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n200080 [17]
3 years ago
11

An object with a mass of 6 kg is dropped from a height of 80 m. Consider the air resistance to be zero and g to be 10 m/s². What

is the kinetic energy of the object when it has fallen 60 m?
A. 1,200 J
B. 3,000 J
C. 3,600 J
D. 4,800 J
Physics
1 answer:
max2010maxim [7]3 years ago
5 0

Answer:

STOP BEING PRESSED

Explanation:

ANSWER  B

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Until a train is a safe distance from the station it must travel at 5 m/s. Once the train is on open track it can speed up
kirza4 [7]

Answer:

I believe the answer is b

Explanation:

5 0
3 years ago
A car has an engine which delivers a constant power. It accelerates from rest at time t = 0, and at t = t0 its acceleration is a
olga2289 [7]
Here is the answer of the given problem above.
Use this formula: <span>P = FV = ma*at = ma^2 t
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then, solve for x which is the acceleration at 2t.
</span>The <span>answer would be a0/sqrt(2).
Hope this answers your question. Thanks for posting.
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7 0
3 years ago
Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m
Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

A=0.0198

From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

5 0
3 years ago
75 pts!
jasenka [17]

again CORRECT ans is 4) light strikes the grooves → different wavelengths of light bend at different angles → diffracted wavelengths reach the eyes → the eyes see different colors.

moderator - plz review the ans as u deleted my right ans n approved the wrong ans :(

6 0
3 years ago
Read 2 more answers
HELP!!!! WILL MARK BRAINLIEST!!!! IF YOU LEAVE AN ANSWER EXPLAIN! THANKS
Travka [436]

Answer:

the answer is C

Explanation:

we know this because if you compare the graphs and look at the direction. it isn't always in the explanation or the few sentences they gave you at the top. also, look at the waves, you can see in Davids drawing that it is directly straight up, A and B do not represent that. A isn't even a valid answer. Notice also in A that the arrow is going in the completely different direction than in Davids drawing. B is also going a different direction even though it is only turned a little bit although if it was straight up like Davids drawing then it would most likely be a correct answer. C does have one arrow going a different direction but look at how it has two, showing in which if the waves were to turn then the arrow is still valid

7 0
3 years ago
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