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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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how much gravitational potential energy do you give a 70 kg person when you lift him up 3 m in the air?
SCORPION-xisa [38]

Given gravitational potential energy when he's lifted is 2058 J.

Kinetic energy is transferred to the person.

Amount of kinetic energy the person has is -2058 J

velocity of person = 7.67 m/s².

<h3>Explanation:</h3>

Given:

Weight of person = 70 kg

Lifted height = 3 m

1. Gravitational potential energy of a lifted person is equal to the work done.

PE_g=W=m\times g\times h\\Acceleration due to gravity = g = 9.8 \ m/s^2 \\PE_g= m = m\times g\times h= 70\times 9.8 \times 3 = 2058\ kg.m/s^2 = 2058\ J

Gravitational potential energy is equal to 2058 Joules.

2. The Gravitational potential energy is converted into kinetic energy. Kinetic energy is being transferred to the person.

3. Kinetic energy gained = Potential energy lost = -PE_g = -2058\ kg.m/s^2

Kinetic energy gained by the person = (-2058 kg.m/s²)

4. Velocity = ?

Kinetic energy magnitude= \frac{1}{2} m\times v^2 = m\times g \times h

Solving for v, we get

v=\sqrt{2gh} =\sqrt{2\times 9.8 \times 3} = \sqrt{58.8} = 7.67 m/s^2

The person will be going at a speed of 7.67 m/s².

4 0
3 years ago
The graphs display velocity data Velocity is on the y-axis (m/s), while time is on the x-axis (s). Based on the graphs, which da
RUDIKE [14]

Answer:

The first graph is showing the constant acceleration (1 m/s)

Explanation:

The second graph showing the flexible velocity therefore a in the graph is different at t1, t2, t3, t4

The last graph is showing constant velocity therefore there is no acceleration (a = 0)

5 0
3 years ago
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6. What are the four chambers of the heart? A. atrium, artery, ventricle, vein B. artery, blood vessel, capillary, vein C. blood
neonofarm [45]

Answer:

The answer is D. right atrium, left atrium, right ventricle, left ventricle

Explanation:

The right atrium receives oxygen-poor blood from the body and pumps it to the right ventricle.The right ventricle pumps the oxygen-poor blood to the lungs.The left atrium receives oxygen-rich blood from the lungs and pumps it to the left ventricle.The left ventricle pumps the oxygen-rich blood to the body.

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3 years ago
A hockey puck with a mass of 0.159 kg slides over the ice. The puck initially slides with a speed of 4.75 m/s, but it comes to a
Anton [14]

Answer:

The energy dissipated as the puck slides over the rough patch is 1.355 J

Explanation:

Given;

mass of the hockey puck, m = 0.159 kg

initial speed of the puck, u = 4.75 m/s

final speed of the puck, v = 2.35 m/s

The energy dissipated as the puck slides over the rough patch is given by;

ΔE = ¹/₂m(v² - u²)

ΔE = ¹/₂ x 0.159 (2.35² - 4.75²)

ΔE = -1.355 J

the lost energy is 1.355 J

Therefore, the energy dissipated as the puck slides over the rough patch is 1.355 J

5 0
3 years ago
A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing throu
Setler79 [48]
<h2>Given :</h2>

  • total charge = 9.0 mC = 0.009 C

Each electron has a charge of :

1.6 \times 10 {}^{ - 19} \:  C

For producing 1 Cuolomb charge we need :

  • \mathrm{\dfrac{1}{1.6 \times 10 {}^{ - 19} } }

  • \dfrac{10 {}^{19} }{1.6}

  • \dfrac{10\times 10 {}^{19} }{16}

  • \dfrac{100 \times 10 {}^{18} }{16}

  • \mathrm{6.24 \times 10 {}^{18}  \:  \: electrons}

Now, for producing 0.009 C of charge, the number of electrons required is :

  • 0.009 \times 6.24 \times  {10}^{18}

  • 0.05616 \times 10 {}^{18}

  • \mathrm{5.616 \times 10 {}^{16}  \:  \: electons}

_____________________________

So, Number of electrons passing through the cross section in 3.6 seconds is :

\mathrm{5.616 \times 10 {}^{16} \:  \: electrons}

Number of electrons passing through it in 1 Second is :

  • \dfrac{5.616 \times  {10}^{16} }{3.6}

  • \mathrm{1.56 \times 10 {}^{16}  \:  \: electrons}

Now, in 10 seconds the number of electrons passing through it is :

  • 10 \times  \mathrm{1.56 \times 10 {}^{16}  \:  \: }

  • \mathrm{1.56 \times 10 {}^{17}  \:  \: electrons}

_____________________________

\mathrm{ \#TeeNForeveR}

6 0
2 years ago
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