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laiz [17]
3 years ago
11

An Earth satellite is in a circular orbit at an altitude of 500 km.

Physics
1 answer:
Dafna1 [17]3 years ago
8 0

Answer:

Speed Unchanged.

Explanation:

As work is defined as a product of force over a distance. If the distance in altitude is constant = 500km, there's 0 change in distance and force, no work would be done by the gravitational force.

Since potential energy of the satellite is unchanged, unless there's additional internal energy source, the kinetic energy would remain unchanged, so would its speed.

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What is the main organ of the circulatory system?
KengaRu [80]

It is the heart

if not im sorry

8 0
3 years ago
Read 2 more answers
a farmer lifts a bale of hay that is 12m off the ground . If he uses a force of 350 n, how much work does the farmer do?
hjlf

Answer:

4200 Joules

Explanation:

Work done =force x distance

From the question , we’re given f =350N ,

d = 12m

Using the above formula, we have

Workdone = 350 x 12

= 4200 Joules

5 0
3 years ago
The velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in
vivado [14]

The force the escaping gas exerts of the rocket is 10.42 N.

<h3>Force escaping gas exerts</h3>

The force the escaping gas exerts of the rocket is calculated as follows;

F = m(v - u)/t

where;

  • m is mass of the rocket
  • v is the final velocity of the rocket
  • u is the initial velocity of the rocket
  • t is time of motion

F = (0.25)(40 - 15)/0.6

F = 10.42 N

Thus, the force the escaping gas exerts of the rocket is 10.42 N.

Learn more about force here: brainly.com/question/12970081

#SPJ1

7 0
2 years ago
on a very muddy football field, a 120 kg linebacker tackles an 75 kg halfback. immediately before the collision, the linebacker
Aleksandr-060686 [28]
B4 the tackle: 

<span>The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north </span>

<span>and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east </span>


<span>After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle </span>

<span>The vector triangle is right angled: </span>

<span>magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s </span>

<span>so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] </span>

<span>v(f) = 5.6 m/s (to 2 sig figs) </span>


<span>direction of v(f) is the same as the direction of the final momentum </span>

<span>so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) </span>


<span>so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E </span>




<span>btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it</span>
4 0
3 years ago
A mover loads a 100 kg box into the back of a moving truck by
NeX [460]

Answer:

2.7

Explanation:

The following data were obtained from the question:

Mass (m) of box = 100 Kg

Length (L) of ramp = 4 m

Height (H) of ramp = 1.5 m

Mechanical advantage (MA) of ramp =?

Mechanical advantage of a ramp is simply defined as the ratio of the length of the ramp to the height of the ramp. Mathematically, it is given by:

Mechanical Advantage = Lenght / height

MA= L/H

With the above formula, we can obtain the mechanical advantage of the ramp as follow:

Length (L) of ramp = 4 m

Height (H) of ramp = 1.5 m

Mechanical advantage (MA) of ramp =?

MA = 4/1.5

MA = 2.7

Therefore, the mechanical advantage of the ramp is 2.7

3 0
3 years ago
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