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Natali [406]
3 years ago
10

What is the magnitude of the orbital velocity of the earth in m/s?

Physics
1 answer:
Lana71 [14]3 years ago
8 0
Circumference C=2πr 
<span>C=2π(1.5x10^8)=9.42x10^8 </span>

<span>In 365 Days there are 8760hr </span>

<span>V=distance/time </span>

<span>V=(9.42x10^8)/8760=107534.2km/hr </span>
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You stand on a merry-go-round which is spinning at f = 0:25 revolutions per second. You are R = 200 cm from the center. (a) Find
ivanzaharov [21]

Answer:

(a) ω = 1.57 rad/s

(b) ac = 4.92 m/s²

(c) μs = 0.5

Explanation:

(a)

The angular speed of the merry go-round can be found as follows:

ω = 2πf

where,

ω = angular speed = ?

f = frequency = 0.25 rev/s

Therefore,

ω = (2π)(0.25 rev/s)

<u>ω = 1.57 rad/s </u>

(b)

The centripetal acceleration can be found as:

ac = v²/R

but,

v = Rω

Therefore,

ac = (Rω)²/R

ac = Rω²

therefore,

ac = (2 m)(1.57 rad/s)²

<u>ac = 4.92 m/s² </u>

(c)

In order to avoid slipping the centripetal force must not exceed the frictional force between shoes and floor:

Centripetal Force = Frictional Force

m*ac = μs*R = μs*W

m*ac = μs*mg

ac = μs*g

μs = ac/g

μs = (4.92 m/s²)/(9.8 m/s²)

<u>μs = 0.5</u>

7 0
3 years ago
On a cold winters day if you left a cup of water sitting outside it could freeze heat is transferred out of the water describe t
valkas [14]
The water molecules would slow down, and as they slow down, the heat created from their movement would cease.
4 0
3 years ago
A rocket powered sled accelerates a jet pilot in training straight forward from rest to 270 km/h in 12.1 seconds. Find:
Ilia_Sergeevich [38]

Answer:

  1. 6.198 m/s²
  2. 4.48 s
  3. 453.77 m

Explanation:

5 0
3 years ago
A 65-kg ice skater stands facing a wall with his arms bent and then pushes away from the wall by straightening his arms. At the
Marrrta [24]

Our values can be defined like this,

m = 65kg

v = 3.5m / s

d = 0.55m

The problem can be solved for part A, through the Work Theorem that says the following,

W = \Delta KE

Where

KE = Kinetic energy,

Given things like that and replacing we have that the work is given by

W = Fd

and kinetic energy by

\frac {1} {2} mv ^ 2

So,

Fd = \frac {1} {2} m ^ 2

Clearing F,

F = \frac {mv ^ 2} {2d}

Replacing the values

F = \frac {(65) (3.5)} {2 * 0.55}

F = 723.9N

B) The work done by the wall is zero since there was no displacement of the wall, that is d = 0.

6 0
3 years ago
What is not changed when work is done by a machine?
Irina18 [472]
B) The amount of work done
8 0
3 years ago
Read 2 more answers
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