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3 years ago

A bowling ball of 35.2kg, generates 218 kg.m/s units of momentum. What is the velocity of the bowling ball?

Physics

1 answer:

brilliants [131]3 years ago

8 0

Answer:

$6.18\:\mathrm{m/s}$

Explanation:

The momentum of an object is given by $p=mv$ .

Plugging in values, we have:

$218=35.2v,\\v=\frac{218}{35.2}=\fbox{$6.18\:\mathrm{m/s}$}$ .

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A student practicing for a track meet ran 3500 m in 500 s. What was her average speed? a. 4,000 m/s b. 0.14 m/s c. 7 m/s d. 1,75

gregori [183]

Answer:

c. 7 m/s

Explanation:

3500 m / 500 s = 7 m/s

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What are some of the phenotypic plant traits that scientists are investigating? Why do you think these traits were chosen?

vodomira [7]

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<u>Explanation</u>:

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3 years ago

A paintball is fired horizontally from a tower 45 m above the ground. If the paintball gun fires at 90 m/s… How long does it tak

ankoles [38]

Answer:

The time taken for the paint ball to hit the ground is $t = 2.143 \ s$

The distance of the landing point from the tower is $d = 192.86 \ m$

Explanation:

From the question we are told that

The height of the tower is $h = 45 \ m$

The speed of the paintball in the horizontal direction is $v_x = 90 \ m/s$

Generally from kinematic equation we have that

$h = ut + \frac{1}{2} at^2$

Here u is the initial velocity of the paintball in the vertical direction and the value is 0 m/s , this because the ball was fired horizontally

a is equivalent to $g = 9.8\ m/s^2$

t is the time taken for the paintball to hit the ground

$45 = 0* t + \frac{1}{2} 9.8 * t^2$

=> $t = 2.143 \ s$

Generally the distance of its landing position from the tower is

$d = v * t$

=> $d = 90 * 2.143$

=> $d = 192.86 \ m$

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3 years ago

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Butoxors [25]

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4 years ago