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4 years ago

HELP! 10 PTS! Two objects of different mass have equal, non-zero kinetic energies. Which object has the greater momentum?

1 answer:

4 0

The heavier one. The explanation for this is that <span>momentum is directly proportional to mass. The more mass there is, the more momentum.

This is as simply as I could explain it, hope I helped :) </span>

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4 years ago

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At what speed, as a fraction of c, will a moving rod have a length 65% that of an identical rod at rest

dezoksy [38]

Answer:

v/c = 0.76

Explanation:

Formula for Length contraction is given by;

L = L_o(√(1 - (v²/c²))

Where;

L is the length of the object at a moving speed v

L_o is the length of the object at rest

v is the speed of the object

c is speed of light

Now, we are given; L = 65%L_o = 0.65L_o, since L_o is the length at rest.

Thus;

0.65L_o = L_o[√(1 - (v²/c²))]

Dividing both sides by L_o gives;

0.65 = √(1 - (v²/c²))

Squaring both sides, we have;

0.65² = (1 - (v²/c²))

v²/c² = 1 - 0.65²

v²/c² = 0.5775

Taking square root of both sides gives;

v/c = 0.76

6 0

3 years ago

If 270 wats of power is used in 42 seconds how much work is done ?

pentagon [3]

Answer:

power is given by work done divide time taken

P=w/t

therefore Pt=w

270x42=11,340.J

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4 years ago

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio

Gnesinka [82]

Answer:

a). $a_p=-2.39x10^{-12} rad/s^2$

b). $t=1016298.8 years$

c). $T_i=80.58x10^{-3}s$

Explanation:

a).

The acceleration for definition is the derive of the velocity so:

$a_p=\frac{dw}{dt}$

$w=\frac{2\pi}{t}$

$a_p=\frac{dw}{dt}=-\frac{2\pi}{t^2}*\frac{dT}{dt}$

$dT=0.0808s$

$dt=1 year*\frac{365d}{1year} \frac{24hr}{1d} \frac{60minute}{1hr} \frac{60s}{1minute}=31.536x10^{6}s$

Replacing

$a_p=-\frac{2\pi}{0.082s^2}*\frac{9.84x10^{-7}}{31.536x10^{6}s}= -2.39x10^{-12} rad/s^2$

b).

If the pulsar will continue to decelerate at this rate, it will stop rotating at time:

$t=\frac{w}{a_p}$

$w=\frac{2\pi }{t}=\frac{2\pi }{0.0820s}=76.62 rad/s$

$t=\frac{76.62 rad/s}{2.39x10^{-12}rad/s^2}= 3.2058x10^{13}s$

$t=1016298.8 years$

c).

582 years ago to 2019

1437

$T_i=0.0820-9.84x10^{-7}*1437)=80.58x10^{-3}s$

5 0

4 years ago

Which factors affect the strength of the electric force between two objects

iris [78.8K]

By definition, the electric force is given by:

$f = k * \frac{q1q2}{d ^ 2}$

Where,

q1: electric charge of object number 1.

q2: electric charge of object number 2.

d: distance between both objects

k: proportionality constant

Therefore, the magnitude of the electric force is affected by:

1) The product of the charges of the objects

2) The distance between objects

Answer:

The factors that affect strength of the electric force between two objects are:

1) The product of the charges of the objects

2) The distance between objects

5 0

4 years ago

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