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3 years ago

Select all the correct answers.

Physics

2 answers:

EleoNora [17]3 years ago

8 0

Correct me if I’m wrong, but I believe it’s A and D, though I’m not fully sure.

Mila [183]3 years ago

7 0

A and d i’m sure
good luck

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A test charge of -3x10^-7 C is located 7 cm to the right of a charge of -9x10^-6 C and 20 cm to the left of a charge of +10x10^-

musickatia [10]

Answer:

5.634 N rightwards

Explanation:

qo = - 3 x 10^-7 C

q1 = - 9 x 10^-6 C

q2 = 10 x 10^-6 C

r1 = 7 cm = 0.07 m

r2 = 20 cm = 0.2 m

The force on test charge due to q1 is F1 which is acting towards right

According to the Coulomb's law

$F_{1}=\frac{Kq_{1}q_{0}}{r_{1}^{2}}$

F1 = (9 x 10^9 x 9 x 10^-6 x 3 x 10^-7) / (0.07 x 0.07)

F1 = 4.959 N rightwards

The force on test charge due to q2 is F1 which is acting towards right

According to the Coulomb's law

$F_{2}=\frac{Kq_{2}q_{0}}{r_{2}^{2}}$

F2 = (9 x 10^9 x 10 x 10^-6 x 3 x 10^-7) / (0.2 x 0.2)

F2 = 0.675 N rightwards

Net force on the test charge

F = F1 + F2 = 4.959 + 0.675 = 5.634 N rightwards

3 0

3 years ago

The force measured to move a sled was 10 N and the sled was moved 10 meters,

sergey [27]

Answer:

work done would be I don't know

3 0

3 years ago

True or false money difficult to carry around

Nina [5.8K]

Answer

yeard

Explanation:

cause it is

6 0

3 years ago

Read 2 more answers

Two horizontal curves on a bobsled run are banked at the same angle, but one has twice the radius of the other. The safe speed (

iren [92.7K]

Answer:

safe speed for the larger radius track u= √2 v

Explanation:

The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.

Also given that r_1= smaller radius

r_2= larger radius curve

r_2= 2r_1..............i

let u be the speed of larger radius curve

now, $\sum F = \frac{mv^2}{r_1} =\frac{mu^2}{r_2}$ ................ii

form i and ii we can write

$v^2= \frac{1}{2} u^2$

⇒u= √2 v

therefore, safe speed for the larger radius track u= √2 v

4 0

4 years ago

1. A 14-cm tall object is placed 26 cm from a converging lens that has a focal length of 13 cm.

AURORKA [14]

Answer:

a) Please find attached the required drawing of light passing through the lens

By the use of similar triangles;

The image distance from the lens = 26 cm

The height of the image = 14 cm

c) The image distance from the lens = 26 cm

The height of the image = 14 cm

Explanation:

Question;

a) Determine the image distance and the height of the image

b) Calculate the image position and height

The given parameters are;

The height of the object, h = 14 cm

The distance of the object from the mirror, u = 26 cm

The focal length of the mirror, f = 13 cm

The location of the object = 2 × The focal length

Therefore, given that the center of curvature ≈ 2 × The focal length, we have;

The location of the object ≈ The center of curvature of the lens

The diagram of the object, lens and image created with MS Visio is attached

From the diagram, it can be observed, using similar triangles, that the image distance from the lens = The object distance from the lens = 26 m

The height of the image = The height of the object - 14 cm

b) The lens equation is used for finding the image distance from the lens as follows;

$\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$

Where;

v = The image distance from the lens

We get;

$v = \dfrac{u \times f}{u - f}$

Therefore;

$v = \dfrac{26 \times 13}{26 - 13} = \dfrac{26 \times 13}{13} = 26$

The distance of the image from the lens, v = 26 cm

The magnification, M =v/u

∴ M = 26/26 = 1, therefore, the object and the image are the same size

Therefore;

The height of the image = The height of the object = 14 cm.

5 0

3 years ago