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ivolga24 [154]
3 years ago
10

Sound travels through air at 343 m/s at 20 °C. A bat emits an ultrasonic squeak and hears the echo 0.05 second later . How far a

way was the object that reflected it?
Physics
1 answer:
Rashid [163]3 years ago
7 0

Answer:

Distance of the object is 8.6 m

Explanation:

As we know that the speed of sound at t degree C is given as

v = 332 + 0.6 t

here we know that the temperature is

t = 20 degree C

so we have

v = 332 + 0.6(20)

v = 344 m/s

now we know that bat heard the echo of sound in 0.05 s

so the to and fro distance of the object is d + d = 2d

so we have

2 d = v\times t

2d = 344(0.05)

d = 8.6 m

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an object has a displacement of + 12 m in 6 seconds if the average velocity is found using the equation change in position / ela
rosijanka [135]

Answer:

+2m/s

Explanation:

average   velocity = displacement traveled / total time taken

                           = +12m/ 6s

                           = +2 m/s

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3 years ago
4: Question 2 - 10.0 pts possible
Orlov [11]

Answer:

5. All of the statements are true; non is false

Explanation:

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2 years ago
A container is filled to a depth of 25.0 cm with water. On top of the water floats a 32.0-cm-thick layer of oil with specific gr
den301095 [7]

Answer:

the answer is 32.0-25.0=7

Explanation:

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6 0
2 years ago
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.
Anettt [7]

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>

6 0
2 years ago
A 5.6 cm diameter parallel-plate capacitor has a 0.58 mm gap. What is the displacement current in the capacitor if the potential
BARSIC [14]

Answer:

1.88\cdot 10^{-5} A

Explanation:

The capacitance of a parallel plate capacitor is given by:

C=\frac{\epsilon_0 A}{d} (1)

where

\epsilon_0 is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

The charge stored on the capacitor is given by

Q=CV (2)

where C is the capacitance and V is the voltage across the capacitor.

The displacement current in the capacitor is given by

J=\frac{Q}{t} (3)

where t is the time elapsed

Substituting (1) and (2) into (3), we find an expression for the displacement current:

J=\frac{CV}{t}=\frac{\epsilon_0 A}{d} \frac{V}{t}

where we have

A=\pi (\frac{d}{2})^2=\pi (\frac{0.056 m}{2})^2=2.46\cdot 10^{-3} m^2

d = 0.58 mm = 5.8\cdot 10^{-4} m

\frac{V}{t}=500,000 V/s

Substituting into the equation, we find

J=\frac{(8.85\cdot 10^{-12} F/m)(2.46\cdot 10^{-3} m^2)}{5.8\cdot 10^{-4}m}(500,000 V/s)=1.88\cdot 10^{-5} A

6 0
3 years ago
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