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1 year ago

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A 1.7-kg block of wood rests on a rough surface. A 0.011-kg bullet strikes the block with a speed of 670 m/s and embeds itself.

The bullet-block system slides forward 2.4 m before coming to rest.

1 answer:

aksik [14]1 year ago

7 0

What is the coefficient of kinetic friction between the block and the surface? Express your answer using two significant figures.

Answer:

0.39

Explanation:

Given information

m1=1.7 Kg

m2=0.011 Kg

v2=670 m/s

d=2.4 m

m2v2=(m1+m2)v hence $v=\frac {m2v2}{m1+m2}$ and also $a=\frac {v^{2}}{2d}$

The deceleration due to friction is given by

$F=\mu_k N=\mu_k W=\mu_k (m1+m2)g$

$F=(m1+m2)a=\mu_k (m1+m2)g$ therefore $a=\mu_k g$

Therefore, $a=\frac {v^{2}}{2d}=\mu_k g$

$\mu_k=\frac {1}{2dg}(\frac {m2v}{m1+m2})^{2}=\frac {1}{2*2.4*9.81}\times (\frac {0.011*670}{1.7+0.011})^{2}$

$\mu_k=0.39$

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A particular cylindrical bucket has a height of 36.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is em

Sergeeva-Olga [200]

Answer:

a. 0.000002 m

b. 0.00000182 m

Explanation:

36 cm = 0.36 m

15 cm = 0.15 m

a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:

$P = \ro g h = 1000 * 10 * 20 = 200000 Pa$

Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same

$P_1V_1 = P_2V_2$

Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:

$V_1 = Ah$

As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:

$P_1Ah_1 = P_2Ah_2$

$h_2 = h_1\frac{P_1}{P_2} = 0.36\frac{1.105}{200000} = 0.000002 m$

b) If the temperatures changes, we can still reuse the ideal gas equation above:

$\frac{P_1Ah_1}{T_1} = \frac{P_2Ah_2}{T_2}$

$h_2 = h_1\frac{P_1T_2}{P_2T_1} = 0.36\frac{1.105 * 275}{200000*300} =0.00000182 m$

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2 years ago

A 2.70 kg cat is sitting on a windowsill. The cat is sleeping peacefully until a dog barks at him. Startled, the cat falls from

Alchen [17]

Answer:

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem, we have that initial potential gravitational energy of the cat ( $U_{g}$ ), in joules, is equal to the sum of the final translational kinetic energy ( $K$ ), in joules, and work losses due to air resistance ( $W_{l}$ ), in joules:

$U_{g} = K +W_{l}$ (1)

By definition of potential gravitational energy, translational kinetic energy and work, we expand the equation presented above:

$m \cdot g\cdot h = \frac{1}{2}\cdot m \cdot v^{2}+W_{l}$ (2)

Where:

$m$ - Mass of the cat, in kilograms.

$g$ - Gravitational acceleration, in meters per square second.

$h$ - Initial height of the cat, in meters.

$v$ - Final speed of the cat, in meters per second.

If we know that $m = 2.70\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h = 5.20\,m$ and $W_{l} = 120\,J$ , then the final speed of the cat is:

$v = \sqrt{\frac{2\cdot (m\cdot g\cdot h-W_{l})}{m} }$

$v = \sqrt{2\cdot g\cdot h-\frac{W_{l}}{m} }$

$v \approx 7.586\,\frac{m}{s}$

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

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2 years ago

four children pull on the same stuffed toy at the same time , yet there is no net force on the toy.how is this possible?

Kay [80]

There was no net force on the stuffed toy, because the kids might have the same strength, The same force is on both sides of it. T<span>hey cancel each other out. They exert a force on the stuffed toy equal in strength but opposite in direction. The forces are balanced and the stuffed toy does not move. </span>Its like a game of tug-o-war, but you and I have the same strength. the rope would be still and not moving.

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2 years ago

Read 2 more answers

A long-distance swimmer is able to swim through still water at 4.0 km/h. She wishes to try to swim from Port Angeles, Washington

Roman55 [17]

Let $\theta$ be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

$\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)$

The current has velocity vector (relative to the Earth)

$\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath$

The swimmer's resultant velocity (her velocity relative to the Earth) is then

$\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}$

$\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath$

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

$\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ$

which is approximately 41º west of north.

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2 years ago

In the circuit shown in the figure above, suppose that the value of R1 is 100 k ohms and the value of R2 is 470 k ohms. At which

Nesterboy [21]

Answer:

A

Explanation:

voltage between A and C is equal battery's voltage.

7 0

2 years ago