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Andreas93 [3]
3 years ago
10

When the north pole of a magnet is moved towards the center of a loop of wire containing a galvanometer, the needle of the galva

nometer flicks to the right. What would also cause the needle on the galvanometer to move to the right?
Physics
1 answer:
andriy [413]3 years ago
6 0

Answer:

Moving the magnet away from the center of the loop with its south pole facing the center of the loop.

Explanation:

Electromagnetic induction is due to a rapidly changing magnetic field, or loop area. The poles of the magnet induce current in the loop but in the opposite direction, depending on the direction of their relative motion. An approaching north pole will induce an anticlockwise current in the loop, while an approaching south pole will do the reverse. To get the galvanometer to flicker in the same direction as of that when the north pole was approaching, we move the magnet away from the center of the loop with its south pole facing the center of the loop.

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The moon Phobos orbits Mars
shepuryov [24]

27.9816 \times 10^{3} s is the period of orbit.

<u>Explanation: </u>

The equation that is useful in describing satellites motion is Newton form after Kepler's Third Law. The period of the satellite (T) and the average distance to the central body (R) are related as the following equation:

                  \frac{T^{2}}{R^{3}}=\frac{4 \times \pi^{2}}{G \times M_{c e n t r a l}}

Where,

T is the period of the orbit

R is the average radius of orbit

G is gravitational constant  6.673 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}

Here, given data

M=6.23 \times 10^{23} \mathrm{kg}

R=9.38 \times 10^{6} \mathrm{m}

Substitute the given values, we get T as

      \frac{T^{2}}{\left(9.38 \times 10^{6}\right)^{3}}=\frac{4 \times(3.14)^{2}}{\left(6.673 \times 10^{-11}\right) \times 6.23 \times 10^{23}}

      T^{2}=\frac{4 \times 9.8596 \times 825.29 \times 10^{18}}{41.57 \times 10^{12}}

      T^{2}=\frac{32548.12 \times 10^{18-12}}{41.57}=782.97 \times 10^{6}

Taking square root, we get

       T=27.9816 \times 10^{3} s

4 0
3 years ago
The current supplied by a battery slowly decreases as the battery runs down. Suppose that the current as a function of time is:
ludmilkaskok [199]

Answer: 8.1 x 10^24

Explanation:

I(t) = (0.6 A) e^(-t/6 hr)

I'll leave out units for neatness: I(t) = 0.6e^(-t/6)

If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).

For neatness let k = 1/(6x3600) = 4.63x10^-5, then:

I(t) = 0.6e^(-kt)

Providing t is in seconds, total charge Q in coulombs is

Q= ∫ I(t).dt evaluated from t=0 to t=∞.

Q = ∫(0.6e^(-kt)

= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.

= -(0.6/k)[e^-∞ - e^-0]

= -0.6/k[0 - 1]

= 0.6/k

= 0.6/(4.63x10^-5)

= 12958 C

Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.

5 0
4 years ago
What is the period of a spring that oscillates with a frequency of 2.09 Hz​
Sergio [31]

Frequency, υ = 2.09 Hz

We know that time period, T = 1/υ

∴ T = 1 ÷ 2.09

⇒ T = 0.4784 seconds

5 0
4 years ago
Read 2 more answers
A 66.5kg astronaut is doing a repair in space on the orbiting space station. She throws a 2.30kg tool away from her at 3.10m/s r
MrRa [10]

Answer:

Explanation:

mass of astronaut, M = 66.5 kg

mass of tool, m = 2.3 kg

velocity of tool, v = 3.10 m/s

Let the velocity of astronaut is V.

(A) According to the conservation of moemntum

Momentum of astronaut = Momentum of tool

M x V = m x v

66.5 x V = 2.3 x 3.10

V = 0.107 m/s

(B) The direction of motion of astronaut is opposite to the direction of motion of tool.

6 0
3 years ago
What's a solution that feels slippery
larisa86 [58]
What are you exactly asking for? A solution that feels slippery? That could mean anything. Please explain a little bit more.
4 0
3 years ago
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