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AlexFokin [52]
3 years ago
15

Which would take more force to stop in 10 seconds: an 8.0-kilogram ball rolling in a straight line at a speed of 0.2 m/sec or a

5.0-kilogram ball rolling along the same path at a speed of 1.0 m/sec?​
Physics
1 answer:
den301095 [7]3 years ago
8 0

Answer:

Higher force would be required to stop 5 kg of ball

Explanation:

The force required to stop a moving object depends on its momentum.

The momentum for 8 kg of rolling ball is = 8 kg * 0.2 m/s = 1.6 kgm/s

The momentum for 5 kg of rolling ball is = 8 kg * 1.0 m/s = 5 kgm/s

Hence, momentum of 5 Kg ball is higher than the momentum of 8 Kg of ball

Hence, higher force would be required to stop 5 kg of ball

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Buses travel at 25 m/s then decelerate steadily. It comes to a complete stop in 40 second. What is its acceleration?
AnnZ [28]
Given is the velocity of the bus at 25 meters per second. Then it decelerates steadily. It stops after 40 seconds. To calculate for the acceleration, use this formula:

a = (vf - vo) / t

vf = 0 m/s (complete stop)
vo = 25 m/s 
t = 40 s

Therefore,

a = (0 - 25) m/s / 40 s

a = - 0.625 m/s^2

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7 0
3 years ago
Heat gained or lost is mass times specific heat times change in temperature.
BlackZzzverrR [31]

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6 0
3 years ago
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A 20.6 g marble sliding to the right at 59.6 cm/s overtakes and collides elastically with a 10.3 g marble moving in the same dir
Likurg_2 [28]

Answer:29.8 cm/s

Explanation:

Given

mass of marble A is m_a=20.6\ gm

Initial velocity of marble A is u_a=59.6\ cm/s

mass of marble B is m_b=10.3\ gm

Initial velocity of marble B u_b=14.9\ cm/s

After collision marble B moves with velocity of v_b=74.5\ cm/s

Conserving momentum

m_au_a+m_bu_b=m_av_a+m_bv_b

20.6\times 59.6+10.3\times 14.9=20.6\times v_a+10.3\times 74.5

v_a=\frac{20.6\times 59.6-10.3(74.5-14.9)}{20.6}

v_a=29.8\ cm/s                                      

4 0
4 years ago
A ramp is often used to raise an object. if 90 joules of energy are needed to lift an object to a platform but 100 joules are us
vova2212 [387]
100 Joules have been used to roll the object up the ramp, but only 90 Joules were actually needed to lift the object, this means that 10 Joules have been wasted in the form of heat due to the friction between the object and the ramp, and this corresponds to
\frac{10 J}{100 J} \cdot 100 = 10 \%
so, 10 % of the energy.
8 0
3 years ago
Hey , please help me im totally stuck?! :)
Setler79 [48]
D thymine
if this is high school than i am tottaly right
3 0
3 years ago
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