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2 years ago

The light waves transfer their oscillations and energy to other object in what way?

Physics

1 answer:

denis23 [38]2 years ago

7 0

Answer: b the energy of light...

Explanation:

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A sample of a substance has a high density, yet a low particle motion. This sample must be a

avanturin [10]

The most possible answer is letter B) Liquid.
If we are going to base the answer to the motion of particles, liquid is the best answer since solid's particle cannot move for they are tightly packed while the particles of gas and plasma (ionized gas) are free and move at high speeds.

3 0

2 years ago

Read 2 more answers

An ion is an atom that has

liraira [26]

Answer:

a charge

Explanation:

it is charged, because the number of electrons does not equal the number of protons in the core of the atom.

8 0

1 year ago

An electron in a TV picture tube is accelerated through a potential difference of 10 kV before it hits the screen. What is the k

laiz [17]

Answer:

10,000 eV

Explanation:

According to the law of conservation of energy, the kinetic energy gained by the electron is equal to its change in electric potential energy:

$K=\Delta U=q \Delta V$

where:

K is the kinetic energy of the electron

$q=1 e$ is the magnitude of the charge of the electron

$\Delta V$ is the potential difference through which the electron has been accelerated

For this electron in the TV, we have

$\Delta V=10 kV=10000 V$

Therefore, the kinetic energy of the electron in electronvolts is

$K=(1 e)(10000 V)=10000 eV$

7 0

3 years ago

A car traveling 34 mi/h accelerates uniformly for 4 s, covering 615 ft in this time. What was its acceleration? Round your answe

Contact [7]

Answer:

51.94 ft/s²

257.63 ft/s

Explanation:

t = Time taken = 4 s

u = Initial velocity = 34 mi/h

v = Final velocity

s = Displacement = 615 ft

a = Acceleration

Converting velocity to ft/s

$34\ mi/h=\frac{34\times 5280}{3600}=49.87\ ft/s$

Equation of motion

$s=ut+\frac{1}{2}at^2\\\Rightarrow a=2\frac{s-ut}{t^2}\\\Rightarrow a=2\left(\frac{615-49.87\times 4}{4^2}\right)\\\Rightarrow a=51.94\ ft/s^2$

Acceleration is 51.94 ft/s²

$v=u+at\\\Rightarrow v=49.87+51.94\times 4\\\Rightarrow v=257.63\ ft/s$

Final velocity at this time is 257.63 ft/s

5 0

2 years ago

Calculate the amount of heat liberated (in kj) from 411 g of mercury when it cools from 88.0°c to 12.0°c.

Katen [24]

You should note that the melting point of mercury is -38.83°C, while the boiling point is at 356.7°C. Then, that means that there is no latent heat involved here. We only compute for the sensible heat.

ΔH = mCpΔT
The Cp of mercury is 0.14 J/g·°C
Thus,
ΔH = (411 g)(0.14 J/g·°C)(88 - 12°C)
<em>ΔH = 4,373.04 J</em>

5 0

2 years ago