Answer:
Gravity changes with altitude. as we know The gravitational force is proportional to 1/R2, where R is your distance from the center of the Earth.
eg. The radius of the Earth at the equator is 6400 kilometers.
Let's say you were in a jet at the equator that was 40 kilometers high above the earth's surface.
may be helpfull
<h2>
The answer got is reasonable.</h2>
Explanation:
We have equation of motion v² = u² + 2as
Initial velocity, u = 300 m/s
Acceleration, a = ?
Final velocity, v = 400 m/s
Displacement,s = 4 km = 4000 m
Substituting
v² = u² + 2as
400² = 300² + 2 x a x 4000
a = 8.75 m/s² = 8.8 m/s²
The acceleration is 8.8 m/s²
The answer got is reasonable.
Answer:
Explanation:
Project mass m=3.8 kg
Initial speed vi= 0m/s
Final speed vf= 9.3×10³ m/s
Force F=9.3×10⁵N
To find
Time t
Solution
From Newtons second law we know that
∑F=ma
Where m is mass
a is acceleration
We can write this equation as
∑F=m(Δv/Δt)
Rearrange this equation to find time t
So
Substitute the given values
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.
We can use the equation for Newton's Law of Gravitation
Fg = (Gm₁m₂)/r²
Where gravitational constant = G = 6.674 x 10⁻¹¹ N · m²/kg²
mass m₁ = 0.145 kg
mass m₂ = 6.8 kg
distance between centers of masses = r = 0.5 m
Substitute these values into...
Fg = (Gm₁m₂)/r²
Fg = ((6.674 x 10⁻¹¹)(0.145)(6.8)) / (0.5)²
Fg = 2.63 x 10⁻¹⁰ N
Therefore, your answer should be <span>2.6 × 10–10</span>
