the answer to the (a) F=m(a−g) and (b) ║F║= 440. 19 N. to solve this we need to use newton's law of motion as follows:
a) Our task is to write an equation giving the vertical force that must apply to the pole.
From the Newton's second law of the motions, the net force [tex](F_{net})[/tex] is
[tex](F_{net})[/tex] = ∑[tex]F= F_{1}[/tex] + [tex]F_{2}[/tex] + [tex]F_{3}[/tex] ..... = m.a
When we accord into equation (1), we get.
F-W = m.a
F−m⋅g = m.a
The vertical force from the equation (3) will be.
Fj= m.aj - m⋅gj
When we take the common factor, final equation will be.
F=m(a−g)
b) Our task is to determine the magnitude of the force.
║F║= m(a - g)
calculations
║F║ = 90kg [tex]( 5.0 \frac{m}{s_{2}} - 9.6 \frac{m}{s_{2}} )[/tex]
= 90 kg ( -4.81 [tex]m/s_{2}[/tex])
║F║= 440. 19 N
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How far will a 50 N force stretch the same spring?
The force required to stretch the spring up to 0.5 m is 20 N. Then, the spring constant of the spring is 40 N/m. Then, a force of 50 N will stretch the same spring up to 1.2 m.
What is spring constant ?The force required to stretch an elastic material is directly proportional to the displacement x of the material.
hence, F = -kx
Where, the proportionality constant k is called spring constant.
Given that, the force required to stretch the spring up to 0.5 m is 20N.
then, spring constant = force/displacement
k = 20 N/0.5 m
= 40 N/m.
Then, the displacement made by a force of 50 N applied on the same spring is:
x = F/k
= 50 N/40 N/m
= 1.2 m.
Therefore, the force of 50 N will stretch the spring up to 1.2 m.
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Your question is incomplete. But your complete question probably was:
A 20 N force stretches a spring to 0.5 m. How far will a 50 N force stretch the same spring?
the three forces are applied to the bracket. determine the range of values for the magnitude of force p so that the resultant of the three forces does not exceed 2400 n. Force P is always directed to the right.
Range of values for given condition will depend on magnitude and directions of the three given forces. [tex]|R| < = 2400 N[/tex]
To find the range of values for the magnitude of force P so that the resultant of the three forces does not exceed 2400 N, find magnitude and direction.
We can do this by using vector addition. Adding the three forces together, we get:
R = F1 + F2 + F3
where F1, F2, and F3 are the magnitudes and directions of the three given forces, and R: magnitude with resultant force direction.
To ensure that the resultant force does not exceed 2400 N, we must have:
|R| <= 2400 N
Therefore, we need to find the range of values for the magnitude of force P that satisfy this inequality.
The magnitudes and directions of the three given forces are not specified in the problem, so we cannot provide a specific numerical answer. However, we can provide a general method for solving the problem.
To find the range of values for the magnitude of force P, we can first find the maximum and minimum values of the magnitude of the resultant force for different values of P. We can then find the range of values of P that satisfy the inequality above.
This can be done numerically by using vector addition and trigonometry to find the magnitude and direction of the resultant force for different values of P. Alternatively, we can use graphical methods such as force polygons or vector diagrams to visualize the resultant force and find the range of values of P that satisfy the inequality.
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The charged objects represented by the gray circles below all have identical masses, but different charges. Which will make it to the finish line first, assuming they are all released from rest at the same time? The black arrows represent a uniform electric field. A. AB. BC. CD. There will be a tie for first place E. None will reach the finish line
B will make it to the finish line first assuming they are all charged objects released from rest at the same time.
This is because the acceleration of the charged object is proportional to the magnitude of the electric field and inversely proportional to the mass of the object. Since all the charged objects have the same mass, the acceleration will depend only on the magnitude of the electric field.
In this case, the electric field is uniform and in the direction of the finish line, which means that the magnitude of the electric field is the same for all the charged objects. Therefore, the acceleration will be the same for all the charged objects.
However, since the magnitude of the force on each object is different due to their different charges, the velocity of the objects will differ. The charged object with the smallest charge, which is object B, will have the highest velocity and make it to the finish line first.
Objects A and D will have the same velocity and reach the finish line simultaneously. Object C will be the slowest and reach the finish line last. Therefore, the correct answer is B.
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the materials that made up the solar nebula can be categorized into these four general types. rank these materials from left to right based on the temperature at which each would condense into a solid, from highest to lowest. note: for a substance that does not condense at all, rank it as very low temperature.- Rock
- Hydrogen and Helium Gas
- Metals
- Hydrogen Compounds
A solar system's planets, sun, and other things are created in the solar nebula, a huge disc-shaped mass of gas and dust. Nebula is a Latin term that means "cloud." The
Describe solar energy.
Visit the Article History. Solar energy is another name for it. Solar energy is the name given to the Sun's radiation that can ignite chemical reactions, produce heat, or create electricity. The total solar energy incidence on Earth is far greater than the global energy needs at the moment and in the future. This is extremely dispersed if properly harnessed.
What advantages does solar energy offer?
When combined with storage, solar energy may supply backup power for nights and outages, lower electricity costs, help build a more robust electrical grid, promote economic growth, create jobs, and operate at equivalent effectiveness on both small and big sizes. There are numerous types and sizes of solar energy systems.
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what are the effects of cooling on the physical properties of a substance
When the liquid cools down, it loses heat energy.
What is the physical effect of cooling on liquid?As the liquid cools, it loses heat energy. As a result, its particles slow down in movement and come closer to one another. Attractive forces begin to hold particles and the crystals of a solid form.
If water is cooled, it can change into ice. If ice is warmed, it can change into a liquid state. Heating a substance makes the molecules move very fast whereas cooling a substance makes the molecules move very slowly.
Heating a liquid increases the speed of the molecules present in it. An increase in the molecule's speed competes with the attraction between molecules and results in the molecules moving apart whereas Cooling a liquid decreases the movement of the molecules.
So we can conclude that the liquid cools down when it loses heat energy.
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The mass is 18 kg. The velocity is 4.7 m/s. What is the kinetic energy?
The kinetic energy is 4400 joules. The mass is 29 kg. What is the speed?
Answer:
Explanation:
Kinetic energy = 1/2*m*v^2
The kinetic energy is 4400 joules. The mass is 29 kg. What is the speed
4400 = 1/2*29*v^2
v^2 = 303.44
v=17.42m/s^2
The mass is 18 kg. The velocity is 4.7 m/s. What is the kinetic energy?
KE= 1/2*18*4.7*4.7=198.81J
A gas is enclosed within a chamber that is fitted with a frictionless piston. The piston is then pushed in, thereby compressing the gas. Which statement below regarding this process is consistent with the first law of thermodynamics?
(a) The internal energy of the gas will increase.
(b) The internal energy of the gas will decrease.
(c) The internal energy of the gas will not change.
(d) The internal energy of the gas may increase, decrease, or remain the same, depending on the amount of heat that the gas gains or loses.
The statement consistent with the first law of thermodynamics is the internal energy of the gas will increase. Option a is correct.
The statement consistent with the first law of thermodynamics is (a) The internal energy of the gas will increase. This is because the work done on the gas is being converted into an increase in internal energy.
Option (b) (c) The internal energy of the gas will decrease, or will not change is incorrect, because work is being done on the gas which will increase its internal energy.
Option (d) does not address the fact that work is being done on the gas, which is also contributing to changes in its internal energy.
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A hot dog cooker heats hot dogs by connecting them to 120 V household electricity. A typical hot dog has a mass of 70 g and a resistance of 160 Ω.
Part A
How long will it take for the cooker to raise the temperature of the hot dog from 20∘C to 85 ∘C? The specific heat of a hot dog is approximately 2500 J/kg⋅K
It will take about 233.34 seconds for the hot dog cooker to raise the temperature of the hot dog from 20∘C to 85∘C.
What is specific heat?Specific heat is the amount of heat energy required to raise the temperature of a substance by one degree Celsius or one Kelvin per unit mass.
Here,
To solve this problem, we can use the formula for the amount of heat required to change the temperature of an object,
Q = mcΔT
First, we need to calculate the amount of heat required to raise the temperature of the hot dog from 20∘C to 85∘C:
Q = (0.07 kg)(2500 J/kg⋅K)(85∘C - 20∘C)
Q = 1058.5 J
Next, we can use the formula for electrical power,
P = IV
We can rearrange this formula to solve for the current:
I = P/V
The power required to heat the hot dog can be calculated using the formula for electrical power:
P = V²/R
Substituting the given values, we get:
P = (120 V)²/160 Ω
P = 90 W
I = 90 W/120 V
I = 0.75 A
Finally, we can use the formula for the amount of time required to transfer a certain amount of heat:
t = Q/(IΔT)
Substituting the values we calculated, we get:
t = 1058.5 J/(0.75 A)(65∘C)
t = 233.34 s
Therefore, it will take about 233.34 seconds for the hot dog cooker to raise the temperature of the hot dog from 20∘C to 85∘C.
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A stone is dropped into a river from a bridge 43.9m above thewater. Another stone is thrown vertically down 1.00s afterthe first is dropped. Both stones strike the water at thesame time. What is the initial speed of the secondstone?
Answer:
Below
Explanation:
Find time of first stone to strike water ....second stone take 1 s less
First stone
d = 1/2 a t^2
43.9 = 1/2 (9.81)(t^2) shows t = ~ 3 seconds
Second stone
d = vo t + 1/2 a t^2
43.9 = vo (t) + 1/2 (9.81) t^2 t = 3 -1 = 2 seconds
43.9 = vo (2) + 4.905 (2)^2
shows vo = 12.1 m/s
A fishing boat in the ocean is moving at a speed of 20.0 km/h and heading in a direction of 40.0° east of north. A lighthouse spots the fishing boat at a distance of 24.0 km from the lighthouse and in a direction of 15.0° east of north. At the moment the fishing boat is spotted, a speedboat launches from a dock adjacent to the lighthouse. The speedboat travels at a speed of 44.0 km/h and heads in a straight line such that it will intercept the fishing boat.
(a)How much time, in minutes, does the speedboat take to travel from the dock to the point where it intercepts the fishing boat?
(b)In what direction does the speedboat travel? Express the direction as a compass bearing with respect to due north.
In order to reach the fishing boat in the smallest amount of time and distance, the speedboat's pilot should point it 15.0 + 13.7 = 28.7° east of true north from the lighthouse.
What is the fishing boat?Typically speaking, if I see the initial line of sight from the lighthouse as being straight north, this is the simplest for me to solve.
The fishing vessel is thus traveling 40.0 – 15.0 = 25.0° east of north.
The fishing boat's eastward speed is 29.0sin25.0, or 12.3 kilometers per hour.
The fishing boat's northward speed is 29.0cos25.0, or 26.3 kilometers per hour.
If the speedboat matches the fishing boat's eastward speed such that the two boats stay on a line that is exactly north/south of one another, the speedboat will go the smallest distance in the quickest time.
The speedboat makes northward progress at
√(52.0² - 12.3²) = 53.4 km/hr
The net northward speed difference is
53.4 – 26.3 = 27.1 km/hr
so the gap between them closes in
16 km/27.1km/hr = 0.6 hr (or 36 min)
The speed boat will be traveling in the direction θ to the right of our artificial north
sinθ = 12.3/52.0
θ = 13.7°
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Finding an unknown weight
As shown, a mass is hung from the pulley. This mass causes a tensile force of 16.0 N in the cable and the pulley to hang h=130.0 cm from the ceiling. Assume that the pulley has no mass. What is the weight of the mass?
Found W to be 27.7 N
b) Finding the mass of the pulley
As shown, an object with mass m=5.8 kg is hung from a pulley and spring system. When the object is hung, the tension in the cable is 41.8 N and the pulley is h=147.2 cm below the ceiling.
Because the tensile force is greater than the object's weight, the pulley cannot be massless as assumed. Find the mass of the pulley. For this problem, use g
=
9.81
m
/
s
2
. Not 1.95 kg
The unknown weight is 16 N and the mass of the pulley is 4.26 kg.
What is the unknown weight?
The weight of the mass can be found using the formula;
W = T
where;
T is the tension in the cable,so W = 16.0 N.
To find the mass of the pulley, we need to take into account its weight and the tension in the cable. Since the tension is greater than the weight of the object, we know that the tension is also supporting the weight of the pulley.
T - m_pulley x g = 0
where;
m_pulley is the mass of the pulley and g is the acceleration due to gravity.m_pulley = T/g
= 41.8 N/9.81 m/s^2 = 4.26 kg
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Two wooden members of uniform rectangular cross section are joined by the simple glued scarf splice shown. Knowing that P= 18 kN, determine the normal and shearing stresses in the glued splice. 150 mm P 75 mm kPa. The normal stress is ... kPaThe shearing stress is ... kPa.
1.12 MPa is the shear stress in the glued joint.
What is class 10 cross section?If you think of an object as a 2D object, its cross section area is one of its areas. Consider a perfectly spherical ball as an illustration. A circle with a radius equal to the ball can be seen if you view the ball as a 2D object. Consider a cone for another illustration.
The axial or normal stress, σ, in the glued joint can be calculated as:
σ = P/A
A = 2 × b × h
A = 2 × 75 mm × 150 mm = 22,500 mm²
Substituting the values of P and A, we get:
σ = 18 kN / 22,500 mm² = 0.8 MPa
So the normal stress in the glued joint is 0.8 MPa.
The shear stress, τ, in the glued joint can be calculated as:
τ = VQ/It
V = P/2 = 9 kN
The first moment of area, Q, can be calculated as:
Q = b×h2/4
When we change the values of b and h, we obtain:
Q = 75 mm × (150 mm)2 / 4 = 1,406,250 mm³
Calculating the second instant of area, I, is as follows:
I = b×h3/12
The result of substituting the values of b and h is:
I = 75 mm × (150 mm)3 / 12 = 1,054,687.5 mm⁴
Substituting the values of V, Q, I, and t, we get:
τ = 9 kN × 1,406,250 mm³/ (1,054,687.5 mm⁴ × 75 mm) = 1.12 MPa
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List the homologous series
The organic compounds in the homologous series have similar chemical properties. The simplest example of homologous series is the first four hydrocarbons; methane, ethane, propane and butane.
What is homologous series?The homologous series is known as the group of organic compounds that differ from each other by a methylene group. They are series of compounds with the same functional group and similar chemical properties.
The compounds of carbon in homologous series have different number of carbon atoms. But they contain the same functional group. Alkanes, alkenes and alkynes form the homologous series.
Thus all the alkanes, alkenes and alkynes form homologous series.
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ntroduction The flow of geophysical fluids (i.e., the Earth’s ocean and atmosphere, and the atmospheres of gas giants planets such as Jupiter and Saturn) is complicated, involving a vast number of processes and interactions among them on scales ranging from centimeters to the planet’s size, and timescales going from seconds to millennia. Two effects mainly constrain the flow of geophysical fluids: the planet’s rotation and stratification. In this lab we will deal with the first of the aforementioned effects. We will learn how the unusual properties of rotating fluids manifest themselves in, and profoundly influence, the circulation of the Earth’s ocean and planetary atmospheres. The planet’s rotation makes these fluids more similar than one might expect. On what scales might the atmosphere, ocean, or our laboratory experiment, “feel” the effect of rotation? Suppose that U is a typical horizontal current speed, and the typical distance over which the currents varies is L. Then the timescale of the motion (Tmotion) is L/U. Compare this with the period of rotation Trot, define a nondimensional number (the Rossby number):
Ro := Trot/Tmotion = Trot × U/L. If Ro is much greater than one, then the timescale of motion is short relative to a rotation period, and rotation will not significantly influence the motion. If Ro is much less than one, then the motion will be aware of rotation. Let us estimate Ro for large-scale flow in the atmosphere and ocean.
• Amosphere: L ∼ 5000 Km, U ∼ 10 m/s, and T = 1 day, giving Ro = 0.2, which suggest the rotation will be important.
• Ocean: L ∼ 1000 Km, U ∼ 0.1 m/s, giving Ro = 0.01, and rotation will be a controlling factor. Pre Lab 1.
It is clear from the Ro estimations above, that rotation is very important in shaping the patterns of air and ocean currents on sufficiently large scales. How can we study this effect on an small rotating tank (L ∼ 30cm)? If we generate a current in the tank of U ∼ 0.1 cm/s, what would be an appropriate rotation period?
Answer:
Explanation:
To study the effect of rotation on a small rotating tank, we want the Rossby number to be much less than one, so that rotation will be a controlling factor in shaping the patterns of flow.
Based on the given information, the length scale of the tank is L = 30 cm and the current speed is U = 0.1 cm/s. To calculate the timescale of the motion, Tmotion, we can use the formula Tmotion = L/U, which gives us:
Tmotion = 30 cm / 0.1 cm/s = 300 s
Next, we need to estimate an appropriate rotation period, Trot, so that the Rossby number Ro = Trot / Tmotion will be much less than one. We can use the formula Ro = Trot * U / L, rearrange it to solve for Trot:
Trot = Ro * L / U
If we take Ro to be 0.1 (for example), then we have:
Trot = 0.1 * 30 cm / 0.1 cm/s = 30 s
So, with a rotation period of 30 s and a current speed of 0.1 cm/s, we should expect the rotation to have a significant influence on the patterns of flow in the small rotating tank.
A piece of wire is cut into two pieces, A and B, which are then tightly stretched and moun rigid walls. A and B have the same stretched lengths, but A is stretched more tightly following quantities will always be larger for waves on A than for waves on B? a) amplitude of the wave b) frequency of the first harmonic c) wave velocity d) wavelength of the first harmonic e) both b and c
The correct answer is e) both b and c.
The quantities that will always be larger for waves on A than for waves on B are the frequency of the first harmonic and the wave velocity.
The frequency of the first harmonic is determined by the tension in the wire. Since A is stretched more tightly than B, the frequency of the first harmonic will be larger for waves on A than for waves on B.
The wave velocity is also determined by the tension in the wire. A higher tension results in a higher wave velocity. Therefore, the wave velocity will also be larger for waves on A than for waves on B.
The amplitude and wavelength of the first harmonic are not affected by the tension in the wire, so they will not be larger for waves on A than for waves on B.
In conclusion, the frequency of the first harmonic and the wave velocity will always be larger for waves on A than for waves on B.
Therefore, the correct answer for quantities that will be larger for waves on A than for waves on B is option e) both b and c.
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An ant crawls on the sidewalk. It first moves south a distance of 10 cm. It then turns southwest and crawls 6 cm. What is the magnitude of the ant’s displacement?
Explanation:
the displacement is the length of direct line of sight from the starting point to the end point. in other word the baseline of the created triangle.
the legs of that triangle are 10 cm and 6 cm.
the angle between both legs is 90 + 45 = 135°.
because 90° is the angle between South and West. and for southwest we have to add 45°
the law of cosine :
c² = a² + b² - 2ab×cos(C)
c is a side, C is the opposite angle, a and b are the other 2 sides.
so, we have in our case
displacement² = 10² + 6² - 2×10×6×cos(135) =
= 100 + 36 - 120×cos(135) =
= 136 - 120×-0.707106781... =
= 136 - 120×-sqrt(2)/2 =
= 136 + 60×sqrt(2) =
= 220.8528137...
displacement = 14.86111751... cm ≈ 15 cm
A jet of water emerging from a hole in the side of a tanke of water covers horizontal distance R before Striking the ground. If the depth of water in the tank is h and the height of the hole from bottom of the tank is yo formula for R for an Identical the derive a fank on the moon where Jm = ¼ де R? Show that the maximum range of jet of water is Rmax = hand is achieved when =h/2 y what is the
The maximum range of the water jet is[tex]Rmax = h(sqrt(2)),[/tex] and it is achieved when the height of the hole is [tex]yo = h/2.[/tex]
What is range?We can use the equations of motion for an object under constant acceleration to derive the formula for the horizontal distance R that a jet of water will travel before striking the ground. The acceleration of the water jet is due to gravity, and it is constant and equal to the acceleration due to gravity, g.
Let t be the time it takes for the water jet to hit the ground after leaving the hole. We can use the equation of motion for the vertical direction:
[tex]y = yo + voy t - (1/2)gt^2[/tex]
where y is the vertical displacement of the water jet, yo is the initial vertical displacement (the height of the hole from the bottom of the tank), voy is the initial vertical velocity (which is zero), and g is the acceleration due to gravity.
Solving for t, we get:
[tex]t = sqrt((2(yo - y))/g)[/tex]
Now we can use the equation of motion for the horizontal direction:
[tex]x = v_{0} x t[/tex]
where x is the horizontal displacement (which is R), and vox is the initial horizontal velocity (which is constant and equal to the velocity of the water jet as it emerges from the hole).
We can express [tex]v_{0} x[/tex] in terms of the vertical displacement y and the time t:
[tex]v_{0} x = R/t = R(sqrt(g/(2(yo - y))))[/tex]
Substituting for t and simplifying, we get:
[tex]v_{0} x = R(sqrt(2g/h))[/tex]
Now we can express the range R in terms of the tank height h and the height of the hole yo:
[tex]R = (v_{0} x^2/h) = 2h(sqrt(yo/h))[/tex]
To derive the formula for an identical tank on the moon where the acceleration due to gravity is Jm = 1/4 of the acceleration due to gravity on Earth (g), we can substitute g/4 for g in the equation for the horizontal velocity vox. This gives:
[tex]vox = R(sqrt(g/(8h)))[/tex]
Substituting into the equation for R, we get:
[tex]R = (vox^2/h) = 8h(sqrt(yo/h))[/tex]
To show that the maximum range of the water jet is achieved when yo = h/2, we can differentiate R with respect to yo and set the result equal to zero:
[tex]dR/dyo = (4/h)(sqrt(yo/h))(h/2 - yo)[/tex]
Setting this equal to zero and solving for yo, we get:
[tex]yo = h/2[/tex]
To find the maximum range Rmax, we substitute yo = h/2 into the equation for R:
[tex]Rmax = 2h(sqrt(h/2h)) = h(sqrt(2))[/tex]
Therefore, the maximum range of the water jet is [tex]Rmax = h(sqrt(2))[/tex], and it is achieved when the height of the hole is yo = h/2.
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why is the world so small compared to the sun and jupider
The Sun appears smaller than the Earth from here on Earth, but that is only because the Earth is considerably closer to you than the Sun is. Jupiter due to its rapid revolution, which increases its diameter in the midsection.
What are the bodies of the solar system?Our solar system consists of the star, Sun, planets: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune, and small planets such as Pluto.
While the Sun is 150 million kilometers away from where you are, you are on the surface of the Earth.
The planet is an oblate spheroid due to its rapid revolution, which increases its diameter in the midsection.
Therefore, Jupiter due to its rapid revolution increases its diameter in the midsection making it bigger as compared to the world, so it is believed small compared to the Sun and Jupiter.
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What is the Reynolds’ number if the average flow speed of blood through the coronary artery is 15 mL/s, the density of the blood is 50 kg/m3 and the vessel has a diameter of 0.2 m?
experimental resistance is calculated using measured values. calculated resistance is determined using equations from your textbook or the background information link. question 1) how is the total resistance related to the individual resistances? question 2) total current to the individual currents? question 3) total voltage to the individual voltages? be sure to show your calculations for the series circuit.
Answer:
Explanation:
edge 2023 b
the origin: 8. A thin rod of length and uniform charge per unit length SA lies along the x axis as shown in Figure P23.8. (a) Show that the electric field at P, a distance d from the rod along its perpendicular bisector, has no x component and is given by E = 2k i sin 0,/d. (b) What If? Using your result to part (a), show that the field of a rod of infinite length is E = 2kX/d. -- -- -- --- F
For solution of a) of the particular question where a thin rod and uniform change lies along the x axis=
λ = linear charge density
Consider a small length "dx" at distance "x" from the origin
dq = small charge on the small length = λ dx
r = distance of small length from point P = sqrt(x2 + d2)
small electric field at P due to small length is given as
dE = k dq/r2
dE = k λ dx /(sqrt(x2 + d2))2
dE = k λ dx /(x2 + d2)
From the diagram , we see that "dE Sinθ" are equal and opposite, hence x-components cancel out.
Net electric field at P is given as
E = ∫ 2 dE Cosθ
E = ∫ 2 (k λ dx /(x2 + d2)) (d/sqrt(x2 + d2))
E = ∫ 2 (k λ d dx /(x2 + d2)3/2)
E = ∫ _{0}^{l/2} 2 (k λ d dx /(x2 + d2)3/2)
E = (2 k λ d) ∫_{0}^{l/2}dx /(x2 + d2)3/2
E = (2 k λ d) ((l/2)/ (d2 sqrt(d2 + (l/2)2))
E = (2 k λ d) (Sinθ _{o}/ d2 )
Since
Sinθ _{o} = (l/2) /sqrt(d2 + (l/2)2)
E = 2 k λ Sinθ _{o}/ d
For solution of b)
for infinite length , θ _{o}= 90
E = 2 k λ Sin90/ d
E = 2 k λ / d
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10 pennies, each with a mass of 3.11 grams, are heated to a temperature of 112 degrees celsius and then dropped into a beaker of water
containing 200 grams of water at 25 degrees celsius. To what temperature will the water heat up to?
C copper = 387 J/kg C
C water = 4186 J/kg C
A-26C
B-29C
C-33C
D-39C
Answer:
Explanation:a
Two identical conducting small spheres are placed with their centers 0.300m apart. One is given a charge of 12.0nC and the other a charge of −18.0nC.
(a) Find the electric force exerted by one sphere on the other.
(b) What If? The spheres are connected by a conducting wire. Find the electric force each exerts on the other after they have come to equilibrium.
Electric force by one sphere on other is [tex]-4.80 * 10^(-3) N[/tex] and after they have come to equilibrium is [tex]1.08 * (10^-3) N[/tex]
(a) The electric force exerted by one sphere on the other can be calculated using Coulomb's law:
F = k * (q1 * q2) / r^2
where F is the force, k is Coulomb's constant (9.0 x 10^9 Nm^2/C^2), q1 and q2 are the charges on the two spheres (12.0nC and -18.0nC), and r is the distance between their centers (0.300m).
Plugging in the values, we get:
F = 9.0 x 10^9 * (12.0 x 10^-9) * (-18.0 x 10^-9) / (0.300)^2 = -4.80 x 10^-3 N
The negative sign indicates that the force is attractive.
(b) When the spheres are connected by a conducting wire, they will share charge until they reach equilibrium. At equilibrium, the net force on each sphere will be zero. The electric force each sphere exerts on the other will be the same in magnitude, but opposite in direction.
To find the magnitude of the force, we can use the fact that the potential of each sphere will be the same at equilibrium. The potential can be calculated using:
V = k * q / r
where V is the potential, k is Coulomb's constant, q is the charge on the sphere, and r is the distance from the sphere's center.
At equilibrium, the potential of each sphere will be the same, so:
k * q1 / r1 = k * q2 / r2
Solving for q1 and q2, we get:
q1 = -q2 * r1 / r2
Plugging in the values, we get:
q1 =[tex]-(-18.0 * 10^(-9)) * 0.150m / 0.150m = 18.0 * 10^(-9) C[/tex]
q2 = -q1 = [tex]-18.0 * 10^-(9) C[/tex]
The electric force each sphere exerts on the other can be calculated using Coulomb's law with the new charges:
F = k * (q1 * q2) / r^2
Plugging in the values, we get:
F =[tex]9.0 * 10^(-9) * (18.0 * 10^(-9)^2 / (0.300)^2 = 1.08 * 10^(-3) N[/tex]
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five coulombs (5 c) of charge pass through the element from point a to point b. if the energy absorbed by the element is 120 j, determine the voltage across the element.
The voltage across the element is 24 V where if the energy absorbed by the element is 120 j.
Given data as per the question:
Charges = 5 coulomb
Energy absorbed by the element = 120 J
As per the formula we have,
Energy = Voltage X Charge
120= Voltage X 5
Voltage = 120/5 = 24 V
The voltage in the whole process will be negative because the energy is absorbed.
In a series circuit, the current is the same for all the components. While the circuit reaches its steady state, the capacitor charges and the voltage across its plates increases until it reaches the one on the terminals, and at that point it is in the steady state.
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If the energy absorbed by the element is 120 j, So 5*V1 =120 , V1 =24 Volts.
The voltage distinction be V1 Volts.
When 5C of charge moves from A to B, its energy increments by 120J.
So 5*V1 =120
V1 =24 Volts.
The voltage distinction is hence 24 Volts.
At the point when the charge moves from higher potential to lower, it loses energy and when it moves from lower potential to higher, it retains energy. The energy ingested (or lost) is relative to the potential (voltage) contrast between the two focuses.
The voltage in the entire cycle will be negative in light of the fact that the energy is retained.
In a series circuit, the current is no different for every one of the parts. While the circuit arrives at its consistent express, the capacitor charges and the voltage across its plates increments until it arrives at the one on the terminals, and by then it is in the consistent state.
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problem 11.143 a race car enters the circular portion of a track that has a radius of 70 m. when the car enters the curve at point p, it is travelling with a speed of 120 km/h that is increasing at 5 m/s2 . three seconds later, determine (a) the total acceleration of the car in xy components, (b) the linear velocity of the car in xy components.
(a) The total acceleration of the car in xy components is 16.73 m/s^2.
(b) The linear velocity of the car in xy components is 33.3 m/s.
What is total acceleration of the car?To find the total acceleration of the car in xy components, we need to find the sum of the centripetal and tangential accelerations.
The centripetal acceleration is given by a = v^2/r
where;
v is the velocity of the car and r is the radius of the circular trackAt point P, the velocity of the car is 120 km/h = 33.33 m/s. Therefore, the centripetal acceleration is:
a_c = v^2/r
a_c = (33.33 m/s)^2 / 70 m
a_c = 15.88 m/s^2
The tangential acceleration is given by a_t = dv/dt
where;
v is the velocity of the car and
t is time
The rate of change of velocity is given as 5 m/s^2. Therefore, the tangential acceleration is:
a_t = 5 m/s^2
The total acceleration of the car in xy components is the vector sum of a_c and a_t. Since the two accelerations are perpendicular, we can use the Pythagorean theorem to find the magnitude of the total acceleration:
a_total = √(a_c^2 + a_t^2)
a = √((15.88 m/s^2)^2 + (5 m/s^2)^2)
a = 16.73 m/s^2
(b) The linear velocity of the car in xy components can be found using the formula;
v = rω
where;
r is the radius of the circular track and ω is the angular velocity of the car.The angular velocity is related to the linear velocity by the formula ω = v/r. At point P, the linear velocity of the car is 33.33 m/s. Therefore, the angular velocity is:
ω = v/r = 33.33 m/s / 70 m
ω = 0.476 rad/s
The linear velocity in the x-direction is v_x = v cos(θ),
where;
θ is the angle between the velocity vector and the x-axis.Since the car is entering the circular portion of the track, the angle θ is 0. Therefore, v_x = v cos(0) = 33.33 m/s.
The linear velocity in the y-direction is v_y = v sin(θ). Since the car is entering the circular portion of the track, the angle θ is 90 degrees. Therefore, v_y = v sin(90°) = 33.33 m/s.
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find the work done by moving an object 3 feet from (0,0) to (3,0) by a force of 15lbs in the direction of (4, 1)
The force of 15 lbs in the direction of (4,1) is: 41.3 ft-lb.
The work done by a force of 15lbs moving an object from (0,0) to (3,0) in the direction of (4,1) is given by the formula:
W = Fdcos(θ), where F is the force, d is the distance, and θ is the angle between the force and the direction of motion.
In this case, θ = arctan(1/4) = 14.036 degrees, so the work done is W = 153cos(14.036) = 41.3 ft-lb.
Force is a physical quantity that is defined as the rate of change of momentum. It is a vector, meaning that it has both magnitude and direction. Force is usually represented by the symbol F and is measured in newtons (N).
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If a car is pushed with a force of 18N for 8m, how much work has been done?
According to the question, for a car with a force of 18 N and displacement of 8 m, the work done by a car is calculated as 144Nm.
What is force?Force may be defined as a process of pushing or pulling on an object that significantly produces acceleration in the body on which it acts. It is an external agent capable of changing a body's state of rest or motion. It has a magnitude and a direction.
According to the question,
The force applied on a car = 18 N
The displacement made by a car = 8m.
Now, the work done is calculated with the help of the given formula:
Work done = Force × Displacement.= 18 N × 8m = 144Nm.
Therefore, the work done by a car is calculated as 144Nm.
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as fleet and bright as a lodestar he wheeled toward guitar and it did not matter which one of them would give up his ghost in the killing arms of his brother. for now he knew what shalimar knew: if you surrendered to the air, you could ride it
This is a quote from the novel Song of Solomon which has been authored by Tomi Morrison.
The narrative comes to an ambiguous end with the vision of flying. Readers are left in the dark regarding the fate of Milkman and Guitar even though they both appear to fly in the conclusion. However, Milkman's ability to fly indicates that he has beyond the limitations of his physical form and has a connection to his great-grandfather, Jake, whom he has spent his entire life trying to understand. Despite the fact that Pilate is dead, the book finishes mystically and with a hopeful attitude.
Song of Solomon is a 1977 novel by American author Toni Morrison, her third to be published. It follows the life of Macon "Milkman" Dead III, an African-American man living in Michigan, from birth to adulthood.
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two square plates eachw ith a side length of 9.32 cm, have equal and opposite charges of magnitude 3.152 calculate the electric field nbetween two fileds
When a charge Q is given to a thin metallic plate then the charge gets equally distributed between the two plates. Thus we have an electric field produced by one side of the plate is
E=σ2ϵo
where σ=QA and A is the area.
A)
Electric field, E = Q/(A e0)
E = (5.952 x 10^-6)/(pi x 0.0522^2 x 8.85 x 10^-12)
E = 7.86 x 10^7 N/C
B)
Stays the same
An electric field (sometimes E-field) is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. It also refers to the physical field for a system of charged particles.
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Calculate the velocity a spherical rain drop would achieve falling (taking downward as positive) from 4.4km in the following situations. (h=4.4km; l=3.8mm; d=1.16kg/m3
a. Calculate the velocity in the absence of air drag in m/s. ____
b. Calculate the velocity wiith air drag in m/s Take the size across of the frop to be 3.8mm, the dnesity of air to be 1.16kg/m3 , he density of water to be 1000kg/m3 , the surface area to be\pir2, and the drage coefficient to be 1.0. _____
To calculate the velocity of a spherical raindrop falling from 4.4 km without air drag, we can use the equations of motion:
v² = u² + 2as
How to calculate velocity?where v is the final velocity, u is the initial velocity (which is 0 in this case), a is the acceleration due to gravity, s is the displacement (which is equal to the height of the fall), and we assume that downward is positive. We can use the acceleration due to gravity as -9.81 m/s². Therefore, we get:v² = 0 + 2(-9.81 m/s²)(4.4 km) = -2(9.81 m/s²)(4400 m) = -86,140 m²/s²
Since the velocity cannot be negative, we take the square root of the magnitude to get the final velocity:v = sqrt(86,140 m²/s²) = 293.9 m/s
Therefore, the velocity of the raindrop falling from 4.4 km without air drag is approximately 293.9 m/s.
b. To calculate the velocity of the raindrop falling with air drag, we can use the following equation:
F_drag = 1/2 * rho * v^2 * A * C_d
where F_drag is the drag force, rho is the density of air, v is the velocity of the raindrop, A is the cross-sectional area of the raindrop (which is pi*(d/2)^2), and C_d is the drag coefficient. We can assume that the weight of the raindrop is balanced by the upward force due to air resistance, so we can write:F_drag = m * g
where m is the mass of the raindrop (which is (4/3)pi(l/2)^3*1000 kg/m³), and g is the acceleration due to gravity. We can rearrange the two equations to get:m * g = 1/2 * rho * v^2 * A * C_d
Solving for v, we get:
v = sqrt((2 * m * g) / (rho * A * C_d))
Substituting the values, we get:
v = sqrt((2 * (4/3) * pi * (3.8/2)^3 * 1000 kg/m³ * 9.81 m/s²) / (1.16 kg/m³ * pi * (3.8/2)^2 * 1.0))
Simplifying, we get:
v = 9.7 m/s
Therefore, the velocity of the raindrop falling from 4.4 km with air drag is approximately 9.7 m/s.
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