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lisabon 2012 [21]
2 years ago
5

A cheetah can run 113 km/h in short busts. How far can a cheetah run in 0.25 hours (15 minutes)?

Physics
1 answer:
Fofino [41]2 years ago
4 0
The cheetah can run 28,25 km
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How are objects in space able to “fall” into orbit?
Zolol [24]

Answer:

MRCORRECT has answered the question

Explanation:

Newton realized that the reason the planets orbit the Sun is related to why objects fall to Earth when we drop them. The Sun's gravity pullson the planets, just as Earth's gravity pulls down anything that is not held up by some other force and keeps you and me on the ground.

7 0
2 years ago
In a discussion person A is talking 1.2 dB louder than person B, and person C is talking 3.2 dB louder than person A. What is th
liberstina [14]

Answer: 3.84dB

Explanation:

Since person A is talking 1.2dB louder than B, we will have

A = 1.2B... (1)

Similarly, person C is talking 3.2 dB louder than person A, we have

C = 3.2A... (2)

From equation 1, B = A/1.2... (3)

To get the ratio of the sound intensity of person C to the sound intensity of person B, we will divide equation 2 by 3 to give

C/B = 3.2A/{A/1.2}

C/B = 3.2A×1.2/A

C/B = 3.2×1.2

C/B = 3.84dB

3 0
3 years ago
A 12 kg box is at rest on your kitchen counter, which your cat is pawing at with a horizontal force of 40 N. If the coefficient
murzikaleks [220]

Answer; I think it's False.

8 0
3 years ago
A planet is 10 light years away from Earth. What speed would you need to go for a trip to the planet and back to take only 5 yea
viva [34]

Answer:

a. speed, v = 0.97 c

b. time, t' = 20.56 years

Given:

t' = 5 years

distance of the planet from the earth, d = 10 light years = 10 c

Solution:

(a) Distance travelled in a round trip, d' = 2d = 20 c = L'

Now, using Length contraction formula of relativity theory:

L'' = L'\sqrt{1 - \frac{v^{2}}{c^{2}}}                           (1)

time taken = 5 years

We know that :

time = \frac{distance}{speed}

5 = \frac{L''}{v}                                                      (2)

Dividing eqn (1) by v on both the sides and substituting eqn (2) in eqn (1):

\frac{L'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5

\frac{20'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5

Squaring both the sides and Solving above eqution, we get:

v = 0.97 c

(b) Time observed from Earth:

Using time dilation:

t'' = \frac{t'}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}

t'' = \frac{5}{\sqrt{1 - \frac{(0.97c)^{2}}{c^{2}}}}

Solving the above eqn:

t'' = 20.56 years

4 0
3 years ago
How much does 1 liter of water weigh
Zinaida [17]

Weight = (mass) x (gravity)

On Earth ...

Weight = (1 kg) x (9.8 m/s^2)

Weight = 9.8 Newtons

8 0
2 years ago
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