Answer:
424088766.068 m
Explanation:
Radius of the circular orbit that the satellite is 2.6 Earth radii (r) = 2.6 R
R = Radius of earth = 6371000 m (mean radius)
In order to find the distance that the satellite travels in 5.89 hours to complete one complete revolution is the circumference of the circular orbit
Circumference of a circle = 2×π×r
⇒Distance travelled in 5.89 hours = 2×π×2.6 R
⇒Distance travelled in 5.89 hours = 2×π×2.6×6371000
⇒Distance travelled in 5.89 hours = 104078451.3393m
Distance travelled in 1 hour = 104078451.3393/5.89 = 17670365.252 m
∴ Distance travelled in 24 hours = 17670365.252×24 = 424088766.068 m
<span>2. mass of both objects doubled?
Hope it helped!</span>
<span>if we assume the origin is at the dropping point and the object is merely dropped and not thrown up or down then y0 = 0 and v0 = 0. The equation reduces to </span>
<span>y = 0 + 0t + ½gt² </span>
<span>y = ½gt² </span>
<span>t = √(2y/g) </span>
<span>in the ft - lb - s system </span>
<span>y = -100 ft </span>
<span>g = -32.2 ft / s² </span>
<span>t = √(2y/g) </span>
<span>t = √(2(-100) / (-32.2)) </span>
<span>t = 2.5 s</span>
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,
KE1 = KE2
The kinetic energy of the system before the collision is solved below.
KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²
This value should also be equal to KE2, which can be calculated using the conditions after the collision.
KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)
The value of x from the equation is 17.16 cm/s.
Hence, the answer is 17.16 cm/s.
Answer:
5.1*10^3 J/m^3
Explanation:
Using E = q/A*eo
And
q =75*10^-6 C
A = 0.25
eo = 8.85*10^-12
Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]
= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]
= 5.1*10^3 J/m^3
