Answer:
Yes, this is <em><u>true</u></em>. Hope this helps :)
Answer: b) A positive charge is moved toward a positive charge.
Answer:
117.72 N
Explanation:
The given parameters are;
The mass m₁ = 2.0 × 10³ kg
The mass m₂ = 4.4 × 10² kg
The mass of the man, m₃ = 6.0 × 10 kg
The condition of the interaction of the surfaces = Frictionless surfaces
The
The tension in the string = The downward force = The weight of (m₂ + m₃) = (m₂ + m₃) × g
Let <em>a</em> represent the acceleration of the connected masses due to the weight of m₂, and m₃, we have;
(m₁ + m₂ + m₃) × a = (m₂ + m₃) × g
∴ a = (m₂ + m₃) × g/(m₁ + m₂ + m₃)
Which gives;
a = (4.4 × 10²+ 6.0 × 10) × 9.81/(2.0 × 10³+ 4.4 × 10²+ 6.0 × 10) = 1.962
The downward acceleration, a = 1.962 m/s²
The apparent weight of the man = The mass of the man, m₃ × The acceleration, <em>a</em>
∴ The apparent weight of the man = 6.0×10 kg ×1.962 m/s² = 117.72 N
<h3>
Answer:</h3>
5.395 × 10^8 Watts
<h3>
Explanation:</h3>
<u>We are given;</u>
- Rate of flow is 1.1 × 10^6 kg/s
- Distance is 50.0 m
- Gravitational acceleration is 9.8 m/s²
We are required to calculate the power that is generated by the falling water
- Power is the rate of work done
- It is given by dividing the energy or work done by time
But; work done = Force × distance
Therefore;
Power = (F × d) ÷ time
The rate is 1.1 × 10^ 6 Kg/s
But, 1 kg = 9.81 N
Therefore, the rate is equivalent to 1.079 × 10^7 N/s
Thus,
Power = Rate (N/s) × distance
= 1.079 × 10^7 N/s × 50.0 m
= 5.395 × 10^8 Watts
The power generated from the falling water is 5.395 × 10^8 Watts