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3 years ago

Name one contact force and one force that acts through a force field

1 answer:

5 0

A contact force is a type of force which act on an object by coming in contact with the object. Examples of contact force that acts through a force field are: applied force, frictional force, air resistance force, tension, spring force, etc.
Examples of forces that act through a force field are gravitational force, electromagnetic force, the weak interaction and the strong interaction.

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Two identical conducting spheres, A and B, carry equal charge. They are separated by a distance much larger than their diameters

Vesnalui [34]

Answer:

C. $\frac{3F}{8}$

Explanation:

Let initial charges on both spheres be, $q$

$F=\frac{Kq^2}{d^2} \ \ \ \ \ \ \ \ \ \ \_i$

When the sphere C is touched by A, the final charges on both will be, $\frac{q}{2}$

#Now, when C is touched by B, the final charges on both of them will be:

$q_c=q_d=\frac{q/2+q}{2}\\\\=\frac{3q}{4}\\$

Now the force between A and B is calculated as:

$F\prime=\frac{k\times\frac{q}{2}\times \frac{3q}{4}}{d^2}\\F\prime=\frac{3F}{8}$

Hence the electrostatic force becomes 3F/8

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2 years ago

A car with a mass of 833 kg rounds an unbanked curve in the road at a speed of 28.0 m/s. If the

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2 years ago

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D, Mercury as a weaker gravitational pull! Due to mercury being farther from the sun and it being a smaller planet it has a weaker pull

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2 years ago

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Suppose a gliding 2-kg cart bumps into, and sticks to, a stationary 5-kg cart. If the speed of the gliding cart before the colli

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Answer:

Therefore,

Final velocity of the coupled carts after the collision is

$v_{f}=4\ m/s$

Explanation:

Given:

Mass of Glidding Cart = m₁ = 2 kg

Mass of Stationary Cart = m₂ = 5 kg

Initial velocity of Glidding Cart = u₁ = 14 m/s

Initial velocity of Stationary Cart = u₂ = 0 m/s

To Find:

Final velocity of the coupled carts after the collision = $v_{f}=?$

Solution:

Law of Conservation of Momentum:

For a collision occurring between two objects in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

It is denoted by "p" and given by

Momentum = p = mass × velocity

Hence by law of Conservation of Momentum we hame

Momentum before collision = Momentum after collision

Here after collision both are stuck together so both will have same final velocity,

$m_{1}\times u_{1}+m_{2}\times u_{2}=(m_{1}+m_{2})\times v_{f}$

Substituting the values we get

$2\times 14 + 5\times 0 =(2+5)\times v_{f}$

$v_{f}=\dfrac{28}{7}=4\ m/s$

Therefore,

Final velocity of the coupled carts after the collision is

$v_{f}=4\ m/s$

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3 years ago

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3 years ago

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