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Zielflug [23.3K]
2 years ago
14

In which situation would a space probe most likely experience centripetal acceleration?

Physics
1 answer:
tester [92]2 years ago
8 0

Question: In which situation would a space probe most likely experience centripetal acceleration?

as it revolves around a planet

as it flies straight past a moon

as it is pulled in a line toward the Sun

as it lifts off from Earth

Answer:

When "space probe revolves around a planet"  most likely to experience centripetal acceleration

Explanation:

Centripetal acceleration defined as the rate in change of tangential velocity. Also, as per Newton's second law, any kind of force will be directly proportional to the acceleration attained by the object. So, for centripetal acceleration, the force will be the centripetal force. The centripetal force will be acting on an object rotating in a circular motion with its direction of force towards the center. Thus, centripetal acceleration will be experienced by an object or a space probe when it is in a circular motion. It means the space probe is revolving around a planet.

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A 52 N sled is pulled across a cement sidewalk at constant speed. A horizontal force of 36 N is exerted. What is the coefficient
Andre45 [30]

Answer:

μ = 0.692

Explanation:

In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.

Attached is an image with the respective forces:

A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.

Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.

The frictional force is defined as the product of the coefficient of friction by the normal force. In this way, we can calculate the coefficient of friction.

The process of solving this problem can be seen in the attached image.

5 0
3 years ago
Jet fuel is sold by the pound. Your jet needs 1,500 gallons of fuel. Jet fuel weighs 7.0 pounds per gallon and costs $12.00 per
galben [10]

Answer:

$ 57,272.73

Explanation:

Create a lengthy equation to convert the units you are given to the units you need to answer the question.

Your starting with gallons and pounds and you need your answer to be in Dollars.

$ 12.00         1 kg          7.0 lb       1500 gal

------------ X ----------- X ------------ X -------------- =  $ 57,272.73

   1 kg           2.2 lb        1 gal              1

5 0
2 years ago
51. A balloon starts rising from the ground, vertically upwards, uniformly at the rate of 1 m s-1. At the end of 4 seconds, a bo
strojnjashka [21]

Answer:

4s

Explanation:

when a bOdy rises into the air,the time it takes to reach a particular height is the same as the time it will take the body to fall from that height to the ground.

5 0
1 year ago
After the slab is charged, and the electrophorus is placed on the slab, the student briefly touches the electrophorus, effective
inna [77]

Answer:

By induction method

Explanation:

Induction method involves charging an electrically neutral body by bringing it in contact with an electrically charged body.

For the electrophus, a charge opposite that on the slab is induced on the side in contact with the slab; driving the opposite charge (this will be the same as that on the slab) to the other end of the elctrophus. Touching the electrophus removes the charge opposite the charge induced on the electrophus by the charged slab either by drawing up charge from the earth or taking the charge to earth (depends on the charge. A negative charge is drawn to earth while a positive charge draws up electrons from the earth)

5 0
3 years ago
A spring is hung vertically with a 425g mass attached to it. The mass is at rest. If the mass causes the spring to stretch 0.67
egoroff_w [7]

Answer:

6.22 N/m

Explanation:

From Hooke's law we deduce that F=kx where F is the applied force and k is the spring constant while x is the extension or compression of the spring. Making k the subject of the above formula then

k=\frac {F}{x}

We also know that the force F is equal to mg where m is the mass of an object and g is acceleration due to gravity hence substituting F with mg we get that

k=\frac {mg}{x}

Substituting m with 425 g which is equivalent to 0.425 kg and g with 9.81 then 0.67 for x we get that

k=\frac {mg}{x}=\frac {0.425\times 9.81}{0.67}=6.222761194 N/m\approx 6.22\ N/m

Therefore, the spring constant is approximately 6.22 N/m

3 0
3 years ago
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