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3 years ago

Calculate the electric current that flows through a electric heater of power 1000watt when collected with 220 volt ac supply

Physics

1 answer:

Anit [1.1K]3 years ago

5 0

Answer:

4.545 A

Explanation:

From the question,

Using,

P = VI....................... Equation 1

Where P = power of the heater, V = Voltage supply, I = Electric current that flows through the heater

make I the subject of the equation above

I = P/V.................. Equation 2

Given: P = 1000 watt, V = 220 volt.

Substitute these values into equation 2

I = 1000/220

I = 4.545 A

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At the equator, the radius of the Earth is approximately 6370 km. A plane flies at a very low altitude at a constant speed of v

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To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From there we will define the distance as the circumference of the earth (approximate as a sphere). With the speed given in the statement we will simply clear the equations below and find the time.

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$v = 239m/s$

$a = 16.5m/s^2$

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$\phi = 2\pi R$

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Here $x = \phi$ , then

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3 years ago

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

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Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

<u>Conservation of kinetic energy:</u>

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

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Cody ...

Everything on this page is solved with the SAME formula !

Distance = (speed) x (time) .

Before I get into how to solve each problem, we need to notice that
this whole sheet deals with speed, NOT velocity.

'Velocity' is speed AND THE DIRECTION OF THE MOTION.
Nothing on this page ever mentions direction, so there's no velocity
anywhere on the page.

Your teacher may not be happy if you talk about this on your homework,
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Say "speed", and if the teacher complains about that, then it's time to
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1). Speed = (distance covered) / (time to cover the distance)

2). Speed = (distance covered) / (time to cover the distance)

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4). Time to cover the distance = (distance) / (speed)

5). Car's speed = (distance the car covered) / (time the car took)
Sprinter speed = (distance the sprinter covered) / (time the sprinter took)

Calculate the car's speed.
Calculate the sprinter's speed.

... Look at the two speeds.
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... Subtract the slower one from the faster one.
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divided by
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3 years ago