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Talja [164]
3 years ago
5

Suppose that you hear a clap of thunder 16 s after seeing the associated lightning strike. How far are you from the lightning st

rike? The speed of sound waves in air is 336 m/s and the speed of light in air is 3 × 108 m/s. Answer in units of km
Physics
1 answer:
riadik2000 [5.3K]3 years ago
4 0

Answer:

d=5.376km

Explanation:

Since <em>light is so fast</em> we can assume no time passes between the lightning strikes and we observe it. We want to know then how far away did the strike occur for the sound to take 16s to reach our ears. Since the definition of velocity tells us that v=d/t, we can write d=vt=(336m/s)(16s)=5376m=5.376km

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A bowling ball with a mass of 8 kg is moving at a speed of 5 m/s. What is its<br> kinetic energy?
natali 33 [55]

Answer:

100J

Explanation:

Kinetic energy=1/2mv^2

Kinetic energy=(1/2 x 8)x5^2

Kinetic energy=4x25

Kinetic energy=100

100J

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a worker uses a board that is 7 m long to pry up a bolder A small rock is used for the fulcrum and is placed 2.5 m from the resi
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3 years ago
A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 65 m horizontally
aivan3 [116]

Answer:

option B

Explanation:

given,

height of building = 0.1 km

ball strikes horizontally to ground at = 65 m    

speed at which the ball strike = ?

vertical velocity  = 0 m/s

time at which the ball strike

s = \dfrac{1}{2}gt^2

t = \sqrt{\dfrac{2s}{g}}

t = \sqrt{\dfrac{2\times 100}{9.8}}

t = 4.53 s

vertical velocity at the time  4.53 s = g × t = 9.8 × 4.53 = 44.39 m/s

horizontal velocity = \dfrac{65}{4.53} =14.35 m/s

speed of the ball = \sqrt{44.39^2+14.35^2}

                             = 46.65 m/s

hence, the speed of the ball just before it strike the ground = 47 m/s

The correct answer is option B

5 0
3 years ago
A sample of an ideal gas has a volume of 2.37 L at 2.80×102 K and 1.15 atm. Calculate the pressure when the volume is 1.68 L and
iogann1982 [59]

Answer:

p_2 = 1.76 atm

Explanation:

given data:

v_1 = 2.37 L

v_2 = 1.68 L

p_1 =1.15 atm

p_2 = ?

t_1 = 280 K

t_2 = 304 K

from Gas Law Equation

, WE HAVE

\frac{p_1 v_1}{t_1} =\frac{p_2 v_2}{t_2}

Putting the values

\frac{1.15*2.37}{280}  =\frac{p_2 *1.68}{304}

9.733*10^{-3} = \frac{p_2 *1.68}{304}

9.733*10^{-3}*304 = p_2*1.68

\frac{9.733*10^{-3}*304}{1.68} =p_2

p_2= 1.76 atm

7 0
3 years ago
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