Answer: we embrace and are simultaneously respecting their rational autonomy
Explanation:
Ursula is a student of Kantian philosophy who argues that we should never use people as a means to an end unless we embrace and are simultaneously respecting their rational autonomy
Do charges exit a circuit with less energy than they had when they entered the circuit
Two tiny conducting spheres are identical and carry charges of - 20.0 µC and +50.0 µC. They are separated by a distance of 2.50 cm. (a) What is the magnitude of the force that each sphere experiences, and is the force attractive or repulsive? (b) The spheres are brought into contact and then separated to a distance of 2.50 cm. Determine the magnitude of the force that each sphere now experiences, and state whether the force is attractive of repulsive.
Answer:
a
The force experience by the two spheres is [tex]F_1 = 1.44*10^4 N[/tex]
This force is attractive cause the charge are unlike charges
b
The force experienced by the two spheres is [tex]F_2 = 3.24*10^{3} \ N[/tex]
The force is repulsive because the two charges are like charges
Explanation:
From the question we are told that
The charge on the first sphere is [tex]q_1 = -20.0 \mu C = 20 *10^{-6} C[/tex]
The charge on the second sphere is [tex]q_2 = 50 \mu C = 50*10^{-6} C[/tex]
The distance of separation is [tex]d = 2.50 \ cm = \frac{2.50}{100} = 0.025 \ m[/tex]
The electrostatic force experienced by the two spheres is mathematically represented as
[tex]F_1 = \frac{k q_1 q_2}{d^2}[/tex]
Now k is the coulomb’s constant with a value of [tex]k = 9*10^9 N \cdot m^2 /C^2[/tex]
So
[tex]F_1 = \frac{9*10^9 * 20 *10^{-6} * 50*10^{-6}}{0.025}[/tex]
[tex]F_1 = 1.44*10^4 N[/tex]
When the sphere are brought together the charge on each sphere would be the average of the total charge and this can be mathematically evaluated as
[tex]q = \frac{q_1 + q_2 }{2}[/tex]
[tex]q = \frac{(-20 + 50)*10^{-6} }{2}[/tex]
[tex]q = 15 \mu C[/tex]
So when they are seperated the electrostatic force experienced is
[tex]F_2 = \frac{kq^2}{d^2}[/tex]
[tex]F_2 = \frac{ 9*10^9 * (15 *10^{-6})}{0.025}[/tex]
[tex]F_2 = 3.24*10^{3} \ N[/tex]
A standard mercury thermometer consists of a hollow glass cylinder, the stem, attached to a bulb filled with mercury. As the temperature of the thermometer changes, the mercury expands (or contracts) and the height of the mercury column in the stem changes. Marks are made on the stem to denote the height of the mercury column at different temperatures such as the freezing point (0∘C ) and the boiling point (100∘C ) of water. Other temperature markings are interpolated between these two points. Due to concerns about the toxic properties of mercury,many thermometers are made with other liquids. Consider drainingthe mercury from the above thermometer and replacing it withanother, such as alcohol. Alcohol has a coefficient of volumeexpansion 5.6 times greater than that of mercury. The amount ofalcohol is adjusted such that when placed in ice water, thethermometer accurately records 0 C. No other changes are made tothe thermometer.
Required:
a. When the alcohol thermometer is placed in 22 C water,what temperature will the thermometer record?
b. When the alcohol thermometer is placed in a -12 Csubstance, what temperature will the thermometer record?
c. If you want to design a thermometer with the same spacingbetween temperature markings as a mercury thermometer, how must thediameter of the inner hollow cylinder of the stem of the alcoholthermometer compare to that of the mercury thermometer? Assume thatthe bulb has a much larger volume than the stem.
1. 5.6 times wider
2. √5.6 times wider
3. the same diameter but different bulbsize
4. √5.6 times smaller
5. 5.6 times smaller
Answer:
a) T_alcohol = 123.2ºC , b) T_alcohol = -67.2ºC and c) r_alcohol =√5,6 r_mercury
Explanation:
We can solve this problem using the thermal expansion equation
ΔV = β Vo ΔT
indicates that alcohol has a coefficient of expansion 5 times that of mercury
β_alcohol = 5 β_mercury
the amount of alcohol used from a correct reading at zero degrees, let's substitute in the equation for alcohol
ΔV = β_ alcohol ΔT
ΔV = 5.6 β_mercury ΔT
we see that the expansion is 5 times the expansion of mercury
T_alcohol = 5.6 T_mercury
T_alcohol = 5.6 22
T_alcohol = 123.2ºC
b) in the case of T = -12ºC
T_alcohol = 5.6 (-12)
T_alcohol = -67.2ºC
c) In this case we want the division to give the same value for the two thermometers
let's use the volume ratio
V = A L
where A is the area of the circle and L the length that the alcohol travels
We know that the volume of alcohol is 5.6 times the volume of mercury
V_alcohol = 5.6 V_mercury
A_alcohol V = 5.6 A_mercury L
A_alcohol = 5.6 A_mercury
if the area of a plum is A = π r², when substituting in this equation
π r_alcohol² = 5.6 π r_mercury²
r_alcohol =√5,6 r_mercury
the consequence the radius of the alcohol thermometer must be √5.6 times the radius of the Mercury thermometer; the correct answer is 2
Which is not a difference between chemical reactions and nuclear reactions?
A.The subatomic particles involved in the reaction is different.
B.The amount of energy absorbed during the reaction is different.
C.The masses of the reactants and products differ.
D.The amount of energy released during the reaction is different.
Answer:
B.The amount of energy absorbed during the reaction is different
Explanation:
I just took the test
The 0.100 kg sphere in (Figure 1) is released from rest at the position shown in the sketch, with its center 0.400 m from the center of the 5.00 kg mass. Assume that the only forces on the 0.100 kg sphere are the gravitational forces exerted by the other two spheres and that the 5.00 kg and 10.0 kg spheres are held in place at their initial positions.
Answer:
the speed of 0.10 sphere when it is moved to 0.20 kg to the left is : [tex]3.3337*10^{-5} \ m/s[/tex]
Explanation:
Using the expression of the Change in Gravitational Potential Energy:
[tex]U= -(\frac{ Gm_1m_2 }{r_2} - \frac{ Gm_1m_2 }{r_1}) \\ \\ U=Gm_1m_2 (\frac{ 1 }{r_2} - \frac{ 1 }{r_1})[/tex]
So when the sphere exerts just 10 kg mass; the change in the gravitational potential energy is :
[tex]U_1=Gm_1m_2 (\frac{ 1 }{r_2} - \frac{ 1 }{r_1})[/tex]
[tex]U_1=6.67*10^{-11}*0.1 \ kg*10 \ kg(\frac{ 1 }{0.6 \ m} - \frac{ 1 }{0.8 \ m})[/tex]
[tex]U_1 = 2.778*10^{-11} J[/tex]
the change in the gravitational potential energy when the sphere exerts just 5 kg mass is ;
[tex]U_2=Gm_1m_2 (\frac{ 1 }{r_2} - \frac{ 1 }{r_1})[/tex]
[tex]U_2=6.67*10^{-11}*0.1 \ kg*5 \ kg(\frac{ 1 }{0.4 \ m} - \frac{ 1 }{0.2 \ m})[/tex]
[tex]U_2 = -8.335*10^{-11} J[/tex]
The net total change is:
[tex]U_{total } = U_1 +U_2[/tex]
[tex]U_{total} = 2.778*10^{-11} + (-8.335*10^{-11})[/tex]
[tex]U_{total} = -5.557*10^{-11}[/tex]
We all know that for there to be a balance ;loss of gravitational potential energy must be equal to the gain in kinetic energy .
SO;
K.E = [tex]5.557*10^{-11}[/tex]
[tex]\frac{1}{2}mv^2 = 5.557*10^{-11}[/tex]
[tex]v^2 = \frac{2*5.557*10^{-11} \ J}{m_1}[/tex]
[tex]v=\sqrt{ \frac{2*5.557*10^{-11} \ J}{0. 1 \kg}[/tex]
v = [tex]3.3337*10^{-5} \ m/s[/tex]
Thus, the speed of 0.10 sphere when it is moved to 0.20 kg to the left is : [tex]3.3337*10^{-5} \ m/s[/tex]
The speed of 0.10 sphere when it is moved to 0.20 kg to the left is :3.3337 *10^-5 m/s
What is speed ?The speed of any object is defined as the movement of any object with respect to the time.
By using the expression of the Change in Gravitational Potential Energy:
[tex]U=-Gm_1m_2(\dfrac{1}{r_2}-\dfrac{1}{r_1})[/tex][tex]U_2=-Gm_1m_2(\dfrac{1}{r_2}-\dfrac{1}{r_1})[/tex]
So when the sphere exerts just 10 kg mass; the change in the gravitational potential energy is :
[tex]U_1=-Gm_1m_2(\dfrac{1}{r_2}-\dfrac{1}{r_1})[/tex]
[tex]U_1=- 6.67\times 10^{-11}\times 0.1\times 10(\dfrac{1}{0.6}-\dfrac{1}{0.8})[/tex]
[tex]U_1=2.778\times 10^{-11][/tex]
The change in the gravitational potential energy when the sphere exerts just 5 kg mass is ;
[tex]U_2=-Gm_1m_2(\dfrac{1}{r_2}-\dfrac{1}{r_1})[/tex]
[tex]U_2=- 6.67\times 10^{-11}\times 0.1\times 5(\dfrac{1}{0.4}-\dfrac{1}{0.2})[/tex]
[tex]U_2=-8.335\times 10^{-11}\ J[/tex]
The net total change is:
[tex]U_{total}=U_1+U_2[/tex]
[tex]U_{total}=2.778\times 10^{-11}+(-8.335\times 10^{-11})[/tex]
[tex]U_{tot\leq al}=-5.557\times 10^{-11}[/tex]
We all know that for there to be a balance ;loss of gravitational potential energy must be equal to the gain in kinetic energy .
SO;
[tex]\rm KE=5.557\times 10^{-11}[/tex]
[tex]\dfrac{1}{2}mv^2=5.557\times 10^{-11}[/tex]
[tex]v^2=\dfrac{2\times 5.557\times 10^{-11}}{m}[/tex]
[tex]v=\sqrt{\dfrac{2\times 5.557\times 10^{-11}}{m}}[/tex]
[tex]v=3.337\times 10^{-5}\ \frac{m}{s}[/tex]
Thus, the speed of 0.10 sphere when it is moved to 0.20 kg to the left is : [tex]v=3.337\times 10^{-5}\ \frac{m}{s}[/tex]
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To understand the experiment that led to the discovery of the photoelectric effect.
In 1887, Heinrich Hertz investigated the phenomenon of light striking a metal surface, causing the ejection of electrons from the metal. The classical theory of electromagnetism predicted that the energy of the electrons ejected should have been proportional to the intensity of the light. However, Hertz observed that the energy of the electrons was independent of the intensity of the light. Furthermore, for low enough frequencies, no electrons were ejected, no matter how great the intensity of the light became. The following problem outlines the methods used to investigate this new finding in physics: the photoelectric effect.
Suppose there is a potential difference between the metal that ejects the electrons and the detection device, such that the detector is at a lower potential than the metal. The electrons slow down as they go from higher to lower electric potential; since they must overcome this potential difference to reach the detector, this potential is known as the stopping potential. To reach the detector, the initial kinetic energy of an ejected electron must be greater than or equal to the amount of energy it will lose by moving through the potential difference.
(A) If there is a potential difference V between the metal and the detector, what is the minimum energy E_min that an electron must have so that it will reach the detector?
Express your answer in terms of V and the magnitude of the charge on the electron, e.
(B) Suppose that the light carries energy E_light. What is the maximum stopping potential V_0 that can be applied while still allowing electrons to reach the detector?
Express your answer in terms e, E_light, and Φ.
Answer:
A) Emin = eV
B) Vo = (E_light - Φ) ÷ e
Explanation:
A)
Energy of electron is the product of electron charge and the applied potential difference.
The energy of an electron in this electric field with potential difference V will be eV. Since this is the least energy that the electron must reach to break out, then the minimum energy required by this electron will be;
Emin = eV
B)
The maximum stopping potential energy is eVo,
The energy of the electron due to the light is E_light.
If the minimum energy electron must posses is Φ, then the minimum energy electron must have to reach the detectors will be equal to the energy of the light minus the maximum stopping potential energy
Φ = E_light - eVo
Therefore,
eVo = E_light - Φ
Vo = (E_light - Φ) ÷ e
4
Select the correct answer.
Which of the following is evidence of the past existence of glaciers?
A.
rift valleys
B. fjords
C.
estuaries
D.
oceanic ridges
Answer:
B. fjords
explanation :
Fjords were created by glaciers. In the Earth's last ice age, glaciers covered just about everything. Glaciers move very slowly over time, and can greatly alter the landscape once they have moved through an area. This process is called glaciation.The fjords are one of the glaciers that existed in the past. They are one of the many glacial relief forms that can give us a insight into the size and power of the glaciers.
Answer:
B. fjords
I hope this helps
A block of mass m = 2.5 kg is attached to a spring with spring constant k = 730 N/m. It is initially at rest on an inclined plane that is at an angle of θ = 21° with respect to the horizontal, and the coefficient of kinetic friction between the block and the plane is μk = 0.19. In the initial position, where the spring is compressed by a distance of d = 0.11 m, the mass is at its lowest position and the spring is compressed the maximum amount. Take the initial gravitational energy of the block as zero.
A) What is the block's initial mechanical energy?
B) If the spring pushes the block up the incline, what distance, L, in meters will the block travel before coming to rest? The spring remains attached to both the block and the fixed wall throughout its motion.
Answer:
A) Em = 4.41 J
B) L = 0.33m
Explanation:
A) The total mechanical energy of the block is the elastic potential energy due to the compressed spring. The gravitational energy is zero. Then you have:
[tex]E_m=\frac{1}{2}k(\Delta x)^2[/tex]
k: constant's spring = 730 N/m
Δx: distance of the compression = 0.11m
You replace the values of k and Δx:
[tex]E_m=\frac{1}{2}(730N/m)(0.11m)^2=4.41\ J[/tex]
B) To find the distance L traveled by the block you take into account that the total mechanical energy of the block is countered by the work done by the friction force, and also by the work done by the gravitational energy.
Then, you have:
[tex]E_m-W_f-W_g=0\\\\W_f=(\mu Mg cos\theta)L\\\\W_g=(Mgsin\theta)L[/tex]
μ: coefficient of kinetic friction = 0.19
g: gravitational acceleration = 9.8m/s^2
M: mass of the block = 2.5kg
θ: angle of the inclined plane = 21°
You replace the values of all parameters:
[tex]E_m-W_f-W_g=0\\\\4.41-(0.19)(2.5kg)(9.8m/s^2)(cos21\°)L-(2.5kg)(9.8m/s^2)(sin21\°)L=0\\\\4.41-4.34L-8.78L=0\\\\4.41-13.12L=0\\\\L=0.33m[/tex]
hence, the distance L in which the block stops is 0.33m
(a) The block's initial mechanical energy on the given position is 4.42 J.
(b) The distance traveled by the block when it is pushed by the spring before coming to rest is 1.02 m.
The given parameters;
mass of the block, m = 2.5 kgspring constant, k = 730 N/mangle of inclination, θ = 21°coefficient of friction, μ = 0.19compression of the spring, x = 0.11 mThe block's initial mechanical energy is calculated as follows;
[tex]E = K.E _i + P.E_i\\\\E = \frac{1}{2} mv^2 \ + \ \frac{1}{2} kx^2\\\\E = \frac{1}{2} m (0)^2 \ + \ \frac{1}{2} \times 730 \times (0.11)^2\\\\E = 4.42 \ J[/tex]
The block will travel up if the energy applied by the spring is greater than the work-done by frictional force on the block.
The work-done on the block by the frictional force is calculated as follows;
[tex]W_f = F_k \times d\\\\W_f= \mu_k F_n \times d\\\\W_f = \mu_k mgcos(\theta) \times d\\\\W_f = (0.19)(2.5)(9.8)cos(21) \times d \\\\W_f = 4.346 d[/tex]
Apply work-energy theorem;
[tex]4.346d = 4.42\\\\d = \frac{4.42}{4.346} = 1.02 \ m[/tex]
Thus, the distance traveled by the block when it is pushed by the spring before coming to rest is 1.02 m.
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Find Density of a cylinder with diameter of 5.0 cm and height of 12.0cm and mass of 600.0g
Answer: 2.55 g/cm^3
Explanation:
density is defined as:
Density = mass/volume
Now, the mass of the cylinder is 600g
and the volume of a cylinder is:
V = pi*r^2*h
where r is the radius (half of the diameter), here r = (5/2)cm and h is the height, here 12 cm
So the volume is:
V = 3.14*(2.5cm)^2*12cm = 235.5cm^3
then the density is:
D = 600g/235.5cm^3 = 2.55 g/cm^3
If you have two spoons of the same
size, one silver and one stainless
steel, there is a quick test to tell
which is which. Hold the end of a
spoon in each hand, then lower them
both into a cup of very hot water.
One spoon will feel hot first. Is that
the silver spoon or the stainless steel
spoon? Explain.
Answer: Silver
Explanation:Silver has ahigh thermal conductivity of 429 watts/meter-k. Stainless steel on the other hand, has low thermal conductivity of 160 watts/meter-k.
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(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)(^o^)
Find the equivalent resistance Re for R1=1Ω ,R2=2Ω , R3=3Ω IS connected in series
Answer:
Just add all the resistors together
1+2+3=6
6Ω is the total resistance
Explanation:
Photons with shorter wavelengths have larger
o
A. color
B. amplitude
c. energy
D. speed
SUBMIT
i think photons with shorter wavelengths have larger speed.
I apologise if its incorrect.
Answer:
energy
Explanation:
A 0.050 kg bullet strikes a 5.0 kg wooden block with a velocity of 909 m/s and embeds itself in the block which fies off its stand. what was the final velocity of the bullet?
Answer:
The final velocity of the bullet is 9 m/s.
Explanation:
We have,
Mass of a bullet is, m = 0.05 kg
Mass of wooden block is, M = 5 kg
Initial speed of bullet, v = 909 m/s
The bullet embeds itself in the block which flies off its stand. Let V is the final velocity of the bullet. The this case, momentum of the system remains conserved. So,
[tex]mv=(m+M)V\\\\V=\dfrac{mv}{m+M}\\\\V=\dfrac{0.05\times 909}{0.050+5}\\\\V=9\ m/s[/tex]
So, the final velocity of the bullet is 9 m/s.
A car with mass mc = 1225 kg is traveling west through an intersection at a magnitude of velocity of vc = 9.5 m/s when a truck of mass mt = 1654 kg traveling south at vt = 8.6 m/s fails to yield and collides with the car. The vehicles become stuck together and slide on the asphalt, which has a coefficient of friction of μk = 0.5.
A) Write an expression for the velocity of the system after the collision, in terms of the variables given in the problem statement and the unit vectors i and j.
B) How far, in meters, will the vehicles slide after the collision?
Answer:
a) v(f) = -4i - 5j
b) 4.18 m
Explanation:
The equation to be used for this question is
v(c)m(c) + v(t)m(t) = [m(c) + m(t)] v(f)
if we rearrange and make v(f) subject of formula, then
v(f) = v(c)m(c) + v(t)m(t) / [m(c) + m(t)]
One vehicle is headed towards south and the other vehicle, west when they collide they will travel together in a southwestern direction. This means that both vehicles are traveling in the negative direction taking a standard frame of reference. Thus, we can write the equation in component form by substituting the values as
v(f) = 1225(-9.5i) + 1654(-8.6j) / 1225 + 1654
v(f) = -11637.5i - 14224.4j / 2879
v(f) = -4i - 5j m/s
From the answer,
v(f) = √(4² + 5²)
v(f) = √41
v(f) = 6.4 m/s
And we know that
KE = ½mv²
Fd = umgd
And, KE = Fd, so
½mv² = umgd
½v² = ugd
Making d the subject of formula,
d = v²/2ug
d = 6.4² / 2 * 0.5 * 9.8
d = 41 / 9.8
d = 4.18 m
(a) The velocity of the system after collision is 4.04 i + 4.9 j.
(b)The distance traveled by the vehicles after collision is 1.73 m.
The given parameters;
mass of the car, Mc = 1225 kgvelocity of the car, Vc = 9.5 m/smass of the truck, Mt = 1654 kgvelocity of the truck, Vt = 8.6 m/sApply the principle of conservation of linear momentum to determine the velocity of the system after collision;
[tex]m_1u_x_1 + m_2 u_y_2 = V(m_1 + m_2)\\\\V = \frac{(1225\times 9.5)_x \ + \ (1654 \times 8.6)_y}{m_1 + m_2} \\\\V = \frac{(1225\times 9.5)_x \ + \ (1654 \times 8.6)_y}{1225+ 1654} \\\\V= \frac{(11,637.5)_x \ + \ (14,224.4)_y}{2879} \\\\V = 4.04x \ + 4.94y\\\\V = 4.04i \ + 4.9 j[/tex]
The magnitude of the final velocity of the system is calculated as;
[tex]V = \sqrt{v_x^2 + v_y^2} \\\\V = \sqrt{(4.04)^2 + (4.9)^2} \\\\V = 6.35 \ m/s[/tex]
The change in the mechanical energy of the system;
[tex]\Delta K.E = K.E_f - K.E_i\\\\[/tex]
The initial kinetic energy of the cars before collision is calculated as;
[tex]K.E_i = \frac{1}{2} m_1u_1_x^2 \ + \frac{1}{2} m_1u_2_y^2 \\\\K.E_i = \frac{1}{2} (1225)(9.5)^2\ + \frac{1}{2} (1654)(8.6)^2\\\\K.E_i = 55,278.13_x \ + \ 61,164.92_y\\\\K.E_i = \sqrt{55,278.13^2 \ + \ 61,164.92^2} \\\\K.E_i = \sqrt{6,796,819,094.9} \\\\K.E_i = 82,442.82 \ J[/tex]
The final kinetic energy of the system;
[tex]K.E_f = \frac{1}{2} (m_1 + m_2)V^2\\\\K.E_f = \frac{1}{2} (1225 + 1654)(6.35)^2\\\\K.E_f = 58,044.24 \ J[/tex]
The change in kinetic energy is calculated as;
[tex]\Delta K.E = K.E_f -K.E_i\\\\\Delta K.E= (58,044.24) - (82,442.82)\\\\\Delta K.E = -24,398.58 \ J[/tex]
Apply the principle of work-energy theorem, to determine the distance traveled by the vehicles after collision;
[tex]W = \Delta K.E\\\\- \mu Fd = - 24,398.58\\\\\mu mgd= 24,398.58\\\\d = \frac{24,398.58}{\mu mg} \\\\d = \frac{24,398.58}{0.5 \times 9.8(1225 + 1654)} \\\\d = 1.73 \ m[/tex]
Thus, the distance traveled by the vehicles after collision is 1.73 m.
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A small particle with positive charge q = +4.25 x 10^-4C and mass m = 5.00 x 10^-5 kg is moving in a region of uniform electric and magnetic fields. The magnetic field is B=4.00 T in the +z-direction. The electric field is also in the +z-direction and has magnitude E = 60.0 N/C. At time t = 0 the particle is on the y-axis at y = +1.00 m and has velocity v = 30.0 m/s in the +x-direction. Neglect gravity.
a) What are the x-, y-, and z-coordinates of the particle at t = 0.0200 s?
b) What is the speed of the particle at t = 0.0200 s?
Answer:
a) r = (0.6 i- 2039 j ^ + 0.102 k⁾ m and b) vₓ = 30.0 m / s , v_{y} = 2.04 10⁵ m / s c) v_{z} = 1.02 10⁻¹m / s
Explanation:
a) To find the position of the particle at a given moment we must know the approximation of the body, use Newton's second law to find the acceleration
Fe + Fm = m a
a = (Fe + Fm) / m
the electric force is
Fe = q E k ^
Fe = 4.25 10-4 60 k ^
Fe = 2.55 10-2 k ^
the magnetic force is
Fm = q v x B
Fm = 4.25 10⁻⁴ [tex]\left[\begin{array}{ccc}i&j&k\\30&0&0\\0&0&49\end{array}\right][/tex]
fm = 4.25 10⁻⁴ (-j ^ 30 4)
fm 0 = ^ -5,10 10⁻² j
We look for every component of acceleration
X axis
aₓ = 0
there is no force
Axis y
ay = -5.10 10²/5 10⁻⁵ j ^
ay = -1.02 107 j ^ / s2
z axis
az = 2.55 10⁻² / 5 10⁻⁵ k ^
az = 5.1 10² k ^ m / s²
Having the acceleration in each axis we can encocoar the position using kinematics
X axis
the initial velocity is vo = 30 m / s and an initial position xo = 0
x = vo t + ½ aₓ t₂2
x = 30 0.02 + 0
x = 0.6m
Axis y
acceleration is ay = -1.02 10⁷ m / s², a starting position of i = 1m
y = I + go t + ½ ay t²
y = 1 + 0 + ½ (-1.02 10⁷) 0.02²
y = 1 - 2.04 10³
y = -2039 m j ^
z axis
acceleration is aza = 5.1 10² m / s², the position and initial speed are zero
z = zo + v₀ t + ½ az t²
z = 0 + 0 + ½ 5.1 10² 0.02²
z = 1.02 10⁻¹ m k ^
therefore the position of the bodies is
r = (0.6 i- 2039 j ^ + 0.102 k⁾ m
b) x axis
since there is no acceleration the speed remains constant
vₓ = 30.0 m / s
Axis y
let's use the equation v = v₀ + [tex]a_{y}[/tex] t
[tex]v_{y}[/tex] = 0 + -1.02 10⁷ 0.02
v_{y} = 2.04 10⁵ m / s
z axis
v_{z} = vo + az t
v_{z} = 0 + 5.1 10² 0.02
v_{z} = 1.02 10⁻¹m / s
A) The coordinates of the particle in x, y and z axis is written as :
( 0.6 i - 2039 j + 0.102 k )m ( i.e. x = 0.6, y = 2039, z = 0.102 )
B) The speed of the particle at t = 0.0200 s
x-axis = 30 m/s y-axis = 2.04 * 10⁵ m/s z-axis = 1.02 * 10⁻¹ m/sA ) Determine the coordinates of the particle in x, y and z axis
applying Newton's second law ; a = ( Fe + Fm ) / m
where : Fe = 2.55 * 10⁻² k
The magnetic force ( Fm ) = qv * B = [tex]\left[\begin{array}{ccc}i&j&k\\30&0&0\\0&0&49\end{array}\right][/tex]
∴ Fm = 0
Acceleration in each axis ( x , y and z )
Ax = 0 because there is no force
Ay = - 1.02 * 10⁷ j/s²
Az = 5.1 * 10² km/s²
i) For the x- axis
x = Vot + ½ * Ax * t²
Vo = 30 m/s
t = 0.02
Ax = 0
∴ x - coordinate of the particle = 0.6m
ii) For the y-axis
y = I + Vo*t + ½ *Ay* t²
y = - 2039 m
iii) For the z-axis
z = zo + v₀* t + ½ * Az* t²
z = 0.102 m .
B ) Determine the speed of the particle in all three axis
Vx = 30 m/s
Vy = v₀ + Ay * t
= 30 + - 1.02 * 10⁷ * 0.020
= 2.04 * 10⁵ m/s
Vz = Vo + Az* t
= 1.02 * 10⁻¹
Hence we can conclude that The coordinates of the particle in x, y and z axis is written as : ( 0.6 i - 2039 j + 0.102 k )m ( i.e. x = 0.6, y = 2039, z = 0.102 )The speed of the particle at t = 0.0200 s
x-axis = 30 m/s y-axis = 2.04 * 10⁵ m/s z-axis = 1.02 * 10⁻¹ m/sLearn more : https://brainly.com/question/19915685
floating airboat conditions
Answer:
They are used in icy conditions.
Explanation:
They can also go over grassy plains and mucky swamps.
Which orbit has the highest energy?
n = 1
n = 2
n = 3
Answer:
3
Explanation:
The closer an orbit is to the nucleus the fewer energy
A car starts its motion from rest and accelerates with an acceleration of 3m/s2 speed reaches to 20.0m/sec. Find time interval during this motion.
Answer:
The time interval is [tex]t = 6.667 \ s[/tex]
Explanation:
From the question we are told that
The acceleration of the car is [tex]a = 3 m/s^2[/tex]
The velocity of the car is [tex]v= 20.0 \ m/s[/tex]
The final velocity of the car can be mathematically represented as
[tex]v =u + at[/tex]
Now since the car start from rest u = 0 So
[tex]v = 0 + at[/tex]
[tex]v = at[/tex]
substituting values
[tex]20 = 3 t[/tex]
[tex]t = \frac{20}{3}[/tex]
[tex]t = 6.667 \ s[/tex]
A charge of +80 µC is placed on the x axis at x = 0. A second charge of –50 µC is placed on the x axis at x = 50 cm. What is the magnitude of the electrostatic force on a third charge of 4.0 µC placed on the x axis at x = 30 cm?
Answer:
Explanation:
77
The magnitude of the electrostatic force on a third charge of 4.0 µC placed on the x-axis at x = 30 cm is equal to 77 N.
What is coulomb's law?According to Coulomb’s law, the force of repulsion or attraction between two charged bodies is the multiplication of their charges and is inversely proportional to the square of the distance between them.
The magnitude of the electric force can be written as follows:
[tex]F =k \frac{q_1q_2}{r^2}[/tex]
where k has a value of 9 × 10⁹ N.m²/C².
Given, the first charge, q₁ = + 80 ×10⁻⁶ C
The second charge , q₂ = - 50 × 10⁻⁶ C
The third charge, q₃ = + 4 ×10⁻⁶ C
The distance between these two charges (q₁ and q₃) , r = 0.30 m
The distance between these two charges (q₂ and q₃) , r = 0.20m
The magnitude of the electrostatic force on the third charges will be:
[tex]F =9\times 10^9 (\frac{80\times 10^{-6}C\times 4 \times 10^{-6}C}{(0.30)^2} ) +9\times 10^9 (\frac{50\times 10^{-6}C\times 4 \times 10^{-6}C}{(0.20)^2} )[/tex]
F = 77 N
Therefore, the magnitude of the electrostatic force is equal to 77 N.
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make sure your response is 3-5 sentences.
What are some of the extreme conditions in space that challenge manned space exploration?
Answer: please see below
Explanation:
A manned space exploration is defined as the exploration of individuals --- astronauts in space using a spacecraft as a vehicle and are responsible for operating its controls
The extreme conditions in space that challenge manned space exploration is as follows.
1. extreme loud sound waves cause by the launch of spacecraft which can shatter the spacecraft
2. extreme Temperatures in space ranging from extreme hot temperatures (near the sun) to extreme cold temperatures ( below freezing point out of space.
3.micrometeorite showers responsible for sandblasting can damage spacecraft.
4.Ultra violet Radiation which can alter the control unit of the spacecraft
Because of theses extreme conditions that pose challenges to space explorations, necessary precautions should be taken into consideration to be able to overcome such challenges. These precautions include building the spacecraft and the control unit in such a way that can resist these harmful conditions, also taking in mind safe escape routes for the astronauts in case of failures.
5.Calculate the entropy changes for the following processes:(a)Melting of one mole of tin at its melting point, 213 ᵒC; ΔHfus = 7.029 kJ/mol(b)Evaporation of one mole of liquid carbon dioxide at its boiling point, 216.6 K. ΔHvap = 15.326 kJ/mol
Answer:
a) ΔS = 14.46 J/k
b) ΔS = 70.76 J/k
Explanation:
The general formula to calculate the entropy change accompanied with a process is:
ΔS = ΔQ/T
where,
ΔS = entropy change for the process
ΔQ = Heat Transfer during the process
T = Absolute Temperature during the process
a)
In this case the heat transfer will be given as:
ΔQ = (ΔHfus)(N)
where,
ΔHfus = Molar Heat of Fusion of Tin = 7.029 KJ/mol
N = No. of moles of tin = 1 mol
Therefore,
ΔQ = (7.029 KJ/mol)(1 mol)
ΔQ = 7.029 KJ = 7029 J
and the absolute temperature is:
T = 213°C +273 = 486 k
using these values in the entropy formula, we get:
ΔS = 7029 J/486 k
ΔS = 14.46 J/k
b)
In this case the heat transfer will be given as:
ΔQ = (ΔHvap)(N)
where,
ΔHvap = Molar Heat of Vaporization of Carbon Dioxide = 15.326 KJ/mol
N = No. of moles of Carbon Dioxide = 1 mol
Therefore,
ΔQ = (15.326 KJ/mol)(1 mol)
ΔQ = 15.326 KJ = 15326 J
and the absolute temperature is:
T = 216.6 k
using these values in the entropy formula, we get:
ΔS = 15326 J/216.6 k
ΔS = 70.76 J/k
a. If the object is not moving relative to the surface it's in contact with. the friction force is static friction. Draw a free-body diagram of the object. The direction of the friction force is such as to oppose sliding of the object relative to the surface.
b. If the object is slicing relative to the surface, then kinetic friction is acting. From Newton's second law, find the normal force n. The friction force is then directed opposite to the motion, and its magnitude is fk= µkn.
c. If the object is rolling along the surface, then rolling friction is acting. From Newton's second law. find the normal force n. The friction force is then directed opposite to the motion. and its magnitude is fr= µrn.
Answer:
a. The free body diagram for this object has been attached. It shows all the forces acting on the body at rest, including the friction force in the opposite direction to sliding of the object (assume it's left to right).
b. Since the object is in contact with the surface, there is a normal force acting on both of them and is equal to the weight exerted by each. This perpendicular force is defined by Newton's second law of motion.
c. The force of friction always acts in a direction opposite to the direction of motion of the body. F = mg ('a' for acceleration is replaced by 'g' gravity because acceleration in this case is just gravity).
Hope that answers the question, have a great day!
A metallic circular plate with radius r is fixed to a tabletop. An identical circular plate supported from above by a cable is fixed in place a distance d above the first plate. Assume that dd is much smaller than r. The two plates are attached by wires to a battery that supplies voltage V.
A)What is the tension in the cable? Neglect the weight of the plate.
Express your answer in terms of the variables d, r, V, and constants ϵ0, π.
B)The upper plate is slowly raised to a new height 2d. Determine the work done by the cable by integrating ∫(from d to 2d) F(z)dz, where F(z) is the cable tension when the plates are separated by a distance z.
Express your answer in terms of the variables d, r, V, and constants ϵ0, π.
C)Compute the energy stored in the electric field before the top plate was raised.
Express your answer in terms of the variables d, r, V, and constants ϵ0, π.
D)Compute the energy stored in the electric field after the top plate was raised.
Express your answer in terms of the variables d, r, V, and constants ϵ0, π.
E)Is the work done by the cable equal to the change in the stored electrical energy? If not, why not?
a)The work done in separating the plates is equal to energy change in the plates.
b)The work done in separating the plates is equal to the magnitude of the energy change in the plates. This does not mean that the work done is equal to the change in the energy stored in the plates. The work done on the plates is positive but the plates lose energy. The plates are connected to the battery, so the potential difference across them remains constant as they are separated. Therefore charge is forced off of the plates through the battery, which does work on the battery.
Answer:
the tension in the cable is [tex]\mathbf{F = \frac{\pi E_o v^2r^2}{2d^2}}[/tex]
the work done by the cable is [tex]\mathbf{W= \frac{\pi E_ov^2r^2}{4d}}[/tex]
Explanation:
A)
If we have two circular plate supported by a cable at a fixed distance, then the electric field formed between the two plate of the capacitor can be represented by the equation.
[tex]\mathbf{E = \frac{voltage \ \ V}{distance \ \ d}}[/tex]
However; the net electric field i.e the sum of the electric filed produced is represented as:
[tex]\mathbf{E' = \frac{E}{2}} \\ \\ \mathbf{E' = \frac{V}{2d}}[/tex]
So, if we assume that the lower plate and the upper plate possess the charge +q and -q respectively. Then, the tension of the cable which is the same as Force F can be written as:
[tex]\mathbf{F = q* E'}[/tex]
[tex]\mathbf{F = \frac{q*v}{2d}}[/tex] ----- equation (1)
Also ; we know that
[tex]\mathbf{C = \frac{q}{v}= \frac{E_oA}{d}}[/tex]
[tex]\mathbf{\frac{q}{v}= \frac{E_o \pi r^2}{d}} \ \ \ \ \ \mathbf{since \ A = \pi r^2}[/tex]
[tex]\mathbf{{q}= \frac{\pi E_o {v} r^2}{d}}[/tex] ----- equation (2)
Replacing equation 3 into equation (2); we have:
[tex]\mathbf{F = \frac{\pi E_o vr^2}{d}* \frac{v}{2d}}[/tex]
[tex]\mathbf{F = \frac{\pi E_o v^2r^2}{2d^2}}[/tex]
Therefore, the tension in the cable is [tex]\mathbf{F = \frac{\pi E_o v^2r^2}{2d^2}}[/tex]
B)
Assume that the upper plate is displaced by dz in an upward direction ; Then we can express the workdone by the tension as :
[tex]\mathbf{dW = T *dz} \\ \\ \mathbf{dW = F*dz} \\ \\ \mathbf{dW = \frac{\pi E_o v^2r^2}{2z^2}dz }[/tex]
The net workdone to raise the plate from separation d to 2d is:
[tex]\mathbf{W = \int\limits^{2d}_{2zd} {dw} = \frac{\pi E_ov^2r^2}{2} \int\limits^{2d}_d \frac{dz}{z^2} }[/tex]
[tex]\mathbf{W= \frac{\pi E_ov^2r^2}{2} [-\frac{1}{z}]^{2d}_d }[/tex]
[tex]\mathbf{W= - \frac{\pi E_ov^2r^2}{2} [\frac{1}{2d}-\frac{1}{d}]}[/tex]
[tex]\mathbf{W= - \frac{\pi E_ov^2r^2}{2} [\frac{-1}{2d}]}[/tex]
[tex]\mathbf{W= \frac{\pi E_ov^2r^2}{4d}}[/tex]
the work done by the cable is [tex]\mathbf{W= \frac{\pi E_ov^2r^2}{4d}}[/tex]
C) To calculate the energy stored in the Electrical energy Capacitor before the top plate is raised ; we have:
[tex]\mathbf{U_i = \frac{1}{2}Cv^2} \\ \\ \mathbf{U_i = \frac{1}{2}(\frac{E_oA}{d})v^2} \\ \\ \mathbf{U_i = \frac{1}{2}(\frac{E_o \pi r^2}{d})v^2} \\ \\ \mathbf{U_i = \frac{E_o \pi r^2 v^2}{2d}} }[/tex]
D) The energy stored in the plate after the the top plate was raised is as follows:
[tex]\mathbf{U_f = \frac{1}{2}C'v^2} \\ \\ \mathbf{U_f = \frac{1}{2}(\frac{E_oA}{2d})v^2} \\ \\ \mathbf{U_f = \frac{1}{2}(\frac{E_o \pi r^2}{2d})v^2} \\ \\ \mathbf{U_f = \frac{E_o \pi r^2 v^2}{4d}} }[/tex]
E) Yes, work done by the cable equal to the change in the stored electrical energy. The Difference in energy stored before and after the top plate is raised:
[tex]\mathbf{U_i-U_f} = \mathbf{\frac{E_o \pi r^2 v^2}{2d}} }} - \mathbf {\frac{E_o \pi r^2 v^2}{4d}} }}[/tex]
[tex]\mathbf{U_i-U_f}= \mathbf {\frac{E_o \pi r^2 v^2}{4d}} }}[/tex]
Thus;
b)The work done in separating the plates is equal to the magnitude of the energy change in the plates. This does not mean that the work done is equal to the change in the energy stored in the plates.
What do you mean by peltier coefficient?
Answer:
The Peltier coefficient is a measure of the amount of heat carried by electrons or holes
Explanation:
Answer:
The Peltier coefficient is a measure of the amount of heat carried by electrons or holes.
Explanation:
A framed picture hangs from two cords attached to the ceiling.
A picture of a picture frame hanging by two cables at the center of the frame at the same length and angle from the vertical.
Which shows the correct free body diagram of the hanging picture?
A free body diagram with two force vectors, the first pointing downward labeled F Subscript g Baseline, the second pointing upward labeled F Subscript N Baseline.
A free body diagram with three force vectors, the first pointing south labeled F Subscript p Baseline, the second pointing northeast labeled F Subscript T Baseline, and the third pointing northwest labeled F Subscript N.
A free body diagram with three force vectors, the first pointing south labeled F Subscript g Baseline, the second pointing northeast labeled F Subscript T Baseline and the third pointing northwest labeled F Subscript T.
Answer:
Answer is C
Explanation:
I just did it E2020
A free body diagram with three force vectors, the first pointing south labeled F Subscript g Baseline, the second pointing northeast labeled F Subscript T Baseline and the third pointing northwest labeled F Subscript T. Thus option C is correct.
What is free body diagram?Free body diagram is defined as a drawing of an interesting object with all of its surroundings removed and the forces at work on the subject's body clearly depicted.
It is also defined as a diagram that reduces the loads and moments acting on a component or system of components.
Free body diagrams are streamlined depictions of an object and the force vectors operating on it in a problem. Since the diagram will depict this body without its surroundings, it is "free" of its environment.
Thus, a free body diagram with three force vectors, the first pointing south labeled F Subscript g Baseline, the second pointing northeast labeled F Subscript T Baseline and the third pointing northwest labeled F Subscript T. Thus option C is correct.
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Mark puts an object that has a mass of 58.2kg in a cart that has a mass of 73.00 kg.
Answer:
131.2 kg, to have the same precision as 58.2 kg
Explanation: