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Irina18 [472]
3 years ago
12

9.5 g bullet has a speed of 1.5 km/s what is the kinetic energy of the bullet

Physics
2 answers:
kicyunya [14]3 years ago
8 0

Answer:

The kinetic energy of the bullet is approximately 10688 Joules, or about 10.7kJ.

Explanation:

Use the kinetic energy formula:

E_k=\frac{1}{2}mv^2= \frac{1}{2}9.5\cdot10^{-3}kg\cdot 1.5^2\cdot 10^6 \frac{m^2}{s^2}\approx 10688J

The kinetic energy of the bullet is approximately 10688 Joules, or about 10.7kJ.

Amiraneli [1.4K]3 years ago
3 0

The answer is  0.000824653J

You need to use the formula Mass * Velocity^2 over 2


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An L-R-C series circuit, R = 160 Ω , L = 0.790 H , and C = 1.30×10−2 μF . The source has a voltage amplitude of 140 V and a freq
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Replacing in the expression for power factor, we have:

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7 0
3 years ago
When the speed of the bottle is 2 m/s, the KE is kg m2/s2. When the speed of the bottle is 3 m/s, the KE is kg m2/s2. When the s
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mass of the bottle in each case is M = 0.250 kg

now as per given speeds we can use the formula of kinetic energy to find it

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2) when speed is 3 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(3)^2 = 1.125 J

3) when speed is 4 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(4)^2 = 2 J

4) when speed is 5 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(5)^2 = 3.125 J

5) when speed is 6 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(6)^2 = 4.5 J

3 0
3 years ago
Read 2 more answers
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