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nekit [7.7K]
3 years ago
5

A 2.00-kilogram mud ball drops from rest at a height of 17.0 m. If the impact between the ball and the ground lasts 0.46 s, what

is the magnitude of the average force exerted by the ball on the ground
Physics
1 answer:
Softa [21]3 years ago
5 0

Answer: 79.35 N

Explanation: according to the impulse momentum theorem,

Impulse = change in momentum

Where impulse = force × time and change in momentum = m ( v - u).

The object was initially at rest, hence it initial velocity is zero.

To get the final velocity, we use the formulae below

v² = u² + 2gh

Where h = height of the cliff = 17m

v² = 2 × 9.8 × 17

v² = 333.2

v = √333.2

v = 18.25 m/s

At t = 0.46s and v = 18.25 m/s, we can get the average force of impact

F×0.46 = 2 (18.25 - 0)

F × 0.46 = 2 (18.25)

F × 0.46 = 36.5

F = 36.5 /0.46 = 79.35 N

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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)

t/RC = ln(2)

t = RC x ln(2)

t = (12 x 10⁻⁶ x 265) x ln(2)

t = 2.2 x 10⁻³ s

t = 2.2 ms

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

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

U = ¹/₂Q²/C

(Ue-^(t/RC))²/2C = Q₀²/2Ce

e^(2t/RC) = e

2t/RC = 1

t = RC/2

t = (265 x 12 x 10⁻⁶)/2

t = 1.59 x 10⁻³ s

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

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6 0
1 year ago
Meg walks with a velocity of 0.9 m/s west. She does so while riding on a train that is traveling with a velocity of 2.7 m/s east
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<span>Velocities are vectors so we can add them!

Let's let +x be East and -x be West.

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a 2000 kg truck is moving eastward at 25 m/s. it collides inelastically with a 1500 kg truck traveling southward at 30 m/s. they
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According to the given statement Final velocity when they stick together is 8.735i^ + 11.25j^​

<h3>What is collision and momentum?</h3>

The unit of momentum is kg ms -1. Momentum is a vector parameter that is influenced by the object's direction. During collisions involving objects, momentum is a relevant concept. The final velocity before a collision between two objects equals the total motion after the impact (in the absence of external forces).

<h3>Briefing:</h3>

From conservation of momentum

Initial momentum = final momentum

m u +M U =(m+M) V

2000×25 i^ +1500×30 j^​ =(2000+1500) V

V = 8.735i^ + 11.25j^​

Final velocity when they stick together is 8.735i^ + 11.25j^​

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The complete question is -

A 2000 kg truck is moving eastward at 25 m/s. it collides inelastically with a 1500 kg truck traveling southward at 30 m/s. they collide at the intersection. Find the direction and magnitude of velocity of the wreckage after the collision, assuming the vehicles stick together after the collision.

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Answer:

sun is the main source while the other object reflect light on the sun

Explanation:

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