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2 years ago

How many joules of electrical energy is transferred per second by 6V 0.5 A lamp?

1 answer:

4 0

When you ask for "joules per second", you're asking for "watts".
The rate of energy "transfer" is 'power'. In this case, the light bulb
transfers energy out of the electrical circuit and into the space
around it, in the form of light and heat radiation.

Electrical power = (voltage) x (current) =

(6 volts) x (0.5 ampere) =

3 watts = 3 joules per second.

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Can someone help me ASAP

Phantasy [73]

Well I don't know. Let's actually LOOK at the picture and see if that helps.

A, B, C, and D all have the same TOTAL length, but A has the most waves crammed into that same total length.

By golly, that means the length of each wave in A must be shorter than each wave in B, C, or D.

The correct choice is A . Looking at the picture did the trick !

7 0

2 years ago

Find the slit separation of a double-slit arrangement that will produce interference fringes 0.018 rad apart on a distant screen

inessss [21]

Answer:

1.64 * 10^(-5) m

Explanation:

Parameters given:

Angular separation, θ = 0.018 rad

Wavelength, λ = 589 nm = 5.89 * 10^(-7) m

The angular separation when there are 2 slots is given as

θ = λ/2d

where d = separation between slits

d = λ/2θ

d = (589 * 10^(-9))/(2 * 0.018)

d = 1.64 * 10^(-5) m

5 0

2 years ago

What is meant by the following statement? "Acceleration is inversely proportional to mass." Acceleration decreases as mass decre

Vladimir79 [104]

It means, Acceleration increases as mass decreases.

So, option C is your answer.

Hope this helps!

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3 years ago

Read 2 more answers

i just said something super cringey to someone and now i wish i had a time machine because i cant forget it :((( SOME HELP PLEAS

Llana [10]

Answer:

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Explanation:

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2 years ago

Read 2 more answers

Water (density = 1x10^3 kg/m^3) flows at 15.5 m/s through a pipe with radius 0.040 m. The pipe goes up to the second floor of th

RUDIKE [14]

Answer:

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

Explanation:

By assuming that fluid is incompressible and there are no heat and work interaction through the line of current corresponding to the pipe, we can calculate the speed of the water floor in the pipe on the second floor by Bernoulli's Principle, whose model is:

$P_{1} + \frac{\rho\cdot v_{1}^{2}}{2}+\rho\cdot g\cdot z_{1} = P_{2} + \frac{\rho\cdot v_{2}^{2}}{2}+\rho\cdot g\cdot z_{2}$ (1)

Where:

$P_{1}$ , $P_{2}$ - Pressures of the water on the first and second floors, measured in pascals.

$\rho$ - Density of water, measured in kilograms per cubic meter.

$v_{1}$ , $v_{2}$ - Speed of the water on the first and second floors, measured in meters per second.

$z_{1}$ , $z_{2}$ - Heights of the water on the first and second floors, measured in meters.

Now we clear the final speed of the water flow:

$\frac{\rho\cdot v_{2}^{2}}{2} = P_{1}-P_{2}+\rho \cdot \left[\frac{v_{1}^{2}}{2}+g\cdot (z_{1}-z_{2}) \right]$

$\rho\cdot v_{2}^{2} = 2\cdot (P_{1}-P_{2})+\rho\cdot [v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})]$

$v_{2}^{2}= \frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})$

$v_{2} = \sqrt{\frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2}) }$ (2)

If we know that $P_{1}-P_{2} = 0\,Pa$ , $\rho=1000\,\frac{kg}{m^{3}}$ , $v_{1} = 15.5\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $z_{1}-z_{2} = -3.5\,m$ , then the speed of the water flow in the pipe on the second floor is:

$v_{2}=\sqrt{\left(15.5\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (-3.5\,m)}$

$v_{2} \approx 13.100\,\frac{m}{s}$

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

4 0

2 years ago