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Mashutka [201]
2 years ago
10

For a 99 kg person standing at the equator, what is the magnitude of the angular momentum about Earth's center due to Earth's ro

tation?
Physics
1 answer:
Fofino [41]2 years ago
6 0

Answer:

L = 4.58 x 10⁴ kg.m²/s

Explanation:

The angular momentum is given by the formula:

L = mvr

but, v = rω

Therefore,

L = mr²ω

where,

L = Angular Momentum of the person = ?

m = mass of person = 99 kg

r = radius of earth = 6.37 x 10⁶ m

ω = Angular Speed of Earth's Rotation = θ/t

Since, earth completes 1 rotation in 1 day. Hence,

ω = (2π rad/1 day)(1 day/24 h)(1 h/3600 s)

ω =  7.27 x 10⁻⁵ rad/s

Therefore,

L = (99 kg)(6.37 x 10⁶ m)²(7.27 x 10⁻⁵ rad/s)

<u>L = 4.58 x 10⁴ kg.m²/s</u>

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The force exerted on the board by the karate master given the data is -4500 N

<h3>Data obtained from the question </h3>
  • Initial velocity (u) = 10 m/s
  • Final velocity (v) = 1 m/s
  • Time (t) = 0.002 s
  • Mass (m) = 1 Kg
  • Force (F) = ?
<h3>How to determine the force</h3>

The force exerted can be obtained as illustrated below:

F = m(v - u) / t

F = 1 (1 - 10) / 0.002

F = (1 × -9) / 0.002

F = -4500 N

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The number of magnetic field force lines passing through the given surface determines: 1. magnetic flux 2. magnetic induction 3.
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A new ski area has opened that emphasizes the extreme nature of the skiing possible on its slopes. Suppose an ad intones "Free-f
Allushta [10]

Answer:

48.6°

Explanation:

The forward force, F equals the component of the weight along the slope.

So mgsinθ = ma where a = acceleration and θ = angle between the slope and the horizontal.

So a = gsinθ

Since we are given that a = 75%g = 0.75g,

0.75g = gsinθ

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You punt a ball straight up at 20 m/s. What is the balls hangtime
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The hang time of the ball is 4.08 s

Explanation:

The ball is in free fall motion: this means that it is acted upon gravity only, so its acceleration is the acceleration of gravity,

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downward (the negative sign refers to the downward direction).

Since this is a uniformly accelerated motion, we can solve the problem by using the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

First we calculate the time it takes for the ball to reach the maximum height, where the velocity is zero:

v = 0

Substituting:

u = +20 m/s

a=-9.8 m/s^2

we find t

t=\frac{v-u}{a}=\frac{0-20}{-9.8}=2.04 s

The motion of the ball is symmetrical, so the total time of flight is just twice the time needed to reach the maximum height, therefore:

T=2t=2(2.04)=4.08 s

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