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Amiraneli [1.4K]
3 years ago
11

This is 30% of my overall grade

Physics
2 answers:
Natalija [7]3 years ago
8 0

Answer: C - The moon will take a larger loop around the earth.

Explanation:

algol [13]3 years ago
6 0

Answer: C

Explanation:

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Determine the values of mm and nn when the following average magnetic field strength of the Earth is written in scientific notat
joja [24]

Answer:

m = 4.51 and n = -5              

Explanation:

The average magnetic field strength of the Earth is 0.0000451 T. We need to write the value of magnetic field strength of the Earth in scientific notation. In scientific notation, any value is given by :

N=m\times 10^n.........(1)

Where

m is a real number

n is any integer

Here,

N=4.51\times 10^{-5}.......(2)

On comparing equation (1) and (2), we get the values as :

m = 4.51 and n = -5

Hence, this is the required solution.

6 0
3 years ago
Which performers can benefit from a protein-rich diet?
GrogVix [38]

Answer:

in my opinion its A

Explanation:

7 0
3 years ago
Two balloons are charged with an identical quantity and type of charge: -4 nc They are held apart at a separation distance of 70
blagie [28]

Answer:

29.4 uN

Explanation:

The electric force between two charges can be calculated using Coulomb's Law. According to this law the force between two point charges is given as:

F=k\frac{q_{1} q_{2} }{r^{2}}

where k is a proportionality constant known as the Coulomb's law constant. Its value is 9 \times 10^{9} Nm²/C²

r = distance between charges = 70 cm = 0.7 m

q1 = q2 = 4nC = 4 \times 10^{-9} C

The negative sign indicates that the charges are negative. In the formula we will only use the magnitude of the charges.

Using these values in the formula, we get:

F=9 \times 10^{9} \times \frac{4 \times 10^{-9} \times 4 \times 10^{-9}}{0.7^{2}}\\\\ F=2.94\times10^{-7} N\\\\ F=29.4 \times 10^{-6} N\\\\ F=29.4 \mu N

Therefore, the magnitude of repulsive force between the given charges will be 29.4 uN

8 0
4 years ago
21 of 35 Review A friend of yours is loudly singing a single note at 401 HzHz while racing toward you at 23.0 m/sm/s on a day wh
Ipatiy [6.2K]

Answer:

The frequency of the sound note as heard = 429 Hz

Explanation:

The frequency of sound waves as heard from a distance for a sound wave coming towards one at v₀ m/s and whose real frequency is f₀ is given by

+f = f₀/[1 - (v₀/v)]

+f = frequency of sound as heard from the distance away = ?

f₀ = real frequency of sound = 401 Hz

v₀ = velocity at which the sound source is moving towards the reference point = 23.0 m/s

v = velocity of sound waves = 343 m/s

+f = 401/[1 - (23/343)]

+f = 429 Hz

5 0
4 years ago
Synonym of an applied force
tatuchka [14]

Answer:

Coerce, compel, constrain

Explanation:

7 0
3 years ago
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