Answer:
Stretch can be obtained using the Elastic potential energy formula.
The expression to find the stretch (x) is 
Explanation:
Given:
Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.
To find: Elongation in the spring (x).
We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).
The formula to find EPE is given as:

Rewriting the above expression in terms of 'x', we get:

Example:
If EPE = 100 J and spring constant, k = 2 N/m.
Elongation or stretch is given as:

Therefore, the stretch in the spring is 10 m.
So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.
The weight of the box is <em>w</em> = <em>mg</em>, where <em>m</em> is the mass. So
<em>m</em> = <em>w</em>/<em>g</em> = (3893.40 N) / (9.80 m/s²) ≈ 397 kg
Then the box has density
(397 kg)/(4.60 m³) ≈ 86.4 kg/m³
which is less than the density of the given liquid, so the box will float.
Answer:10
Explanation:
You have to do speed divided by time so your answers 10
Answer:
C. 0.25J
Explanation:
Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;
L is the inductance
I is the current flowing in the inductor
Given parameters
L = 20mH = 20×10^-3H
I = 5A
Required
Energy stored in the magnetic field.
E = 1/2 × 20×10^-3 × 5²
E = 1/2 × 20×10^-3 × 25
E = 10×10^-3 × 25
E = 0.01 × 25
E = 0.25Joules.
Hence the energy stored in the magnetic field of this inductor is 0.25Joules
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