Answer:
4.44 rpm
Explanation:
= Angular speed
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
r = Radius of Europa = 
R = Radius of arm = 6 m
The acceleration due to gravity is given by
Here the centripetal acceleration of the arm and acceleration due to gravity are equal
Converting to rpm
The angular speed of the arm is 4.44 rpm
Answer: 20.2 m/s
Explanation:
From the question above, we have the following data;
M1 = 800kg
M2 = 1200kg
V1 = 13m/s
V2 = 25m/s
U (common velocity) =?
M1V1 + M2V2 = (M1 + M2). U
(800*13) + (1200*25) = (800+1200) * U
10400 + 30000 = 2000u
40400 = 2000u
U = 40400 / 2000
U = 20.2 m/s
Wave power can be regarded as a reliable source of energy because the ocean currents are always moving.
<h3>What can be the challenges of wave power?</h3>
Wave power is a device that can be used to convert the mechanical energy of the ocean waves into electrical energy based on the principle of conservation of energy.
The major challenges that face the use of wave power in electricity generation is the unreliability of the waves which leads to uncertainty in the quantity of power generated Also, the wave direction and direction of ocean currents all limit the amount of power generated by this method. However, in spite of challenges, it can be regarded as a reliable source of energy because the ocean currents are always moving.
Learn more about wave power:brainly.com/question/1362067
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Answer:
The correct answer is B
Explanation:
To calculate the acceleration we must use Newton's second law
F = m a
a = F / m
To calculate the force we use the defined pressure and the radiation pressure for an absorbent surface
P = I / c absorbent surface
P = F / A
F / A = I / c
F = I A / c
The area of area of a circle is
A = π r²
We replace
F = I π r² / c
Let's calculate
F = 8.0 10⁻³ π (1.0 10⁻⁶)²/3 10⁸
F = 8.375 10⁻²³ N
Density is
ρ = m / V
m = ρ V
m = ρ (4/3 π r³)
m = 4500 (4/3 π (1 10⁻⁶)³)
m = 1,885 10⁻¹⁴ kg
Let's calculate the acceleration
a = 8.375 10⁻²³ / 1.885 10⁻¹⁴
a = 4.44 10⁻⁹ m/s² absorbent surface
The correct answer is B
That's what scientists and other technical people call the object's "<em>volume</em>".
