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NARA [144]
3 years ago
14

A bowling ball traveling with constant speed hits the pins at the end of a bowling lane 16.5 m long. The bowler hears the sound

of the ball hitting the pins 2.80 s after the ball is released from his hands. What is the speed of the ball, assuming the speed of sound is 340 m????s?
Physics
1 answer:
Strike441 [17]3 years ago
4 0

Answer:

5.997m/s

Explanation:

We were told to calculate the speed of the ball,

Given speed of sound as 340 m

And we know that the sound of the ball hitting the pins is at 2.80 s after the ball is released from his hands.

Speed of ball = distance traveled/(time of hearing - time the sound travels).

Speed= S/t

Where S= distance traveled

t= time of hearing - time the sound travels

time=time for ball to roll+timefor sound to come back.

time of sound=16.5/340

=0.048529secs

solving for speedof ball

Then,Speed of ball = distance traveled/(time of hearing - time the sound travels).

=16.5/(2.80-0.048529) m/s = 5.997m/s

Therefore, the speed of the ball is

5.997m/s

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41.1 ÷ 40.0

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Explanation:

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7 0
2 years ago
Two objects 5 kg and 7 kg are attached to the end of inextensible string which passes over frictionless pulley
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Answer:

ans:

tenson(T) = 20 N

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Explanation:

T + mg = Mg

T = Mg - mg

T = g( M - m )

T = 10× ( 7-5 )

T = 20 N

again;

T = 20

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a = 20 / 7

= 2.86 m/s

3 0
2 years ago
La tensión en newtons necesaria para que una onda transversal cuya longitud de onda es 3.33 cm vibre a razón de 625 ciclos por s
NemiM [27]

Answer:

9.34 N

Explanation:

First of all, we can calculate the speed of the wave in the string. This is given by the wave equation:

v=f \lambda

where

f is the frequency of the wave

\lambda is the wavelength

For the waves in this string we have:

f=625 Hz, since it completes 625 cycles per second

\lambda=3.33 cm = 0.033 m is the wavelength

So the speed of the wave is

v=(625)(0.0333)=20.6 m/s

The speed of the waves in a string is related to the tension in the string by

v=\sqrt{\frac{T}{\mu}} (1)

where

T is the tension in the string

\mu=\frac{m}{L} is the linear density

In this problem:

m=16.5 g = 16.5\cdot 10^{-3} kg is the mass of the string

L = 0.75 m is the its length

Solving the equation (1) for T, we find the tension:

T=\mu v^2 = \frac{m}{L} v^2 = \frac{16.5\cdot 10^{-3}}{0.75}(20.6)^2=9.34 N

8 0
3 years ago
4.39 moles of gas in a box has a pressure of 2.25 atm at temperature of 385K. What is the volume of the box?
bazaltina [42]

Answer:

0.0619 m^3

Explanation

number of moles = n = 4.39 mol

pressure = P = 2.25 atm =2.25×1.01×10^5 Pa= 2.27×10^5 Pa

Molar gas constant =R = 8.31 J/(mol K)

Temperature T= 385K

volume of gas = V =?

BY GENERAL GAS LAW WE HAVE

PV = nRT

or V = nRT/P

or V = (4.39×8.31×385)/(2.27×10^5)

V = 0.0618728

V =  0.0619 m^3

5 0
3 years ago
Read 2 more answers
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