All categories

4 years ago

Give an explaination to the following formula p=I^2R

Physics

1 answer:

Art [367]4 years ago

4 0

Answer:

Explanation:

This is the equation that defines the power consumed by an electrical circuit. It is normally defined as the product of the voltage applied (V) times the current (I) that runs through it. And in the case of a resistive circuit (resistors in the circuit totaling a certain resistance R), one can use Ohm's law to replace the voltage by the product of the current (I) times the resistance (R), obtaining then the square of the current:

$P=V*I=(I*R)*I=I^2 * R$

You might be interested in

Why are there two fluoride ions in magnesium fluoride but only one fluoride ion in lithium fluoride

postnew [5]

This would be more of a chemistry question. Remember magnesium has a charge of 2+, and would need to hand off its two extra electrons. Fluorine can only take one electron at a time, so there needs to be two fluorines to take one magnesium's 2 electrons.

With lithium, it has a +1 charge, so it has one extra electron, which it can hand off to just 1 fluorine atom.

Another way of looking at this is: $Mg^{2+}$ + 2 $F^{-}$ = MgF2 (the charges must balance out to zero)

$Li^{+}$ + $F^{-}$ = LiF (the charges balance out to zero)

5 0

3 years ago

Read 2 more answers

The coefficient of linear expansion of copper is 17 x 10^-6 K-1. A block of copper 30 cm wide, 45 cm long, and 10 cm thick is he

schepotkina [342]

Answer:

The change in volume is $6.885\times 10^{- 5}\$

Solution:

As per the question:

Coefficient of linear expansion of Copper, $\alpha = 17\times 10^{- 6}\ K^{- 1}$

Initial Temperature, T = $0^{\circ}$ = 273 K

Final Temperature, T' = $100^{\circ}$ = 273 + 100 = 373 K

Now,

Initial Volume of the block, V = $30\times 45\times 10\times 10^{- 6}\ m^{3} = 0.0135\ m^{3}$

$V' = V(1 + \gamma \Delta T)$

$\gamma = 3\alpha$

$V' = V(1 + 3\alpha \Delta T)$

where

V' = Final volume

$V' - V= 0.0135\times 17\times 10^{- 6} \times (T' - T))$

$\Delta V= 0.0135\times 3\times 17\times 10^{- 6} \times (373 - 273)) = 6.885\times 10^{- 5}\$

7 0

3 years ago

Bill throws a tennis ball to his dog. He throws the ball at a speed of 15 m/s at an angle of 30° to the horizontal. Assume he th

Sidana [21]

1a) Bill and the dog must have a speed of 13.0 m/s

1b) The speed of the dog must be 22.5 m/s

2a) The ball passes over the outfielder's head at 3.33 s

2b) The ball passes 1.2 m above the glove

2c) The player can jump after 2.10 s or 3.13 s after the ball has been hit

2d) One solution is when the player is jumping up, the other solution is when the player is falling down

Explanation:

1a)

The motion of the ball in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

In part a), we want to know at what speed Bill and the dog have to run in order to intercept the ball as it lands on the ground: this means that Bill and the dog must have the same velocity as the horizontal velocity of the ball.

The ball's initial speed is

u = 15 m/s

And the angle of projection is

$\theta=30^{\circ}$

So, the ball's horizontal velocity is

$v_x = u cos \theta = (15)(cos 30)=13.0 m/s$

And therefore, Bill and the dog must have this speed.

1b)

For this part, we have to consider the vertical motion of the ball first.

The vertical position of the ball at time t is given by

$y=u_yt+\frac{1}{2}at^2$

where

$u_y = u sin \theta = (15)(sin 30) = 7.5 m/s$ is the initial vertical velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity

The ball is at a position of y = 2 m above the ground when:

$2=7.5t + \frac{1}{2}(-9.8)t^2\\4.9t^2-7.5t+2=0$

Which has two solutions: t=0.34 s and t=1.19 s. We are told that the ball is falling to the ground, so we have to consider the second solution, t = 1.19 s.

The horizontal distance covered by the ball during this time is

$d=v_x t =(13.0)(1.19)=15.5 m$

The dog must be there 0.5 s before, so at a time

t' = t - 0.5 = 0.69 s

So, the speed of the dog must be

$v_x' = \frac{d}{t'}=\frac{15.5}{0.69}=22.5 m/s$

2a)

Here we just need to consider the horizontal motion of the ball.

The horizontal distance covered is

$d=98 m$

while the horizontal velocity of the ball is

$v_x = u cos \theta = (34)(cos 30)=29.4 m/s$

where u = 34 m/s is the initial speed.

So, the time taken for the ball to cover this distance is

$t=\frac{d}{v_x}=\frac{98}{29.4}=3.33 s$

2b)

Here we need to calculate the vertical position of the ball at t = 3.33 s.

The vertical position is given by

$y= h + u_y t + \frac{1}{2}at^2$

where

h = 1.2 m is the initial height

$u_y = u sin \theta = (34)(sin 30)=17.0 m/s$ is the initial vertical velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity

Substituting t = 3.33 s,

$y=1.2+(17)(3.33)+\frac{1}{2}(-9.8)(3.33)^2=3.5 m$

And sinc the glove is at a height of y' = 2.3 m, the difference in height is

y - y' = 3.5 - 2.3 = 1.2 m

2c)

In order to intercept the ball, he jumps upward at a vertical speed of

$u_y' = 7 m/s$

So its position of the glove at time t' is

$y'= h' + u_y' t' + \frac{1}{2}at'^2$

where h' = 2.3 m is the initial height of the glove, and t' is the time from the moment when he jumps. To catch the ball, the height must be

y' = y = 3.5 m (the height of the ball)

Substituting and solving for t', we find

$3.5 = 2.3 + 7t' -4.9t'^2\\4.9t'^2-7t'+12 = 0$

Which has two solutions: t' = 0.20 s, t' = 1.23 s. But this is the time t' that the player takes to reach the same height of the ball: so the corresponding time after the ball has been hit is

$t'' = t -t'$

So we have two solutions:

$t'' = 3.33 s - 0.20 s = 3.13 s\\t'' = 3.33 s - 1.23 s = 2.10 s$

So, the player can jump after 2.10 s or after 3.13 s.

2d)

The reason for the two solutions is the following: the motion of the player is a free fall motion, so initially he jump upwards, then because of gravity he is accelerated downward, and therefore eventually he reaches a maximum height and then he falls down.

Therefore, the two solutions corresponds to the two different part of the motion.

The first solution, t'' = 2.10 s, is the time at which the player catches the ball while he is in motion upward.

On the other hand, the second solution t'' = 3.13 s, is the time at which the player catches the ball while falling down.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0

3 years ago

How much force is needed to keep the 750000 Newton Space Shuttle moving at a constant speed of 28000 km/h, in a straight line?

Alika [10]

The force needed to keep the space shuttle moving at constant speed is 0.

The given parameters;

<em>weight of the space shuttle, F = 750,000 N</em>
<em>constant speed of the space shuttle, v = 28,000 km/h</em>

The mass of the space shuttle is calculated as follows;

$W = mg\\\\m = \frac{W}{g} \\\\m = \frac{750,000}{9.8} \\\\m = 76,530.61 \ kg$

The force needed to keep the space shuttle moving at constant speed is calculated as follows;

$F = ma$

$F = 76,530.61 \times a$

where;

a is the acceleration of the space shuttle

At a constant speed, acceleration is zero.

F = 76,530.61 x 0

F = 0

Thus, the force needed to keep the space shuttle moving at constant speed is 0.

Learn more here:brainly.com/question/16374764

6 0

2 years ago

An object moves from the position +16m to the position +43m in 12s. What us the total displacement

NeX [460]

First method

initial distance = 16m

final distance= 43 m

total distance covered= final -initial

=43m -16m

=27m

Second method

Si= 16m

Sf =43 m

t= 12 s

first we will find V

V = (Sf-Si)/ t

V =( 43- 16)/ 12

V = 27/12 ⇒ V= 9/4

V= distance / time

distance= V×time

distance = (9/4) ×12

distance =27

3 0

3 years ago