Answer:
3.6*10^18s
Explanation:
To find the period of the satellite
We need to apply kephler's third law
Which is
MP² = (4π²/G) d³
d=semi-major axis which is the distance from center of moon = 98km+1740km = 1838km
where M= mass of the moon = 7.3x10^22kg
P=period
G=newtonian gravatational constant= 6.67x10^-11
To find the Period solve for P
P = √[(4π²/G M)xd³]
P=√(4 π²/6.67x10^-22*7.3x10^22kg) x (1.838x10^6m)³]
= 3.6*10^18s
<span>When water freezes to form ice, its volume expands. However, we know from conservation of mass that the mass of the ice is the same as the mass of the water. Since density is defined fundamentally as mass / volume, and we have an expanding volume at a constant mass, the denominator of the equation grows, and thus the density of ice is lower than that of liquid water.</span>
Answer:
1.) U = 39.2 m/s
2.) t = 4s
Explanation: Given that the
height H = 78.4m
The projectile is fired vertically upwards under the acceleration due to gravity g = 9.8 m/s^2
Let's assume that the maximum height = 78.4m. And at maximum height, final velocity V = 0
Velocity of projections can be achieved by using the formula
V^2 = U^2 - 2gH
g will be negative as the object is moving against the gravity
0 = U^2 - 2 × 9.8 × 78.4
U^2 = 1536.64
U = sqrt( 1536.64 )
U = 39.2 m/s
The time it takes to reach its highest point can be calculated by using the formula;
V = U - gt
Where V = 0
Substitute U and t into the formula
0 = 39.2 - 9.8 × t
9.8t = 39.2
t = 39.2/9.8
t = 4 seconds.
A) 0.189 N
The weight of the person on the asteroid is equal to the gravitational force exerted by the asteroid on the person, at a location on the surface of the asteroid:
where
G is the gravitational constant
8.7×10^13 kg is the mass of the asteroid
m = 130 kg is the mass of the man
R = 2.0 km = 2000 m is the radius of the asteroid
Substituting into the equation, we find
B) 2.41 m/s
In order to orbit just above the surface of the asteroid (r=R), the centripetal force that keeps the astronaut in orbit must be equal to the gravitational force acting on the astronaut:
where
v is the speed of the astronaut
Solving the formula for v, we find the minimum speed at which the astronaut should launch himself and then orbit the asteroid just above the surface:
