Answer:
23376 days
Explanation:
The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

where k is a constant.
From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

Let the orbital period of the earth be
and its mean distance of from the sun be
.
Also let the orbital period of the planet be
and its mean distance from the sun be
.
Equation (2) therefore implies the following;

We make the period of the planet
the subject of formula as follows;

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

Substituting equation (5) into (4), we obtain the following;

cancels out and we are left with the following;

Recall that the orbital period of the earth is about 365.25 days, hence;

The applicable relationship is N1/N2 = V1/V2, meaning the ratio of primary voltage to secondary voltage is equal to the ratio of primary turns to secondary turns.
Here N1 = 1000, V1 = 250, V2 = 400V and N2 = TBD.
Rewriting the above relationship, N2 = N1 V2/V1 = 1000 x 400/250 = 1600 turns.
Answer:
F = 200 N
Explanation:
Given that,
The mass suspended from the rope, m = 20 kg
We need to find the resultant force acting on the rope. The resultant force on the rope is equal to its weight such that,
F = mg
Where
g is acceleration due to gravity
Put all the values,
F = 20 kg × 10 m/s²
F = 200 N
So, the resultant force on the mass is 200 N.
Answer:
Explanation:
side of the square loop, a = 7 cm
distance of the nearest side from long wire, r = 2 cm = 0.02 m
di/dt = 9 A/s
Integrate on both the sides

i = 9t
(a) The magnetic field due to the current carrying wire at a distance r is given by


(b)
Magnetic flux,





(c)
R = 3 ohm

magnitude of voltage is
e = 1.89 x 10^-7 V
induced current, i = e / R = (1.89 x 10^-7) / 3
i = 6.3 x 10^-8 A
Answer: The bottom of the ladder is moving at 3.464ft/sec
Explanation:
The question defines a right angle triangle. Therefore using pythagorean
h^2 + l^2 = 10^2 = 100 ...eq1
dh/dt = -2ft/sec
dl/ dt = ?
Taking derivatives of time in eq 1 on both sides
2hdh/dt + 2ldl/dt = 0 ....eq2
Putting l = 5ft in eq2
h^ + 5^2 = 100
h^2 = 25 = 100
h Sqrt(75)
h = 8.66 ft
Put h = 8.66ft in eq2
2 × 8.66 × (-2) + 2 ×5 dl/dt
dl/dt = 17.32 / 5
dl/dt = 3.464ft/sec