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Eva8 [605]
3 years ago
7

A "570-W" electric heater is designed to operate from 120-V lines. What current does it draw? If the line voltage drops to 110 V

, what power does the heater take? (Assume that the resistance is constant. Actually, it will change because of the change in temperature.)
Physics
2 answers:
Andreas93 [3]3 years ago
5 0
<h2>Answer:</h2>

4.75A, 479W

<h2>Explanation:</h2>

The power P, taken by an appliance is related to the current I, drawn by the appliance and the voltage V, at which it is operating as follows;

P = I x V          -------------------(i)

<em>From the question, the appliance is the electric heater and;</em>

P = 570W

V = 120V

<em>Substitute these values into equation (i) as follows;</em>

570 = I x 120

i = \frac{570}{120}

I = 4.75

Therefore, the current that it draws is 4.75A

(b) First, lets find the resistance, R, in the heater as follows;

P = \frac{V^{2} }{R}   -----------------------(ii)

Where;

P = Power

V = Voltage

<em>Substitute V=120V and P=570W into equation (ii) as follows;</em>

570 = \frac{120^{2} }{R}

<em>Cross multiply and solve for R;</em>

570R = 120²

570R = 14400

R = \frac{14400}{570}

R =  25.26Ω

The resistance of the heater is 25.26Ω

Now, if the voltage drops to 110V and resistance is assumed to be constant, then using equation (ii);

P = \frac{V^{2} }{R}  

...we can calculate the power the heater takes by substituting V=110V and R = 25.26Ω as follows;

P = \frac{110^{2} }{25.26}

P = 479

Therefore, the power the heater takes is 479W

Viefleur [7K]3 years ago
4 0

Answer:

Watts=Volt*Amps

So A=570/120=4.75amps

If voltage drops to 110V We get A=570/110=5.(18...)amps

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worty [1.4K]

If the ball is dropped with no initial velocity, then its velocity <em>v</em> at time <em>t</em> before it hits the ground is

<em>v</em> = -<em>g t</em>

where <em>g</em> = 9.80 m/s² is the magnitude of acceleration due to gravity.

Its height <em>y</em> is

<em>y</em> = 40 m - 1/2 <em>g</em> <em>t</em>²

The ball is dropped from a 40 m height, so that it takes

0 = 40 m - 1/2 <em>g</em> <em>t</em>²

==>  <em>t</em> = √(80/<em>g</em>) s ≈ 2.86 s

for it to reach the ground, after which time it attains a velocity of

<em>v</em> = -<em>g</em> (√(80/<em>g</em>) s)

==>  <em>v</em> = -√(80<em>g</em>) m/s ≈ -28.0 m/s

During the next bounce, the ball's speed is halved, so its height is given by

<em>y</em> = (14 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

Solve <em>y</em> = 0 for <em>t</em> to see how long it's airborne during this bounce:

0 = (14 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

0 = <em>t</em> (14 m/s - 1/2 <em>g</em> <em>t</em>)

==>  <em>t</em> = 28/<em>g</em> s ≈ 2.86 s

So the ball completes 2 bounces within approximately 5.72 s, which means that after 5 s the ball has a height of

<em>y</em> = (14 m/s) (5 s - 2.86 s) - 1/2 <em>g</em> (5 s - 2.86 s)²

==>  (i) <em>y</em> ≈ 7.5 m

(ii) The ball will technically keep bouncing forever, since the speed of the ball is only getting halved each time it bounces. But <em>y</em> will converge to 0 as <em>t</em> gets arbitrarily larger. We can't realistically answer this question without being given some threshold for deciding when the ball is perfectly still.

During the first bounce, the ball starts with velocity 14 m/s, so the second bounce begins with 7 m/s, and the third with 3.5 m/s. The ball's height during this bounce is

<em>y</em> = (3.5 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

Solve <em>y</em> = 0 for <em>t</em> :

0 = (3.5 m/s) <em>t</em> - 1/2 <em>g t</em>²

0 = <em>t</em> (3.5 m/s - 1/2 <em>g</em> <em>t</em>)

==>  (iii) <em>t</em> = 7/<em>g</em> m/s ≈ 0.714 s

As we showed earlier, the ball is in the air for 2.86 s before hitting the ground for the first time, then in the air for another 2.86 s (total 5.72 s) before bouncing a second time. At the point, the ball starts with an initial velocity of 7 m/s, so its velocity at time <em>t</em> after 5.72 s (but before reaching the ground again) would be

<em>v</em> = 7 m/s - <em>g t</em>

At 6 s, the ball has velocity

(iv) <em>v</em> = 7 m/s - <em>g</em> (6 s - 5.72 s) ≈ 4.26 m/s

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Eduardwww [97]

Answer:

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Explanation:

The statement being made is completely true. This layer of rock is called a Sedimentary Rock level and is slowly formed over millions of years with minerals and organic remains from the bottom of the Oceans that may no longer be covered in water anymore. Since it is made up of all these minerals and remains, it is studied widely by Geologists and Archeologists to better understand the Earth's past.

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Hi there! :)

Reference the diagram below for clarification.

1.

We must begin by knowing the following rules for resistors in series and parallel.

In series:
R_T = R_1 + R_2 + ... + R_n

In parallel:
\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}

We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules.

\frac{1}{R_{T, parallel}} = \frac{1}{10} + \frac{1}{10}\\\\\frac{1}{R_{T, parallel}} = \frac{2}{10} = \frac{1}{5}\\\\R_{T, parallel} = 5\Omega

Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:
R_T = 15 + 5\\\\\boxed{R_T = 20 \Omega}

2.

We can use Ohm's law to solve for the current in the circuit.

i = \frac{V}{R_T}\\\\i = \frac{120}{20} = \boxed{6 A}

3.

For resistors in series, both resistors receive the SAME current.

Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.

In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>.

4.

Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.

V = iR\\\\V = (6)(15) = \boxed{90 V}

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