Answer:
a) 6636 km
b) 0.0154
Explanation:
The height above the earth at its furthest point is 368 km
The height above the earth at its closest point is 164 km
Radius of the Earth is 6370 km
The distance of the satellite from the center of the earth to the furthest point is 6370 + 368 km = 6738 km
The distance of the satellite from the center of the earth to the closest point is 6370 + 164 = 6534 km
If we add together the sum of the distance of the satellite from the furthest and its closest distance, it is equal to the 2 major semi axis.
Basically,
2a = R + r
a = (R + r) / 2
a = (6738 + 6534) / 2
a = 13272 / 2
a = 6636 km
Eccentricity, e = (a - r) / a
Eccentricity, e = (6636 - 6534) / 6636
Eccentricity, e = 102 / 6636
Eccentricity, e = 0.0154
Answer:
649kg/m^3
Explanation:
Let p be the density of this particular object.
Formula for density:

We can substitute the givenmass and volume to find density of the object.

Therefore the density of this object is 649kg/m^3.
Answer:

Explanation:
For answer this we will use the law of the conservation of the angular momentum.

so:

where
is the moment of inertia of the merry-go-round,
is the initial angular velocity of the merry-go-round,
is the moment of inertia of the merry-go-round and the child together and
is the final angular velocity.
First, we will find the moment of inertia of the merry-go-round using:
I = 
I = 
I = 359.375 kg*m^2
Where
is the mass and R is the radio of the merry-go-round
Second, we will change the initial angular velocity to rad/s as:
W = 0.520*2
rad/s
W = 3.2672 rad/s
Third, we will find the moment of inertia of both after the collision:



Finally we replace all the data:

Solving for
:

Answer:
Explanation:
Density = Mass / Volume = 850 / 40*10*5 = 0.425 g /cm^3
F=nmv
where;
n=no. of bullets = 1
m=mass of bullets=2g *10^-3
V=velocity of bullets200m/sec
F=1
loss in Kinetic energy=gain in heat energy
1/2MV^2=MS∆t
let M council M
=1/2V^2=S∆t
M=2g
K.E=MV^2/2
=(2*10^-3)(200)^2/2
2 councils 2
2*10^-3*4*10/2
K.E=40Js
H=mv∆t
(40/4.2)
40Js=40/4.2=mc∆t
40/4.2=2*0.03*∆t
=158.73°C