3 years ago

How long must a 400w electrical engine work in order to produce 300kj of work

Physics

1 answer:

vladimir1956 [14]3 years ago

8 0

Answer:

300000 J / 400 J/s = 750 s = 12.5 minutes

Explanation:

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A point charge of -3.0 x 10-C is placed at the origin of coordinates. Find the clectric field at the point 13. X= 5.0 m on the x

mezya [45]

Answer:

-1.0778×10⁻¹⁰ N/C

Explanation:

Applying,

E = kq/r²................ equation 1

Where E = elctric field, q = charge, r = distance, k = coulomb's law

From the question,

Given: q = -3.0×10 C, r = 5.0 m

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values in equation 1

E = (-3.0×10)(8.98×10⁹)/5²

E = -1.0778×10⁻¹⁰ N/C

Hence the electric field on the x-axis is -1.0778×10⁻¹⁰ N/C

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3 years ago

Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m

Oksi-84 [34.3K]

Answer:

Observed time, t = 5.58 s

Explanation:

Given that,

Speed of light in a vacuum has the hypothetical value of, c = 18 m/s

Speed of car, v = 14 m/s along a straight road.

A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s.

We need to find the time the driver of the car measure for his trip between the poles. The relation between real and observed time is given by :

$T=\dfrac{t}{\sqrt{1-\dfrac{v^2}{c^2}} }$

t is observed time.

$t=T\times \sqrt{1-\dfrac{v^2}{c^2}} \\\\t=8.89\times \sqrt{1-\dfrac{14^2}{18^2}} \\\\t=5.58\ s$

So, the time observed by the driver of the car measure for his trip between the poles is 5.58 seconds.

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