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3 years ago

Mass of water= 357g density water= 1.0g/cm3

1 answer:

8 0

$m=357g\\\\\rho=1.0\ \dfrac{g}{cm^3}\\\\\rho=\dfrac{m}{V}\to V=\dfrac{m}{\rho}\\\\\text{substitute}\\\\V=\dfrac{357g}{1.0\frac{g}{cm^3}}=375g\cdot1.0\dfrac{cm^3}{g}=375\ cm^3$

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Answer:

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3 years ago

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Answer:

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3 years ago

A speed boat moving at a velocity of 25 m/s runs out of gas and drifts to a stop 3 minutes later 100 meters away. What is its ra

mr_godi [17]

<h2>Answer:</h2>

The rate of deceleration is -0.14 $m/s^{2}$

<h2>Explanation:</h2>

Using one of the equations of motion;

v = u + at

where;

v = final velocity of the boat = 0m/s (since the boat decelerates to a stop)

u = initial velocity of the boat = 25m/s

a = acceleration of the boat

t = time taken for the boat to accelerate/decelerate from u to v = 3 minutes

<em>Convert the time t = 3 minutes to seconds;</em>

=> 3 minutes = 3 x 60 seconds = 180seconds.

<em>Substitute the values of v, u, t into the equation above. We have;</em>

v = u + at

=> 0 = 25 + a(180)

=> 0 = 25 + 180a

<em>Make a the subject of the formula;</em>

=> 180a = 0 - 25

=> 180a = -25

=> a = -25/180

=> a = -0.14 $m/s^{2}$

The negative value of a shows that the boat is decelerating.

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4 years ago

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baherus [9]

Answer:

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Explanation:

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3 years ago

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earnstyle [38]

Answer:

120 kg m/s

Explanation:

The magnitude of the momentum of an object is given by

$p=mv$

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m is the mass of the object

v is its speed

For the block in this problem,

m = 10.0 kg (mass of the block)

v = 12.0 m/s (speed of the block)

Therefore the magnitude of the block's momentum is

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3 years ago