The distance traveled by the wood after the bullet emerges is 0.16 m.
The given parameters;
- <em>mass of the bullet, m = 23 g = 0.023 g</em>
- <em>speed of the bullet, u = 230 m/s</em>
- <em>mass of the wood, m = 2 kg</em>
- <em>final speed of the bullet, v = 170 m/s</em>
- <em>coefficient of friction, μ = 0.15</em>
The final velocity of the wood after the bullet hits is calculated as follows;
The acceleration of the wood is calculated as follows;
The distance traveled by the wood after the bullet emerges is calculated as follows;
Thus, the distance traveled by the wood after the bullet emerges is 0.16 m.
Learn more here:brainly.com/question/15244782
Answer: 1000 Hz · 0,5 m = 500 m/s
Explanation: speed = frequency · wavelength
Answer:
fibrous =potato
taproot =radish
stilt =maize and sugar cane
Answer:
Explanation:
A mass of 700 kg will exert a force of
700 x 9.8
= 6860 N.
Amount of compression x = 4 cm
= 4 x 10⁻² m
Force constant K = force of compression / compression
= 6860 / 4 x 10⁻²
= 1715 x 10² Nm⁻¹.
Let us take compression of r at any moment
Restoring force by spring
= k r
Force required to compress = kr
Let it is compressed by small length dr during which force will remain constant.
Work done
dW = Force x displacement
= -kr -dr
= kr dr
Work done to compress by length d
for it r ranges from 0 to -d
Integrating on both sides
W = 
= [ kr²/2]₀^-4
= 1/2 kX16X10⁻⁴
= .5 x 1715 x 10² x 16 x 10⁻⁴
= 137.20 J
