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2 years ago

The electrical force on a 2-c charge is 60 n. the electric field where the charge is located is

1 answer:

5 0

The electrical force acting on a charge q immersed in an electric field is equal to
$F=qE$
where
q is the charge
E is the strength of the electric field

In our problem, the charge is q=2 C, and the force experienced by it is
F=60 N
so we can re-arrange the previous formula to find the intensity of the electric field at the point where the charge is located:
$E= \frac{F}{q}= \frac{60 N}{2 C}=30 N/C$

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two cars are moving at constant speeds in a straight line along a major highway. The first is travelling at 20ms^-1 and the seco

Reika [66]

Answer:

$t=750s$

Explanation:

The two cars are under an uniform linear motion. So, the distance traveled by them is given by:

$\Delta x=vt\\x_f-x_0=vt\\x_f=x_0+vt$

$x_f$ is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have:

$x_0_1=6km\\x_0_2=0$

We have $x_f_1=x_f_2$ . Thus, solving for t:

$x_0_1+v_1t=x_0_2+v_2t\\x_0_1=t(v_2-v_1)\\t=\frac{x_0_1}{v_2-v_1}\\t=\frac{6*10^3m}{28\frac{m}{s}-20\frac{m}{s}}\\t=750s$

8 0

2 years ago

Karla Ayala pulls a sled on an icy road (dangerous!). Because of Karla's pull, the tension force is 151 N, and the rope makes a

skelet666 [1.2K]

Answer:

W = 1418.9 J = 1.418 KJ

Explanation:

In order to find the work done by the pull force applied by Karla, we need to can use the formula of work done. This formula tells us that work done on a body is the product of the distance covered by the object with the component of force applied in the direction of that displacement:

W = F.d

W = Fd Cosθ

where,

W = Work Done = ?

F = Force = 151 N

d = distance covered = 10 m

θ = Angle with horizontal = 20°

Therefore,

W = (151 N)(10 m) Cos 20°

6 0

3 years ago

In a tug of war between Mrs. Brenneman and Mr. Schroedl seems at a standstill. Then Mrs. Brenneman tugs hard giving a force of 4

sergiy2304 [10]

He feels a 10 N to the left force moves. Yes ,he moves.

4 0

3 years ago

A planet exerts a gravitational force of magnitude 4e22 N on a star. If the planet were 3 times closer to the star (that is, if

Alex_Xolod [135]

Answer:

$3.6\times10^{23} N$

Explanation:

$F=\frac{GmM}{r^2}=4\times10^{22} N$

$F'=\frac{GmM}{(r/3)^2}=9\frac{GmM}{r^2}=9\times4\times10^{22}=3.6\times10^{23} N$

7 0

2 years ago

Complete the following:

masha68 [24]

When light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.

To find the answer, we have to know about the rules followed by drawing ray-diagram.

<h3>What are the rules obeyed by light rays?</h3>

If the incident ray is parallel to the principal axis, the refracted ray will pass through the opposite side's focus.
The refracted ray becomes parallel to the major axis if the incident ray passes through the focus.
The refracted ray follows the same path if the incident light passes through the center of the curve.

Thus, we can conclude that, when light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.

Learn more about refraction by a lens here:

brainly.com/question/13095658

#SPJ1

8 0

1 year ago