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3 years ago

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

Physics

1 answer:

tester [92]3 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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Experiment 1: A study is done to determine which of two fuel mixtures allows a rocket to travel farther over a period of time. R

leva [86]

Answer: D

Experiment 1 has a confounding variable related to the mass of the rockets. Any variation in mass may cause a discrepancy in the distance traveled.

This is the answer to the question because:

Both experiments do have a confounding variable.
Experiment 1 doesn't have to stay constant.
A double-blind experiment will not do anything to the placebo.
High blood pressure people will not make the results confusing.

The answer has to be the option D. Hope this helps you!

3 0

4 years ago

You look in the sky and see two jetliners that you know are of equal size, yet one appears to be much larger. Because of your kn

Ksju [112]

Answer:

Because of the knowledge of <u>relative size</u>, it will be assumed that the smaller jetliner is farther away.

Explanation:

According to the theory of relative size, the distance that an object has to the viewing individual affects the perception of the individual regarding the size of the object.

As stated in this case, one of the jetliners is farther away from the other. Therefore, even if the jets are of equal size, the one that is at a greater distance is perceived to be smaller as it is at a greater viewing range. The one that is nearer to the individual seems bigger in comparison to the one farther away due to a closer viewing range.

Therefore, the jet that is nearer appears larger.

To know more about relative size, refer to:

brainly.com/question/19998265

#SPJ4

8 0

2 years ago

Imagine that two charged objects are the system of interest. When the objects are infinitely far from each other, the electric p

podryga [215]

Answer: Option (c) is the correct answer.

Explanation:

It is known that the expression for potential energy related to charges and distance between their separation is as follows.

$U_{q} = \frac{kq_{1}q_{2}}{r}$

where, $q_{1}$ and $q_{2}$ are two charges

r = distance of separation between the charges

k = electrostatic constant

So, when both the charges are carry the same charge and r is small then the value of potential energy will be positive in nature.

Whereas if the distance of separation between the charges is infinitely large then the potential energy calculated will be zero everywhere.

Thus, we can conclude that the statement one object is negatively charged and the other one is positively charged, is incorrect.

5 0

3 years ago

If a 950 kg car has 30,400 J of kinetic energy, how fast is it moving?

abruzzese [7]

Hello

The kinetic energy K of a moving object is:
$K= \frac{1}{2} m v^2$
where m is the mass and v the velocity of the object.
Using this formula, we can calculate v for this problem:
$v= \sqrt{ \frac{2E}{m} }= \sqrt{ \frac{2\cdot30400~J}{950~Kg} } = 8~m/s$

6 0

3 years ago

Richard is driving home to visit his parents. 150 mi of the trip are on the interstate highway where the speed limit is 65 mph .

Serggg [28]

Answer:

t = 25.5 min

Explanation:

To know how many minutes does Richard save, you first calculate the time that Richard takes with both velocities v1 = 65mph and v2 = 80mph.

$t_1=\frac{x}{v_1}=\frac{150mi}{65mph}=2.30h\\\\t_2=\frac{x}{v_2}=\frac{150mi}{80mph}=1.875h$

Next, you calculate the difference between both times t1 and t2:

$\Delta t=t_1-t_2=2.30h-1.875h=0.425h$

This is the time that Richard saves when he drives with a speed of 80mph. Finally, you convert the result to minutes:

$0.425h*\frac{60min}{1h}=25.5min=25\ min\ \ 30 s$

hence, Richard saves 25.5 min (25 min and 30 s) when he drives with a speed of 80mph

3 0

3 years ago