Answer:
(orbital speed of the satellite) V₀ = 3.818 km
Time (t) = 4.5 × 10⁴s
Explanation:
Given that:
The radius of the Earth is 6.37 × 10⁶ m; &
the acceleration of gravity at the satellite’s altitude is 0.532655 m/s
We can calculate the orbital speed of the satellite by using the formula:
Orbital Speed (V₀) = √(r × g)
radius of the orbit (r) = 21000 km + 6.37 × 10⁶ m
= (2.1 × 10⁷ + 6.37 × 10⁶) m
= 27370000
= 2.737 × 10⁷m
Orbital Speed (V₀) = √(r × g)
Orbital Speed (V₀) = √(2.737 × 10⁷ × 0.532655 )
= 3818.215
= 3.818 × 10³
= 3.818 Km
To find the time it takes to complete one orbit around the Earth; we use the formula:
Time (t) = 2 π × 
= 2 × 3.14 × 
= 45019.28
= 4.5 × 10 ⁴ s
Assuming that all the bullet’s energy heats the paraffin, its final temperature is 27.1 degree C. The correct option is D.
<h3>What is temperature?</h3>
Temperature is the degree of hotness and coldness of the material.
The energy of the bullet E = 1/2 mv²
E = 1/2 x 10 x 10⁻³ x (2000)²
E = 2 x 10⁴ J
This heat is used in heating the paraffin
E = m x c ΔT = m x c (Tfinal -Tinitial)
2 x 10⁴ J = 1 x 2.8 x 10³ x (Tfinal -20)
Tfinal = 27.1°C
Thus, the final temperature is 27.1 degree C. The correct option is D.
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The gravitational attraction between two planets is 4905.95 N
<h3>What is gravitational attraction?</h3>
When two objects with masses are placed at a distance, there will an attractive force acting between them.
According to the Newton's law of gravitation, gravitational force is
F = Gm₁m₂ /r²
where r is the distance between the masses m₁ and m₂ and G is the gravitational constant G = 6.67 x 10⁻¹¹ N-m²/kg²
Substitute the values into the expression, we get
F = 6.67 x 10⁻¹¹ x 2.25 x 10²⁰ x 6.20 x 10¹⁸ / (435,500 x 1000)²
F= 4905.95 N
Thus, the gravitational attraction between two planets is 4905.95 N.
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Answer:
The strength of the magnetic field is
.
Explanation:
Given that,
Length of the rod, L = 1.01 m
Speed with which the rod is moving, v = 3.47 m/s
We need to find the strength of the magnetic field that is perpendicular to both the rod and your direction of motion and that induces an EMF of 0.265 mV across the rod. When the rod is moving with some speed, an emf gets induced and it is given by :

B is magnetic field

So, the strength of the magnetic field is
.