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4 years ago

The speed of a bullet of mass 20 g is 216 kilometre-per-hour what is kinetic energy in joules

Physics

1 answer:

chubhunter [2.5K]4 years ago

3 0

Answer:

2160joules

Explanation:

M = 20g = 0.02kg

V = 216km/h

K.E = 1/2 m (v)^2 = 1/2 × 0.02 × 216

K.E = 2.16kj

1kj = 1000joules

2.16kj = 2160joules

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3 years ago

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A 1.20-m cylindrical rod of diameter 0.570 cm is connected to a power supply that maintains a constant potential difference of 1

nasty-shy [4]

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

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$R=\rho \frac{L}{A}$ (1)

where

$\rho$ is the material resistivity

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A is the cross-sectional area

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

$A=\pi r^2=\pi (2.85\cdot 10^{-3} m)^2=2.55\cdot 10^{-5} m^2$

The resistance at 20°C can be found by using Ohm's law. In fact, we know:

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- The current at this temperature is I = 18.6 A

So, the resistance is

$R=\frac{V}{I}=\frac{15.0 V}{18.6 A}=0.81 \Omega$

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

$\rho=\frac{RA}{L}=\frac{(0.81 \Omega)(2.55\cdot 10^{-5} m^2)}{1.20 m}=1.72\cdot 10^{-5} \Omega m$

(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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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)

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$T_0 = 20^{\circ}$

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The speed of a bullet of mass 20 g is 216 kilometre-per-hour what is kinetic energy in joules​

The speed of a bullet of mass 20 g is 216 kilometre-per-hour what is kinetic energy in joules