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Verizon [17]
3 years ago
15

What is the velocity of an object that has been in free fall for 1.5s?

Physics
1 answer:
Crank3 years ago
8 0

Answer:

D. 15 m/s downward

Explanation:

v = at + v₀

v = (-9.8 m/s²) (1.5 s) + (0 m/s)

v = -14.7 m/s

Rounded to two significant figures, the answer is D, 15 m/s downward.

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If Bill threw a ball straight up on the Moon (g=1.6 m/s2) with a starting velocity of 22m/s from a cliff and it fell past him an
kap26 [50]

Answer:

Write two important of physical state

5 0
2 years ago
Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radiu
madam [21]

The speed of the satellite in a circular orbit around the Earth is 1.32 x 10⁵ m/s.

<h3>Speed of the satellite</h3>

v = √(GM/r)

where;

  • G is universal gravitation constant
  • M is mass of Earth
  • r is radius of the satellite

v = √(6.67 x 10⁻¹¹ x 5.98 x 10²⁴/3.57 x 6.37x 10³)

v = 1.32 x 10⁵ m/s

Thus, the speed of the satellite in a circular orbit around the Earth is 1.32 x 10⁵ m/s.

Learn more about speed of satellite here: brainly.com/question/22247460

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7 0
2 years ago
A boy on rollerskates is travelling along at 8 m/s. He has a mass of 60 kg and is carrying his
Brrunno [24]

Answer:

6m/s

Explanation:

the original momentum = mass x velocity = 8x (60+10) = 560

momentum after = mass x velocity of the school bag + mass x velocity of the boy = 10x20 + 60x A

200+60A = 560

A=6

5 0
3 years ago
What is the new current?
nikdorinn [45]
Voltage = current x resistance
since R is doubled, current must reduce by half.
So,
new current = 120/2 = 60mA
3 0
3 years ago
Please someone help, I’m very confused and it’s due soon, thanks
Anit [1.1K]

Answer:

  1. 1 s
  2. 19.6 m
  3. 2 s
  4. 0.8 m/s^2
  5. 28 m/s
  6. 79 m/s
  7. 0.37 s
  8. 26 m/s
  9. 242 m/s
  10. 19,930 m

Explanation:

In physics, many of the relationships between speed, distance, and acceleration are tied up in the equations for potential and kinetic energy. For an object of mass M* at height h in a gravity field with acceleration g, the potential energy is

  PE = Mgh

At velocity v, the kinetic energy of the object is ...

  KE = 1/2Mv^2

When an object is dropped or launched from rest, the height and velocity are related by the fact that kinetic energy gets translated to potential energy, or vice versa. This gives rise to ...

  PE = KE

  Mgh = (1/2)Mv^2

The mass (M) can be factored out of this, so we have ...

  2gh = v^2

This can be solved for height:

  h = v^2/(2g) . . . . [eq1]

or for velocity:

  v = √(2gh) . . . . [eq2]

__

When acceleration is constant, as assumed here, the velocity changes linearly (to/from 0). So, over the time of travel, the average velocity is half the final velocity. That is,

  t = 2h/v

Depending on whether you start with h or with v, this resolves to two more equations:

  t = 2(v^2/(2g))/v = v/g . . . . [eq3]

  t = 2h/(√(2gh)) = √(4h^2/(2gh)) = √(2h/g) . . . . [eq4]

The last of these can be rearranged to give distance as a function of time:

  h = gt^2/2 . . . . [eq5]

or acceleration as a function of time and distance:

  g = 2h/t^2 . . . . [eq6]

__

These 6 equations can be used to solve the problems posed. Just "plug and chug." For problems in Earth's gravity, we use g=9.8 m/s^2. (You may want to keep these equations handy. Be aware of the assumptions they make.)

_____

* M is used for mass in these equations so as not to get confused with m, which is used for meters.

_____

1) Use [eq4]: t = √(2·6 m/(9.8 m/s^2)) ≈ 1.107 s ≈ 1 s

__

2) Use [eq5]: h = (9.8 m/s^2)(2 s)^2/2 = 19.6 m

__

3) Use [eq4]: t = √(25 m/(4.9 m/s^2)) ≈ 2.259 s ≈ 2 s

__

4) Use [eq6]: g = 2(10 m)/(5 s)^2 = 0.8 m/s^2

__

5) Use [eq2]: v = √(2·9.8 m/s^2·40 m) = 28 m/s

__

6) Use [eq2]: v = √(2·9.8 m/s^2·321 m) ≈ 79.32 m/s ≈ 79 m/s

__

7) Using equation [eq3], we will find the time until Tina reaches her maximum height. Her actual off-the-ground total time is double this value. Using [eq3]: t = v/g = (1.8 m/s)/(9.8 m/s^2) = 9/49 s. Tina is in the air for double this time:

  2(9/49 s) ≈ 0.37 s

__

8) Use [eq2]: v = √(2·9.8 m/s^2·33.5 m) ≈ 25.624 m/s ≈ 26 m/s

__

9) Use [eq2]: v = √(2·9.8·3000) m/s ≈ 242.49 m/s ≈ 242 m/s

(Note: the terminal velocity in air is a lot lower than this for an object like a house.)

__

10) Use [eq1]: h = (625 m/s)^2/(2·9.8 m/s^2) ≈ 19,930 m

_____

<em>Additional comment</em>

Since all these questions make use of the same equation development, I have elected to answer them. Your questions are more likely to be answered if you restrict your posts to 3 or fewer questions each.

5 0
3 years ago
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