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3 years ago

All satellite eventually lose their orbital energy and plummet to earth because of______.

Physics

1 answer:

Softa [21]3 years ago

8 0

Answer:

gravity

Explanation:

Even when satellites are thousands of miles away, Earth's gravity still tugs on them. Gravity—combined with the satellite's momentum from its launch into space—cause the satellite to go into orbit above Earth, instead of falling back down to the ground.

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james throws a baseball with a horizontal component velocity of 30 m/s east and a vertical component velocity of 40m/s North.How

olga2289 [7]

Force vectors, use Pythagoras to get resultant force: sqrt ((30^2 + (40^2)) = 50m/s

6 0

3 years ago

Charlie pulls horizontally to the right on a wagon with a force of 37.2 N. Sara pulls horizontally to the left with a force of 2

Sidana [21]

Answer:

The work done on the wagon is 37 joules.

Explanation:

Given that,

The force applied by Charlie to the right, F = 37.2 N

The force applied by Sara to the left, F' = 22.4 N

We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :

$F_n=F-F'$

$F_n=37.2-22.4$

$F_n=14.8\ N$

Work done on the wagon is given by the product of net force and displacement. It is given by :

$W=F_n\times d$

$W=14.8\ N\times 2.5\ m$

W = 37 Joules

So, the work done on the wagon is 37 joules. Hence, this is the required solution.

7 0

3 years ago

Read 2 more answers

Select the volume units that are greater than one liter.

Andreas93 [3]

A.) kiloliter. 1 kiloliter = 1,000 liters
c.) megaliter. 1 megaliter = 1,000,000 liters

hope this helps

5 0

3 years ago

Read 2 more answers

A Hooke's law bowstring is stretched x meters until a force of f newtons is applied, and then held. By what factor will the elas

Liono4ka [1.6K]

The elastic potential energy increases by a factor of 9

Explanation:

The elastic potential energy of a bowstring is given by

$E=\frac{1}{2}kx^2$ (1)

where

k is the spring constant

x is the elongation of the bowstring

Hooke's law states the relationship between the force applied and the elongation of an elastic object:

$F=kx$

where

F is the force applied

x is the elongation

We can rewrite it as

$x=\frac{F}{k}$

And substituting into (1),

$E=\frac{1}{2}k(\frac{F}{k})^2=\frac{F^2}{2k}$

In this problem, the force applied to the bowstring is tripled,

F' = 3F

So the final elastic potential energy is:

$E'=\frac{(3F)^2}{2k}=9(\frac{F^2}{2k})=9E$

So, the elastic potential energy increases by a factor of 9.

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

7 0

3 years ago

In a new scenario, the block only makes it (exactly) half-way through the rough spot. How far was the spring compressed from its

Mila [183]

Answer: hello below is the missing part of your question

A mass m = 10 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5029 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is μk = 0.49. The mass leaves the spring at a speed v = 3.4 m/s.

answer

x = 0.0962 m

Explanation:

<em>First step :</em>

Determine the length of the rough patch/spot

F = Uₓ (mg)

and w = F.d = Uₓ (mg) * d

hence;

d( length of rough patch) = w / Uₓ (mg) = 46.55 / (0.49 * 10 * 9.8) = 0.9694 m

<em>next : </em>

work done on unstretched spring length

Given that block travels halfway i.e. d = 0.9694 / 2 = 0.4847 m

w' = Uₓ (mg) * d

= 0.49 * 10 * 9.81 * 0.4847 = 23.27 J

also given that the Elastic energy of spring = work done ( w')

1/2 * kx^2 = 23.27 J

x = $\sqrt{\frac{2*23.27}{5029} }$ = 0.0962 m

4 0

3 years ago