1) 0.0011 rad/s
2) 7667 m/s
Explanation:
1)
The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

where
is the angular displacement of the object
t is the time elapsed
is the angular velocity
In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is
rad
And the time taken is

Therefore, the angular velocity of the telescope is

2)
For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

where
v is the linear velocity
is the angular velocity
r is the radius of the circular orbit
In this problem:
is the angular velocity of the Hubble telescope
The telescope is at an altitude of
h = 600 km
over the Earth's surface, which has a radius of
R = 6370 km
So the actual radius of the Hubble's orbit is

Therefore, the linear velocity of the telescope is:

Answer:
Explanation:
Normal length of spring = 28.3 cm
stretched length of spring = 38.2 cm
length of extension = 38.2 - 28.3 = 9.9 cm
= 9.9 x 10⁻² m
force applied to stretch = .55 x 9.8 ( mg )
= 5.39 N
Force constant = force applied / extension
= 5.39 / 9.9 x 10⁻²
= .5444 x 10² N /m
= 54.44 N/m
Answer:
A(many people think that no energy or matter exists outside the universe)
Explanation:
The law of conservation of energy<span>, a fundamental concept of physics, states that the total amount of </span>energy<span> remains constant in an isolated system. It implies that </span>energy<span> can neither be created nor destroyed, but can be change from one form to another.</span>
Answer:
Explanation:
a ) Let let the frictional force needed be F
Work done by frictional force = kinetic energy of car
F x 107 = 1/2 x 1400 x 35²
F = 8014 N
b )
maximum possible static friction
= μ mg
where μ is coefficient of static friction
= .5 x 1400 x 9.8
= 6860 N
c )
work done by friction for μ = .4
= .4 x 1400 x 9.8 x 107
= 587216 J
Initial Kinetic energy
= .5 x 1400 x 35 x 35
= 857500 J
Kinetic energy at the at of collision
= 857500 - 587216
= 270284 J
So , if v be the velocity at the time of collision
1/2 mv² = 270284
v = 19.65 m /s
d ) centripetal force required
= mv₀² / d which will be provided by frictional force
= (1400 x 35 x 35) / 107
= 16028 N
Maximum frictional force possible
= μmg
= .5 x 1400 x 9.8
= 6860 N
So this is not possible.