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Strike441 [17]
3 years ago
7

This problem uses the same concepts as Multiple-Concept Example 17, except that kinetic, rather than static, friction is involve

d. A crate is sliding down a ramp that is inclined at an angle 30.6 ° above the horizontal. The coefficient of kinetic friction between the crate and the ramp is 0.360. Find the acceleration of the moving crate.

Physics
1 answer:
coldgirl [10]3 years ago
8 0

Answer:1.94 m/s^2

Explanation:

Given

inclination \theta =30.6^{\circ}

Coefficient of kinetic Friction \mu =0.36

Forces on crate is

mg\sin \theta -f_r=ma

f_r=\mu mg\cos \theta

mg\sin \theta -\mu mg \cos \theta =ma

a=g\sin \theta -\mu g\cos \theta

a=1.94 m/s^2  

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The accompanying table shows measurements of the Hall voltage and corresponding magnetic field for a probe used to measure magne
aalyn [17]

0.125 mm . is the thickness of the sample.

<h3>What do you mean by hall voltage ?</h3>

The Hall effect is the creation of a voltage difference (the Hall voltage) across an electrical conductor, which is transverse to an applied magnetic field perpendicular to the current and an electric current in the conductor. Edwin Hall made the discovery in 1879.

We need to know the material's current, magnetic field, length, number of charge carriers, and area in order to calculate the Hall voltage. The Hall voltage is computed using the formula: v=IBlneA=(100A)(1.5T)(1.0102m)(5.91028/m3)(1.61019C)(2.0105m2)=7.9106V.

lof4

First we have to plot those point Then we can use some computer program to fit those point linearly to get slope

of that graph a and interception b. We already know, from theory, that Hall's voltage AVH and magnitude of

magnetic field B are connected as

ΔV_{H} =\frac{I}{nqt} B

where I is current trough probe, n is concentration of charge carriers, q = 1.6 • 10¯19 C is charge of charge

carries and t is thickness of the material. We have put the data from the problem on a graph and fitted linearly and

got

a = 100 μ\frac{V}{T}

b = —0.02  μV.

As we can see, our result are in agreement with theoretical assumptions because interception b is almost O, and a

is asked relation between Hall's voltage A VH and magnitude of magnetic field B. Then we can write

ΔVH =100X10^{-6} V/TB

(4) Then we can use result (4) and numbers from the textbook to calculate the thickness of the sample as

a=\frac{I}{nqt} \\t=\frac{I}{anq} \\t=\frac{.200A}{100X10^{-6}X 1.6 X10^{-19}X10^{26}  } \\t=0.125mm

To learn more about the hall voltage , Visit: brainly.com/question/19130911

#SPJ4

8 0
1 year ago
What is a successful result of science
Vlada [557]
I think the answer is discovery.
7 0
2 years ago
If the body temperature of a person is recorded as 37°C, what is the person’s body temperature in K? Round your answer to the ne
melisa1 [442]
K= 37°C+273.15
K= 310.15
Round to the nearest whole number
310K
6 0
2 years ago
A 4.2-kg block of ice originally at 263 K is placed in thermal contact with a 13.7-kg block of silver (cAg = 233 J/kg-K) which i
Natalka [10]

Answer:

an example was provided below. switch out my number of 13.4 and plug in 13.7kg and solve

Explanation:

so your equation would be (4.2)(2030)(10)+(4.2)(3.33x10^5) + (4.2)(4186)(T-273)=(13.7)(233)(1075-T)

3 0
3 years ago
A 1.6-kg block is attached to the end of a 2.0-m string to form a pendulum. The pendulum is released from rest when the string i
irinina [24]

Answer:

1002.2688 m/s

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

h = The length of a string = 2 m

m = Mass of block = 1.6 kg

m_2 = Mass of bullet = 0.01 kg

Here, the potential energy of the fall will balance the kinetic energy of the bullet

mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 2}\\\Rightarrow v=6.26418\ m/s

Velocity of block is 6.26418 m/s

As the momentum of system is conserved we have

mv=m_2u\\\Rightarrow u=\dfrac{mv}{m_2}\\\Rightarrow u=\dfrac{1.6\times 6.26418}{0.01}\\\Rightarrow u=1002.2688\ m/s

The magnitude of velocity just before hitting the block is 1002.2688 m/s

7 0
3 years ago
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