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salantis [7]
3 years ago
8

A table lamp draws a current of 200 mA when it is connected to a 120-V source. What is the resistance of the table lamp?

Physics
1 answer:
Nitella [24]3 years ago
7 0

Answer: 600Ω

Explanation:

Voltage of source (V) = 120 Volts

Current drawn by table lamp (I) = 200mA

(Since 1 mA = 1 x 10^-3A

200mA = 200 x 10^-3A)

Resistance of table lamp (R)= ?

Then, apply the formula for ohms law

Voltage = Current x resistance

i.e V = I x R

120V = 200 x 10^-3A x R

R = 120V / 200 x 10^-3A

R = 600Ω (Ω is the symbol for ohms)

Thus, the resistance of the table lamp is 600Ω

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The most common liquid on planet earth is water
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Which switch configuration creates a short circuit?
lara31 [8.8K]
I think the answer would be Switch 1 open; switches 2, 3, and 4 closed. It is letter D. This will happen when there is a low resistance and a high circuit so from the choices above, when two bulbs are open and one bulb is close, so there is a low resistance and high voltage.
6 0
3 years ago
Find the ratio of the lengths of the two mathematical pendulums, if the ratio of periods is 1.5​
juin [17]

Answer:

The ratio of lengths of the two mathematical pendulums is 9:4.

Explanation:

It is given that,

The ratio of periods of two pendulums is 1.5

Let the lengths be L₁ and L₂.

The time period of a simple pendulum is given by :

T=2\pi \sqrt{\dfrac{l}{g}}

or

T^2=4\pi^2\dfrac{l}{g}\\\\l=\dfrac{T^2g}{4\pi^2}

Where

l is length of the pendulum

l\propto T^2

or

\dfrac{l_1}{l_2}=(\dfrac{T_1}{T_2})^2 ....(1)

ATQ,

\dfrac{T_1}{T_2}=1.5

Put in equation (1)

\dfrac{l_1}{l_2}=(1.5)^2\\\\=\dfrac{9}{4}

So, the ratio of lengths of the two mathematical pendulums is 9:4.

3 0
2 years ago
An unknown radioactive sample is observed to decrease in activity by a factor of two in a one hour period. What is its half-life
elena55 [62]

Answer:

The half-life is t_{1/2} = 1.005 h

Explanation:

Using the decay equation we have:

A=A_{0}e^{-\lambda t}

Where:

  • λ is the decay constant
  • A(0) the initial activity
  • A is the activity at time t

We know the activity decrease by a factor of two in a one hour period (t = 1 h), it means that A = \frac{A_{0}}{2}

\frac{A_{0}}{2}=A_{0}e^{-\lambda*1 h}

0.5=e^{-\lambda*1 h}

Taking the natural logarithm on each side we have:

ln(0.5)=-\lambda

\lambda=0.69 h^{-1}

Now, the relationship between the decay constant λ and the half-life t(1/2) is:

\lambda = \frac{ln(2)}{t_{1/2}}

t_{1/2} = \frac{ln(2)}{\lambda}

t_{1/2} = \frac{ln(2)}{0.69}

t_{1/2} = 1.005 h

I hope it helps you!

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3 years ago
What is the unit used for work?
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Joules is used for work. 
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3 years ago
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