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statuscvo [17]
3 years ago
13

A test charge is introduced into the electric field of a charge. It feels a force of 2 F where the electric feild is 4 E. What w

ould the force on the test charge be where the electric feild is 1 E
Physics
1 answer:
Elza [17]3 years ago
3 0

Answer:

0.5 F

Explanation:

since electric field is defined as force per charge!

so, force on test charge is directly proportional to electric field

here new electric field becomes 1/4 of old one

i.e, 4E-->1E

so force is also will become 1/4 of old one

i.e, 2F--> 1/4 × 2F =0.5 F

✌️:)

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Three light bulbs are connected to a battery in a series circuit.How will the bulbs behave if the circuit is closed
Drupady [299]

The answer is

<em>All the three light bulbs will glow</em>

hope this helps

6 0
3 years ago
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Heart cells must contract simultaneously to move blood.
krek1111 [17]

Answer:

<u>A</u>

Explanation:

The heart cells must contract simultaneously to move blood.

This means that it needs to act fast and efficiently.

Therefore, the connections among heart cells are characterized by :

  • having many branches
  • having many communicating junctions

The correct option should be <u>A</u>

4 0
2 years ago
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What is the frequency of a wave that has a wavelength of 20 cm and a speed of 10 m/s
jarptica [38.1K]
You would probably have a low frequency due to how much the wavelength is spread out.
8 0
3 years ago
how fast will and in what direction will a 20kg object accelerate if one force pushes at a 30 degree angle and another pushes at
DiKsa [7]

Answer:

|a|=2.83\ m/s^2

\theta=75^o

Explanation:

<u>Net Force And Acceleration </u>

The Newton's second law relates the net force applied on an object of mass m and the acceleration it aquires by

\vec F_n=m\vec a

The net force is the vector sum of all forces. In this problem, we are not given the magnitude of each force, only their angles. For the sake of solving the problem and giving a good guide on how to proceed with similar problems, we'll assume both forces have equal magnitudes of F=40 N

The components of the first force are

\vec F_1=

\vec F_1=\ N

The components of the second force are

\vec F_2=

\vec F_2=\ N

The net force is

\vec F_n=

\vec F_n=\ N

The magnitude of the net force is

|F_n|=\sqrt{14.64^2+54.64^2}

|F_n|=\sqrt{3200}=56.57\ N

The acceleration has a magnitude of

\displaystyle |a|=\frac{|F_n|}{m}

\displaystyle |a|=\frac{56.57}{20}

|a|=2.83\ m/s^2

The direction of the acceleration is the same as the net force:

\displaystyle tan\theta=\frac{54.64}{14.64}

\theta=75^o

5 0
3 years ago
A 55 newton force applied on an object moves the object 10 meters in the same direction as the force. What is the value of work
kifflom [539]

Answer: Option D: 5.5×10²Joules

Explanation:

Work done is the product of applied force and displacement of the object in the direction of force.

W = F.s = F s cosθ

It is given that the force applied is, F = 55 N

The displacement in the direction of force, s = 10 m

The angle between force and displacement, θ = 0°

Thus, work done on the object:

W = 55 N × 10 m × cos 0° = 550 J = 5.5 × 10² J

Hence, the correct option is D.

3 0
3 years ago
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