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Zielflug [23.3K]
3 years ago
14

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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Why smartphone and computer are called modern technology?​
topjm [15]

Answer:

Hey

I am sure from what i know that your answer should be the following:

Modern technology is deeply tied to efficiancy and speed. Think of the knew 5G network it gives us very fast internet, or perhaps our very nessisitys such as our phone or cumputer. On a phone you have so much (thats to say you can acces alot easly); pictures videos dictionarys video games etc etc.

6 0
3 years ago
What is a force that attracts all matter to each other?
Temka [501]
The answer is gravity. I hope this helps. 
5 0
3 years ago
Read 2 more answers
A test car starts from rest on a horizontal circular track of 115-m radius and increases its speed at a uniform rate to reach 90
Wewaii [24]

Answer:

a= 3.49 m/s^2

Explanation:

magnitude of total acceleration = sqrt{radial acceleration^2+tangential acceleration^2}.

we know that tangential acceleration a_t= change in velocity /time taken

now 90 km/h = 25 m/s

a_t = 25/17 = 1.47 m/s^2.

radial acceleration a_r = v^2/r

v= a_t×t = 1.47×13 = 19.11 m/s

a_r = 19.11^2/115=  3.175

now,

a= \sqrt{a_t^2+a_r^2}

a= \sqrt{1.47^2+3.175^2}

a= 3.49 m/s^2

3 0
3 years ago
A worker drives a 0.562 kg spike into a rail tie with a 2.26 kg sledgehammer. The hammer hits the spike with a speed of 64.4 m/s
zubka84 [21]

Answer:

Explanation:

Given that,

Mass of sledge hammer;

Mh =2.26 kg

Hammer speed;

Vh = 64.4 m/s

The expression fot the kinetic energy of the hammer is,

K.E(hammer) = ½Mh•Vh²

K.E(hammer) = ½ × 2.26 × 64.4²

K.E ( hammer) = 4686.52 J

If one forth of the kinetic energy is converted into internal energy, then

ΔU = ¼ × K.E(hammer)

∆U = ¼ × 4686.52

∆U = 1171.63 J

Thus, the increase in total internal energy will be 1171.63 J.

4 0
3 years ago
In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a fur
Margaret [11]

Answer:

D = -4/7 = - 0.57

C = 17/7 = 2.43

Explanation:

We have the following two equations:

3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)

First, we isolate C from equation (2):

2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)

using this value of C from equation (3) in equation (1):

3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}

<u>D = - 0.57</u>

Put this value in equation (3), we get:

C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C =  \frac{17}{7}\\

<u>C = 2.43</u>

5 0
3 years ago
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