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3 years ago

Electric field strength of a point charge is E. What is the electric potential at a point where electric field will be E/4?

Physics

1 answer:

AlladinOne [14]3 years ago

3 0

Answer:

Explanation:

We shall find first the distance where electric field is E/4 .

Let the charge be Q and distance be d where electric field is E . From the coulomb's Law

E = k Q / d²

Let distance be d₁ where field is E/4

E/4 = kQ / d₁²

Dividing the two equation

4 = d₁² / d²

d₁ = 2d

We shall have to find Potential at d₁ which is equal to 2 d .

Potential at d₁

V = k Q / 2d

= kQ d / 2d²

= E d / 2 . where d is distance of the point where field is E .

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The magnitude of the momentum of an object is the product of its mass m and speed v. If m = 3 kg and v = 1.5 m/s, what is the ma

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Answer:

Momentum, p = 5 kg-m/s

Explanation:

The magnitude of the momentum of an object is the product of its mass m and speed v i.e.

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Mass, m = 3 kg

Velocity, v = 1.5 m/s

So, momentum of this object is given by :

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p = 4.5 kg-m/s

p = 5 kg-m/s

So, the magnitude of momentum is 5 kg-m/s. Hence, this is the required solution.

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A 2kg mass is initially at rest on a frictionless surface. A 45N force acts at an angle of 300

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Answer:

a) The work done by the force is 136.400 joules.

b) The speed of the mass at the end of the 3.5 meters is approximately 11.679 meters per second.

Explanation:

The correct statement is shown below:

A 2-kg mass is initially at reat on a frictionless surface. A 45 N force acts at an angle of 30º to the horizontal for a distance of 3.5 meters:

a) Find the work done by the force.

b) Find the speed of the mass at the end of the 3.5 meters.

a) Given that a external constant force is acting on the mass on a frictionless surface, the work done by the force ( $W_{F}$ ), measured in joules, is:

$W_{F} = F\cdot \Delta s \cdot \cos \theta$ (1)

Where:

$F$ - External constant force exerted on the mass, measured in newtons.

$\Delta s$ - Horizontal travelled distance, measured in meters.

$\theta$ - Direction of the external force regarding the horizontal, measured in sexagesimal degrees.

If we know that $F = 45\,N$ , $\Delta s = 3.5\,m$ and $\theta = 30^{\circ}$ , then the work done by the force is:

$W_{F} = (45\,N)\cdot (3.5\,m)\cdot \cos 30^{\circ}$

$W_{F} = 136.400\,J$

The work done by the force is 136.400 joules.

b) The final speed of the mass at the end ot the 3.5 meter is calculated by means of the Work-Energy Theorem, which means that the work done on the mass is transformed into a translational kinetic energy. That is:

$W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})$ (2)

Where:

$m$ - Mass, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speeds of the mass, measured in meters per second.

If we know that $W_{F} = 136.400\,J$ , $m = 2\,kg$ and $v_{1} = 0\,\frac{m}{s}$ , then the final speed of the mass is:

$\frac{2\cdot W_{F}}{m} = v_{2}^{2}-v_{1}^{2}$

$v_{2}^{2} =v_{1}^{2}+\frac{2\cdot W_{F}}{m}$

$v_{2} =\sqrt{v_{1}^{2}+\frac{2\cdot W_{F}}{m} }$

$v_{2} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+\frac{2\cdot (136.400\,J)}{2\,kg} }$

$v_{2} \approx 11.679\,\frac{m}{s}$

The speed of the mass at the end of the 3.5 meters is approximately 11.679 meters per second.

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3 years ago

"A parcel moving in a horizontal direction with speed v0 = 13 m/s breaks into two fragments of weights 1.4 N and 1.9 N, respecti

Lady bird [3.3K]

Answer:

the smaller particle moves with speed of 8.706 m/s in the opposite direction to the bigger particle.

Explanation:

Speed of the original particle = 13 m/s

We designate particles as A and B

The final weights of the component particles are

Particle A = 1.4 N

particle B = 1.9 N

The speed of the larger piece (particle B) = 29 m/s

We know that weight is the product of a body's mass and acceleration due to gravity g which is equal to 9.81 m/s^2, therefore, masses of the particles are

particle A = 1.4/9.81 = 0.143 kg

Particle B = 1.9/9.81 = 0.194 kg

The momentum of a body is the product of its mass and its velocity i.e

P = mv

This means that the mass of the particle before splitting is

0.143 kg + 0.194 kg = 0.337 kg

Momentum of the initial whole particle = mv

==> 0.337 x 13 = 4.381 kg-m/s

The bigger particle B remains horizontal, and has a momentum of

mv = 0.194 x 29 = 5.626 kg-m/s

According to the conservation of momentum, the total initial momentum of a system must be equal tot the total final momentum of the system.

Initial total momentum of the system = 4.381 kg-m/s (momentum of original particle before splitting)

Final total momentum of the system = Total momentum of the particles after splitting = 5.626 kg-m/s + ( 0.143 kg x $V_{B}$ )

where $V_{B}$ is the velocity of smaller particle A

final total momentum of the system = 5.626 + 0.143 $V_{B}$

Equating the two momenta of the system, we'll have

4.381 = 5.626 + 0.143 $V_{B}$

4.381 - 5.626 = 0.143 $V_{B}$

-1.245 = 0.143 $V_{B}$

$V_{B}$ = -1.245/0.143 = -8.706 m/s

The negative sign indicates that the smaller particle moves in the opposite direction to the bigger particle

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3 years ago