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scZoUnD [109]
3 years ago
6

A rocket achieves a lift-off velocity of 500.0 m/s from rest in 30.0 seconds. Calculate the average acceleration of the rocket.

Physics
1 answer:
Anettt [7]3 years ago
6 0

Answer:

The average acceleration is 16.6 m/s² ⇒ 1st answer

Explanation:

A rocket achieves a lift-off velocity of 500.0 m/s from rest in

30.0 seconds

The given is:

→ The initial velocity = 0

→ The final velocity = 500 meters per seconds

→ The time is 30 seconds

Acceleration is the rate of change of velocity of the rocket

→ a=\frac{v-u}{t}

where a is the acceleration, v is the final velocity, u is the initial velocity

and t is the time

→ u = 0 , v = 500 m/s , t = 30 s

Substitute these values in the rule

→ a=\frac{500-0}{30}=\frac{500}{30}=16.6 m/s²

<em>The average acceleration is 16.6 m/s²</em>

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Answer:

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Explanation:

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3 years ago
A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and whirling it aroun
cluponka [151]

Answer:Highest frequency  =618.89Hz

Lowest frequency=582.22Hz

Explanation:

 The linear velocity of a sound generator  is related to angular velocity and is given as

Vs = rω  where

r = the radius of circular path = 1.0 m

ω is the angular velocity of the sound generator. = 100 rpm

1 rev/min = 0.10472 rad/s

100rpm =10.472 rad/ s

Vs = rω

= 1m x 10.472rad/ s=  10.472m/s

A) Highest frequency  heard by a student in the classroom = Maximum frequency. Using the Doppler effect formulae,

f max = (v/ v-vs) fs

Where , v is the speed of the sound in air at 20 degrees celcius =

343 metres per second

vs is the linear velocity of the sound generator=10.472m/s

fs is the frequency of the sound generator= 600 Hz

f max = (343/ 343 - 10.472) x 600

=343/332.528) x600

=618.89Hz

B) Lowest frequency  heard by a student in the classroom = Minimum frequency

f min = (v/ v+vs) fs

(343/ 343 + 10.472) x 600

=343/353.472) x 600

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5 0
3 years ago
A parallel-plate capacitor is charged by connecting it to a battery. If the battery is disconnected and then the separation betw
TEA [102]

Answer:

The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)

Explanation:

The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. Q, the amount of charge stored in this capacitor, will stay the same.

The formula \displaystyle Q = C\, V relates the electric potential across a capacitor to:

  • Q, the charge stored in the capacitor, and
  • C, the capacitance of this capacitor.

While Q stays the same, moving the two plates apart could affect the potential V by changing the capacitance C of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:

\displaystyle C = \frac{\epsilon\, A}{d},

where

  • \epsilon is the permittivity of the material between the two plates.
  • A is the area of each of the two plates.
  • d is the distance between the two plates.

Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of \epsilon. Neither will that change the area of the two plates.

However, as d (the distance between the two plates) increases, the value of \displaystyle C = \frac{\epsilon\, A}{d} will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.

On the other hand, the formula \displaystyle Q = C\, V can be rewritten as:

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The value of Q (charge stored in this capacitor) stays the same. As the value of C becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.  

3 0
3 years ago
Can someone pls help, thank you in advance!
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A) Geothermal
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4 0
3 years ago
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