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labwork [276]
3 years ago
14

Like magnetic poles Always.

Physics
2 answers:
natka813 [3]3 years ago
8 0
It's A they repel eachother
muminat3 years ago
5 0
Repel each other.

When it deals with magnetic poles think of it this way:
Opposites attract, likes repel.
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HELP PLS!!!! Light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is th
maw [93]

You've already told us the speed in ft/s .  It's right there in the question.  You said that light travels about  982,080,000 ft/s.

We don't know how accurate that number is, but for purposes of THIS question, that's the number we're going with.

In scientific notation, it's written . . . <em>9.8208 x 10⁸ ft/s .</em>

We don't know where you were going with the number of seconds in a year.  But to answer the question that you eventually asked, it turned out that we don't even need it.

6 0
3 years ago
At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . It moves at constan
Jet001 [13]

The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.

At time t_1, the acceleration vector has direction \theta_1 such that

\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ

which indicates the particle is situated at a point on the lower left half of the circle, while at time t_2 the acceleration has direction \theta_2 such that

\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ

which indicates the particle lies on the upper left half of the circle.

Notice that \theta_1-\theta_2=90^\circ. That is, the measure of the major arc between the particle's positions at t_1 and t_2 is 270 degrees, which means that t_2-t_1 is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

Recall that

\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}

where R is the radius of the circle and T is the period. We have

t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s

and the magnitude of the particle's acceleration toward the center of the circle is

\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}

So we find that the path has a radius R of

7.21\dfrac{\rm m}{\mathrm s^2}=\dfrac{4\pi^2R}{(4.00\,\mathrm s)^2}\implies\boxed{R=2.92\,\mathrm m}

8 0
3 years ago
How is the Sun similar to the stars that can be seen at night?
prisoha [69]

Answer:

A

Explanation:

4 0
3 years ago
Read 2 more answers
A weightlifter lifts a set of weights a vertical distance of 2 meters. If a constant net force of 350 N is exerted on the weight
o-na [289]

Answer:

The work done by the weightlifter, W = 700 J

The power of the weightlifter, P = 350 watts

Explanation:

A weightlifter lifts a set of weights a vertical distance, s = 2 m

The force exerted to lift the weight, F = 350 N

The work done by the body is defined as the product of the force applied by the body to the displacement it caused.

                            W = F x s

                                = 350 N x 2 m

                                = 700 J

The work done by the weightlifter, W = 700 J

The time taken by the weightlifter to lift the weight, t = 2 s

The power is defined as the rate of body to do work. It is given by the equation,

                            P = W / t

                               = 700 J / 2 s

                               = 350 watts

Hence, the power of the weightlifter, P = 350 watts

6 0
3 years ago
If an athlete executing a long jump leaves the ground at a 30° angle and travels 7.80 m, what was the takeoff speed
Mumz [18]
R=u^2 sin2x / g
7.8=u^2 sin2*30 /10
u=9.5m/s
6 0
3 years ago
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