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makkiz [27]
2 years ago
15

a wall switch is connected to a wall outlet. A lamp is plugged into the same outlet. You turn the switch on, but the lamp stays

off. Give three reasons why this might happen.
Physics
1 answer:
jolli1 [7]2 years ago
5 0
1:   only half  the outlet is switched  and the lamp is in the  other half
2:  the lamp is turned off.
3:   The  light bulb  is burned out
4:  the switch might be broken
5:   the fuse   might be blown
6: the electricity might be off
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You go to the hardware store to buy a new 50 ft garden hose. You find you can choose between hoses of ½ inch and 5/8 inch inner
omeli [17]

To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,

Q = V*A

Where,

A= Cross-sectional Area

V = Velocity

The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,

Q_1 = Q_2

V_1A_1=V_2A_2

Our values are given as,

A_1=\frac{1}{2}^2*\pi=0.785 in^2

A_2=\frac{5}{8}^2*\pi=1.227 in^2

Re-arrange the equation to find the first ratio of rates we have:

\frac{V_1}{V_2}=\frac{A_2}{A_1}

\frac{V_1}{V_2}=\frac{1.227}{0.785}

\frac{V_1}{V_2}=1.56

The second ratio of rates is

\frac{V2}{V1}=\frac{A_1}{A2}

\frac{V2}{V1}=\frac{0.785}{1.227}

\frac{V2}{V1}=0.640

3 0
3 years ago
One gram of Uranium averages release 1.01 KJ (10^7) of energy. How much mass could be converted to energy to release this much e
frutty [35]

Answer:

The amount of mass that needs to be converted to release that amount of energy is 1.122 X 10^{-7}  kg

Explanation:

From Albert Einstein's Energy equation, we can understand that mass can get converted to energy, using the formula

E= \Delta mc^{2}

where \Delta m = change in mass

c = speed of light = 3 \times 10 ^{8}m/s

Making m the subject of the formula, we can find the change in mass to be

\Delta m = \frac{E}{c^{2}}= \frac{1.01 \times 10^{3} \times 10^{7}}{(3 \times 10^{8})^{2}}= 1.122 \times 10 ^{-7}kg

There fore, the amount of mass that needs to be converted to release that amount of energy is 1.122 X 10 ^-7 kg

5 0
3 years ago
Identify some common fuels
kozerog [31]
On a worldwide scale, the most common fuels are wood, grass, peat, coal, and animal fats and oils.
4 0
3 years ago
Read 2 more answers
What happens when gasoline is used to power a vehicle?
mojhsa [17]
<span>Energy is neither lost nor gained as it transforms from chemical, to heat, to mechanical energy.</span>
5 0
3 years ago
A bicycle racer inflates their tires to 7.1 atm on a warm autumn afternoon when temperatures reached 27 °C. By morning the tempe
natulia [17]

Answer:

The required pressure is 6.4866 atm.

Explanation:

The given data : -

In the afternoon.

Initial pressure of tire ( p₁ ) = 7 atm = 7 * 101.325 Kpa =  709.275 Kpa

Initial temperature ( T₁ ) = 27°C = (27 + 273) K = 300 K

In the morning .

Final temperature ( T₂ ) = 5°C = ( 5 + 273 ) K = 278 K

Given that volume remains constant.

To find final pressure ( p₂ ).

Applying the ideal gas equation.

p * v = m * R * T

\frac{p}{T}  = constant

\frac{p_{1} }{T_{1} }  = \frac{p_{2} }{T_{2} }

p_{2}  = \frac{T_{2} }{T_{1} } *p_{1}  

p_{2}  = \frac{278}{300}  * 709.275  = 657.2615 Kpa = 6.486 atm

8 0
2 years ago
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