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2 years ago

It velocity of light scalar or vector equality

Physics

1 answer:

suter [353]2 years ago

4 0

Answer:

velocity is a vector quantity

Explanation:

velocity is a vector quantity because it has a mass and a direction

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A bead slides without friction around a loopthe-loop. The bead is released from a height 21.9 m from the bottom of the loop-the-

wariber [46]

Answer:

Part a)

$v = 12.45 m/s$

Part B)

$F_n = 0.05 N$

Explanation:

Part A)

As we know that the point A lies on the top of the loop

so we will have by energy conservation

$mgH = \frac{1}{2}mv^2 + mg(2R)$

so the speed at point A is given as

$mg(H - 2R) = \frac{1}{2}mv^2$

$v = \sqrt{2g(H - 2R)}$

$v = \sqrt{2(9.81)(21.9 - 2\times 7)}$

$v = 12.45 m/s$

Part B)

Now the force equation at point A is given as

$F_n + mg = \frac{mv^2}{R}$

$F_n = \frac{mv^2}{R} - mg$ [/tex]

$F_n = 0.004(\frac{12.45^2}{7} - 9.81)$

$F_n = 0.05 N$

6 0

3 years ago

Which of the following is not a type of system?

swat32

Social Classical is not a type of system.

5 0

4 years ago

A ball of mass 0.160 kg is dropped from a height of 2.25 m. When it hits the ground it compresses 0.087 m.

Studentka2010 [4]

A) 6.64 m/s downward

B) 0.026 s

C) -40.9 N

Explanation:

We can solve this problem by using the law of conservation of energy.

In fact, since the total mechanical energy of the ball must be conserved, this means that the initial gravitational potential energy of the ball before the fall is entirely converted into kinetic energy just before it reaches the floor.

So we can write:

$PE=KE\\mgh = \frac{1}{2}mv^2$

where

m = 0.160 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 2.25 m is the initial height of the ball

v is the final velocity of the ball before hitting the ground

Solving for v, we find:

$v=\sqrt{2gh}=\sqrt{2(9.8)(2.25)}=6.64 m/s$

And the direction of the velocity is downward.

The motion of the ballduring the collision is a uniformly accelerated motion (= with constant acceleration), so the time of impact can be found by using a suvat equation:

$s=(\frac{u+v}{2})t$

where:

v is the final velocity

u is the initial velocity

s is the displacement of the ball during the impact

t is the time

Here we have:

u = 6.64 m/s is the velocity of the ball before the impact

v = 0 m/s is the final velocity after the impact (assuming it comes to a stop)

s = 0.087 m is the displacement, as the ball compresses by 0.087 m

Therefore, the time of the impact is:

$t=\frac{2s}{u+v}=\frac{2(0.087)}{0+6.64}=0.026 s$

The force exerted by the floor on the ball can be found using the equation:

$F=\frac{\Delta p}{t}$

where

$\Delta p$ is the change in momentum of the ball

t is the time of the impact

The change in momentum can be written as

$\Delta p = m(v-u)$

So the equation can be rewritten as

$F=\frac{m(v-u)}{t}$

Here we have:

m = 0.160 kg is the mass of the ball

v = 0 is the final velocity

u = 6.64 m/s is the initial velocity

t = 0.026 s is the time of impact

Substituting, we find the force:

$F=\frac{(0.160)(0-6.64)}{0.026}=-40.9 N$

And the sign indicates that the direction of the force is opposite to the direction of motion of the ball.

4 0

3 years ago

The Coulomb force between two charges q1 and q2 at separation r in the air is 10N. If half of the separation is filled with medi

adoni [48]

Answer:

The value of new coulomb force is 1.43 N.

Explanation:

Given;

Coulomb's force in vacuum (air), $F_v$ = 10 N

dielectric constant, K = 7

The Coulomb's force between two charges separated by a distance r in a vacuum is given as;

$F_v = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2}$

The Coulomb's force between two charges separated by a distance r in a medium with dielectric constant is given as

$F_m = \frac{1}{4\pi K\epsilon_0} \frac{q_1q_2}{r^2}$

Take the ratio of the two forces;

$\frac{F_v}{F_m} = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2} \ \times \ \frac{4\pi K\epsilon_0 r^2}{q_1q_2} = K\\\\\frac{F_v}{F_m} = K\\\\\frac{10}{F_m} = 7\\\\F_m = \frac{10}{7} \\\\F_m = 1.43 \ N$

Therefore, the value of new coulomb force is 1.43 N.

5 0

3 years ago

A brick falls to the ground. if the time for the collision of the brick and the ground is increased by a factor of 4, the force

melomori [17]

Answer:

By a factor of 1/4.

Explanation:

The impulse force that applies to an object undergoing rapid deceleration just before coming to a stop on the ground is given by the following formula,

$\\\begin{aligned}\\\small F &=\small \frac{\Delta (mV)}{\Delta T}\end{aligned}$

in which $\small \Delta (mV)$ , $\small \Delta t$ represent the change in momentum and the time taken for that change.

If one increases the time that is taken for the momentum change (which remains constant for this situation) by a factor 4 and if that new force is represented by $\small F_1$ , the following manipulation confirms the answer to this question.

$\begin{aligned}\\\small F_1 &=\small \frac{\Delta (mV)}{4\Delta t}\\\\&=\small \frac{1}{4}\times\bigg[\frac{\Delta (mV)}{\Delta t}\bigg]\\\\&=\small \frac{1}{4}F\end{aligned}$

Here $\small F$ is the force that was applied to the object previously.

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4 0

2 years ago

It velocity of light scalar or vector equality ​

It velocity of light scalar or vector equality