Answer:
Explanation:
Orbital radius of satellite A , Ra = 6370 + 6370 = 12740 km
Orbital radius of satellite B , Rb = 6370 + 19110 = 25480 km
Orbital potential energy of a satellite = - GMm / r where G is gravitational constant , M is mass of the earth and m is mass of the satellite
Orbital potential energy of a satellite A = - GMm / Ra
Orbital potential energy of a satellite B = - GMm / Rb
PE of satellite B /PE of satellite A
= Ra / Rb
= 12740 / 25480
= 1 / 2
b ) Kinetic energy of a satellite is half the potential energy with positive value , so ratio of their kinetic energy will also be same
KE of satellite B /KE of satellite A
= 1 / 2
c ) Total energy will be as follows
Total energy = - PE + KE
- P E + PE/2
= - PE /2
Total energy of satellite B / Total energy of A
= 1 / 2
Satellite B will have greater total energy because its negative value is less.
The distance traveled by the particle at the given time interval is 0.28 m.
<h3>
Position of the particle at time, t = 0</h3>
The position of the particle at the given time is calculated as follows;
x = 2 sin2(t)
y = 2 cos2(t)
x(0) = 2 sin2(0) = 0
y(0) = 2 cos2(0) = 2(1) = 2
<h3>
Position of the particle at time, t = 4</h3>
x = 2 sin2(t)
y = 2 cos2(t)
x(4) = 2 sin2(4) = 0.28
y(4) = 2 cos2(4) = 2(1) = 1.98
<h3>Distance traveled by the particle at the given time interval</h3>
d = √[(x₄ - x₀)² + (y₄ - y₀)²]
d = √[(0.28 - 0)² + (1.98 - 2)²]
d = 0.28 m
Thus, the distance traveled by the particle at the given time interval is 0.28 m.
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Answer:
3.4 mT
Explanation:
L = 0.53 m
i = 7.5 A
Theta = 19 degree
F = 4.4 × 10^-3 N
Let B be the strength of magnetic field.
Force on a current carrying conductor placed in a magnetic field.
F = i × L × B × Sin theta
4.4 × 10^-3 = 7.5 × 0.53 × B × Sin 19
B = 3.4 × 10^-3 Tesla
B = 3.4 mT
Answer:
h' = 603.08 m
Explanation:
First, we will calculate the initial velocity of the pellet on the surface of Earth by using third equation of motion:
2gh = Vf² - Vi²
where,
g = acceleration due to gravity on the surface of earth = - 9.8 m/s² (negative sign due to upward motion)
h = height of pellet = 100 m
Vf = final velocity of pellet = 0 m/s (since, pellet will momentarily stop at highest point)
Vi = Initial Velocity of Pellet = ?
Therefore,
(2)(-9.8 m/s²)(100 m) = (0 m/s)² - Vi²
Vi = √(1960 m²/s²)
Vi = 44.27 m/s
Now, we use this equation at the surface of moon with same initial velocity:
2g'h' = Vf² - Vi²
where,
g' = acceleration due to gravity on the surface of moon = 1.625 m/s²
h' = maximum height gained by pellet on moon = ?
Therefore,
2(1.625 m/s²)h' = (44.27 m/s)² - (0 m/s)²
h' = (1960 m²/s²)/(3.25 m/s²)
<u>h' = 603.08 m</u>
The correct unit for the speed of light is [ m s⁻¹ ].
Time = (distance) / (speed)
Time = (9.3 x 10^7 miles) x (1609 m/mile) / (3 x 10^8 m/s) = 498.8 seconds .
That would be <em>8.31 minutes</em>.
