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blsea [12.9K]
3 years ago
14

A 0.39 kg apple is dropped,Earths exerts a force that accelerates at 9.8 m/s^2. If the mass of the earth is 5.98x10^24 kg,what i

s the magnitude of the earths acceleration
Physics
1 answer:
Jobisdone [24]3 years ago
4 0

The magnitude of the Earth's acceleration is 6.39\cdot 10^{-25} m/s^2.

Explanation:

We have to use different concepts to answer this question.

First of all, we have to find the gravitational force exerted by the Earth on the apple: this is known as weight, and it is calculated as

F=mg

where

m = 0.39 kg is the mass of the apple

g=9.8 m/s^2 is hte acceleration of gravity

Substituting, we find the force exerted on the apple by the Earth:

F=(0.39)(9.8)=3.82 N

For the second part, we have to use Newton's third law of motion, which states that:

"When an object A exerts a force on an object B, object B exerts an equal and opposite force on object A"

Therefore in this situation, if we identify object A with the apple and object B as the Earth, we can say that the force exerted by the Earth on the apple is equal (and opposite) to the force exerted by the apple on the Earth. Therefore, also the Earth experiences a force of

F = 3.82 N

Now we can finally calculate the Earth's acceleration by using Newton's second law of motion:

F=Ma

where

F = 3.82 N is the net force on the Earth

M=5.98\cdot 10^{24} kg is the mass of the Earth

a is the Earth's acceleration

And solving for a, we find

a=\frac{F}{M}=\frac{3.82}{5.98\cdot 10^{24}}=6.39\cdot 10^{-25} m/s^2

Learn more about Newton's second law here:

brainly.com/question/3820012

About forces and weight:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

About Newton's third law:

brainly.com/question/11411375

#LearnwithBrainly

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Maru [420]

The drag force acting on the rocket is 80N.

<h3>Give an explanation of drag force?</h3>

The divergence in velocity between the fluid and the item, also known as drag, exerts a force on it. Between the liquid and the solid object, there should be motion. Drag is absent in the absence of motion.

The air molecules are more compressed (pushed together) on the surfaces that are facing the front while being more dispersed (spread out) on the surfaces facing the back. Turbulent flow, which occurs when air layers split from the surface and start to swirl, is what causes this.

The drag force acting on the rocket F = ma

Given,

m = 4kg, a = 20ftm/s²

Substituting m and a values in the above formula,

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What kind of reaction (endothermic or exothermic)
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A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a
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Answer:

The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

Explanation:

Given that

q₁ = 5 μ C

q₂ = - 4 μ C

The distance between charges = 50 cm

d= 50 cm

Lets take at distance x from the charge μ C ,the electrical field is zero.

That is why the distance from the charge - 4 μ C =  50 - x cm

We know that ,electric field is given as

E=K\dfrac{q}{r^2}

K\dfrac{5\ \mu}{x^2}=K\dfrac{4\mu }{(50-x)^2}\\\\\dfrac{5}{x^2}=\dfrac{4 }{(50-x)^2}\\\\\\5(50-x)^2=4x^2\\(50-x)^2=0.8x^2\\\\50-x =0.89x\\\ x=\dfrac{50}{1.89}\ cm\\\\\\x=26.45\ cm\\

Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

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3 years ago
A ski gondola is connected to the top of a hill by a steel cable of length 620 m and diameter 1.5 cm. As the gondola comes to th
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Answer:

(a) 89 m/s

(b) 11000 N

Explanation:

Note that answers are given to 2 significant figures which is what we have in the values in the question.

(a) Speed is given by the ratio of distance to time. In the question, the time given was the time it took the pulse to travel the length of the cable twice. Thus, the distance travelled is twice the length of the cable.

v=\dfrac{2\times 620 \text{ m}}{14\text{ s}} = \dfrac{1240\text{ m}}{14\text{ s}}=88.571428\ldots \text{ m/s}= 89\text{ m/s}

(b) The tension, T, is given by

v =\sqrt{\dfrac{T}{\mu}}

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Hence,

T = \mu\cdot v^{2}

To determine \mu, we need to know the mass of the cable. We use the density formula:

\rho = \dfrac{m}{V}

where m is the mass and V is the volume.

m=\rho\cdot V

If the length is denoted by l, then

\mu = \dfrac{m}{l} = \dfrac{\rho\cdot V}{l}

T = \dfrac{\rho\cdot V}{l} v^{2}

The density of steel = 8050 kg/m3

The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is

V = \pi \dfrac{d^{2}}{4} l

T = \dfrac{\rho\cdot\pi d^2 l}{4l}v^2 = \dfrac{\rho\cdot\pi d^2}{4}v^2

T = \dfrac{8050\times\pi\times0.015^2}{4} \times 88.57^2

T = 11159.4186\ldots \text{ N} = 11000 \text{ N}

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