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tangare [24]
3 years ago
12

PLEASE ANSWER, I NEED HELP

Physics
1 answer:
Scorpion4ik [409]3 years ago
6 0

1) The gravitational force between Ellen and the moon is 1.56\cdot 10^{-3} N

2) The two forces are equal, while the acceleration of the bus is smaller than the acceleration of the bicycle.

Explanation:

1)

The magnitude of the gravitational force between two objects is given by

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m_1, m_2 are the masses of the two objects

r is the separation between them

In this problem, we have:

m_1 = 47 kg is the mass of Ellen

m_2 = 7.35\cdot 10^{22} kg is the mass of the moon

r=3.84\cdot 10^8 m is the distance between Ellen and the moon

Substituting, we find the gravitational force between Ellen and the moon:

F=(6.67\cdot 10^{-11})\frac{(47)(7.35\cdot 10^{22})}{(3.84\cdot 10^8)^2}=1.56\cdot 10^{-3} N

2)

We can analyze the forces acting in the collision between the bus and the bicycle by using Newton's third law of motion, which states that:

"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"

Applied to our problem, this means that the force exerted by the bus on the bicycle during the collision (action force) is equal (and opposite) to the force exerted by the bicycle on the bus (reaction force).

Now let's analyze the accelerations of the two vehicles. We can find the acceleration of each vehicle by using Newton's second law:

a=\frac{F}{m}

where

a is the acceleration

F is the force exerted on the vehicle

m is the mass of the vehicle

As we said previously, the force F exerted on each of the two vehicles: so, the acceleration only depends on the mass. In particular, the acceleration is inversely proportional to the mass: therefore, the larger the mass of the vehicle, the smaller the acceleration. This means that the acceleration of the bus is smaller than the acceleration of the bicycle.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

And about Newton's third law:

brainly.com/question/11411375

#LearnwithBrainly

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The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

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The final area of the copper wire;

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v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s

The final drift velocity of the electrons is calculated as;

v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7}  \ m/s

The change in the mean drift velocity is calculated as;

\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s

The time of motion of electrons for the initial wire diameter is calculated as;

t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s

The time of motion of electrons for the final wire diameter is calculated as;

t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s

The average acceleration of the electrons is calculated as;

a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

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