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gizmo_the_mogwai [7]
2 years ago
6

Which of the following can be correct units for acceleration?

Physics
1 answer:
SVETLANKA909090 [29]2 years ago
6 0

Answer:

B. Km/s/hr

Explanation:

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Answer:

D)

Explanation:

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3 0
2 years ago
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Match the terms to the correct descriptions. Question 1 options:
faust18 [17]
1) Refraction
2)Reflection
3)Concave
4)Convex

I took the test and got this right so you can believe me :)
Hope this helps
4 0
3 years ago
Read 2 more answers
1. An object in free fall will have an initial velocity equal to zero when: a. It is thrown vertically down
Ne4ueva [31]

Answer:

b. It is dropped

Explanation:

If the initial velocity is zero, the object move from rest. That happens if the object is dropped

6 0
3 years ago
If earth increase the distance from the sun, what will happen to the period of orbi t(the time it takes to complete one revoluti
Mandarinka [93]

The period of the orbit would increase as well

Explanation:

We can answer this question by applying Kepler's third law, which states that:

"The square of the orbital period of a planet around the Sun is proportional to the cube of the semi-major axis of its orbit"

Mathematically,

\frac{T^2}{a^3}=const.

Where

T is the orbital period

a is the semi-major axis of the orbit

In this problem, the question asks what happens if the distance of the Earth from the Sun increases. Increasing this distance means increasing the semi-major axis of the orbit, a: but as we saw from the previous equation, the orbital period of the Earth is proportional to a, therefore as a increases, T increases as well.

Therefore, the period of the orbit would increase.

Learn more about Kepler's third law:

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5 0
3 years ago
Two power lines run parallel for a distance of 222 m and are separated by a distance of 40.0 cm. if the current in each of the t
earnstyle [38]
1) Magnitude of the force:

The magnetic field generated by a current-carrying wire is
B= \frac{\mu_0I}{2 \pi r}
where
\mu_0 is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated

Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance 
r=40.0 cm=0.40 m, 
which is the distance at which the other wire is located:
B= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} N/A^2)(135 A) }{2 \pi (0.40 m)}=6.75 \cdot 10^{-5} T

Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
F=ILB=(135 A)(222 m)(6.75 \cdot 10^{-5}T)=2.03 N

2) direction of the force: 
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)

Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive

A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.
6 0
3 years ago
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