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Andrei [34K]
3 years ago
9

Two 0.006 Kg bullets are fired with speeds of 20.0 m/s and 50.0 m/s respectively. What are their kinetic energies? Which bullet

has more kinetic energy? What is the difference of their kinetic energies?
Physics
1 answer:
SashulF [63]3 years ago
5 0

Answer:

a) Kinetic energies

K₁ = 1.2 J

K₂ = 7.5 J

b) The bullet that has the highest kinetic energy is the one with the highest speed , v = 50 m/s , K₂ = 7.5 J

c) K₂ -K₁  = 6.3 J

Explanation:

The kinetic energy (K) is that due to the movement of a body and is calculated as follows:

K = (1/2) m*v²  (J)

Where :

m : the mass of the body ( kg)

v is the speed of the body (m/s)

Data

m₁ = m₂ = 0.006 Kg

v₁ = 20 m/s

v₂ = 50 m/s

a)Calculation of the kinetic energy

K₁ = (1/2) (m₁)*(v₁)²

K₁ = (1/2) (0.006)*(20)²

K₁ = 1.2 J

K₂= (1/2) (m₂)*(v₂)²

K₂ = (1/2) (0.006)*(50)²

K₂ = 7.5 J

b) K₂ ˃ K₁

The bullet that has the highest kinetic energy is the one with the highest speed , v = 50 m/s, K₂ = 7.5 J

c) Difference of their kinetic energies (K₂ -K₁)

K₂ -K₁  = 7.5 J - 1.2 J = 6,3 J

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3 years ago
A 12.0 μF capacitor is charged to a potential of 50.0 V and then discharged through a 265 Ω resistor. A)How long does the capaci
larisa [96]

(a) The time for the capacitor to loose half its charge is 2.2 ms.

(b) The time for the capacitor to loose half its energy is 1.59 ms.

<h3>Time taken to loose half of its charge</h3>

q(t) = q₀e-^(t/RC)

q(t)/q₀ = e-^(t/RC)

0.5q₀/q₀ = e-^(t/RC)

0.5 = e-^(t/RC)

1/2 =  e-^(t/RC)

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t = 2.2 ms

<h3>Time taken to loose half of its stored energy</h3>

U(t) = Ue-^(t/RC)

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(Ue-^(t/RC))²/2C = Q₀²/2Ce

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t = 1.59 ms

Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.

Learn more about energy stored in capacitor here: brainly.com/question/14811408

#SPJ1

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