Select all the correct locations on the image.

a particle with charge Q is on the y axis a distance a from the origin and a particle with charge qi is on the x axis at a dista

vova2212 [387]

Answer:

Explanation:

Force on Q due to q₁

= k Q.q₁ / r² , r is distance between two.

X component

= k Q.q₁ / r² x cos θ , θ is angle r makes with x - axis.

= k Q.q₁ / r² x cos θ

r = a / sinθ

X component

= k Q.q₁ sin²θ . cosθ/ a²

= k Q.q₁ / 2 a² x sin2θ . sinθ

for maximum value of it ,

sin2θ = 1 , θ = 45°

a / d = tan45

a = d .

3 0

3 years ago

A ball is dropped from a building of height h. Derive an equation for its velocity just before it hits the ground. Assume no air

swat32

Answer:

The final velocity of the ball just hits the ground is $\sqrt{2gh}$ .

Explanation:

A ball is dropped from the height h. As the ball is dropped means there is no initial velocity given to the ball.

GIVEN:

The distance traveled by the ball is equal to h.

Initial velocity of the ball is zero.

Concept:

Two type of force comes into play in dropping of ball.

1. Gravitational force

Due to gravitational force ball forced to fall downward towards the surface.

2. Air resistance

Air develops drag force on the ball in opposite direction of the gravitational force.

Assumption:

There is no air resistance in falling of ball. So, there will be no drag force on the ball.

Calculation:

It is given that a ball is dropped from a building of height h.

The Newton’s equations of motion are as follows:

$V^{2}-u^{2}=2as$

$s=ut+\frac{1}{2}at^{2}$

$v=u+at$

STEP 1

Newton’s equation of motion is expressed as follows:

$V^{2}-u^{2}=2as$

Here, V is final velocity of the ball just before the ball hits the ground, a is acceleration due to gravity, u is the initial velocity of the ball and s is the distance traveled by the ball.

Here, initial velocity is zero. So, u = 0 m/s.

Vertical distance traveled by the ball is h. So, s= h.

Only gravitational force is responsible for motion as well as for acceleration. Thus, the acceleration due to gravity acts on the ball. Acceleration due to gravity is denoted by small g.

Acceleration due to gravity is constant for different planate. The acceleration due to gravity for the earth is 9.8 m/s². Here, no planet is defined. So the acceleration due to gravity is a = g.

Substitute 0 for m/s, g for a and h for s in above equation as follows:

$V^{2}-u^{2}=2as$

$V^{2}-(0)^{2}=2gh$

$V^{2}=2gh$

$v=\sqrt{2gh}$

Thus, the final velocity of the ball just hits the ground is $\sqrt{2gh}$ .

5 0

3 years ago

Can anyone answer this question its a science question NO LINKS !!!!

liubo4ka [24]

Answer: Change

Explanation:

5 0

3 years ago

What help in reversing direction of current of current

Novay_Z [31]

Answer:

To reverse the direction of an electric current, we simply reverse the voltage either automatically with the help of some switching circuitry or manually by changing the voltage source terminals connection.

Explanation:

For electric current to flow, there must be a potential difference, usually referred to as the voltage. The electric current flow is analogous to the flow of water under the action of a pump, through a series of pipe connections. The voltage is similar to the driving action of the pump, and current flows the same way water flows. The resistance due to drag on the pipe wall is equivalent to electric resistance. For current to flow in the reverse direction, the voltage or rather, the potential difference is changed, causing the current to flow in the opposite direction. This can be done by switching the terminals of the voltage source, or by automatic means. The automatic switching can be done with a transistor based circuitry.

3 0

4 years ago

Calculate the amount of heat needed to raise the temperature of 200g of ice from-30°C 50C water. (C = .5 for ice, C 1 for water,

Nitella [24]

Answer: The amount of heat needed is 29000 Cal.

Explanation:

The process involved in this problem are:

$(1):H_2O(s)(-30^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(50^oC)$