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trasher [3.6K]
1 year ago
13

If you determined the electric field intensity in a field using a test charge of 1.0×10−6 C and then repeated the process with a

test charge of 2.0×10−6 C, would the forces on the charges be the same? Would you find the value for E?
Physics
1 answer:
uranmaximum [27]1 year ago
4 0

The force F exerted by an electric field on a test charge q is given by:

F=qE

From this equation we notice that if we change the test charge then the force will change as well.

Therefore, the forces on the charges will not be the same.

We can't determine the electric field without further information on the problem; we need the force on one of the test charges to determine the magnitude of the field.

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Definition of definite​
KIM [24]

Google definition: a statement of the exact meaning of a word, especially in a dictionary.

the degree of distinctness in outline of an object, image, or sound, especially of an image in a photograph or on a screen.

My definition: The meaning or deintion of any word.

5 0
4 years ago
Light from a lamp is shining on a surface. How can you increase the intensity of the light on the surface? Light from a lamp is
Kay [80]

Answer:

the correct option is C

Explanation:

The intensity of a lamp depends on the power of the lamp that is provided by the current flowing over it, therefore the intensity would increase if we raise the current.

Another way to increase the intensity is to decrease the area with a focusing lens, as the intensity is power over area, decreasing the area increases the power.

When we see the possibilities we see that the correct option is C

3 0
3 years ago
A hiker walks 11 km due north from camp and then turns and walks 11 km due east. What is the total distance walked by the hiker?
AnnyKZ [126]

Answer:

22 km

Explanation:

11km + 11km= 22km

3 0
3 years ago
Read 2 more answers
PLZZZ HELPP ASAP!!!!
Tamiku [17]

Answer:

1.68N

Explanation:

W=F*d

KE=1/2*M*V^2

M=32kg

V= 5.1m/s

D=50m

0.5(32)(5.1^2)= 416 J

Initial energy=500 J

final energy= 416 J

lost energy= 500-416

heat lost to friction=84 J

84=F*50

84/50=F*50/50

F=1.68N

3 0
3 years ago
An athlete does one push-up. In the process, she moves half of her body weight, 250 newtons, a distance of 20 centimeters. This
s2008m [1.1K]
We know that an athlete moves half of her body weight 250 N a distance of 20 cm.
So:  F = 250 N,  d = 20 cm = 0.2 m
The work formula:
W = F * d
W = 250 N * 0.2 m
W = 50 J
Answer: Her work after one push-up is 50 J.
6 0
3 years ago
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