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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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If your birthday was celebrated already then you would be  23, but if your birthday isn't celebrated then you would be 22

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Answer:

Option B is correct

Explanation:

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A moving curling stone, A, collides head on with stationary stone, B. Stone B has a larger mass than stone A. If friction is neg
Kitty [74]

Answer:

The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.

Explanation:

Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively

and the mass of big rock be 'M'

Initial momentum of the system equals

p_i=mv+0=mv

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'

Thus the final momentum of the system is

p_f=mv'+MV'

Equating initial and the final momenta we get

mv=mv'+MV'\\\\m(v-v')=MV'.....i

Now since the surface is frictionless thus the energy is also conserved thus

E_i=\frac{1}{2}mv^2

Similarly the final energy becomes

E_f=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\

Equating initial and final energies we get

\frac{1}{2}mv^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\\\\mv^2=mv'^2+MV'^2\\\\m(v^2-v'^2)=MV'^2\\\\m(v-v')(v+v')=MV'^2......(ii)

Solving i and ii we get

v+v'=V'

Using this in equation i we get

v'=\frac{v(m-M)}{(M-m)}=-v

Thus putting v = -v' in equation i  we get V' = 0

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4 0
3 years ago
2. In these diagrams, the sunlight is coming from the left, as shown by the arrows. Which
RideAnS [48]

Answer:

Explanation:

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3 0
3 years ago
An engineer weighs a sample of mercury (ρ = 13.6 × 103 kg/m3 ) and finds that the weight of the sample is 6.0 n. what is the sam
Amanda [17]
Given:
ρ = 13.6 x 10³ kg/m³, density of mercury
W = 6.0 N, weight of the mercury sample
g = 9.81 m/s², acceleration due to gravity.

Let V =  the volume of the sample.
Then
W = ρVg
or
V =  W/(ρg)
   = (6.0 N)/[(13.6 x 10³ kg/m³)*(9.81 m/s²)]
   = 4.4972 x 10⁻⁵ m³

Answer: The volume is 44.972 x 10⁻⁶ m³
5 0
3 years ago
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