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Nezavi [6.7K]
3 years ago
5

An electric oven has a resistance of 1.00 kΩ and a voltage of 220 v. how much current does it draw?

Physics
2 answers:
lorasvet [3.4K]3 years ago
7 0

Answer:

0.22 A

Explanation:

According to Ohm's law, at constant temperature, the current flowing through a conductor is directly proportional to the potential difference applied across the ends of the conductor.

V = IR

Where, I is the current and R be the resistance.

So, Here, V = 220 V and R = 1 Kilo ohm = 1000 ohm

By using Ohm's law

I = \frac{V}{R}

I = \frac{220}{1000}

I = 0.22 A

Semenov [28]3 years ago
3 0
Assume that this is a direct current circuit.
V = IR
220 = I(1k)
I = .22 A
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In an inelastic collision, as compared to an elastic collision, what is to be expected?.
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Two golf balls are hit from the same point on a flat field. Both are hit at an angle of 55 degree above the horizontal. Ball 2 h
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Answer:

d_2 = 4d_1

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= (2U)²sin2θ/g

= 4U²sin2θ/g

= 4d_1   (since d_1 = U²sin2θ/g)

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A ball is thrown upward. what can be said about the system ?
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Read 2 more answers
Suppose that a person gets hit by a bus moving at 30 mi/h with a 58,000 lbs of force in the direction of motion. If the mass of
alexandr402 [8]

The impulse of a force is due to the change in the motion of an object

A. The persons speed after impact is approximately 59.38 mi/h

B. The expected speed is <u>29.89 mi/h</u> which is less than the findings

Reason:

Known parameters are;

The speed of the bus, v = 30 mi/h

The force with which the person was hit, F = 58,000 lbs

Mass of the bus, M = 40,000 lbs

Mass of the person, m = 150 lbs

Duration of the impact, Δt = 0.007 seconds

A. The speed of the person at the end of the impact, <em>v</em>, is given as follows;

The impulse of the force = F × Δt = m × Δv

For the person, we get;

58,000 lbf ≈ 1866094.816 lb·ft./s²

58,000 lbf × 0.007 s = 150 lbs × Δv

1,866,094.816 lb·ft./s²

\Delta v = \dfrac{1,866,094.816\ lbs \times 0.007 \, s}{150 \, lbs} \approx  87.084  \ ft./s

Δv = v₂ - v₁

The initial speed of the person at the instant, can be as v₁ = 0

The final speed, v₂ = Δv - v₁

∴ v₂ ≈  87.084 ft./s - 0 = 87.084 ft./s

≈ <u>87.084 ft./s</u>

<u />v_2 \approx \dfrac{87.084 \ ft./s}{y} \times\dfrac{1 \ mi}{5280 \ ft.} \times \dfrac{3,600 \ s}{1 \, hour} \approx 59.38 \ mi/h<u />

The speed of the person at the end of the impact, v₂ ≈ <u>59.38 mi/h</u>

B. Where the momentum is conserved, we have;

m₁·v₁ + m₂v₂ = (m₁ + m₂)·v

v = \dfrac{m_1 \cdot v_1 + m_2 \cdot v_2}{m_2 + m_1}

v = \dfrac{40,000 \times 30  + 150 \times 0}{40,000 + 150} \approx 29.89

The expected speed of the person at the end of the impact is 29.89 mi/h, and therefore, <u>the findings does not agree with the expectation</u>

Learn more here:

brainly.com/question/18326789

3 0
3 years ago
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