WheN a car is driven some distance the air pressure in the tyre increases,why?

1 answer:

8 0

Over the course of a long drive, as a result of alternating between braking and acceleration and of course steering, tires get heated up. When this happens, the air within the tires gets heated up, causing the air within the tire to expand. This expanding air exerts force on the tire from inside thus increasing tire pressure.

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Read 2 more answers

A bullet with a mass m b = 11.9 mb=11.9 g is fired into a block of wood at velocity v b = 261 m/s. vb=261 m/s. The block is atta

fredd [130]

Answer:

0.372 kg

Explanation:

The collision between the bullet and the block is inelastic, so only the total momentum of the system is conserved. So we can write:

$mu=(M+m)v$ (1)

where

$m=11.9 g = 11.9\cdot 10^{-3}kg$ is the mass of the bullet

$u=261 m/s$ is the initial velocity of the bullet

$M$ is the mass of the block

$v$ is the velocity at which the bullet and the block travels after the collision

We also know that the block is attached to a spring, and that the surface over which the block slides after the collision is frictionless. This means that the energy is conserved: so, the total kinetic energy of the block+bullet system just after the collision will entirely convert into elastic potential energy of the spring when the system comes to rest. So we can write

$\frac{1}{2}(M+m)v^2 = \frac{1}{2}kx^2$ (2)

where

k = 205 N/m is the spring constant

x = 35.0 cm = 0.35 m is the compression of the spring

From eq(1) we get

$v=\frac{mu}{M+m}$

And substituting into eq(2), we can solve to find the mass of the block:

$(M+m) \frac{(mu)^2}{(M+m)^2}=kx^2\\\frac{(mu)^2}{M+m}=kx^2\\M+m=\frac{(mu)^2}{kx^2}\\M=\frac{(mu)^2}{kx^2}-m=\frac{(11.9\cdot 10^{-3}\cdot 261)^2}{(205)(0.35)^2}-11.9\cdot 10^{-3}=0.372 kg$