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BARSIC [14]
3 years ago
5

Which of the following statements is true?A. Radiant energy is the same as sound waves. B. Radiant energy is a form of potential

energy only. C. Radiant energy does not serve any useful purpose. D. Radiant energy does not require a medium through which to travel. Reset
Physics
2 answers:
svetoff [14.1K]3 years ago
5 0
D.Radiant energy does not require a medium through which to travel.
svetlana [45]3 years ago
5 0

Answer:

D. Radiant energy does not require a medium through which to travel.

Explanation:

Radiant energy is the energy of electromagnetic waves. It is a form of energy that can travel through space. For example, we receive the heat from the sun, which is located very far from the earth via radiation. The sun's heat is not transmitted through any solid medium, but through a vacuum. Welcome.

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Two objects with masses of 2.75 kg and 4.45 kg are connected by a light string that passes over a light frictionless pulley to f
Llana [10]

Answer:

Explanation:

F = ma

4.45g - 2.75g = (4.45 + 2.75)a

a = 9.81(4.45 - 2.75) / (4.45 + 2.75) = 2.31625 ≈ 2.32 m/s²

a)

T = 2.75(9.81 + 2.32) = 33.3 N

or

T = 4.45(9.81 - 2.32) = 33.3 N

b) 2.32 m/s² upward for 2.75 kg mass

    2.32 m/s² downward for 4.45 kg mass

c) y = ½at² = ½(2.31625/3)1² = 1.158125 ≈ 1.16 m

4 0
2 years ago
Read the given list of organisms.
Nadya [2.5K]
They are all herbivores
3 0
3 years ago
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What weather would create the weakest surfing waves?
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Its B hope I could help
5 0
3 years ago
A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she
lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

f_{(1  \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}}  }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}  = 1.3225 \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz  = 740.6 \  Hz

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, L_{new} = 2/3·L

The value of the fundamental frequency will then be given as follows;

f_{(1  \, new)} =  \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \times \dfrac{2 \times L}{3} }  =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz =  840 \ Hz

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0
3 years ago
A swan on a lake gets airborne by flapping its wings and running on top of the water. If the swan must reach a velocity of 6.00
ira [324]

Answer:

A. 51.42 m.

B. 17.14 s.

Explanation:

Using equations of motion:

vf^2 = vi^2 + 2 * aS

Where,

vf = final velocity

a = acceleration

S = distance to which swan traveled

vi = 0 m/s

6.00^2 = 2 * 0.350S

S = 36/0.7

= 51.42 m.

B.

vf = vi + at

6 = 0 + 0.35t

t = 6/0.35

= 17.14 s.

5 0
3 years ago
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