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suter [353]
3 years ago
15

PLZZZ HELP

Physics
1 answer:
matrenka [14]3 years ago
4 0

B. A sandbar is formed by water. A sand dune is formed by wind.

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A child pulls a sled up a snow covered hill. If the child does 504J of work on the sled while pulling the sled 23m up the hill t
zhenek [66]

Explanation:

ans is equal to 504j* 23 m* 10 ms

7 0
2 years ago
How does a physicist answer a scientific question?
lara31 [8.8K]

Answer:

b

Explanation:

the answer is B

3 0
2 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:

V = \displaystyle \frac{Q}{C}.

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
A wave has a frequency of 6 Hz and a speed of 30 m/s. What is the wavelength of the wave?
Helen [10]

Answer:

D. 5m

Explanation:

fλ = c, where f is frequency, λ is wavelength and c is speed.

6λ=30

λ=30/6=5

5 0
3 years ago
You are standing on a sidewalk. There is a car in the distance. The horn on the car sounds. You hear it at a pitch that correspo
AnnyKZ [126]

Answer: The answer: The car is moving away from you.

Both A and C are true as Car can be moving in line away from you or has component of velocity in opposite direction.

Explanation:The decrease in the frequency of the sound is the result of Doppler's effect. A/c to Doppler's effect the frequency of received sound of source is changed if it is moving relative to the receiver, i.e. the distance between them is changing due to motion.

The general formula of Doppler's Effect is attached as the picture.

In this formula v_D is the velocity of Detector i.e the receiver relative to wind. While v_s is the velocity of source relative to wind and v is the velocity of sound.

The Doppler's effect is not effected by the velocity of wind as the wind itself could not change the distance between the two objects i.e. you and the car. Wind velocity can change the speed of sound and its wavelength but the change does not effect the frequency.

Hence if we assume the car to be moving with velocity v_c and you are stationary

f'=f_s*\frac{v}{v-v_c}

hence the frequency is reduced.

4 0
3 years ago
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