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Zielflug [23.3K]
2 years ago
8

YO IF YOU ARE GOOD AS SCIENCE PLEASE ANSWER THIS QUESTION IT WOULD HELP A LOT NO LINKS!!!

Physics
2 answers:
zhuklara [117]2 years ago
7 0

Answer:

1  and 5 and 3

Explanation:

hope this helps

viva [34]2 years ago
3 0

Answer:

1.) the study of matter and energy (I think that's the only answer)

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During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0
oksian1 [2.3K]

Answer:

v = 384km/min

Explanation:

In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.

You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

d=\frac{1,280,000,000km}{37,000}=34,594.59km

You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

v=\frac{d}{t}=\frac{34,594.59km}{90min}=384.38\frac{km}{min}\approx384\frac{km}{min}

hence, the speed of the Hubble is approximately 384km/min

5 0
3 years ago
What is the centripetal force that would be required to keep a 4.0 kg mass moving in a horizontal circle with a radius of 0.80 m
KIM [24]

Answer:

D. 1.8 × 102 newtons radially inward

Explanation:

The magnitude of the centripetal force is given by:

F=m\frac{v^2}{r}

where

m is the mass of the object

v is the tangential speed

r is the radius of the circular trajector

In this problem, we have m = 4.0 kg, v = 6.0 m/s and r = 0.80 m, therefore substituting into the equation we get

F=(4.0 kg)\frac{(6.0 m/s)^2}{0.80 m}=180 N

The centripetal force is the force that keeps the object in a circular trajectory, so it is a force that is always directed inward (towards the centre of the circular path) and radially. Therefore, the correct answer is

D. 1.8 × 102 newtons radially inward

4 0
3 years ago
Read 2 more answers
If a 42kg rolling object slows from 11.5m/s to 3.33m/s how much work did friction do
geniusboy [140]

Answer: 1608.39 J

Explanation: Given that the

mass M = 42kg

U = 11.5m/s

V = 3.33m/s

how much work did friction do

Work done = Force × distance

Work done = Ma × distance

But acceleration a = V/t

Work done = M × V/t × d

Work done = M × V × d/t

Where d/t = velocity

Therefore,

Work done = M × U × V

Work done = 42 × 11.5 × 3.33

Work done = 1608.39 J

8 0
3 years ago
Which of the following is a part of the digestive system
BlackZzzverrR [31]

The Pancreas

<em>Is an inflammation organ that lies in the lower part of the stomach which plays a big part in the digestive system.</em>

3 0
3 years ago
A sled with mass 8.00kg moves in a straight line on a frictionless horizontal surface. At one point in it's path, it's speed is
kati45 [8]

We will use two definitions to solve this problem. The first will be given by the conservation of energy, whereby the change in kinetic energy must be equivalent to work. At the same time, work can be defined as the product between the force by the distance traveled. By matching these two expressions and clearing for the Force we can find the desired variable.

W = KE_f-KE_i

Fd = \frac{1}{2}mv_f^2-\frac{1}{2} mv_i^2

Thus the force acting on the sled is,

F = \frac{m}{2s} (v_f^2-v_i^2)

Replacing,

F = \frac{8}{2(2.5)}(6^2-4^2)

F = 32N

Therefore the Force acting on the sled is 32N

8 0
3 years ago
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