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harkovskaia [24]
3 years ago
10

A force is applied to an object at rest with a mass of 100 kg. The same force is applied to an object at rest with a mass of 1 k

g. What differences in their motion would you observe? Explain your reasoning.
Physics
2 answers:
Pani-rosa [81]3 years ago
8 0

Answer: the object with less mas will move faster than the other object.

Explanation: Let's do a simplification of this case, I will only use newton's second law, this law says F = ma

where f is force, m is mass and a is acceleration.

Then, if in bot objects the force is the same, let's call it F, we have:

F = 100kg*a1

and

F = 1kg*a2

now let's isolate the accelerations:

a1 = F/100kg

a2 = F/1kg

here we can see that a2 is bigger than a1, this means that the object with less mass has a bigger acceleration, then the object with less mass will move faster than the object with more mass.

Darya [45]3 years ago
3 0
The object with a mass of 100 kg would move slower than the object with a mass of 1 kg. This is because a greater force is required to move the object with a mass of 100 kg compared to the object of 1 kg.
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Two equally charged tiny spheres of mass 1.0 g are placed 2.0 cm apart. When released, they begin to accelerate away from each o
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Answer:

The magnitude of the charge on each sphere is 0.135 μC

Explanation:

Given that,

Mass = 1.0

Distance = 2.0 cm

Acceleration = 414 m/s²

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Using newton's second law

F= ma

a=\dfrac{F}{m}

Put the value of F

a=\dfrac{kq^2}{mr^2}

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414=\dfrac{9\times10^{9}\times q^2}{1.0\times10^{-3}\times(2.0\times10^{-2})^2}

q^2=\dfrac{414\times1.0\times10^{-3}\times(2.0\times10^{-2})^2}{9\times10^{9}}

q^2=1.84\times10^{-14}

q=0.135\times10^{-6}\ C

q=0.135\ \mu C

Hence, The magnitude of the charge on each sphere is 0.135μC.

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In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m
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(a) -39.4^{\circ}

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

p_y =0

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2

where

m is the mass of each ball

v_1= 4.60 m/s is the velocity of the cue ball after the collision

v_2 = 3.40 m/s is the velocity of the second ball after the collision

\phi_1=28.0^{\circ} is the angle of the cue ball with the x-axis

\phi_2 is the angle of the second ball

Solving for \phi_2, we find the angle between the direction of motion of the second ball and the original direction of motion:

sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m

The initial total momentum along the x-direction as

p_x = m u

where

m is the mass of the cue ball

u is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

mu = 6.69 m\\u = 6.69 m/s

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3 years ago
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