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vekshin1
3 years ago
15

The hubbles telescopes orbit is 5.6 x10 ^5 meters above earths suface. the telescope has a mass os 1.1 x10^4 kilograms. earth ex

erts a gravitational force of 9.1 x10^4 newtons on the telescope. what is the magnitude of earths gravitational field strength at this location
Physics
1 answer:
Andru [333]3 years ago
7 0
(3) 8.3 N/kg. The gravitational field strength at a point is the force per unit mass exerted on a mass placed at that point. So at the point where the Hubble telescope is, it is (9.1 x 10^4)N/(1.1 x 10^4 kg) = 8.3 N/kg

Fam
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Calculate the east component of a resultant 32.5 m/s, 35.0° east of north.
ValentinkaMS [17]

Answer:

East component is: 18.64 m/s

Explanation:

If the resultant is 32.5 m/s directed 35 degrees east of north, then we use the sin(35) projection to find the east component of the velocity:

East component = 32.5 m/s * sin(35) = 18.64 m/s

4 0
3 years ago
A 100 kg bungee jumper leaps from a bridge. The bungee cord has an un-streched equilibrium length of 10 m, and a spring constant
nika2105 [10]

Answer:

11.78meters

Explanation:

Given data

Mass m = 100kg

Length of cord= 10m

Spring constant k= 35N/m

At the greatest vertical distance, the spring potential energy is equal to the gravitational potential energy

That is

Us=Ug

Us= 1/2kx^2

Ug= mgh

1/2kx^2= mgh

0.5*35*10^2= 100*9.81*h

0.5*35*100=981h

1750=981h

h= 1750/981

h= 1.78

Hence the bungee jumper will reach 1.78+10= 11.78meters below the surface of the bridge

6 0
3 years ago
Read 2 more answers
A jetliner has a velocity of 95 m/s. What is the displacement of the jetliner at t=3.0 seconds?
charle [14.2K]
Distance = speed / time

speed = 95 m/s
time = 3 s

distance = 95 / 3 m

displacement = 95/3 m or 32 m (2 s.f.)
5 0
3 years ago
A football player, with a mass of 69.0 kg, slides on the ground after being knocked down. At the start of the slide, the player
White raven [17]

Answer:

(a) -472.305  J

(b) 1 m

Explanation:

(a)

Change in mechanical energy equals change in kinetic energy

Kinetic energy is given by0.5mv^{2}

Initial kinetic energy is 0.5\times 69\times 3.7^{2}=472.305 J

Since he finally comes to rest, final kinetic energy is zero because the final velocity is zero

Change in kinetic energy is given by final kinetic energy- initial kinetic energy hence

0-472.305  J=-472.305  J

(b)

From fundamental kinematic equation

v^{2}=u^{2}+2as

Where v and u are final and initial velocities respectively, a is acceleration, s is distance

Making s the subject we obtain

s=\frac {v^{2}-u^{2}}{-2a} but a=\mu g hence

s=\frac {v^{2}-u^{2}}{-2\mu g}=\frac {0^{2}-3.7^{2}}{-2*0.7*9.81}=0.996796272\approx 1 m

7 0
3 years ago
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Answer:

6

Explanation:

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