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galina1969 [7]
3 years ago
9

A lamp is 10% efficient.How much electrical energy must be supplied to the lamp each second if it produces 20 J of light energy

per second?
Physics
1 answer:
k0ka [10]3 years ago
6 0

If it produces 20J of light energy in a second, then that 20J is the 10% of the supply that becomes useful output.

20 J/s = 10% of Supply

20 J/s = (0.1) x (Supply)

Divide each side by 0.1:

Supply = (20 J/s) / (0.1)

<em>Supply = 200 J/s  </em>(200 watts)

========================

Here's something to think about:  What could you do to make the lamp more efficient ?  Answer:  Use it for a heater !

If you use it for a heater, then the HEAT is the 'useful' part, and the light is the part that you really don't care about.  Suddenly ... bada-boom ... the lamp is 90% efficient !

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Guys please help me ​
Likurg_2 [28]

Answer:

1)t=2.26\: s

2)S=33.9\: m

3)v=26.77\: m/s

4)\alpha=55.92

Explanation:

1)

We can use the following equation:

y_{f}=y_{0}+v_{iy}t-0.5*g*t^{2}

Here, the initial velocity in the y-direction is zero, the final y position is zero and the initial y position is 25 m.

0=25-0.5*9.81*t^{2}

t=2.26\: s

2)

The equation of the motion in the x-direction is:

v_{ix}=\frac{S}{t}

15=\frac{S}{2.26}

S=33.9\: m

3)

The velocity in the y-direction of the stone will be:

v_{fy}=v_{iy}-gt

v_{fy}=0-(9.81*2.26)

v_{fy}=-22.17\: m/s

Now, the velocity in the x-direction is 15 m/s then the velocity will be:

v=\sqrt{v_{x}^{2}+v_{fy}^{2}}=\sqrt{15^{2}+(-22.17)^{2}}

v=26.77\: m/s

4)

The angle of this velocity is:

tan(\alpha)=\frac{22.17}{15}

\alpha=tan^{-1}(\frac{22.17}{15})

\alpha=55.92

Then α=55.92° negative from the x-direction.

I hope it helps you!

6 0
3 years ago
Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1.2 m high. Student 2 pushes an identical box up a 5 m ram
Troyanec [42]

The student who did the most work is student 2 with 2500 Joules.

<u>Given the following data:</u>

  • Force 1 = 500 Newton
  • Distance 1 = 1.2 meter
  • Force 2 = 500 Newton
  • Distance 2 = 5 meter

To determine which of the students did the most work:

Mathematically, the work done by an object is given by the formula;

Work\;done = Force \times distance

<u>For </u><u>student 1</u><u>:</u>

Work\;done = 500 \times 1.2

Work done = 600 Joules

<u>For </u><u>student 2</u><u>:</u>

Work\;done = 500 \times 5

Work done = 2500 Joules.

Therefore, the student who did the most work is student 2 with 2500 Joules.

Read more: Read more: brainly.com/question/13818347

7 0
3 years ago
Read 2 more answers
What is the relationship between Newton's first law of motion and inertia?​
adelina 88 [10]

Explanation: Newton's first law of motion states that a body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force. This is also known as the law of inertia. Inertia is the tendency of an object to remain at rest or remain in motion.

4 0
3 years ago
An astronaut takes an object to the moon where there is less gravity. Explain how the mass and weight of the object on the moon
alexandr1967 [171]
The mass of the object will remain the same rather it's on the moon or on the Earth and even in other places. But the weight will change on the moon, so its weight will be different from the one it had on Earth
3 0
3 years ago
A soccer player pumps air into a soccer ball until no more air can be pushed inside. Describe the air inside the soccer ball com
inessss [21]

Answer:

the filling stops when the pressure of the pump equals the pressure of the interior air plus the pressure of the walls.

Explanation:

This exercise asks to describe the inflation situation of a spherical fultball.

Initially the balloon is deflated, therefore the internal pressure is equal to the pressure of the air outside, atmospheric pressure, when it begins to inflate the balloon with a pump this creates a pressure in the inlet valve and as it is greater than the pressure inside, the air enters it, this is repeated in each filling cycle, manual pump.

When the ball is full we have two forces, the one created by the external walls and the one aired by the pressure of the pump, these forces are directed towards the inside, but the air molecules exert a pressure towards the outside, which translates into a force. When these two forces are equal, the pump is no longer able to continue introducing air into the balloon.

Consequently the filling stops when the pressure of the pump equals the pressure of the interior air plus the pressure of the walls.

4 0
3 years ago
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