Answer:
The temperature rise [tex]\Delta T = 0.3088 \ ^oC[/tex]
Explanation:
From the question we are told that
The speed is [tex]v = 60 \ km /h = 16.67 \ m/s[/tex]
Generally according to the law of energy conservation we have that
The kinetic energy = increase in the internal energy of the car
i,e [tex]\frac{1}{2} * m * v^2 = m * c_p * \Delta T[/tex]
Here [tex]c_p[/tex] of iron is [tex]c_p = 450 \ J/kg K[/tex]
So
[tex]0.5 * 16.67^2 = 450 * \Delta T[/tex]
=> [tex]\Delta T = 0.3088 \ ^oC[/tex]
The temperature increase of each car will be 0.3088 °C.
What is temperature?Temperature directs to the hotness or coldness of a body. In clear terms, it is the method of finding the kinetic energy of particles within an entity. Faster the motion of particles more the temperature.
The given data in the problem is;
v is the velocity= 60 km/h=16.67 m/sec
ΔT is the temperature difference=?
[tex]\rm c_P[/tex] is the specific heat of iron=450 J/KgK
From the law of energy conservation of the energy, the kinetic energy is equal to the increase in the internal energy of the car.
[tex]\rm KE=\triangle E \\\\\ \frac{1}{2} mv^2=mc_p \triangle T \\\\ \triangle T=\frac{v^2}{2C_p} \\\\\ \triangle T=\frac{(16.67)^2}{2\times 450} \\\\ \triangle T=0.3088 ^0\C[/tex]
Hence the temperature increase of each car will be 0.3088 °C.
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A tire swing hanging from a branch reaches nearly to the ground. How could you estimate the height of the branch using only a stopwatch?
Answer:
A stop watch measures time and the number of seconds making an osculation is the period T
Which is T = 2π√(L/g)........solving for L (the length of the pendulum),
Then the L = T²x g/( 4πr²) substitute and then u find length or height of the swing
A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 58.9 cm ( 0.589 m) and the flow speed of the petroleum is 12.1 m/s. At the refinery, the petroleum flows at 6.29 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?
Answer:
Explanation:
The volume rate of flow = a x v where a is cross sectional area of pipe and v is velocity of flow
putting the values
π x .2945² x 12.1
= 3.3 m³ /s
To know the pipe's diameter at the refinery we shall apply the following formula
a₁ v₁ = a₂ v₂
a₁ v₁ and a₂ v₂ are volume rate of flow of liquid which will be constant .
3.3 = a₂ x 6.29
a₂ = .5246 m³
π x r² = .5246
r = .4087 m
= 40.87 cm
diameter
= 81.74 cm
Consider a vacuum-filled parallel plate capacitor (no dielectric material between the plates). What is the ratio of conduction current Jc to displacement current Jd at 10 MHz
Answer: 1798
Explanation:
Given that there is no dielectric material between the plates,
the permittivity of free space = 9 × 10^-12 f/m
The frequency F = 10 MHz
The ratio of conduction current JC to the displacement current Jd is also known as loss tangent.
Please find the attached file for the solution
The x component of vector is 8.7 units, and its y component is -6.5 units. The magnitude of is closest to
Answer:
F = 10.86 units
Explanation:
The magnitude of a vector in terms of the magnitude of its rectangular components is given by the following formula:
F = √(Fₓ² + Fy²)
where,
F = Magnitude of the Vector = ?
Fₓ = magnitude of the x-component of vector = 8.7 units
Fy = magnitude of y component of vector = - 6.5 units
Therefore, using these values in the equation, we get:
F = √[(8.7 units)² + (- 6.5 units)²]
F = √(117.94 units²)
F = 10.86 units
A family made the observation that more expensive brands of popcorn seem to produce more popped kernels than cheaper brands of popcorn. In order to test this hypothesis, the family conducted an experiment in which they followed the procedures below: The family bought 2 differing brands of popcorn to test: Wilbur Bockenreder Popcorn and PopWhisper. They measured 1 cup of kernels and popped 5 bowls of each type of popcorn. They ensured they popped the popcorn for the same amount of time, used the same wattage microwave, and did not have any other ingredients added to the kernels (butter, etc). They counted and recorded the number of unpopped kernels in each of the 10 bowls. As they analyzed the data collected, they were able to conclude that their initial hypothesis was supported. They formed the conclusion that the more expensive brand of popcorn popped more kernels than the cheaper brand of popcorn. From this experiment, let's answer some questions. Identify the independent variable.
Answer:
The different types of corn used.
Explanation:
Independent variable: In research methods, the term "independent variable" is determined as a variable that is being manipulated, changed, or altered in an experiment by the experimenter in order to see its effect on the dependent variable. The changes in the dependent variable in an experiment depends on the independent variable directly.
The independent variable in the popcorn experiment is brand of popcorn used.
INDEPENDENT VARIABLE:Independent variable in an experiment is the variable that the experimenter changes or manipulates in order to bring about a response. According to this question, a family is conducting an experiment to test which brand of popcorn seem to produce more popped kernels. They used two brands of popcorn as follows: Wilbur Bockenreder Popcorn and PopWhisper.However, the brand of popcorn was changed in this experiment, hence, the brand of the popcorn is the independent variable.
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The temperature of a plastic cube is monitored while the cube is pushed 8.6 m across a floor at constant speed by a horizontal force of 19 N. The monitoring reveals that the thermal energy of the cube increases by 120 J. What is the increase in the thermal energy of the floor along which the cube slides
Answer:
Answer:
43.4J
Explanation:
We know that
Work done = total heat energy
But work done is force x distance
=> F = 19 x8.6 = 163.4 J
So the total heat. Will be Heat of cube + heat of floor = 163.4J
So that heat of floor will now be
floor = 163.4 J - 120 J = 43.4 Joules
Explanation:
A monatomic gas is adiabatically compressed to 0.250 of its initial volume. Do each of the following quantities change?
A) What is rms speed?
B) What is the mean free path?
C) What is the thermal energy of gas?
D) What is the molar specific at a constant volume?
Answer:
Given that
V2/V1= 0.25
And we know that in adiabatic process
TV^န-1= constant
So
T1/T2=( V1 /V2)^ န-1
So = ( 1/0.25)^ 0.66= 2.5
Also PV^န= constant
So P1/P2= (V2/V1)^န
= (1/0.25)^1.66 = 9.98
A. RMS speed is
Vrms= √ 3RT/M
But this is also
Vrms 2/Vrms1= (√T2/T1)
Vrms2=√2.5= 1.6vrms1
B.
Lambda=V/4π√2πr²N
So
Lambda 2/lambda 1= V2/V1 = 0.25
So the mean free path can be inferred to be 0.25 times the first mean free path
C. Using
Eth= 3/2KT
So Eth2/Eth1= T2/T1
So
Eth2= 2.5Eth1
D.
Using CV= 3/2R
Cvf= Cvi
So molar specific heat constant does not change
The rms speed of the system is 1.6 while Mean free path is 0.25 times the first mean free path.
Given that
[tex]\bold{\dfrac {V_2}{V_1}= 0.25}[/tex]
In adiabatic process
[tex]\bold {TV^\gamma^-^1}[/tex]= constant
So
[tex]\bold {\dfrac {T1}{T2}=(\dfrac {V1 }{V2})^ \gamma^-^1}}\\[/tex]
So,
[tex]\bold {\dfrac {T_1} {T_2} = (\dfrac { 1}{0.25})^ 0^.^6^6= 2.5}[/tex]
for pressure,
[tex]\bold {PV^\gamma = \ Constant }[/tex]
So
[tex]\bold {\dfrac {P1}{P2}=(\dfrac {V1 }{V2})^ \gamma}}\\\\\bold {\dfrac {P1}{P2}=(\dfrac {1}{0.25})^ 0^.^6^6} = 9.98 } }[/tex]
A. rms speed can be calculated as
[tex]\bold {\dfrac {Vrms 2}{Vrms1}= \sqrt {T2T1})}\\\\\bold {Vrms2 =\sqrt {2.5} = 1.6\ Vrms1 }[/tex]
B. The mean free path can be calculated as
[tex]\bold {\dfrac {\lambda_1 }{\lambda_2} = \dfrac {V_1}{V_2} = 0.25 \\ }[/tex]
Mean free path is 0.25 times the first mean free path.
C.
[tex]\bold {Eth= \dfrac {3}{2}kT}}\\\\\bold {\dfrac {E_t2}{E_t1} = \dfrac {T_2}{T_1}}\\\\\bold {E_t2= 2.5\ E_t1}[/tex]
D. the molar specific at a constant volume can be calculated a using,
[tex]\bold {CV= \dfrac 3{2}R }[/tex]
[tex]\bold {Cv_f= Cv_i }[/tex]
So, molar specific heat constant will not change.
Therefore, the rms speed of the system is 1.6 while Mean free path is 0.25 times the first mean free path.
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Simple Harmonic Motion: The quartz crystal in a digital watch has a frequency of 32.8 kHz. What is its period of oscillation
Answer:
Time Period of Oscillation = 3.04 x 10⁻⁵ s
Explanation:
The time period of oscillation and the the frequency of oscillation are two inter linked quantities. They are actually the reciprocals of each other. So, the time period of oscillation of quartz crystal is given by the following formula:
Time Period of Oscillation = 1/Frequency of Oscillation
where,
Frequency of Oscillation = 32.8 KHz
Frequency of Oscillation = 32800 Hz
Therefore,
Time Period of Oscillation = 1/32800 Hz
Time Period of Oscillation = 3.04 x 10⁻⁵ s
A cheetah can accelerate from rest to 25.0 m/s in 6.22 s. Assuming constant acceleration, how far has the cheetah run in this time
Answer:
The distance covered by the cheetah during the motion is 77.75 m
Explanation:
Given;
initial velocity of cheetah, u = 0
final velocity of cheetah, v = 25 m/s
time of acceleration, t = 6.22 s
Apply kinematic equation;
[tex]s = (\frac{v+u}{2} )t[/tex]
where;
s is the distance covered by the cheetah during the motion
[tex]s = (\frac{v+u}{2} )t\\\\s = (\frac{25+0}{2} )6.22\\\\s = 77.75 \ m[/tex]
Therefore, the distance covered by the cheetah during the motion is 77.75 m
Two point charges totaling 8 μC exert a repulsive force of 0.15 N on one another when separated by 0.5 m. What is the charge on each? A. 4.0x10-6 C 4.0x10-6 C B. 7.4x10-6 C 0.6x10-6 C C. 6.6x10-6 C 1.4x10-6 C D. 5.0x10-6 C 3.0x10-6 C
Answer:
B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C
Explanation:
From Coulomb's Law the electrostatic repulsive force is given by the following formula:
F = kq₁q₂/r²
where,
F = Repulsive Force = 0.15 N
k = Coulomb's Constant = 9 x 10⁹ N.m²/C²
q₁ = Magnitude of 1st Charge = ?
q₂ = Magnitude of 2nd Charge = ?
r = Distance between Charges = 0.5 m
Therefore,
0.15 N = (9 x 10⁹ N.m²/C²)q₁q₂/(0.5 m)²
q₁q₂ = (0.15 N)(0.5 m)²/(9 x 10⁹ N.m²/C²)
q₁q₂ = 4.17 x 10⁻¹²
q₁ = (4.17 x 10⁻¹²)/q₂ -------------------- equation (1)
The sum of charges is given as:
q₁ + q₂ = 8 μC
q₁ + q₂ = 8 x 10⁻⁶
using equation (1):
(4.17 x 10⁻¹²)/q₂ + q₂ = 8 x 10⁻⁶
(4.17 x 10⁻¹²) + q₂² = 8 x 10⁻⁶ q₂
q₂² - (8 x 10⁻⁶) q₂ + (4.17 x 10⁻¹²) = 0
Solving this quadratic equation:
q₂ = 7.4 x 10⁻⁶ C (OR) q₂ = 0.56 x 10⁻⁶ C
q₂ = 7.4 μC (OR) q₂ = 0.6 μC
Therefore,
q₁ = (4.17 x 10⁻¹² C)/(7.4 x 10⁻⁶ C)
q₁ = 0.6μC
Now, if we solve with q₂ = 0.6 μC, we will get q₁ = 7.4 μC.
Therefore, the correct option will be:
B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C
How large an expansion gap should be left between steel railroad rails if they may reach a maximum temperature 36.0°C greater than when they were laid? Their original length is 26.0 m.
Answer:
11 mm
Explanation:
Original length L' = 26 m
Increase in temperature dT = 36.0°C
The gap l =?
The coefficient of linear expansion for steel & = 12 × 10^−6 per °C
The gap will be gotten from the equation of linear expansion.
l = L' x & x dT
substituting, we have
L = 26 x 12 × 10^−6 x 36
L = 0.011 m = 11 mm
How far can a person run in 0.25hr if he or she runs at a average speed of 16km/he?
Answer:
4km
Explanation:
d = st
d=0.25*16
d=4km
You are looking down on a single coil in a constant magnetic field B = 1.2 T which points directly into of the screen. The dimensions of the coil go from a = 8 cm and b = 17 cm, to a* = 16 cm and b* = 22 cm in t = 0.04 seconds. If the coil has resistance that remains constant at 1.2 ohms. What would be the magnitude of the induced current in amperes?
Answer:
The current is [tex]I = 0.5425 \ A[/tex]
Explanation:
From the question we are told that
The magnetic field is [tex]B = 1.2 \ T[/tex]
The first length is [tex]a = 8 \ cm = 0.08 \ m[/tex]
The second length is [tex]a^* = 16 \ cm = 0.16 \ m[/tex]
The first width is [tex]b = 17 \ cm = 0.17 \ m[/tex]
The second width is [tex]b^* = 22 \ cm = 0.22 \ m[/tex]
The time interval is [tex]dt = 0.04 \ s[/tex]
The resistance is [tex]R = 1.2 \ \Omega[/tex]
Generally the first area is
[tex]A = a * b[/tex]
=> [tex]A = 0.08 * 0.17[/tex]
=> [tex]A = 0.0136 \ m^2[/tex]
The second area is
[tex]A^* = a^* * b^*[/tex]
=> [tex]A^* = 0.16 * 0.22[/tex]
=> [tex]A^* = 0.0352 \ m^2[/tex]
Generally the induced emf is mathematically represented as
[tex]\epsilon = - \frac{ B * [A^* - A]}{dt}[/tex]
This negative show that it is moving in the opposite direction of the motion producing it
=> [tex]|\epsilon | = \frac{ 1.2 * [ 0.0352-0.0135]}{0.04}[/tex]
=> [tex]|\epsilon | = 0.651 \ V[/tex]
The induced current is
[tex]I = \frac{|\epsilon|}{R}[/tex]
=> [tex]I = \frac{ 0.651}{1.2}[/tex]
=> [tex]I = 0.5425 \ A[/tex]
When your scalpel gets dull or is broken, it should be disposed of by
a. Placing in the broken glass/sharps container
b. Thrown in the trash
c. Kept in your dissection kit
How many grams of water (H2O) have the same number of oxygen atoms as 7.0 mol of oxygen gas
Answer:
126 g of water, H2O.
Explanation:
First, we'll begin by calculating the number of atoms in 7 mole oxygen gas.
This is illustrated below:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms. This means that 1 mole of O2 also contains 6.02×10²³ atoms.
Now, if 1 mole of O2 contains 6.02×10²³ atoms, then 7 moles of O2 will contain = 7 × 6.02×10²³ = 4.214×10²⁴ atoms.
Finally, we shall determine the mass of H2O that will contain 4.214×10²⁴ atoms.
This is illustrated below:
6.02×10²³ atoms is present in 1 mole of any substance contains
1 mole of H2O = (2x1) + 16 = 18 g
6.02×10²³ atoms is present in 18 g of H2O.
Therefore, 4.214×10²⁴ atoms will be present in = (4.214×10²⁴ × 18)/6.02×10²³
= 126 g of water, H2O.
Therefore, 126 g of water, H2O will contain the same number of oxygen atoms as 7.0 mol of oxygen gas.
7.0 mol of oxygen gas contain the same number of oxygen atoms as 250 g of water.
We want to know the mass of water (H₂O) that has the same number of oxygen atoms as 7.0 mol of oxygen gas (O₂).
First, we will calculate the number of oxygen atoms in 7.0 mol of O₂ considering the following relations.
1 mol of O₂ contains 6.02 × 10²³ molecules of O₂ (Avogadro's number).1 molecule of O₂ contains 2 atoms of O.[tex]7.0 mol O_2 \times \frac{6.02 \times 10^{23}molecule O_2 }{1molO_2} \times \frac{2atomO}{1molecule O_2} = 8.4 \times 10^{24} atomO[/tex]
Now, we want to calculate the mass of H₂O that contains 8.4 × 10²⁴ atoms of O. We will consider the following relations.
1 molecule of H₂O contains 1 atom of O.1 mol of H₂O contains 6.02 × 10²³ molecules of H₂O (Avogadro's number).The molar mass of H₂O is 18.02 g/mol.[tex]8.4 \times 10^{24} atomO \times \frac{1moleculeH_2O}{1atomO} \times \frac{1molH_2O}{6.02 \times 10^{23}moleculeH_2O } \times \frac{18.02gH_2O}{1molH_2O} = 250 gH_2O[/tex]
7.0 mol of oxygen gas contain the same number of oxygen atoms as 250 g of water.
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The velocity of any point on a rigid body is _________ to the relative position vector extending from the IC to the point. Group of answer choices
Answer:
The velocity of any point on a rigid body is ___Always perpendicular______ to the relative position vector extending from the IC to the point
Explanation:
This is because The instantaneous center (IC) of zero velocity for the rigid body is at the point in contact with ground. The velocity direction at any point on the body is always perpendicular to the line connecting the point to the IC.
The velocity of any point on a rigid body is : Perpendicular to the relative position vector extending from the IC to the point.
The IC ( instantaneous center ) is the point of the rigid body in contact with the ground when the rigid body is at zero velocity.
The velocity direction from any point of the rigid body will always be perpendicular to the ( IC ) since the IC is the reference point of the rigid body when the body is at rest ( zero velocity ).
Hence we can conclude that the velocity of any point on a rigid body will be perpendicular.
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Although the options related to your question is missing an accurate answer is provided within the scope of your question
A spring with a mass of 5 Kg has natural length 0.5m. A force of 35.6 N is required to maintain it stretched to a length of 0.5m. If the spring is stretched to a length of 0.5m and released with initial velocity 0, find the position of the mass at any time t. Here damping constant is zero.
Answer:
Explanation:
force constant of spring k = force / extension
= 35.6 / 0.5
k = 71.2 N / m
angular frequency ω of oscillation by spring mass system
[tex]\omega = \sqrt{\frac{k}{m} }[/tex]
where m is mass of the body attached with spring
Putting the values
[tex]\omega = \sqrt{\frac{71.2}{5} }[/tex]
ω = 3.77 radian / s
The oscillation of the mass will be like SHM having amplitude of 0.5 m and angular frequency of 3.77 radian /s . Initial phase will be π / 2
so the equation for displacement from equilibrium position that is middle point can be given as follows
x = .5 sin ( ω t + π / 2 )
= 0.5 cos ω t
= 0.5 cos 3.77 t .
x = 0.5 cos 3.77 t .
What does this picture show?
Answer:
d poor accuracy,poor precision
The diagram represents how much accurate and precise the result data is. It is clear from the image that, the data points have poor accuracy and good precision.
What is accuracy ?Accuracy of a result is the measure of the closeness of the experimental or calculated value to the true value or absolute value of a measurement. For a reproducible experiment, the results for each trials can be differ or close.
The closeness between values of a set of experiments is called precision of the results. Not all accurate value be can be precise. Similarly precision of values does not need to meet accuracy.
Here, the central dot indicates the true value. The green points are a set of experimental values. They are not so close to the true value and hence it is less accurate. However, all the points and closer together. Hence, they are precise. Therefore, option B is correct.
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Capacitor C1 is in series with capacitors C2 and C3 in parallel. Then three capacitor system is connected to battery with V0. Determine the charge stored by C1 when C1 = 20 μF, C2 = 10 μF, C3 = 30 μF, and V0 = 18 V.g
Answer:
Q₁ = 2.4 10⁻⁴ C
Explanation:
We have a circuit with several capacitors, let's find the equivalent capacitor of the parallel
[tex]C_{eq1}[/tex] = C₂ + C₃
C_{eq1} = (10 +30) 10⁻⁶
C_{eq1} = 40 10⁻⁶ F
There remains a series system between C₁ and C_{eq1}, let's find the equivalent capacitor
1/C_{eq2} = 1 / C₁ + 1 / C_{eq1}
1 /C_{eq2} = 1 / 20 10⁻⁶ + 1/40 10⁻⁶
1 / C_{eq2} = 0.075 10⁶
C_{eq2} = 13.33 10⁻⁶ F
let's use the relationship
V = Q / C_{eq2}
Q = V C_{eq2}
Q = 18 13.33 10⁻⁶
Q = 2.4 10⁻⁴ C
In a combination of capacitors in series the charge is constant, so the charge on C₁ is the same
Q₁ = 2.4 10⁻⁴ C
Where are the most reactive elements located on the periodic table.
Answer:
The most reactive elements are at the bottom left corner of the periodic table. Those are the elements that have the most active/most reactive. For example, lithium, sodium, and potassium all react with water!
Explanation:
Don't know what to explain :v
The spaceship Intergalactic landed on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.44 m. She sets the pendulum swinging, and her collaborators carefully count 1.10×102 complete cycles of oscillation during 2.00×102 s. What is the result
Answer:
18808.7 m/s^2
Explanation:
Given
Length of the pendulum L = 1.44 m
Number of complete cycles of oscillation n = 1.10 x 10^2
total time of oscillation t = 2.00 x 10^2 s
The period of the T = n/t
T = (1.10 x 10^2)/(2.00 x 10^2) = 0.55 ^-s
The period of a pendulum is gotten as
T = [tex]2\pi \sqrt{\frac{L}{g} }[/tex]
where g is the acceleration due to gravity
substituting values, we have
0.55 = [tex]2\pi \sqrt{\frac{1.44}{g} }[/tex]
0.0875 = [tex]\sqrt{\frac{1.44}{g} }[/tex]
squaring both sides of the equation, we have
7.656 x 10^-3 = 144/g
g = 144/(7.656 x 10^-3) = 18808.7 m/s^2
A person whose near-point distance is 45.5 cm wears a pair of glasses that are 2.1 cm from her eyes. With the aid of these glasses, she can now focus on objects 25 cm away from her eyes.
Required:
a. Find the focal length of her glasses.
b. Find the refractive power of her glasses.
Answer:
a
[tex]f = 0.4848 \ m[/tex]
b
[tex]p = 2.063 D[/tex]
Explanation:
From the question we are told that
The near point distance is [tex]k = 45.5 \ cm[/tex]
The distance of the glasses from the eye is [tex]y = 2.1 \ cm[/tex]
The distance of an object she can focus with the glass is [tex]i= 25 \ cm[/tex]
Generally the image distance is mathematically evaluated as
[tex]v = -(45.5 - 2.1)[/tex]
[tex]v = -43.4 \ cm[/tex]
Generally the object distance is mathematically represented as
[tex]u = (25 -2.1)[/tex]
[tex]u = 22.9 \ cm[/tex]
The negative sign tells us that the image was formed behind the eye
Generally the lens formula is mathematically represented as
[tex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/tex]
=> [tex]\frac{1}{f} = \frac{1}{22.9} + \frac{1}{ - 43.4}[/tex]
=> [tex]f = 48.48 \ cm[/tex]
converting to meters
[tex]f = 0.4848 \ m[/tex]
Thus the refractive power is mathematically represented as
[tex]p = \frac{1}{f}[/tex]
=> [tex]p = \frac{1}{0.4848 }[/tex]
=> [tex]p = 2.063 D[/tex]
1) A cyclist moves a distance of 2000 meters during a time of 20 minutes. How fast is the cyclist?
2) A car moves for 20 minutes on a road with constant speed. If he traveled 60 kilometers, what is his speed?
Answer:
1) speed of a cyclist = 100 m/min.
2) speed of a car moving = 3 km/min.
Explanation:
1) speed of a cyclist = distance over time
= 2000 m / 20 min
= 100 m/min.
2) speed of car moving = distance over time
= 60 km / 20 min
= 3 km/min.
The mass of the crate can also be adjusted by clicking on the More Features tab and then using the slider bar in the right panel. How does the maximum angle for which the crate can remain at rest on the ramp depend on the mass of the crate
Answer:
The maximum angle does not depend on the mass
Explanation:
This is because In as much as the force acting on the normal which is the the maximum force of static friction increases as the mass increases, the component of the force of gravity parallel to the ramp increases at the same rate. Thus The maximum angle is independent of the mass.
The rotating loop in an AC generator is a square 12.0 cm on a side. It is rotated at 70.0 Hz in a uniform field of 0.800 T. Calculate the following quantities as functions of time t, where t is in seconds. I need help with d and e.
(a) the flux through the loop
11.52cos(140πt) mT·m2
(b) the emf induced in the loop
5.067sin(140πt) V
(c) the current induced in the loop for a loop resistance of 2.50 Ω
2.03sin(140πt) A
(d) the power delivered to the loop
4.12sin2(140πt) W (wrong)
(e) the torque that must be exerted to rotate the loop
mN·m
Answer:
Explanation:
d )
power in an electrical loop = volt x current
= 5.067sin(140πt) x 2.03sin(140πt)
= 10.286 sin²(140πt) .
10.286 ( 1 - cos 280πt ) / 2
= 5.143 ( 1 - cos 280πt )
e )
Torque on a current carrying loop in a magnetic field
= M B sinθ
M is magnetic moment of coil , B is magnetic field , θ is angle between area vector of coil and direction of B .
Here magnetic moment of coil
= A i where A is area of coil and i is current .
= .12² x 2.03sin(140πt)
= .029 sin(140πt)
B = .8 T
MB = .8 x .029 sin(140πt)
= .0232 sin(140πt) x sin(140πt)
=.0232 sin²(140πt).
For a conductor in electrostatic equilibrium, which of the following properties are true?
a) Any excess charge is uniformly spread throughout the volume of the conductor.
b) The electric field inside is zero.
c) The entire conductor is at the same potential.
Answer:
b) The electric field inside is zero.
Explanation:
This is simply because the charges in the field will be in motion, thus this charges kind of spread around to make the conductor field zero.
A 1.5m wire carries a 7 A current when a potential difference of 87 V is applied. What is the resistance of the wire?
Answer:
Ohm's law states that I=V/R (Current=volts divided by resistance). Since we're looking for resistance, we'll rewrite it as R=V/I. Then just plug in the numbers; R=84/9, R= 9 1/3 or 28/3. The resistance of the wire is 9.33... or 9 1/3 ohm's, depending on how you wanna write it.
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Answer:
[tex]R\approx12.43 \,\, \Omega[/tex]
Explanation:
We can use Ohm's Law to find the resistance R of a wire that carries a current I under a given potential difference:
[tex]V=I\,\,R\\R = \frac{V}{I} \\R=\frac{87}{7} \\R\approx12.43 \,\, \Omega[/tex]
A plane moves 100 meters to the right in two seconds. What is its velocity?
0 m/s
O 30 m/s
50 m/s
O 100 m/s
200 m/s
Answer:
50 m/s
Explanation:
[tex]Distance = 100m\\Time = 2 secs\\\\Velocity = \frac{Distance}{Time} \\\\V = \frac{100m}{2s} \\\\= 50m/s[/tex]
Which describes the amplitude of a wave when it carries more ene
O It is higher.
O It is lower.
It is darker.
Oli is lighter
Answer:
It is higher.
Explanation:
Hello,
In this case, the quantity of energy that a wave is able to transfer is directly related its amplitude exhibiting that the higher the carried energy, the higher its amplitude and on the flip side, the lower the energy the lower the amplitude, therefore, the answer is it is higher. You can verify this on the attached picture.
Regards.
Answer:
It is higher.
Explanation:
If your front lawn is 21.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1350 new snowflakes every minute, how much snow, in kilograms, accumulates on your lawn per hour
Answer:
Mass of flake = 67.716 kg
Explanation:
Given:
Length of lawn = 21 ft
Width of lawn = 20 ft
Note:
Each snow flake mass = 1.90 mg
Find:
Weight of total flake in a minute
Computation:
Area of lawn = 21 × 22
Area of lawn = 440 ft²
Amount of flake per minute = 440 × 1350
Amount of flake per minute = 594000 flake/ minutes
Mass of flake = 594000 × 1.90 mg × 60 minutes
Mass of flake = 67,716,000 mg
Mass of flake = 67.716 kg