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1 year ago

At a distance r from a charge e on a particle of mass m the electric field value is

1 answer:

7 0

At a distance r from a charge e on a particle of mass m the electric field value is 8.9876 × 10⁹ N·m²/C². Divide the magnitude of the charge by the square of the distance of the charge from the point. Multiply the value from step 1 with Coulomb's constant.

<h3>what is magnitude ?</h3>

Magnitude can be defined as the maximum extent of size and the direction of an object.

It is used as a common factor in vector and scalar quantities, as we know scalar quantities are those quantities that have magnitude only and vector quantities are those quantities have both magnitude and direction.

There are different ways where magnitude is used Magnitude of earthquake, charge on an electron, force, displacement, Magnitude of gravitational force

For more details regarding magnitude, visit

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If the force exerted by a man is halved and the distance is multiplied by four, by which factor does the work change ?

jeyben [28]

increase factor by 8 thousands

6 0

3 years ago

Which of the following statements are true? Check all that apply.

lidiya [134]

We have 6 levels and we will verify which of them is true.

A ) Electric charge is conserved

According to the conservation of charge, an electric charge can neither be created nor be destroyed.

Therefore this statement is TRUE.

B ) A positive charge and negative charge attract each other

According to the properties of charges, the opposite charges attract each other and similar charges repel each other

Therefore this statement is TRUE

C) Two positive charges attract each other.

Using the previous law, we can observe that this statement is false, because equal charges repel each other.

Therefore this statement is FALSE.

D ) Two negative charges repel each other.

Using the same principle as the two previous statements, two equal charges repel each other.

Therefore this statement is TRUE

E) Electric charge is quantized.

According to the quantization of charge, charges can take values only in integral multiples of a fundamental unit of charge.

Therefore this statement is TRUE.

F) A neutral object contains no charge.

A neutral object contains equal amount of negative and positive charge. Thus, in totality a neutral object contains no charge.

Therefore this statement is TRUE.

8 0

3 years ago

On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s

dsp73

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

= 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2

= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J

4 0

3 years ago

When you view the pendulum’s swing, it shows that at the very top of the swing KE = 0. What does that tell you about the pendulu

nadezda [96]

it tells you that it has a lot of force

5 0

3 years ago

Read 2 more answers

An apple weighs 1.00 N. When you hang it from the end of a long spring of force constant 1.50 N/m and negligible mass, it bounce

Ierofanga [76]

Answer:

2.67 m

Explanation:

k = Spring constant = 1.5 N/m

g = Acceleration due to gravity = 9.81 m/s²

l = Unstretched length

Frequency of SHM motion is given by

$f_s=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

Frequency of pendulum is given by

$f_p=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}$

Given in the question

$f_p=\dfrac{1}{2}f_s$

$\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\\Rightarrow \sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\sqrt{\dfrac{k}{m}}\\\Rightarrow \dfrac{g}{l}=\dfrac{1}{4}\dfrac{k}{m}\\\Rightarrow l=\dfrac{4gm}{k}\\\Rightarrow l=\dfrac{4\times 9.81\times \dfrac{1}{9.81}}{1.5}\\\Rightarrow l=2.67\ m$

The unstretched length of the spring is 2.67 m

6 0

3 years ago