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3 years ago

9.

Physics

1 answer:

mote1985 [20]3 years ago

6 0

Answer:

D. location

Explanation:

The gravitational field strength of Earth is determined by the virtue of the location within the Earth's gravitational field.

That's why all objects regardless of their mass, shape, and size free fall towards the Earth with an acceleration equal to the acceleration at that location in the absence of air resistance.

According to the gravitational force between two bodies, the force experienced by one body due to the other is independent of its own mass.

The gravitational force is given by equation

F = GMm/r²

If F is the force acting on the smaller body of mass 'm', then

F = ma

Therefore, the equation becomes,

ma = GMm/r²

a = GM/r²

The value of 'a' changes with respect to the value of 'r' such that if 'r' is the radius of the Earth, then the acceleration at a height 'h' from Earth surface is given by

a = GM/(r+h)²

Here it is clear that the acceleration at any point is only the inherent property of the Earth itself.

The gravitational field strength of Earth is determined by the virtue of the location within the Earth's gravitational field.

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Karla Ayala pulls a sled on an icy road (dangerous!). Because of Karla's pull, the tension force is 151 N, and the rope makes a

skelet666 [1.2K]

Answer:

W = 1418.9 J = 1.418 KJ

Explanation:

In order to find the work done by the pull force applied by Karla, we need to can use the formula of work done. This formula tells us that work done on a body is the product of the distance covered by the object with the component of force applied in the direction of that displacement:

W = F.d

W = Fd Cosθ

where,

W = Work Done = ?

F = Force = 151 N

d = distance covered = 10 m

θ = Angle with horizontal = 20°

Therefore,

W = (151 N)(10 m) Cos 20°

W = 1418.9 J = 1.418 KJ

6 0

3 years ago

A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-

djverab [1.8K]

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

$\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}$

$\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}$

$\frac{3-y}{y}=\sqrt{\frac{7}{2}}$

3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.

6 0

4 years ago

I need help ASAP I need to get this right plz plz plz!!!!!

Rina8888 [55]

The atomic nucleus consists of protons and neutrons.

Additional information:

Protons are positively charged particle and neutrons are negatively charged particle.

Hope this will help u...:)

3 0

3 years ago

Read 2 more answers

An electron is fired horizontally at 2.5 x 10 m/s between two horizontal parallel plates 7.5 cm long, as shown in the diagram. T

DochEvi [55]

Answer:

the answer to your question is 2.5

Explanation:

8 0

3 years ago

Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass g

djyliett [7]

Answer:

a) m = 69.0 kg

b) release some gas in the opposite direction to the astronaut's movement

Explanation:

a) Let's use Newton's second law

F = m a

m = F / a

m = 60.0 / 0.870

m = 69.0 kg

b) when we exert a force on the astronaut it acquires a momentum po, as the astronaut system plus spacecraft is isolated, the momentum is conserved

p₀ = p_f

m v = M v '

v ’= $\frac{m}{M} \ v$

so we see that the ship is moving backwards, but since the mass of the ship is much greater than the mass of the astronaut, the speed of the ship is very small.

One method to avoid this effect is to release some gas in the opposite direction to the astronaut's movement so that the initial momentum of the astronaut plus the gas is zero and therefore no movement is created in the spacecraft.

3 0

3 years ago