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REY [17]
3 years ago
5

2. A 3.0 kg block rests on a level surface. The coefficient of static friction is µs = 0.50, and the coefficient of kinetic fric

tion is µk = 0.20. A horizontal force F is applied to the block. As F is increased, the block begins moving. Describe how the force of friction varies as F increases from the moment the block is at rest to when it begins moving. Indicate how you could determine the force of friction at each value of F—before the block starts moving, at the point it starts moving, and after it is moving. Show your work.
Physics
1 answer:
yan [13]3 years ago
6 0
Friction force is equal to the applied force F until the block starts moving. Hence, it will increase until the maximum value of 3.0*10*0.50=150 N. 

While it's moving, the friction will be 3.0*10*0.2 = 60 N, constant. The rest of the applied force will accelerate the block.

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A research vessel is mapping the bottom of the ocean using sonar. It emits a short sound pulse called "ping" downward. The frequ
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Answer:

d = 4180.3m

wavelengt of sound is 0.251m

Explanation:

Given that

frequency of the sound is 5920 Hz

v=1485m/s

t=5.63s

let d represent distance from the vessel to the ocean bottom.

an echo travels a distance equivalent to 2d, that is to and fro after it reflects from the obstacle.

velocity=\frac{distance}{time}\\\\ v=\frac{2d}{t} \\\\vt=2d\\\\d=\frac{vt}{2}

d=\frac{1485*5.63}{2}\\d= 4180.3m

wavelengt of sound is \lambda = v/f

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7 0
3 years ago
Block b rests upon a smooth surface. if the coefficients of static and kinetic friction between a and b are μs = 0.4 and μk = 0.
aliina [53]

Given

Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient of kinetic friction µk = 0.3. If a force P is applied to the body, no relative motion will take place until the applied force is equal to the force of friction Ff, which is acting opposite to the direction of motion. Magnitude of static force of friction between block A and block B, Fs = µsN, where N is reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA + mB)a,

 

6 = (20 / 32.2 + 50 / 32.2)a

 

2.173a = 6

 

A = 2.76 ft/s^2

 

To check slipping occurs between block A and block B, consider block A:

P – Ff = mAaA

6 – Ff = 1.71

Ff = 4.29 lb

 

And also,

N = wA. We know static friction,

Fs = µsN

Fs = 0.4 x 20

Fs = 8lb

Frictional force is less than static friction. Ff < Fs

<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of block B aB = 2.76 ft/s^2</span>

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3 years ago
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3 years ago
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Degger [83]

Answer:

7.16 m /s

Explanation:

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#need any question answered within secs/mins, hit me up and I got you!

#branilest

:) <3

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