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3 years ago

8

You are in a spaceship moving in a straight line at constant speed. You cannot see out of the ship. Assuming perfectly uniform m

otion, what can you do to prove you are moving?

1 answer:

Dimas [21]3 years ago

8 0

Answer:

There is no experiment to prove that you are in motion

Explanation:

A frame of reference which has constant velocity is known as an inertial frame of reference. Motion is relative. One can detect one's motion only when one observes change in position with respect to a fixed body.

Thus, if you are in a spaceship moving at a constant speed in a straight line and unable to look outside, you would not be able to prove that you are moving. Everything within the spaceship would have same speed. If you will throw any object within the spaceship, then the parameters measured by you would also not show that the spaceship is in motion.

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two cars are moving at constant speeds in a straight line along a major highway. The first is travelling at 20ms^-1 and the seco

Reika [66]

Answer:

$t=750s$

Explanation:

The two cars are under an uniform linear motion. So, the distance traveled by them is given by:

$\Delta x=vt\\x_f-x_0=vt\\x_f=x_0+vt$

$x_f$ is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have:

$x_0_1=6km\\x_0_2=0$

We have $x_f_1=x_f_2$ . Thus, solving for t:

$x_0_1+v_1t=x_0_2+v_2t\\x_0_1=t(v_2-v_1)\\t=\frac{x_0_1}{v_2-v_1}\\t=\frac{6*10^3m}{28\frac{m}{s}-20\frac{m}{s}}\\t=750s$

8 0

3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

What is the density of an object with mass of 60 g and a volume of 2cm3

Artemon [7]

Answer:

D=mass /volume

=60/0.002

=30000

4 0

2 years ago

Metals can be described as?

luda_lava [24]

They tend to be lustrous, ductile, malleable, and good conductors of electricity, while nonmetals are generally brittle (for solid nonmetals), lack lustre, and are insulators. Use google.

6 0

3 years ago

Which has greater value, a newton of gold on earth or a newton of gold on the moon?

jek_recluse [69]

The moon, because the acceleration due to gravity is less.

4 0

2 years ago