Answer:
e = 10 V
Explanation:
given,
number of the coaxial loops = 10
Cross sectional area = 0.5 m²
magnitude of magnetic field =
B = 3 T + (2 T/s)*t.
B = ( 3+ 2 t ) T
induced potential difference = ?
At time = 2 s
we know,
induced emf

∅ = B . A




e = -10 V
magnitude of induced emf
|e| = |-10 V|
e = 10 V
the induced potential difference in the loop = e = 10 V
The JWST is postioned about 1.5 million kilometers from the earth on the side facing away from the sun
Answer:
When object is placed between the focus (F) and pole (P) of a concave mirror, magnified and erect image of the object is formed on the back of the mirror.
When object is placed between the centre of curvature and the principal focus of a concave mirror, magnified and inverted image is formed in front of the mirror.
Explanation:
Answer:
1716.75 J
Explanation:
<u>Step </u><u>1</u><u>:</u> First check what we are provided with. As per given question we have:
mass (m) = 70 kg, height (h) = 2.5 m and acceleration due to gravity (g) = 9.81 m/s².
<u>Step</u><u> </u><u>2</u><u>:</u> Check what we are asked to find out.
Work done = Change in Potential energy
The stuff required to solve this question is potential energy. Using the formula: P = mgh. Where P is Potential energy, m is mass, g is acceleration due to gravity and h is height.
<u>Step</u><u> </u><u>3</u><u>:</u> Substitute the known values in the above formula.
→ P = 70 × 2.5 × 9.81
→ P = 1716.75 J
Hence, the work done against the force of gravity is 1716.75 J.
I think it’s D) all of the above