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3 years ago

Why do astronauts float aboard the international space station?

1 answer:

8 0

They’re falling toward earth & moving forward at about the same velocity. because the downward and forward forces are nearly equal, the astronauts are not pulled in any specific direction, so they float . <span>
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What is the mass of a school bus if it can accelerate from rest to 15.5 m/s over 8.25

alexandr1967 [171]

The mass of a school bus if it can accelerate from rest to 15.5 m/s over 8.25s with 7,500 N of force is 3989.4kg.

HOW TO CALCULATE MASS:

The mass of an object can be calculated by dividing the force applied to the object by its acceleration.

According to this question, a bus can accelerate from rest to 15.5 m/s over 8.25s. The acceleration can be calculated as follows:

a = (v - u)/t

a = 15.5 - 0/8.25

a = 15.5/8.25

a = 1.88m/s²

The mass of the bus = 7500N ÷ 1.88m/s²

The mass of the bus = 3989.4kg

Therefore, the mass of a school bus if it can accelerate from rest to 15.5 m/s over 8.25s with 7,500 N of force is 3989.4kg.

Learn more about mass at: brainly.com/question/20259048?referrer=searchResults

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3 years ago

A car that increases its speed from 20 km/h to 100 km/h undergoes

Kamila [148]

Answer:

accelaration

Explanation:

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3 years ago

The fraction of nonreflected radiation that is transmitted through a 5-mm thickness of a transparent material is 0.95. if the th

Zielflug [23.3K]

Check the attached file for the answer.

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3 years ago

What happens when a particle of matter meets its corresponding antiparticle of antimatter?

shutvik [7]

Answer:

The combined mass of the two particles is completely transformed into energy (photons). This process is called matter-antimatter annihilation.

Explanation:

6 0

3 years ago

A rocket rises vertically, from rest, with an acceleration of 3.6m/s^2 until it runs out of fuel at an altitude of 1500m. After

Montano1993 [528]

A. 103.9 m/s

The motion of the rocket until it runs out of fuel is an accelerated motion with constant acceleration a = 3.6 m/s^2, so we can use the following equation

$v^2 -u^2 = 2ad$

where

v is the velocity of the rocket when it runs out of fuel

u = 0 is the initial velocity of the rocket

a = 3.6 m/s^2 is the acceleration

d = 1500 m is the distance covered during this first part

Solving for v, we find

$v=\sqrt{u^2+2ad}=\sqrt{(0^2+2(3.6 m/s^2)(1500 m)}=103.9 m/s$

B. 28.9 s

We can calculate the time taken for the rocket to reach this altitude with the formula

$d=\frac{1}{2}at^2$

where

d = 1500 m

a = 3.6 m/s^2

Solving for t, we find

$t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2(1500 m)}{3.6 m/s^2}}=28.9 s$

C. 2050.8 m

We can calculate the maximum altitude reached by the rocket by using the law of conservation of energy. In fact, from the point it runs out of fuel (1500 m above the ground), the rocket experiences the acceleration due to gravity only, so all its kinetic energy at that point is then converted into gravitational potential energy at the point of maximum altitude:

$K_i = U_f\\\frac{1}{2}mu^2 = mgh$

where h is distance covered by the rocket after it runs out of fuel, and v=103.9 m/s is the velocity of the rocket when it starts to decelerate due to gravity. Solving for h,

$h=\frac{v^2}{2g}=\frac{(103.9 m/s)^2}{2(9.8 m/s^2)}=550.8 m$

So the maximum altitude reached by the rocket is

$h' = d+h=1500 m +550.8 m=2050.8 m$

D. 39.5 s

The time needed for the second part of the trip (after the rocket has run out of fuel) can be calculated by

$h=\frac{1}{2}gt^2$

where

h = 550.8 m is the distance covered in the second part of the trip

g = 9.8 m/s^2

Solving for t,

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(550.8 m)}{9.8 m/s^2}}=10.6 s$

So the total time of the trip is

$t'=28.9 s+10.6 s=39.5 s$

E. 200.5 m/s

When the rocket starts moving downward, it is affected by gravity only. So the gravitational potential energy at the point of maximum altitude is all converted into kinetic energy at the instant the rocket hits the ground:

$\frac{1}{2}mv^2 = mgh$