Answer:
Distance traveled will be 5.6307 m
Explanation:
Time t = 3 sec
We have given force F = 25 N
We know that force is given by F = ma
So ma = 25 -----------eqn 1
Weight is given by W = 196 N
We know that weight is given by W = mg
So mg = 196 -----------------eqn 2
From equation 1 and equation 2 
Initial velocity is given as 0 so u = 0 m/sec
From second equation of motion 
Air resistance is the answer
Answer:
18.1 × 10⁻⁶ A = 18.1 μA
Explanation:
The current I in the wire is I = ∫∫J(r)rdrdθ
Since J(r) = Br, in the cylindrical wire. With width of 10.0 μm, dr = 10.0 μm. r = 1.20 mm. We have a differential current dI. We integrate first by integrating dθ from θ = 0 to θ = 2π.
So, dI = J(r)rdrdθ
dI/dr = ∫J(r)rdθ = ∫Br²dθ = Br²∫dθ = 2πBr²
Now I = (dI/dr)dr at r = 1.20 mm = 1.20 × 10⁻³ m and dr = 10.0 μm = 0.010 mm = 0.010 × 10⁻³ m
I = (2πBr²)dr = 2π × 2.00 × 10⁵ A/m³ × (1.20 × 10⁻³ m)² × 0.010 × 10⁻³ m = 0.181 × 10⁻⁴ A = 18.1 × 10⁻⁶ A = 18.1 μA
<h3>
Answer:</h3>
20 seconds
<h3>
Explanation:</h3>
<u>We are given</u>;
- Power of the engine as 400 watts
- Force as 100 N
- Distance the object is lifted up as 80 m
We are required to determine the time taken.
- We need to know that power is the rate of work done
Therefore;
But, work done = Force × distance
Work done = 100 N × 80 m
= 8000 Joules
- Since , Power = Work done ÷ time
Then, time = Work done ÷ Power
Thus;
Time = 8000 J ÷ 400 W
= 20 s
Therefore, the time taken by the engine to lift the object is 20 seconds
Kinetic energy = 1/2 m v²
If we reduce the mass by half > m/2
Kinetic energy = 1/2 m/2 v²
We should know that 1/2 × 1/2 = 1/4
So kinetic energy will be :
1/4 × m × v²
