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Nastasia [14]
3 years ago
15

_____ is the preceived frequency of a sound wave.​

Physics
1 answer:
Lina20 [59]3 years ago
7 0
I guess the answer is pitch
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If a car travels 400m in 20 seconds how fast is it going? 20 m/s
scoundrel [369]

Answer:

20m/s

Explanation:

Speed = distance / time

Speed = 400/20

Speed = 20

5 0
3 years ago
Sarah launches her purse straight up in air with a velocity of 30.2m/s<br> How high will it go?
yaroslaw [1]

Answer: 46.53

Explanation:

6 0
3 years ago
A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at lif
Serggg [28]

Answer:

Height of the rocket be one minute after liftoff is 40.1382 km.

Explanation:

v(t)=-gt-v_e\times \ln \frac{m-rt}{m}

v = velocity of rocket at time t

g = Acceleration due to gravity =9.8 m/s^2

v_e = Constant velocity relative to the rocket = 2,900m/s.

m = Initial mass of the rocket at liftoff = 29000 kg

r = Rate at which fuel is consumed = 170 kg/s

Velocity of the rocket after 1 minute of the liftoff =v

t = 1 minute = 60 seconds'

Substituting all the given values in in the given equation:

v(60)=-9.8 m/s^2\times 60 s-2,900m/s\times \ln (\frac{29,000 kg-170 kg/s\times 60 s}{2,9000 kg})

v(60) = 668.97 m/s

Height of the rocket = h

Velocity=\frac{Displacement}{time}

668.97 m/s=\frac{h}{60 s}

h=668.97 m/s\times 60 s=40,138.2 m = 40.1382 km

Height of the rocket be one minute after liftoff is 40.1382 km.

4 0
3 years ago
A string under a tension of 50.4 N is used to whirl a rock in a horizontal circle of radius 2.51 m at a speed of 21.1 m/s. The s
Leokris [45]

Answer:

619.8 N

Explanation:

The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:

T=m\frac{v^2}{r}

where

T is the tension

m is the mass of the rock

v is the speed

r is the radius of the circular path

At the beginning,

T = 50.4 N

v = 21.1 m/s

r = 2.51 m

So we can use the equation to find the mass of the rock:

m=\frac{Tr}{v^2}=\frac{(50.4)(2.51)}{21.1^2}=0.284 kg

Later, the radius of the string is decreased to

r' = 1.22 m

While the speed is increased to

v' = 51.6 m/s

Substituting these new data into the equation, we find the tension at which the string breaks:

T'=m\frac{v'^2}{r'}=(0.284)\frac{(51.6)^2}{1.22}=619.8 N

5 0
3 years ago
Help with question 16 and 17
storchak [24]
<h3>16.</h3>

Your answer is correct.

___

<h3>17.</h3>

The fractional change in resistance is equal to the given temperature coefficient multiplied by the change in temperature.

  R = R₀×(1 + α×ΔT)

  R = (10.0 Ω)×(1 + 0.004×(65 -20)) = 11.8 Ω

5 0
3 years ago
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