I'm assuming we're applying the standard Integral form of the calculation of work. The solution is provided in the image.
Answer:
R = 2Ω
Explanation:
Potential difference (V) = current (I) * Resistance (R)
V = IR
I = 2.0A
V = 10v
R = ?
V = IR
R = V / I
R = 10 / 2
R = 2Ω
The resistance across the wire is 2Ω
Magnitude of acceleration
Explanation:
We know that acceleration can increase depending in the force applied on an object, any object with a greater mass will apply a greater force. F = M(a).
(a) 5.66 m/s
The flow rate of the water in the pipe is given by

where
Q is the flow rate
A is the cross-sectional area of the pipe
v is the speed of the water
Here we have

the radius of the pipe is
r = 0.260 m
So the cross-sectional area is

So we can re-arrange the equation to find the speed of the water:

(b) 0.326 m
The flow rate along the pipe is conserved, so we can write:

where we have

and where
is the cross-sectional area of the pipe at the second point.
Solving for A2,

And finally we can find the radius of the pipe at that point:

Answer:
45000 kg and 45 tons
Explanation:
The expression in kilograms and tons is shown below;
As we know that
1 gr is 0.001 kg
So, 45000000 = 45,000 kg
And,
1 kg = 0.001 tons
So, 45000 kg = 45 tons
Therefore the same would be considered