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2 years ago

6

A 20-gram bullet traveling at 250 m/s strikes a block of wood that weighs 2 kg. With what velocity will the block and bullet mov

e after the collision? [convert g to kg]

1 answer:

Shkiper50 [21]2 years ago

6 0

Answer:

the velocity of the bullet-wood system after the collision is 2.48 m/s

Explanation:

Given;

mass of the bullet, m₀ = 20 g = 0.02 kg

velocity of the bullet, v₀ = 250 m/s

mass of the wood, m₁ = 2 kg

velocity of the wood, v₁ = 0

Let the velocity of the bullet-wood system after collision = v

Apply the principle of conservation of linear momentum to calculate the final velocity of the system;

Initial momentum = final momentum

m₀v₀ + m₁v₁ = v(m₀ + m₁)

0.02 x 250 + 2 x 0 = v(2 + 0.02)

5 + 0 = v(2.02)

5 = 2.02v

v = 5/2.02

v = 2.48 m/s

Therefore, the velocity of the bullet-wood system after the collision is 2.48 m/s

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2 years ago

Immediately after being struck by a hammer, the nail (mass of 50 g) has a velocity of 50 m/s. The total frictional force is 62.5

Marina86 [1]

Answer:

nail will stop after traveling 1000 m

Explanation:

We have given mass m = 50 gram = 0.05 kg

Frictional force which is used to stop the mass F = 62.5 kN

Initial velocity is given u = 50 m/sec

From newton's law force is equal to F = ma, here m is mass and a is acceleration

So $a=\frac{F}{m}=\frac{62.5\times 10^3}{0.05}=1.25\times 10^6m/sec^2$

As finally nail stops so final velocity v = 0 m/sec

From third equation of motion $v^2=u^2+2as$

So $0^2=50^2-2\times 1.25\times 10^6\times s$

s = 1000 m

So nail will stop after traveling 1000 m

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2 years ago

This is a compound made from a group of covalently bonded atoms.

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3 years ago

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Column A: x-axis, input, domain

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2 years ago

A 5.00 kg rock whose density is 4300 kg/m3 is suspended by a string such that half of the rock's volume is under water. You may

12345 [234]

Answer:

The tension on the string is $T = 43.302 \ N$

Explanation:

From the question we are told that

The mass of the rock is $m_r = 5.00 \ kg = 5000 \ g$

The density of the rock is $\rho = 4300 \ kg/m^3 = 4.3 g/dm^3$

Generally the volume of the rock is mathematically evaluated as

$V = \frac{m_r}{\rho}$

substituting values

$V = \frac{5000}{4.3}$

$V = 1162.7 \ dm^3$

The volume of the rock immersed in water is

$V_w = \frac{V}{2}$

substituting values

$V_w = \frac{1162.7 }{2}$

$V_w = 581.4 \ dm^3$

mass of water been displaced by the this volume is

$m_w = V_w$ According to Archimedes principle

=> $m_w = 581.4 \ g$

$m_w = 0.5814 \ kg$

The weight of the water displace is

$W _w = m_w * g$

$W _w = 0.5814 * 9.8$

$W _w = 5.698 \ N$

The actual weight of the rock is

$W_r = m_r * g$

$W_r = 5.0 * 9.8$

$W_r = 49.0 \ N$

The tension on the string is

$T = W_r - W_w$

substituting values

$T = 49.0 - 5.698$

$T = 43.302 \ N$

4 0

2 years ago