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3 years ago

Calculate the kinetic energy of a 64.g bullet moving at a speed of 411.ms . Round your answer to 2 significant digits.

Physics

1 answer:

Mandarinka [93]3 years ago

8 0

Answer:

The kinetic energy of the bullet is 5.4 × 10³ J

Explanation:

Hi there!

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the bullet.

v = speed of the bullet.

Let´s convert the mass unit to kg so that our result is in Joules:

64 g · ( 1 kg / 1000 g) = 0.064 kg

Then, the kinetic energy will be the following:

KE = 1/2 · 0.064 kg · (411 m/s)²

KE = 5.4 × 10³ J

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4 years ago

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Explanation:

Displacement = 5 km

Converting km/h to m/s,

10 km/h * 1000 m/1 km * 1 h/3600 s

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3 km/h * 1000 m/1 km * 1 h/3600 s

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3 years ago

Over the last several decades, scientists have addressed the problem of nonrenewable natural resources such as fossil fuels. Hum

ExtremeBDS [4]

Answer:

Option (C)

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3 years ago

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allochka39001 [22]

Answer:

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$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

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$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

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$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

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$|E| =-\nabla V$ (6)

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$|E| =-\frac{dV(z)}{dz}$ (7)

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