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Likurg_2 [28]
3 years ago
5

Please help will mark brainleist ASAP

Physics
2 answers:
-Dominant- [34]3 years ago
8 0
B because I took that test
kiruha [24]3 years ago
4 0

Answer:

B - Earth's path around the Sun

Explanation:

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As an object moves, the distance it travels increases with time.<br><br> Agree<br> Disagree
Mnenie [13.5K]
Agree

Hope this helps
4 0
3 years ago
Read 2 more answers
Which is greater, the force exerted by the Earth on the Sun, or the force exerted by the Sun on the Earth? Why?
LekaFEV [45]

Answer:

There is no great force, the force exerted by the Earth on the Sun, and the force exerted by the Sun on the Earth are equal

Explanation:

By definition...

3 0
3 years ago
Help plzzzzzzz i need thissssssssss
frutty [35]

Answer:

The final graph

Explanation:

The graph that curves downwards is negative acceleration. While the position decreases the slop increases.

8 0
3 years ago
Suppose that the sound level of a conversation is initially at an angry 70 db and then drops to a soothing 50 db. assuming that
stepan [7]
Angry sound level = 70 db
Soothing sound level = 50 db
Frequency, f = 500 Hz
Assuming speed of sound = 345 m/s
Density (assumed) = 1.21 kg/m^3
Reference sound intensity, Io = 1*10^-12 w/m^2

Part (a): Initial sound intensity (angry sound)
10log (I/Io) = Sound level
Therefore,
For Ia = 70 db
Ia/(1*10^-12) = 10^(70/10)
Ia = 10^(70/10)*10^-12 = 1*10^-5 W/m^2

Part (b): Final sound intensity (soothing sound)
Is = 50 db
Therefore,
Is = 10^(50/10)*10^-12 = 18*10^-7 W/m^2

Part (c): Initial sound wave amplitude
Now,
I (W/m^2) = 0.5*A^2*density*velocity*4*π^2*frequency^2

Making A the subject;
A = Sqrt [I/(0.5*density*velocity*4π^2*frequency^2)]

Substituting;
A_initial = Sqrt [(1*10^-5)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-8 m = 69.7 nm

Part (d): Final sound wave amplitude
A_final = Sqrt [(1*10^-7)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-9 m = 6.97 nm
4 0
4 years ago
At the Indianapolis 500, you can measure the speed of cars just by listening to the difference in pitch of the engine noise betw
allsm [11]

To develop this problem it is necessary to apply the concepts related to the Dopler effect.

The equation is defined by

f_i = f_0 \frac{c}{c+v}

Where

f_h= Approaching velocities

f_i= Receding velocities

c = Speed of sound

v = Emitter speed

And

f_h = f_0 \frac{c}{c+v}

Therefore using the values given we can find the velocity through,

\frac{f_h}{f_0}=\frac{c-v}{c+v}

v = c(\frac{f_h-f_i}{f_h+f_i})

Assuming the ratio above, we can use any f_h and f_i with the ratio 2.4 to 1

v = 353(\frac{2.4-1}{2.4+1})

v = 145.35m/s

Therefore the cars goes to 145.3m/s

7 0
3 years ago
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