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3 years ago

The current in a toaster is 20 amps, and its voltage is 200 volts. What is the resistance in the toaster?

2 answers:

8 0

<span>Ohm's law deals with the relation between voltage and current in an ideal conductor. It states that: Potential difference across a conductor is proportional to the current that pass through it. It is expressed as V=IR.

V = IR
200 = 20R
R = 10 ohms</span>

Verdich [7]3 years ago

8 0

Answer:

The answer would be

Explanation: B. 10 ohms

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Electric power is to be generated by installing a hydraulic turbine-generator at a site 120 m below the free surface of a large

iogann1982 [59]

Answer:

P = 2234400 Watt

Explanation:

Power generation is given as rate of work done

Here the turbine generator is located at 120 m below the free surface

So here rate of work done is given as rate of potential energy

$Power = \frac{dw}{dt}$

$Power = \frac{mgh}{t}$

so we have

$Power = (\frac{dm}{dt})gh$

now we have

$Power = 1900(9.8)(120) = 2234400 Watt$

so the power generation potential will be 2234400 Watt

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3 years ago

4. A car is accelerating at 40 m/s2, if the car is 400 kg how much force

Ksenya-84 [330]

Sorry I’m only in kindergarten is it 10kg must be supplied???

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3 years ago

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$