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barxatty [35]
2 years ago
14

Calculate the efficiency of a light bulb that gives 40J of light from 200J of electrical energy’s

Physics
1 answer:
nekit [7.7K]2 years ago
7 0

Answer:

This formula is the most common one used. The efficiency is in a %

20%

Explanation:

Efficiency = energy out / energy in * 100 (for a %)

Efficiency = 40 / 200 * 100

Efficiency = 20%

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Atoms that have a positive charge will be attracted to atoms with a
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Answer:

negative

Explanation:

positive charges attract negative charges and vice versa. and are possible to nullify

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Is energy matter? Is gravity matter? Why or why not
Alja [10]

No, because the energy is the capacity for performing work. Gravity is the force that draws everything to the earth's center.

5 0
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The amount of diffraction that a sound wave undergoes depends on
BigorU [14]
The amount of diffraction of sound waves depends on the medium the sound wave travels to and the frequency. Diffraction happens as soon as it has been out of the source.
5 0
3 years ago
If we start with 1.000 g of cobalt-60, 0.675 g will remain after 3.00 yr. this means that the of is _____
Tju [1.3M]
<span>Cobalt-60 is undergoing a radioactivity decay.

The formula of the decay is n=N(1/2)</span>∧(T/t).
<span>Where N </span>⇒ original mass of cobalt
<span>           n </span>⇒ remaining mass of cobalt after 3 years
          T ⇒ decaying period
           t ⇒ half-life of cobalt.

So,
0.675 = 1 × 0.5∧(3/t)
log 0.675 = log 0.5∧(3/t)
3/t = log 0.675 ÷log 0.5
 3/t= 0.567

t = 3÷0.567
  = 5.290626524

the half-life of Cobalt-60 is 5.29 years. 

<span>           
</span><span>
</span>
8 0
2 years ago
A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130
konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(c) W = 0J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

W=Fd      (1)

F: applied force = 130N

d: distance = 5.0m

W=(130N)(5.0m)=650J

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

W_f=F_fd=\mu Mgd       (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

W = 0J

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

\Delta K=W_T         (3)

You calculate the total work:

W_T=Fd-F_fd=(F-F_f)d     (4)

F: applied force = 130N

Ff: friction force

d: distance = 5.00m

The friction force is:

F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N

Next, you replace the values of all parameters in the equation (4):

W_T=(130N-105.84N)(5.00m)=120.80J

\Delta K=120.80J

The change in the kinetic energy of the box is 120.8J

(e) The final speed of the box is calculated by using the equation (3):

W_T=\frac{1}{2}M(v_2^2-v_1^2)       (5)

v2: final speed of the box

v1: initial speed of the box = 0 m/s

You solve the equation (5) for v2:

v_2 = \sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(120.8J)}{36.0kg}}=2.58\frac{m}{s}

The final speed of the box is 2.58m/s

5 0
3 years ago
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