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2 years ago

Which color of visibe light has a lower frequency than orange light

1 answer:

8 0

red having the frequency of 4.62 and at its limit 4.29 has the lowest frequency. It is the lowest before orange which has 5.00.

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In the experiment, a meter is hooked up to a speaker to monitor the amplitude of the received sound. Suppose the background sign

FrozenT [24]

Answer:

The answer is " $15.38\%$ "

Explanation:

Background $= 13 \ mv\\\\$

corrected signal $= 91 \ mv-13\ mv= 78\ mv\\\\$

with attenuator $=25\ mv-13\ mv= 12\ mv\\\\$

$\to \frac{p_t}{p_i}=\frac{12}{78}\times 100= 15.38\%$

4 0

3 years ago

Which of the following statements is/are true?Check all that apply.a. A conservative force permits a two-way conversion between

Eddi Din [679]

Answer:

a). A conservative force permits a two-way conversion between kinetic and potential energies.

TRUE

Because there is no energy loss in presence of conservative forces so energy conversion in two ways are possible.

b). A potential energy function can be specified for a conservative force.

TRUE

negative gradient of potential energy is equal to conservative force

$F = -\frac{dU}{dr}$

c). A non-conservative force permits a two-way conversion between kinetic and potential energies.

FALSE

here energy is lost against non-conservative forces

d). The work done by a conservative force depends on the path taken.

FALSE

work done by conservative force is independent of path

e). The work done by a non-conservative force depends on the path taken.

TRUE

work done by non conservative forces depends on path.

f). A potential energy function can be specified for a non-conservative force.

FALSE

It is not defined for non conservative forces

3 0

3 years ago

Think about the electricity sent to your home from a power plant. How does the voltage of the electricity that leaves the plant

tatiyna

C. High voltage to low voltage

6 0

3 years ago

Two forces (4N and 3N) pull your the left, while a 12N force pulls to the right. What is the net force

Lapatulllka [165]

Answer:

The answer is 11N to the right

Explanation:

Because 4N-3N= 1N

Therefore, 12N-1N=11N

The netforce is 11N to the right, because the greatest force is 12N to the right so it is more likely that the object is being pulled to the right.

8 0

3 years ago

1) A boy drags a wooden crate with a mass of 20 kg, a distance of 12 m, across a rough level floor at a constant speed of 1.5 m/

mojhsa [17]

Answer: a) 49.560 and 21.13 b) i) 50 N, ii) 196 N iii) 196 N iv) 47.685 N

c) i) 594.72 ii) 0 iii) 0 iv) 0

d) 594.72

Explanation: question a)

The force is inclined at an angle of 25° to the horizontal

The horizontal component of force = 50 cos 25° = 49.560 N

The vertical component of force = 50 sin 30°= 21.130N

Question b)

i) according to the question applied force is 50 N

ii) if g = 9.8m/s², w=mg where m = mass of object = 20kg hence weight = 20* 9.8 = 196 N

iii) the normal force is the force the floor exerts on the body as a result of the weight of the object.

Normal reaction R = W = mg, we already deduced that w = mg, hence R = 196 N.

iv) according to newton's laws of motion

F - Fr = ma

F = applied force = horizontal component of force = 49.560 N.

We need to get the acceleration (a) by using Newton laws of motion before we can be able to compute the frictional force..

The body started from rest hence initial velocity u = 0

Final velocity v = 1.5m/s distance covered (s) = 12m

v ² = u² + 2as

But u = 0

v² = 2as

1.5² = 2(a) * 12

2.25 = 24a

a = 2.25/24 = 0.09735m/s²

From F - Fr = ma

49.560 - Fr = 20 * 0.09735

49.560 - Fr = 1.875

Fr = 49.560 - 1.875

Fr = 47.685 N

Question c)

i) The applied force = 49.560 N, distance covered = 12m

Work done = force * distance

Work done = 49.560 * 12

Work done = 594.72 J

ii) the weight of the object does not make the object move a distance, hence work done = 0 ( since distance covered is 0)

iii) the normal force is the same thing as the weight and they did not cover any distance hence work done is zero.

iv) the frictional force does not cover any distance, hence work done is zero.

Question d)

The total work done = work done by applied force + work done by weight + work done by normal reaction + work done by frictional force.

Total work done = 594.72 + 0 + 0 + 0 = 594.72 J

8 0

3 years ago