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Cerrena [4.2K]
2 years ago
12

A car runs along a horizontal path at speed of 20m/s.The driver observes the rain hitting his car at 60 to vertical.If rain is a

ctually falling vertically then what is speed of rain drop
Physics
1 answer:
Mama L [17]2 years ago
7 0

Answer:

The speed of the raindrop is 11.55 m/s.

Explanation:

We need to find the speed of the rain in the "y" direction. We have:

v_{r}: speed of the raindrop at 60° (vertical) =?

v_{c}: speed of the car = 20 m/s (in the "x" direction)

Hence, the speed of the car and the speed of the raindrop are related by:

tan(\theta) = \frac{v_{c}}{v_{r}}        

v_{r} = \frac{v_{c}}{tan(60)} = \frac{20}{tan(60)} = 11.55 m/s

Therefore, the speed of the raindrop is 11.55 m/s.

I hope it helps you!      

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A piece of aluminum has a volume of 1.50 10-3 m3. the coefficient of volume expansion for aluminum is β = 69 ✕ 10-6 (°c)-1. the
Alex17521 [72]

Answer:

W = 3.12 J

Explanation:

Given the volume is 1.50*10^-3  m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:

β = 69*10^-6 (°C)^-1       V = 1.50*10^-3  m^3       ΔT = 298°C

So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):

dV = \beta V_{0} dT\\dV = (69*10^{-6})( C)^{-1} * (1.50*10^{-3})m^{3} * (298)C\\dV = 3.0843*10^-5

So ΔV = 3.0843*10^-5 m^3

Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa

To get work, multiply the air pressure and the volume change.

W = P * dV = (1.01 * 10^5)Pa * (3.0843*10^{-5})m^3\\W = 3.115143J

W = 3.12 J

Hope this helps!

4 0
3 years ago
The electric force between two charged balloons is 0.12 newtons. If the distance between the two balloons is halved, what will b
marshall27 [118]

Answer:

The answer is D, I just took the test

Explanation:

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A gymnast dismounts off the uneven bars in a tuck position with a radius of 0.3m (assume she is a solid sphere) and an angular v
kifflom [539]

Here we will say that there is no external torque on the system so we will have

L_i = L_f

here we know that

L_i = I_1\omega_1

where we know that

I_1 = \frac{2}{5}mr^2

Also we know that

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initial angular speed will be

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now we have

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6 0
2 years ago
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