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lisov135 [29]
3 years ago
14

A 12.0 g bullet was fired horizontally into a 1 kg block of wood. The bullet initially had a speed of 250 m/s. The block of wood

was hanging from a 2 m long piece of (massless) string. After the collision the block/bullet combined object swings upward on the string. Find the height the block/bullet combined object rises.
a. 0.66m
b. 0.45m
c. 0.12 m
d. 0.35 m
e. 0.27m
Physics
1 answer:
Vaselesa [24]3 years ago
4 0

Answer:

So height though which combination of block bullet rises is 0.45 m

Explanation:

We have given  mass of the bullet  m_1=12gram=0.012kg

Velocity of the bullet v_1=250m/sec

Mass of block of wood m_2=1kg

Block is at rest so v_2=0m/sec

From conservation of momentum.

m_1v_1+m_2v_2=(m_1+m_2)v

0.012\times 250+0\times 0=(1+0.012)v

v=2.964m/sec

From third equation of motion

h=\frac{v^2}{2g}=\frac{2.964^2}{2\times .8}=0.45m

So option (b) will be the correct answer.

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If the separation distance is doubled, then the electric field decreases by a factor of 4.

<h3>What is the electric field strength?</h3>

We know that the electric field strength is known to depend on the magnitude of the charge and the distance of separation. We know that the electric field refers to the region in which the influence of a charge is felt. Recall that a charge is a specie that is positively or negatively charged. The charge on a specie must always be shown by its sign.

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