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3 years ago

What is the electric potential difference of 5cm from a point charge of -2.0µc

Physics

1 answer:

vodka [1.7K]3 years ago

7 0

Answer:

V = 360 kV

Explanation:

Given that,

Charge, q = -2µC

We need to find the electric potential difference of 5cm from a point charge.

We know that the formula for the electric potential is given by :

$V=\dfrac{kq}{r}\\\\V=\dfrac{9\times 10^9\times 2\times 10^{-6}}{0.05}\\\\V=360000\ V$

$V=360\ kV$

So, the electric potential difference is 360 kV.

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A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

What is the streak of a mineral used for

ad-work [718]

The streak of a mineral is used for identifying what the mineral may be.

Hope this helps :)

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3 years ago

Read 2 more answers

The value of the normal force exerted on a 2,000 gram block on a table

kobusy [5.1K]

Answer:

N = 19.6 N

Explanation:

Given that,

Mass of a block, m = 2000 g

1 kg = 1000 g

It means, 2000 g = 2kg

We need to find the value of normal force on the block on a table. Normal force is balanced by the weight of the block as follows :

N = mg, g is acceleration due to gravity

N = 2 kg × 9.8 m/s²

N = 19.6 N

So, the normal force acting on the block is 19.6 N.

4 0

3 years ago

A small ferryboat is 3.90 m wide and 6.30 m long. when a loaded truck pulls onto it, the boat sinks an additional 5.00 cm into t

Leya [2.2K]

When the truck's weight is added to the boat, the boat sinks 5 cm deeper,
and displaces additional water whose weight is equal to the weight of the
truck.

The volume of the additional displaced water is

(3.9 m) x (6.3 m) x (5.0 cm)

= (3.9 m) x (6.3 m) x (0.05 m) = 1.2285 m³ .

The weight of that much water is the weight of the truck.

Mass of 1 liter of water = 1 kilogram

1.2285 m³ = 1,228.5 liters = 1,228.5 kg of water.

Weight = (mass) x (gravity)

= (1,228.5 kg) x (9.8 m/s²) = 12,039 Newtons.

(about 2,708 pounds)

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3 years ago

What is the pressure at the bottom of a tank 1-m deep? Take the density of water to be 1000-kg/m^3, and gravitational accelerati

lara31 [8.8K]

To solve this problem we will use the concepts related to hydrostatic pressure. Which determines the pressure of a body at a given depth of a liquid.

Mathematically this can be described as

$P= \rho gh$