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2 years ago

Need Some Help Please :)

2 answers:

6 0

1. The amount of energy carried by the wave is related to the Amplitude of the wave.
2. A mechanical wave requires an initial energy input, Once this initial energy is added the wave travels through the medium until all it's energy is transferred.

yanalaym [24]2 years ago

6 0

Answer:

The higher the wave, the higher the frequency
Energies are transferred in mechanical waves through a medium

Explanation:

<u>How frequency relates to the energy of the wave</u>

Literally, frequency of a wave refers to the number of waves that move past a certain point during a given amount of time, (although it is often measured per second )

So, this means that when more waves moves past the given points, the frequency of the wave will get increased and if other wise, the frequency if the wave will get reduced, this is so because wave frequency is related to wave energy.

Since all that waves really are is traveling energy, the more energy in a wave, the higher its frequency. The lower the frequency is, the less energy in the wave.

<u>How energy is transferred through mechanical wave</u>

Mechanical waves are such that they're not capable of transmitting energy via vacuum. Hence, they need a medium to transmit these energies from one point to another.

All type of mechanical waves need a form of medium in order to transmit.

Take for instance:

A slinky wave requires the coils of the slinky;
A water wave requires water

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An 84% efficient single pulley is used to lift a 230 kg piano 3.5 m. How much work must be input?

Sati [7]

Answer:

35%

Explanation:

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3 years ago

Please help!!!! Will mark brainliest.

julia-pushkina [17]

Answer:

Approximately $8.4 \times 10^{2}\; \rm N$ , assuming that $g = 9.8\; \rm m \cdot s^{-2}$ .

Explanation:

Let $m$ and $a$ denote the mass and acceleration of Spiderman, respectively.

There are two forces on Spiderman:

Downward gravitational attraction from the earth: $W = m \cdot g$ .
Upward tension force from the strand of web $F(\text{tension})$ .

The directions of these two forces are exactly opposite of one another. Besides, because Spiderman is accelerating upwards, the magnitude of $F(\text{tension})$ (which points upwards) should be greater than that of $W$ (which points downwards towards the ground.)

Subtract the smaller force from the larger one to find the net force on Spiderman:

$(\text{Net Force}) = F(\text{tension}) - W$ .

On the other hand, apply Newton's Second Law of motion to find the value of the net force on Spiderman:

$(\text{Net Force}) = m \cdot a$ .

Combine these two equations to get:

$m \cdot a = (\text{Net Force}) = F(\text{tension}) - W$ .

Therefore:

$\begin{aligned}& F(\text{tension})\\ &= m \cdot a + W \\ &= m \cdot (a + g)\\ &= 76\; \rm kg \times \left(1.3\; \rm m \cdot s^{-2} + 9.8\; \rm m \cdot s^{-2}\right)\\ &\approx 8.4\times 10^{2}\; \rm N\end{aligned}$ .

By Newton's Third Law of motion, Spiderman would exert a force of the same size on the strand of web. Hence, the size of the force in the strand of the web should be approximately $8.4\times 10^{2}\; \rm N$ (downwards.)

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3 years ago

Rudik [331]

Answer:

Explanation:

cause logic

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3 years ago

Ah electron is a negatively charged particle that has a charge of magnitude, e - 1.60 x 10-19 C. Which one of the following stat

Advocard [28]

Answer:

The correct statement is "The electric field is directed toward the electron and has a magnitude of $\rm \dfrac{ke}{r^2}$ ".

Explanation:

According to Coulomb's law, the magnitude of the electric field due to a static point charge q at a point r distance away from it is given by

$\rm E = \dfrac{k|q|}{r^2}.$

k is the Coulmob's constant.

The direction of the electric field along the line joining the charge and the point where electric field is to be found and it is directed from positive charge to negative charge.

Conventionally, we assume a positive test charge placed at the point where electric field is to be found, the test charge has very small charge such that its charge does not affect the electric field due to the given charge.

The charge on the electron = -e.

The electric field due to an electron is given by

$\rm E = \dfrac{k|-e|}{r^2}=\dfrac{ke}{r^2}.$

The direction of this electric field is from positive test charge, placed at the point where electric field is to be found, towards the electron along the line joining the two.

Thus, the correct statement is "The electric field is directed toward the electron and has a magnitude of $\rm \dfrac{ke}{r^2}$ ".

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