Answer:
0.625 c
Explanation:
Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.
In the context,
The relative speed of body 2 with respect to body 1 can be expressed as :

Speed of rocket 1 with respect to rocket 2 :



Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c
Answer:
32000joule.
Explanation:
given
mass. (m)=160kg
speed (v)=20m/s
now
kinetic energy =1/2 (mv²)=1/2 ×{160×20²}=32000joule.
The constants could be the day because the experiment is all on the same day or the person throwing the pumpkin because it will always be the same perosn
Answer:
a) 1321.45 N
b) 1321.45 N
c) 2.66 m/s^2
d) 2.21*10^-22 m/s^2
Explanation:
Hello!
First of all, we need to remember the gravitational law:

Were
G = 6.67428*10^-11 N(m/kg)^2
m1 and m2 are the masses of the objects
r is the distance between the objects.
In the present case
m1 = earth's mass = 5.9742*10^24 kg
m2 = 497 kg
r = 1.92 earth radii = 1.92 * (6378140 m) = 1.2246*10^7 m
Replacing all these values on the gravitational law, we get:
F = 1321.45 N
a) and b)
Both bodies will feel a force with the same magnitude 1321.45 N but directed in opposite directions.
The acceleration can be calculated dividing the force by the mass of the object
c)
a_satellite = F/m_satellite = ( 1321.45 N)/(497 kg)
a_satellite = 2.66 m/s^2
d)
a_earth = F/earth's mass = (1321.45 N)/( 5.9742*10^24 kg)
a_earth = 2.21*10^-22 m/s^2
Answer:
.
Explanation:
F = kx so k = 800/((10-5)/100) = 16000 N/m
W = 1/2 kx^2 = 1/2 * 16000 * .05^2 = 20 J.
(sorry if it's wrong)