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2 years ago

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Charge q is a test charge used to measure the electric field created by a source charge Q. What happens to the magnitude of the

electric force on these charges if q is doubled?

1 answer:

Arte-miy333 [17]2 years ago

7 0

Answer:

The electric force will also double.

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hi, Need help with triangle law of vector addition worksheet and a verifying Newtons second law worksheet?

Yuri [45]

Use Newton's second law and the free body diagram to determine the net force and acceleration of an object. In this unit, the forces acting on the object were always directed in one dimension.

The object may have been subjected to both horizontal and vertical forces but there was no single force directed both horizontally and vertically. Moreover, when free-body diagram analysis was performed, the net force was either horizontal or vertical, never both horizontal and vertical.

Times have changed and we are ready for situations involving two-dimensional forces. In this unit, we explore the effects of forces acting at an angle to the horizontal. This makes the force act in two dimensions, horizontal and vertical. In such situations, as always in situations involving one-dimensional network forces, Newton's second law applies.

Learn more about Newton's second law here:-brainly.com/question/25545050

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5 0

3 months ago

A 60 kilogram student jumps down from a laboratory counter. At the instant he lands on the floor hus speed is 3 meters per secon

erastovalidia [21]

As per Newton's law rate of change in momentum is net force

so we can write it as

$F = \frac{dP}{dt}$

$F = \frac{m(v_f - v_i)}{\Delta t}$

now we know that

$m = 60 kg$

$v_f = 3 m/s$

$v_i = 0$

$\Delta t= 0.2 s$

from above equation

$F = \frac{60(3 - 0)}{0.2} = 900 N$

so he will experience 900 N force in above case

5 0

2 years ago

If both mass and speed are doubled, what happens to its momentum?

puteri [66]

The general formula is: Momentum = (mass) x (speed)

I never like to just write a bunch of algebra without explaining it.
But in this particular case, there's really not much to say, and
I think the algebra will pretty well explain itself. I hope so:

Original momentum = (original mass) x (original speed)

New momentum = (2 x original mass) x (2 x original speed)

   = (2) x (original mass) x (2) x (original speed)

   = (2) x (2) x (original mass) x (original speed)

   = (4) x (original mass) x (original speed)

   = (4) x (original momentum).

7 0

2 years ago

A student fills a tank of radius r with water to a height of h1 and pokes a small, 1.0 cm diameter hole at a distance h2 from th

Alik [6]

when a hole is made at the bottom of the container then water will flow out of it

The speed of ejected water can be calculated by help of Bernuolli's equation and Equation of continuity.

By Bernoulli's equation we can write

$Po + \frac{1}{2}\rho v_1^2 + \rho g h = Po + \frac{1}{2}\rho v_2^2 + \rho g *0$

Now by equation of continuity

$A_1v_1 = A_2v_2$

$\pi (0.2)^2 v_1 = \pi (0.01)^2 v_2$

from above equation we can say that speed at the top layer is almost negligible.

$v_1 = 0$

now again by equation of continuity

$\rho g h = \frac{1}{2} \rho v^2$

$v = \sqrt{2 g h}$

here we have

$h = h_1 - h_2$

$h = 0.50 - 0.03 = 0.47m$

now speed is given by

$v = \sqrt{2* 9.8 * 0.47}$

$v = 3.03 m/s$

7 0

2 years ago

A highway curves to the left with radius of curvature of 36 m and is banked at 28 ◦ so that cars can take this curve at higher s

iVinArrow [24]

Answer: 30.34m/s

Explanation:

The sum of forces in the y direction 0 = N cos 28 - μN sin28 - mg

Sum of forces in the x direction

mv²/r = N sin 28 + μN cos 28

mv²/r = N(sin 28 + μcos 28)

Thus,

mv²/r = mg [(sin 28 + μ cos 28)/(cos 28 - μ sin 28)]

v²/r = g [(sin 28 + μ cos 28)/(cos 28 - μ sin 28)]

v²/36 = 9.8 [(0.4695 + 0.87*0.8829) - (0.8829 - 0.87*0.4695)]

v²/36 = 9.8 [(0.4695 + 0.7681) / (0.8829 - 0.4085)]

v²/36 = 9.8 (1.2376/0.4744)

v²/36 = 9.8 * 2.6088

v²/36 = 25.57

v² = 920.52

v = 30.34m/s

5 0

2 years ago