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emmainna [20.7K]
3 years ago
13

A 5 N force is applied to a 3 kg ball to change its velocity from 9 m/s to 3 m/s. What is the impulse on the ball ?

Physics
1 answer:
nika2105 [10]3 years ago
3 0
So, first the formula of Impulse is
I = force * time
We have force but no time.
Then, find time.
Next find acceleration,
F = mass * acceleration
5 = 3 * a
1.67 m/s^2
Next find time,
Acceleration = change in velocity / time
Change in velocity is velocity final - velocity initial
1.67 = 3 - 9 / time
Time = 3.6 s (round to 2 s.f.)
Lastly,
Impulse = force * time
Impulse = 5 * 3.6
Impulse is 18 Ns
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3 0
3 years ago
When you take your 1900-kg car out for a spin, you go around a corner of radius 55 m with a speed of 15 m/s. The coefficient of
pentagon [3]

Answer:

7772.72N

Explanation:

When u draw your FBD, you realize you have 3 forces (ignore the force the car produces), gravity, normal force and static friction. You also realize that gravity and normal force are in our out of the page  (drawn with a frame of reference above the car). So that leaves you with static friction in the centripetal direction.

Now which direction is the static friction, assume that it is pointing inward so

Fc=Fs=mv²/r=1900*15²/55=427500/55=7772.72N

Since the car is not skidding we do not have kinetic friction so there can only be static friction. One reason we do not use μFn is because that is the formula for maximum static friction, and the problem does not state there is maximum static friction.

7 0
3 years ago
A ball is thrown upward. As it passes 5.0 m height it is traveling at 4.0 m/s up. What was its initial upward velocity? (a) 7.0
sveticcg [70]

Answer:

c) 10.7m/s

Explanation:

From the exercise we know that at 5m the ball  is traveling at 4m/s

To calculate its initial velocity we need to solve the following equation:

v_{y}^{2}=v_{oy}^{2}+2g(y-y_{o})

Since the initial height is 0

Solving for v_{o}

v_{oy}=\sqrt{v_{y}^{2}-2gy}=\sqrt{(4m/s)^2-2(-9.8m/s^2)(5m)}=10.7m/s

5 0
3 years ago
A 9.0 kg bowling ball races down the lane at 15 m/s before striking a bowling pin (at rest) with a mass of 0.85 kg. If the 0.85
artcher [175]

Answer:

v = 10.75\,\frac{m}{s}

Explanation:

The system ball-pin is modelled by the Principle of Moment Conservation:

(9\,kg)\cdot (15\,\frac{m}{s} ) + (0.85\,kg)\cdot (0\frac{m}{s} ) = (9\,kg)\cdot v + (0.85\,kg)\cdot (45\,\frac{m}{s} )

The velocity of the bowling ball after the collision is:

v = 10.75\,\frac{m}{s}

8 0
3 years ago
Read 2 more answers
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
Strike441 [17]

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

4 0
3 years ago
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