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3 years ago

PLEASE HELPPPPPPP!!!!!!!!!!

Physics

1 answer:

natka813 [3]3 years ago

8 0

Answer:

F' = 800 N

Explanation:

The electrical force between charges is 400 N.

The electrical force between two charges is given by :

$F=k\dfrac{q_1q_2}{r^2}$

If q₁' = 2q₁, new force becomes,

$F'=k\dfrac{q_1'q_2'}{r^2}\\\\F'=k\dfrac{2q_1\times q_2}{r^2}\\\\F'=2\times F\\\\F'=2\times 400\\\\F'=800\ N$

So, the new force becomes 800 N.

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What is the kinetic energy of a 2995 kg truck when it moving 47 ms

Dafna11 [192]

To find the kinetic energy . You have to use this equation : 1/2m(v^2) .

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3 years ago

A horse does 910 J of work in 380 seconds while pulling a wagon. What is the power output of the horse? Round your answer to two

frutty [35]

Answer: 2.4 watts

Explanation:

Work done by horse = 910J

Time = 380 seconds

Power output of the horse = ?

Recall that power is the rate of doing work. Thus, power is workdone divided by time taken.

i.e Power = (work/time)

Power = 910J/380seconds

Power = 2.395 watts (rounded to two significant figures as 2.4 watts)

Thus, the power output of the horse is 2.4 watts

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3 years ago

Ignoring air resistance and the little friction from the plastic tube, the magnet was a freely-falling object in each trial. If

horsena [70]

Answer:

emf will also be 10 times less as compared to when it has fallen $40 \mathrm{m}$

Explanation:

We know, from faraday's law-

$e m f=-N \frac{\Delta \Phi}{\Delta T}$

and $\Phi=B . A$

So, as the height increases the velocity with which it will cross the ring will also increase. $(v=\sqrt{2 g h})$

Given

$\mathrm{V} 1(\text { Speed at } 40 \mathrm{m})=2 \mathrm{x} \mathrm{V} 2(\text { speed at } 10 \mathrm{m})$

$\sqrt{2 g h_{2}}=2 \times \sqrt{2 g h_{1}}=28.28 \mathrm{m} / \mathrm{s}$

Now, from $40 \mathrm{cm}$

$V_{3}=\sqrt{2 g h_{3}}=\sqrt{2 \times 10 \times 0.4}=2.82 \mathrm{m} / \mathrm{s}$

From equation a and b we see that velocity when dropped from $40 \mathrm{m}$ is 10 times greater when height is 40 $\mathrm{cm}$ so, emf will also be 10 times less as compared to when it has fallen $40 \mathrm{m}$

4 0

3 years ago

A light source shines light consisting of two wavelengths, λ1 = 540 nm (green) and λ2 = 450 nm (blue), on two slits separated by

densk [106]

Answer:

The maximun distance is $z_1 = z_2 = 0.0138m$

Explanation:

From the question we are told that

The wavelength are $\lambda _ 1 = 540nm (green) = 540 *10^{-9}m$

$\lambda_2 = 450nm(blue) = 450 *10^{-9}m$

The distance of seperation of the two slit is $d = 0.180mm = 0.180 *10^{-3}m$

The distance from the screen is $D = 1.53m$

Generally the distance of the bright fringe to the center of the screen is mathematically represented as

$z = \frac{m \lambda D}{d}$

Where m is the order of the fringe

For the first wavelength we have

$z_1 = \frac{m_1 (549 *10^{-9} * (1.53))}{0.180*10^{-3}}$

$z_1=0.00459m_1 m$

$z_1= 4.6*10^{-3}m_1 m ----(1)$

For the second wavelength we have

$z_2 = m_2 \frac{450*10^{-9} * 1.53 }{0.180*10^{-3}}$

$z_2 = 0.003825m_2$

$z_2 = 3.825 *10^{-3} m_2 m$ ----(2)

From the question we are told that the two sides coincides with one another so

$zy_1 =z_2$

$4.6*10^{-3}m_1 m = 3.825 *10^{-3} m_2 m$

$\frac{m_1}{m_2} = \frac{3.825 *10^{-3}}{4.6*10^{-3}}$

Hence for this equation to be solved

$m_1 = 3$

and $m_2 = 4$

Substituting this into the equation

$z_1 = z_2 = 3 * 4.6*10^{-3} = 4* 3.825*10^{-3}$

Hence $z_1 = z_2 = 0.0138m$

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3 years ago

g as measured from the earth, a spacecraft is moving at speed .80c toward a second spacecraft moving at speed .60c back toward t

cupoosta [38]

Answer:

the speed of the first spacecraft as viewed from the second spacecraft is 0.95c

Explanation:

Given that;

speed of the first spacecraft from earth v $_a$ = 0.80c

speed of the second spacecraft from earth v $_b$ = -0.60 c

Using the formula for relative motion in relativistic mechanics

u' = ( v $_a$ - v $_b$ ) / ( 1 - (v $_b$ v $_a$ / c²) )

we substitute

u' = ( 0.80c - ( -0.60c) ) / ( 1 - ( ( 0.80c × -0.60c) / c² ) )

u' = ( 0.80c + 0.60c ) / ( 1 - ( -0.48c² / c² ) )

u' = 1.4c / ( 1 - ( -0.48 ) )

u' = 1.4c / ( 1 + 0.48 )

u' = 1.4c / 1.48

u' = 0.9459c ≈ 0.95c { two decimal places }

Therefore, the speed of the first spacecraft as viewed from the second spacecraft is 0.95c

7 0

3 years ago