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Tom [10]
3 years ago
15

Tom is sitting facing forward in the train as it is getting ready to leave the station. He puts a smooth ball on the train floor

. The train jolts into motion, accelerating rapidly up to speed. According to Newton's first law of motion, what happens to the ball? (1 point)
A) Nothing changes.
B) The ball rolls forward slowly.
C) The ball rolls forward rapidly.
D) The ball rolls backward.
Physics
1 answer:
Vesnalui [34]3 years ago
8 0

According to Newton's first law of motion, what happens to the ball is  the ball rolls backward.

<h3>What is the first law?</h3>

This means that an object at rest or in motion will remain uniformly rectilinear and tend to be in that state if the net force on it is zero.

In this case, we have to think that the ball is at rest and the train is moving with a velocity that way, the reaction of the ball will be to go in the opposite direction to the motion.

See more about  first law at brainly.com/question/3808473

#SPJ1

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An object is acted upon by a constant force which gives it a constant acceleration a. At a certain time t1, having started from
Elenna [48]

Answer:

t_2=\sqrt{2}(t_1)

Explanation:

The kinetic energy of a body is that energy it possesses due to its motion. In classical mechanics, this energy depends only on its mass and speed, as follows:

K=\frac{mv^2}{2}

The speed in an uniformly accelerated motion is given by:

v=at

Replacing this expression in the formula for the kinetic energy, we have:

K=\frac{ma^2t^2}{2}\\

So, if we have K_2=2K_1:

K_1=\frac{ma^2t_1^2}{2}(1)\\K_2=\frac{ma^2t_2^2}{2}\\2K_1=\frac{ma^2t_2^2}{2}\\K_1=\frac{ma^2t_2^2}{4}(2)\\

Equaling (1) and (2) and solving for t_2:

\frac{ma^2t_1^2}{2}=\frac{ma^2t_2^2}{4}\\t_2=\frac{4t_1^2}{2}\\t_2=\sqrt{2t_1^2}\\t_2=\sqrt{2}(t_1)

7 0
4 years ago
How much potential energy is stored on a stone kept on the earth surface? Why?​
olga2289 [7]

Answer:

When the body is kept at the surface there height of the stone is equal to zero. hence, if the height of the stone is zero then Potential energy is equal to zer

6 0
3 years ago
Read 2 more answers
A police car with its 300-Hz siren is moving toward a warehouse at 30 m/s, intending to crash through the door. The sound bounce
Lady bird [3.3K]

Answer: The frequency heard will be f = 275.675Hz

Explanation: When an object emitting sound is moving, it occurs a phenomenon called Doppler shift or Doppler effect. What happens is that the sound gets higher when the moving object comes closer the observer and becomes lower after it passes, This change is due to the quantity of waves that passes through an area in an unit of time.

The formula to calculate the Doppler effect is as follows

f = (\frac{c}{c+Vs}) · f₀

f is the observed frequency;

c is the speed of sound;

Vs is velocity of the source;

f₀ is the emitted frequency of source;

Substituting and calculating,

f = \frac{340}{340+30} · 300

f = 275.675 Hz

Thus, the frequency heard by the police officer is 275.675Hz.

8 0
4 years ago
10 The magnitude J of the current density in a certain lab wire with a circular cross section of radius R = 2.00 mm is given by
rewona [7]

Answer:

I = 0.002593 A = 2.593 mA

Explanation:

Current density = J = (3.00 × 10⁸)r² = Br²

B = (3.00 × 10⁸) (for ease of calculations)

The current through outer section is given by

I = ∫ J dA

The elemental Area for the wire,

dA = 2πr dr

I = ∫ Br² (2πr dr)

I = ∫ 2Bπ r³ dr

I = 2Bπ ∫ r³ dr

I = 2Bπ [r⁴/4] (evaluating this integral from r = 0.900R to r = R]

I = (Bπ/2) [R⁴ - (0.9R)⁴]

I = (Bπ/2) [R⁴ - 0.6561R⁴]

I = (Bπ/2) (0.3439R⁴)

I = (Bπ) (0.17195R⁴)

Recall B = (3.00 × 10⁸)

R = 2.00 mm = 0.002 m

I = (3.00 × 10⁸ × π) [0.17195 × (0.002⁴)]

I = 0.0025929449 A = 0.002593 A = 2.593 mA

Hope this Helps!!!

4 0
3 years ago
A +26.3 uC charge qy is repelled by a force
Musya8 [376]

Answer:

+1.46×10¯⁶ C

Explanation:

From the question given above, the following data were obtained:

Charge 1 (q₁) = +26.3 μC = +26.3×10¯⁶ C

Force (F) = 0.615 N

Distance apart (r) = 0.750 m

Electrical constant (K) = 9×10⁹ Nm²/C²

Charge 2 (q₂) =?

The value of the second charge can be obtained as follow:

F = Kq₁q₂ / r²

0.615 = 9×10⁹ × 26.3×10¯⁶ × q₂ / 0.750²

0.615 = 236700 × q₂ / 0.5625

Cross multiply

236700 × q₂ = 0.615 × 0.5625

Divide both side by 236700

q₂ = (0.615 × 0.5625) / 236700

q₂ = +1.46×10¯⁶ C

NOTE: The force between them is repulsive as stated from the question. This means that both charge has the same sign. Since the first charge has a positive sign, the second charge also has a positive sign. Thus, the value of the second charge is +1.46×10¯⁶ C

5 0
3 years ago
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