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marusya05 [52]
4 years ago
15

Which describes the attraction or repulsion caused by magnets?

Physics
1 answer:
Andrews [41]4 years ago
7 0
I think the correct answer from the choices listed above is the second option. It is magnetic force that describes the attraction or repulsion caused by magnets. This <span>arises between electrically charged particles because of their motion. Hope this answers the question.</span>
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The temperature at the surface of the Sun is approximately 5,300 K, and the temperature at the surface of the Earth is approxima
N76 [4]

Answer:

The entropy change of the Universe that occurs is 19.346 J/K

Explanation:

Given;

temperature of the sun, T_s = 5,300 K

temperature of the Earth, T_E = 293 K

radiation energy transferred by the sun to the earth, E = 6000 J

The sun loses Q of heat and therefore decreases its entropy by the amount

\delta S_{sun} = \frac{-Q}{T_s}

The earth gains Q of heat and therefore increases its entropy by the amount

\delta S_{Earth} = \frac{-Q}{T_E}

The total entropy change is:

\delta S_{Earth} + \delta S_{sun} = \frac{Q}{T_E} -\frac{Q}{T_S} \\\\                                                      = Q(\frac{1}{T_E} -\frac{1}{T_S} )\\\\= 6000(\frac{1}{293} -\frac{1}{5300} )\\\\=6000(0.0032243)\\\\= 19.346 \ J/K

Therefore, the entropy change of the Universe that occurs is 19.346 J/K

4 0
3 years ago
Which type of motion most accurately describes the behavior of a friction -less pendulum?
kaheart [24]

Answer:

B) Periodic Motion

Explanation:

When a pendulum is friction-less, i.e there are no damping forces acting on it, its motion will be periodic, i.e it will bob up and down going from potential energy to kinetic energy and back. Thus, the motion of the pendulum can be best described by the term "period motion", hence choice B.

If however, forces do act on the pendulum, and if they acts as to damp the pendulum, it will oscillate less and less as time goes by, and eventually come to a stop (in the real world this damping force is usually air resistance ). And if the force acts in such a way that it increases the oscillations, the pendulum will swing higher and higher, and the system will go haywire! :)

4 0
4 years ago
Which rope would have the longest wavelength-one with a frequency of 2 hz or one with a frequency of 3 hz?
Elis [28]
The frequency f is related to the wavelength \lambda by the equation
\lambda =  \frac{v}{f}
where v is the speed of the wave in the rope.

We can see from the formula that, if v is kept constant, smaller frequency means larger wavelength. So, the rope with frequency 2 Hz will have the longest wavelength.
3 0
3 years ago
The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla
Vladimir [108]

Answer

given,

Radius of sphere = 6.38 × 10⁶ m

time  = 1 day = 86400 s

\omega = \dfrac{2\pi}{T}

\omega = \dfrac{2\pi}{ 86400 }

\omega = 7.272 \times 10^{-5}\ rad/s

a) at equator

v = R_E \omega

v = 6.38 \times 10^6\times 7.272 \times 10^{-5}

v = 464 m/s

acceleration of the person

a = \omega^2R_E

a = (7.272 \times 10^{-5})^2(6.38 \times 10^6)

a = 0.03374 m/s^2

b) at a latitude of 61.0 ° north of the equator.

R = R_E cos \theta

R = 6.38 \times 10^6\times cos 61^0

R = 3.093 \times 10^6 m

v = R \omega

v = 3.093 \times 10^6 \times 7.272 \times 10^{-5}

v = 225 m/s

acceleration of the person

a = \omega^2R_E

a = (7.272 \times 10^{-5})^2(3.093 \times 10^6 )

a = 0.01635 m/s^2

4 0
3 years ago
A projectile is launched from ground level to the top of a cliff which is 195 m away and 155 m high. If the projectile lands on
muminat

Answer:

66.02m/s

Explanation:

the equation describing the distance covered in the horizontal direction is

x=ucos\alpha t-(1/2)gt^{2} but the acceleration in the horizontal path is zero, hence we have

x=ucos\alpha t

Since the horizontal distance covered is 155m at 7.6secs, we have ucos\alpha =\frac{155}{7.6} \\ucos\alpha =20.38.............equation 1

Also from the vertical path, the distance covered is expressed as

y=usin\alpha t-(1/2)gt^{2}

since the horizontal distance covered in 7.6secs is 195m, then we have

y=usin\alpha t-(1/2)gt^{2}\\y=7.6usin\alpha -4.9(7.6)^{2}\\478.02=7.6usin\alpha \\usin\alpha =62.9...........equation 2

Hence if we divide both equation 1 and 2 we arrive at

\frac{usin\alpha }{ucos\alpha } =\frac{62.9}{20.38} \\tan\alpha =3.08\\\alpha =tan^{-1}(3.08)\\\alpha =72.02^{0}\\

Hence if we substitute the angle into the equation 1 we have

ucos72.02=20.38\\u=66.02m/s

Hence the initial velocity is 66.02m/s

3 0
4 years ago
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