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S_A_V [24]
2 years ago
14

Two events are observed in a frame of reference S to occur at the same space point, with the second event occurring after a time

of 1.70 s . In a second frame S' moving relative to S, the second event is observed to occur after a time of 2.25 s . What is the difference between the positions of the two events as measured in S'?

Physics
1 answer:
Deffense [45]2 years ago
5 0

Answer:

he difference between the positions of the two events as measured in S = 4.42 x 10 8 m

Explanation:

The concept of time dilation is applied to solve the question as shown in the attachment.

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A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.
myrzilka [38]

Answer:

a. The plane speeds up but the cargo does not change speed.

Explanation:

Just to make it clear, the question is as follows from what I understand.

A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.  You can neglect air resistance.

Just after the cargo has fallen out:

a. The plane speeds up but the cargo does not change speed.

b. The cargo slows down but the plane does not change speed.

c. Neither the cargo nor the plane change speed.

d. The plane speeds up and the cargo slows down.

e. Both the cargo and the plane speed up.

And we are requested to choose the right answer under the given conditions. We know the glider has no motor, then it must be in free fall movement, then it is experiencing some force that pulls it to the from due to the gravity effect on it, and a force in general is calculated by

F=m*a, m:= mass of the object, a:= acceleration.

Here we are only considering the horizontal effect of the forces, then since the mass is reduced the acceleration must increase to compensate and maintain  the equilibrium of the forces, then the glider being lighter can travel faster due to the acceleration. On the other hand by the time the cargo left the glider there was no acceleration and the speed it had at the moment he left the plane continues, then the cargo does not change its speed, then horizontally speaking the answer would be a. The plane speeds up but the cargo does not change speed.

5 0
2 years ago
Two climbers are on a mountain. Simon, of mass m, is sitting on a snow covered slope that makes an angle θ with the horizontal.
elena-14-01-66 [18.8K]

Answer:

Explanation:

It is required that the weight of Joe must prevent Simon from being pulled down . That means he is not slipping down but tends to be towed down . So in equilibrium , force of friction will act in upward direction on Simon.

Let in equilibrium , tension in rope be T

For balancing Joe

T = M g

For balancing Simon

friction + T = mgsinθ

μmgcosθ+T = mgsinθ

μmgcosθ+Mg = mgsinθ

M = (msinθ - μmcosθ)

M = m(sinθ - μcosθ)

5 0
2 years ago
On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he
Liula [17]

Answer:

correct is d) a ’= g / 2

Explanation:

For this exercise let's use the kinematics equations

On earth

      v = v₀ - a t

     a = (v₀- v) / T

On planet X

    v = v₀ - a' t’

    a ’= (v₀-v) / 2T

Let's substitute the land values ​​in plot X

     a’= a / 2

Now let's use Newton's second law

       W = ma

      m g = m a

      a = g

We substitute

      a ’= g / 2

So we see that on planet X the acceleration is half the acceleration of Earth's gravity

4 0
1 year ago
In a laboratory, Nakisha mixes a sodium hydroxide solution with an indicator called phenolphthalein. When combined, they create
Gnesinka [82]

by making qualitative observations about the sodium hydroxide and phenolphthalein in solution

by comparing the given solution to other familiar solutions containing phenolphthalein

by designing an experiment to test phenolphthalein with other solutions

8 0
1 year ago
Read 2 more answers
A spaceship of mass 8600 kg is returning to Earth with its engine turned off. Consider only the gravitational field of Earth. Le
Katyanochek1 [597]

Answer:

\Delta KE = 4.20\times 10^{13}\ J

Explanation:

given,

mass of spaceship(m) = 8600 Kg

Mass of earth = 5.972 x 10²⁴ Kg

position of movement of space ship

R₁ = 7300 Km

R₂ = 6700 Km

the kinetic energy of the spaceship increases by = ?

Increase in Kinetic energy = decrease in potential energy

    \Delta KE = GMm (\dfrac{1}{R_2}-\dfrac{1}{R_1})

    \Delta KE = GMm (\dfrac{R_1-R_2}{R_2R_1})

    \Delta KE = 6.67 \times 10^{-11}\times 5.972 \times 10^{24}\times 8600 (\dfrac{7300 - 6700}{7300 \times 6700})

    \Delta KE = 6.67 \times 10^{-11}\times 5.972 \times 10^{24}\times 8600 (\dfrac{600}{48910000})

    \Delta KE = 4.20\times 10^{13}\ J

5 0
2 years ago
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