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Scorpion4ik [409]
3 years ago
13

Which scenario is the best example of matter absorbing energy from a wave?

Physics
2 answers:
julia-pushkina [17]3 years ago
7 0

Answer:

C I think so

Explanation:

ocean waves hitting a concert barrier that reduces their energy

dimulka [17.4K]3 years ago
7 0

Answer:

whats the answer ppl

Explanation:

why haven't no one else try to do this lile come on. this app is for ppl to help each other. like come on.

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Why is a device never 100% efficient? *
oksano4ka [1.4K]
I think it’s c, friction
I’m not sure tho
5 0
3 years ago
A 7450 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25 m/s2 and feels no appreci
sashaice [31]

Answer:

a) 112.5 m

b) 15.81s

Explanation:

a)We can use the following equation of motion to calculate the velocity v of the rocket at s = 500 m at a constant acceleration of a = 2.25 m/s2

v^2 = 2as

v^2 = 2*2.25*500 = 2250

v = \sqrt{2250} = 47.4 m/s

After the engine failure, the rocket is subjected to a constant deceleration of g = -10 m/s2 until it reaches its maximum height where speed is 0. Again if we use the same equation of motion we can calculate the vertical distance h traveled by the rocket after engine failure

0^2 - v^2 = 2gh

-2250 = 2(-10)h

h = 2250/20 = 112.5 m

So the maximum height that the rocket could reach is 112.5 + 500 = 612.5 m

b) Using ground as base 0 reference, we have the following equation of motion in term of time when the rocket loses its engine:

s + vt + gt^2/2 = 0

500 + 47.4t - 10t^2/2 = 0

5t^2 - 47.4t - 500 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{47.4\pm \sqrt{(-47.4)^2 - 4*(5)*(-500)}}{2*(5)}

t= \frac{47.4\pm110.67}{10}

t = 15.81 or t = -6.33

Since t can only be positive we will pick t = 15.81s

7 0
3 years ago
Read 2 more answers
The output power of a solar panel is the rate of transfer of...?
gtnhenbr [62]
Power is the rate of generating, moving, or using energy.
5 0
3 years ago
Two Carnot air conditioners, A and B, are removing heat from different rooms. The outside temperature is the same for both rooms
lakkis [162]

Answer:

(a)  333.77 J

(b) 237.85 J

(c) 4763.77 J

(d) 4667.85 J

Explanation:

Temperature of source, TH = 314 K

Temperature of A, Tc = 292 K

Temperature of B, Tc' = 298 K

heat taken out, Qc = 4430 J

Let the heat deposited outside is QH and QH' by A and B respectively.

\frac{Q_H}{Q_c}=\frac{T_H}{T_c}\\\\Q_H = \frac{4430\times314}{292}=4763.77 J

Now

\frac{Q_H'}{Q_c}=\frac{T_H}{T_c'}\\\\Q_H' = \frac{4430\times314}{298}=4667.85 J

(a) Work done for A

W = QH - QC = 4763.77 - 4430 = 333.77 J

(b) Work done for B

W' = QH' - Qc = 4667.85 - 4430 = 237.85 J

(c) QH = 4763.77 J

(d) QH' = 4667.85 J

4 0
3 years ago
Please Help No Links
ale4655 [162]

Answer:

the moment of inertia

Explanation:

5 0
3 years ago
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