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gogolik [260]
3 years ago
14

A 0.020-kg bullet traveling at a speed of 300 m/s embeds in a 1.0-kg wooden block resting on a horizontal surface. The block sli

des horizontally 4.0 m on a surface before stopping. What is the coefficient of friction between the block and the surface?
Physics
1 answer:
lys-0071 [83]3 years ago
5 0

Answer:

The coefficient of friction between the block and the surface is μk=0.441

Explanation:

Let

mb = bullet mass = 0.02 kg

vb = bullet speed  = 300 m/s

mw = mass of the wood block = 1 kg

d = distance covered by the block plus the bullet = 4 m

u = speed of the bullet plus the block

g = gravity = 9.8 m/s2

μk = coefficient of friction, variable unknown

We will use the following expression to calculate the coefficient of friction:

μk = \frac{(\frac{mb*vb}{mb+mw})^{2}  }{2*g*d}

Replacing known values:

μk=\frac{(\frac{(0.02kg)*(300m/s)}{0.02kg+1kg})^{2}  }{2*9.8m/s^{2}*4m }=0.441

The coefficient of friction is a dimensionless value

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Matter can exist in one of three main states: solid, liquid, or gas. Solid matter is composed of tightly packed particles. A solid will retain its shape; the particles are not free to move around. Liquid matter is made of more loosely packed particles. Hopefully this helps:)

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3 years ago
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A 2.5 kg tribble is placed in a bucket and whirled in a 1.4 m radius vertical circle at a constant tangential speed. If the forc
Over [174]

Given that,

Mass of a tribble, m = 2.5 kg

Radius, r = 1.4 m

The force on the tribble from the bucket does not exceed 10 times its weight.

To find,

The maximum tangential speed.

Solution,

The force acting on the tribble is equal to the centripetal force.

F = 10mg

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F=\dfrac{mv^2}{r}

v is maximum tangential speed

v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{mgr}{m}} \\\\v=\sqrt{{10gr}} \\\\v=\sqrt{10\times 9.8\times 1.4} \\\\v=11.7\ m/s

So, the maximum tangential speed is 11.7 m/s.

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3 years ago
Stan does 178 J of work lifting himself 0.5 m. What is Stan’s mass? The acceleration of gravity is 9.8 m/s 2 . Answer in units o
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3 years ago
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I need help with questions 6-8. Thank you!! Image is attached
Digiron [165]

6) b) 2.7 m/s

7) b) DCA

8) b) B

Explanation:

6)

In a displacement-time plot, the slope of the line is given by

m=\frac{\Delta y}{\Delta x}

where

\Delta y is the change in the y-variable, so it is the displacement

\Delta x is the change in the x-variable, so it is the time elapsed

So, the slope of the line in a displacement-time plot corresponds to the velocity:

v=m=\frac{d}{t}

Therefore, to find the velocity of the object, we have to estimate the slope of its curve.

To estimate the velocity of object B, we have to estimate the slope of the line tangent to curve B at 10 seconds.

By doing an estimate by eye, we see that the displacement of object B changes from -10 m to 0 m when time increases from about 8 s to 12 s, so the velocity is about:

v=\frac{0-(-10)}{12-8}\sim 2.5 m/s

So the closest option is b) 2.7 m/s.

7)

As we said in part A, the velocity of each object is given by the slope of each curve.

Therefore:

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- The less steep the curve, the lower the velocity

From the graph, we observe that, among A, C and D:

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- Curve C is less steep than curve C, so object C has the second largest magnitude of velocity

- Curve A is flat, so the slope is zero, so its velocity is zero

So, from greatest magnitude to lowest magnitude of velocity, we have:

b) DCA

8)

In the graph, the overall displacement of each object is given by the change in the y-variable, \Delta y.

This means that the object with largest displacement is the object whose curve has the largest variation in y.

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|\Delta y| = 25 -(-15)=40 m

Finally, object C has displacement

\Delta y = 20-(-5)=25 m

While object A has displacement zero. Therefore, the correct option is

b) B

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