Answer:
density = 5520 kg/m^3
Explanation:
given that
radius of earth = 6378 km
G = 6.67 x 10⁻¹¹ m³/kg.s²
g = 9.80 m/s²
we know,
mass of earth
M = 5.972 x 10²⁴ kg
density =
V = volume of the earth = 4/3πr³
V = 4/3 x 3.14 x (6378 x 10³)³
V = 1.08 x 10²¹ m³
density = 
density = 5.52 x 10³ kg/m^3
density = 5520 kg/m^3
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Answer:</h2><h2>
The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/
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Explanation:
A meteoroid is in a circular orbit 600 km above the surface of a distant planet.
Mass of the planet = mass of earth = 5.972 x 
Kg
Radius of the earth = 90% of earth radius = 90% 6370 = 5733 km
The acceleration of the meteoroid due to the gravitational force exerted by the planet = ?
By formula, g = 
where g is the acceleration due to the gravity
G is the universal gravitational constant = 6.67 x 
M is the mass of the planet
r is the radius of the planet
Substituting the values, we get
g = 
g = 12.12 m/
The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/
Looks like you simply substitute the length of the femur
Explanation:
Show that the motion of a mass attached to the end of a spring is SHM
Consider a mass "m" attached to the end of an elastic spring. The other end of the spring is fixed
at the a firm support as shown in figure "a". The whole system is placed on a smooth horizontal surface.
If we displace the mass 'm' from its mean position 'O' to point "a" by applying an external force, it is displaced by '+x' to its right, there will be elastic restring force on the mass equal to F in the left side which is applied by the spring.
According to "Hook's Law
F = - Kx ---- (1)
Negative sign indicates that the elastic restoring force is opposite to the displacement.
Where K= Spring Constant
If we release mass 'm' at point 'a', it moves forward to ' O'. At point ' O' it will not stop but moves forward towards point "b" due to inertia and covers the same displacement -x. At point 'b' once again elastic restoring force 'F' acts upon it but now in the right side. In this way it continues its motion
from a to b and then b to a.
According to Newton's 2nd law of motion, force 'F' produces acceleration 'a' in the body which is given by
F = ma ---- (2)
Comparing equation (1) & (2)
ma = -kx
Here k/m is constant term, therefore ,
a = - (Constant)x
or
a a -x
This relation indicates that the acceleration of body attached to the end elastic spring is directly proportional to its displacement. Therefore its motion is Simple Harmonic Motion.
Answer:
N = 337.96 N
Explanation:
∅ = 32º
F = 249 N
m = 21 Kg
N = ?
We can apply:
∑ F = 0 (↑)
- Fy - W + N = 0 ⇒ N = Fy + W
⇒ F*Sin ∅ + m*g = N
⇒ N = (249 N*Sin32º) + (21 Kg*9.81 m/s²)
⇒ N = 337.96 N (↑)
