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2 years ago

In the ballistic pendulum experiment, a bullet of mass 0.06 kg is fired

Physics

1 answer:

Usimov [2.4K]2 years ago

8 0

The velocity of the bullet and the block just after impact is mathematically given as

u=6.647m/s

v=1.534m/s

<h3> Velocity of the bullet and the block</h3>

Question Parameters:

In the ballistic pendulum experiment, a bullet of mass 0.06 kg is fired

horizontally into a wooden block of mass 0.2 kg.

the block is 0.12 m above its initial position.

Generally the equation for the conservation of energy is mathematically given as

0.5(m+n)v^2=(m+n)gh

Therefore

v^2=2gh

$v=\sqrt{2*9.8*0.12}$

v=1.534m/s

Hence

mu+m*0=(m+n)v

0.060*u=(0.060+0.2)*1.534

u=6.647m/s

For more information on mass

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(a) $1.72\cdot 10^{-5} \Omega m$

The resistance of the rod is given by:

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where

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The radius of the rod is half the diameter: $r=0.570 cm/2=0.285 cm=2.85\cdot 10^{-3} m$ , so the cross-sectional area is

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So, the resistance is

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And now we can re-arrange the eq.(1) to solve for the resistivity:

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(b) $8.57\cdot 10^{-4} /{\circ}C$

First of all, let's find the new resistance of the wire at 92.0°C. In this case, the current is

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So the resistance is

$R=\frac{V}{I}=\frac{15.0 V}{17.5 A}=0.86 \Omega$

The equation that gives the change in resistance as a function of the temperature is

$R(T)=R_0 (1+\alpha(T-T_0))$

where

$R(T)=0.86 \Omega$ is the resistance at the new temperature (92.0°C)

$R_0=0.81 \Omega$ is the resistance at the original temperature (20.0°C)

$\alpha$ is the temperature coefficient of resistivity

$T=92^{\circ}C$

$T_0 = 20^{\circ}$

Solving the formula for $\alpha$ , we find

$\alpha=\frac{\frac{R(T)}{R_0}-1}{T-T_0}=\frac{\frac{0.86 \Omega}{0.81 \Omega}-1}{92C-20C}=8.57\cdot 10^{-4} /{\circ}C$

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