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rewona [7]
4 years ago
6

What will happen if an orbiting object is moving too fast?

Physics
1 answer:
Ronch [10]4 years ago
4 0
It will sooner crash
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What happens when light enters a pair of glasses
alexdok [17]

Answer:

It refracts when it hits the glass.

6 0
3 years ago
a car accelerates uniformly from rest to a speed of 51.9mi/h in 9.37s. find the constant acceleration of the car. answer in unit
Darina [25.2K]
Here is my step-by-step-work. Let me know if you have any questions! :)

3 0
3 years ago
A force of 100N is applied to move an object a horizontal distance of 20m to the right. The work done by this force on the objec
horsena [70]
WORKDONE = FORCE * DISPLACEMENT
W=F*S
HERE, THE FORCE = 100N AND DISTANCE = 20M
WORKDONE = 100*20
WORKDONE=2000
ITS S.I UNIT IS JOULE OR J
SO, 2000J
5 0
3 years ago
arzan, who weighs 700 N, swings from a cliff at the end of a convenient vine thatis 20 m long. From the top of the cliff to the
vekshin1

Answer:

= 7.07 m

Explanation:

The Tarzan reaches bottom of swing after descending 2.5 m,

change in his potential energy equals his kinetic energy at bottom of swing

m g h = (1/2) m v²   ,  

hence speed v of Tarzan at bottom of swing is given as  

v = ( 2 g h )1/2

= ( 2 × 9.8 × 2.5 )1/2

= 7 m/s

At the bottom of swing, if the vine breaks, then he is moving with horizontal velocity 7 m/s in gravitational field.  

If vertical distance from ground to bottom of swing is 5 m, then time t for Tarzan to reach ground is given by

S = (1/2)g t2 or   t = (2S/g)1/2

= ( 2 × 5 / 9.8 )1/2

= 1.01 s

Horizontal distance traveled by Tarzan = 1.01 × 7

= 7.07 m

7 0
3 years ago
A car is traveling up a hill that is inclined at an angle θ above the horizontal. Determine the ratio of the magnitude of the no
padilas [110]

Explanation:

The weight of the car is equal to, W_c=m\times g...........(1)

Where

m is the mass of car

g is the acceleration due to gravity

The normal or vertical component of the force is, F_N=mg\ cos\theta

or

F_N=mg\ cos(13).............(2)

The horizontal component of the force is, F_H=mg\ sin\theta

Taking ratio of equation (1) and (2) as :

\dfrac{F_N}{W_c}=\dfrac{mg\ cos\theta}{mg}

\dfrac{F_N}{W_c}=cos(13)

\dfrac{F_N}{W_c}=0.97

or

\dfrac{F_N}{W_c}=\dfrac{97}{100}

Hence, this is the required solution.

5 0
3 years ago
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