The Hubble Space Telescope is a joint ESA/NASA project and was launched in 1990 by the Space Shuttle mission STS-31 into a low-Earth orbit 569 km above the ground. During its lifetime Hubble has become one of the most important science projects ever. Hope this helps! ~ Autumn :)
Answer:
Explanation:
Impulse of a force is measured by force x time or F X t
Impulse also equals change in momentum or
F x t = m v₂ - m v₁
The given case is as follows
in the first case
F x t = mv - o = mv
F = mv / t
in the second case
F₁ x 4 t = mv
F₁ = 1/4 x mv /t
F₁ = F / 4
option a) is correct .
iii )
In the last case
F₂ X t = m v/2 -0
F₂ = 1/2 x mv / t
= 1/2 x F
F₂ = F/2
Option e ) is correct.
Answer:
The Force exerted by the two objects will be same and 4,401,189.49 × 10−11 N
Explanation:
Let m1 be the mass of the first object and m2 be the mass of the second object and the distance between them be d. Then
m1= 181kg
m2= 712 kg
d= 0.442 m
G is the gravitational Constant
According to Newtons Law Gravitational Force is given by
F=G m1 m2 /d²
F= 6.672 × 10−11 ×181×712/(0.442)²
F= 859,576.24 × 10−11/0.195364
F= 4,401,189.49 × 10−11 N
The Force exerted by the two masses on the third mass will be same and 4,401,189.49 × 10−11 N
Answer:
the runner's average kinetic energy during the run is 476.96 J.
Explanation:
Given;
mass of the runner, m = 85 kg
distance covered by the runner, d = 42.2 km = 42,200 m
time to complete the race, t = 3 hours 30 mins = (3 x 3600s) + (30 x 60s)
= 12,600 s
The speed of the runner, v = d/t
v = 42,200 / 12,600
v = 3.35 m/s
The runner's average kinetic energy during the run is calculated as;
K.E = ¹/₂mv²
K.E = ¹/₂ × 85 × (3.35)²
K.E = 476.96 J
Therefore, the runner's average kinetic energy during the run is 476.96 J.
Answer:
the electron only emits electromagnetic radiation when it descends to a lower-energy orbit
Explanation:
