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3 years ago

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A rope is being used to pull a mass of 10 kg vertically upward. Determine the tension on the rope, if starting from rest, the ma

ss acquires a velocity of 4 ms-1 in 8s (g=10ms?)

1 answer:

Sergio [31]3 years ago

6 0

Answer:

mgh= 10 x 8 x 10

= 800

but you can try 10 x 8 x 4^-1 x 10

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2) 1. A vibration of an electric charge.

Anestetic [448]

Explanation:

<h2> Answers</h2>

1.Electromagnetic waves

2.Electromagnetic radiation

3.Electromagneticwaves

5 0

3 years ago

Name three ways that being obese puts a strain on your heart & can lead to serious health problems.

Wittaler [7]

Answer:

Obesity changes in the structure and function of the heart. It increases your risk of heart disease. The more you weigh, the more blood you have flowing through your body. The heart has to work harder to pump the extra blood.

Explanation:

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3 years ago

Read 2 more answers

84. Three resitors each of value 30 respectively are connected in a parallel

kati45 [8]

Answer:

each resistor draws 1/3 of an amp or 0.33333 amps

Explanation:

V = I * R

V = 10 volts

R = 30 ohms

10 = I * 30 Divide by 30

10/30 = I

I = 0.33333

8 0

3 years ago

What is the acceleration of a race car if it has a mass of 1200kg and is moving with an engine force of 400N

meriva

Answer:

0.34 m/s^2

Explanation:

force=mass × acceleration

400 =1200 × acceleration

acceleration=400/1200

=0.34 m/s^2

5 0

2 years ago

what is the rotational kinetic energy of the earth? use the moment of inertia you calculated in part a rather than the actual mo

Ivenika [448]

The Earth's rotational kinetic energy is the kinetic Energy that the Earth

has due to rotation.

The rotational kinetic energy of the Earth is approximately <u>3.331 × 10³⁶ J</u>

Reasons:

<em>The parameters required for the question are; </em>

<em>Mass of the Earth, M = </em><em>5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = </em><em>6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = </em><em>24.0 hrs</em><em>.</em>

$The \ moment \ of \ inertia \ of \ uniform \ sphere \ is \ I = \mathbf{\dfrac{2}{5} \cdot M \cdot R^2}$

Which gives;

$\mathbf{I_{Earth}} = \dfrac{2}{5} \times 5.97 \times 10 ^{24} \cdot \left(6.38 \times 10^6 \right)^2 = 9.7202107 \times 10^{37}$

$\mathrm{The \ rotational \ kinetic \ energy \ is} \ E_{rotational} = \mathbf{\dfrac{1}{2} \cdot I \cdot \omega^2}$

$\mathrm{The \ angular \ speed, \ \omega} = \mathbf{\dfrac{2 \dcdot \pi}{T}}$

Therefore;

$\omega = \dfrac{2 \cdot \pi}{24} = \dfrac{\pi}{24}$

Which gives;

$\mathbf{E_{rotational}} = \dfrac{1}{2} \times 9.7202107 \times 10^{37} \times \left( \dfrac{\pi}{12} \right)^2 = 3.331 \times 10^{36}$

The rotational kinetic energy of the Earth, $E_{rotational}$ = <u>3.331 × 10³⁶ Joules</u>

Learn more here:

brainly.com/question/13623190

<em>The moment of inertia from part A of the question (obtained online) is that of the Earth approximated to a perfect sphere</em>.

<em>Mass of the Earth, M = 5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = 6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = 24.0 hrs</em>

3 0

3 years ago