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vekshin1
3 years ago
15

On a day when the velocity of sound is 340 m/s, a boy drops a stone from the top of a high tower. If the tower is 50 m tall, how

long will it take for the boy to hear the returning sound?
Physics
1 answer:
luda_lava [24]3 years ago
4 0

Answer:

It takes 3.31 s for the boy to hear the returning sound of the stone

Explanation:

Ok, the time after which splash is heard is equal to the time acquired by the stone to reach the ground + time taken by sound to return.

First, u=0 m/s (this is the initial speed with which the stone was thrown), h=50 m (is the height of the tower) and g=10 m/s^{2} (is the gravity)

We can use this equation that relates the free fall of the object:

h=u+\frac{1}{2} gt^{2}

50=0+\frac{1}{2}(10)t^{2}

50=5t^{2}

10=t^{2}

and t=\sqrt{10}=3.16 s

t =3.16 s is the time it takes the stone to fall

Second, h=50 m and v=340 m/s (that it's the speed of sound)

We can use this equation:

h=vt

t=\frac{h}{v}

t=\frac{50}{340}=0,15 s

t=0.15 is the time it takes for the sound to return

Finally, both times are added, obtaining the time in which the boy will hear the returning sound of the stone:

t= 3.16 s + 0.15s=3.31 s

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Ultrasound with a frequency of 4.257 MHz can be used to produce images of the human body. If the speed of sound in the body is t
shutvik [7]

Answer:

2.49 * 10^(-4) m

Explanation:

Parameters given:

Frequency, f = 4.257 MHz = 4.257 * 10^6 Hz

Speed of sound in the body, v = 1.06 km/ = 1060 m/s

The speed of a wave is given as the product of its wavelength and frequency:

v = λf

Where λ = wavelength

This implies that:

λ = v/f

λ = (1060) / (4.257 * 10^6)

λ = 2.49 * 10^(-4) m

The wavelength of the sound in the body is 2.49 * 10^(-4) m.

4 0
3 years ago
A high-jump athlete leaves the ground, lifting her center of mass 1.8 m and crossing the bar with a horizontal velocity of 1.4 m
Romashka [77]

Answer:

The minimum speed when she leave the ground is 6.10 m/s.

Explanation:

Given that,

Horizontal velocity = 1.4 m/s

Height = 1.8 m

We need to calculate the minimum speed must she leave the ground

Using conservation of energy

K.E+P.E=P.E+K.E

\dfrac{1}{2}mv_{1}^2+0=mgh+\dfrac{1}{2}mv_{2}^2

\dfrac{v_{1}^2}{2}=gh+\dfrac{v_{2}^2}{2}

Put the value into the formula

\dfrac{v_{1}^2}{2}=9.8\times1.8+\dfrac{(1.4)^2}{2}

\dfrac{v_{1}^2}{2}=18.62

v_{1}=\sqrt{2\times18.62}

v_{1}=6.10\ m/s

Hence, The minimum speed when she leave the ground is 6.10 m/s.

6 0
3 years ago
A parallel plate capacitor with plates of area A and plate separation d is charged so that the potential difference between its
faust18 [17]

Answer:

Explanation:

Given a parallel plate capacitor of

Area=A

Distance apart =d

Potential difference, =V

If the distance is reduce to d/2

What is p.d

We know that

Q=CV

Then,

V=Q/C

Then this shows that the voltage is inversely proportional to the capacitance

Therefore,

V∝1/C

So, VC=K

Now, the capacitance of a parallel plate capacitor is given as

C= εA/d

When the distance apart is d

Then,

C1=εA/d

When the distance is half d/2

C2= εA/(d/2)

C2= 2εA/d

Then, applying

VC=K

V1 is voltage of the full capacitor V1=V

V2 is the required voltage let say V'

Then,

V1C1=V2C2

V × εA/d=V' × 2εA/d

VεA/d = 2V'εA/d

Then the εA/d cancels on both sides and remains

V=2V'

Then, V'=V/2

The potential difference is half when the distance between the parallel plate capacitor was reduce to d/2

6 0
3 years ago
Read 2 more answers
a 60g tennis ball travelling at 30m/s hi a wall and bounce back at 20m/s.calculate (a)momentum of the ball before impact (b)mome
frez [133]

Explanation:

the formula for momentum is denoted by p=mv where p is momentum, m is mass and v is velocity. thus, the velocity before impact would be 0.060 x 30 = 1.8 kg/ms

the second one would just be 0.060 x 20 0.72kg/ms

I'm not 100 percent sure this is correct but yeah

6 0
3 years ago
Describe the motion of an automobile on an east-west
Tanya [424]

Answer:

1. The automobile is traveling due east and is speeding up.

2. The car is traveling due east and is is slowing down.

3. The automobile is traveling due east at a constant speed.

4. The car is traveling due west and is slowing down.

5. The automobile is traveling due west and is speeding up.

6. The automobile is traveling due west at a constant speed.

7. The automobile is accelerating due east from rest.

8. The automobile is accelerating due west from rest.

Explanation:

The key to understanding this is:

When the acceleration and initial velocity of the automobile have the same sign (positive or negative) then the automobile is speeding up. Explained further, if acceleration and the initial velocity are both positive or they are both negative the automobile is speeding up but whenever they have opposite signs (that is acceleration is positive and initial velocity is negative or vice versa) the automobile is slowing down. When the acceleration is zero the automobile is maintaining a unform motion at a constant speed (the speed is not changing with time). The + or - sign indicates the direction of travel. In this case east is + and west is -. It is my pleasure answering this question. I hope you find it helpful. Thank you.

4 0
3 years ago
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