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3 years ago

Which of these results in kinetic energy of an object? position motion mass volume

Physics

2 answers:

mote1985 [20]3 years ago

7 0

Answer:

motion

Explanation:

when an object moves, it gains kinetic energy. so correct option in motion.

hope it helps!

Dominik [7]3 years ago

7 0

Motion..............…

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Two objects, Object A and Object B, need to be identified. Object A's index of refraction is determined to be 1.77, and Object B

Slav-nsk [51]

The correct answer is

C. Light can pass through Object B faster than it can pass through Object A.

In fact, the index of refraction of a material is defined as:

$n=\frac{c}{v}$

where c is the speed of light in vacuum and v is the speed of light in the material. Rearranging the equation, we can write the speed of light in the material as:

$v=\frac{c}{n}$

So we that, the smaller the refractive index n, the greater the speed of light in the material, v. In this problem, object B has lower refractive index than object A, so light travels faster in object B.

4 0

3 years ago

Read 2 more answers

An electron is initially moving at 1.4 x 107 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N

algol13

Answer:

K.E = 15.57 x 10⁻¹⁷ J

Explanation:

First, we find the acceleration of the electron by using the formula of electric field:

E = F/q

F = Eq

but, from Newton's 2nd Law:

F = ma

Comparing both equations, we get:

ma = Eq

a = Eq/m

where,

E = electric field intensity = 120 N/C

q = charge of electron = 1.6 x 10⁻¹⁹ C

m = Mass of electron = 9.1 x 10⁻³¹ kg

Therefore,

a = (120 N/C)(1.6 x 10⁻¹⁹ C)/(9.1 x 10⁻³¹ kg)

a = 2.11 x 10¹³ m/s²

Now, we need to find the final velocity of the electron. Using 3rd equation of motion:

2as = Vf² - Vi²

where,

Vf = Final Velocity = ?

Vi = Initial Velocity = 1.4 x 10⁷ m/s

s = distance = 3.5 m

Therefore,

(2)(2.11 x 10¹³ m/s²)(3.5 m) = Vf² - (1.4 x 10⁷)²

Vf = √(1.477 x 10¹⁴ m²/s² + 1.96 x 10¹⁴ m²/s²)

Vf = 1.85 x 10⁷ m/s

Now, we find the kinetic energy of electron at the end of the motion:

K.E = (0.5)(m)(Vf)²

K.E = (0.5)(9.1 x 10⁻³¹ kg)(1.85 x 10⁷ m/s)²

4 0

3 years ago

A wave With wavelength 20 m has a frequency of 12 Hz what is the waves speed

pentagon [3]

Answer:

240m/s

Explanation:

The equation to calculate is wavelength= velocity/ frequency so to find the velocity you would have to multiply frequency by wavelength.

7 0

3 years ago

a. How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 25.0 cm in diameter t

Agata [3.3K]

Answer:

Explanation:

The electric field outside the sphere is given as,

E = k Q /r²

here Q = n x 1.6 x 10⁻¹⁹ C

where n is the number of electons

if the dimeter of sphere d= 25 cm= 0.25 m

then the radius r = 0.125 m

we get

n= E r²/ k x 1.6 x 10⁻¹⁹ C

n = 1350N/C x (0.125m)² / (8.99 x 10⁹ N m²/C² x 1.6 x 10⁻¹⁹ C)

n = 14664731646

5 0

3 years ago

A thin soap bubble of index of refraction 1.33 is viewed with light of wavelength 550.0nm and appears very bright. Predict a pos

faust18 [17]

Answer:

The possible thickness of the soap bubble = $1.034\times 10^{-7}\ m.$

Explanation:

<u>Given:</u>

Refractive index of the soap bubble, $\mu=1.33.$
Wavelength of the light taken, $\lambda = 550.0\ nm = 550.0\times 10^{-9}\ m.$

Let the thickness of the soap bubble be $t$ .

It is given that the soap bubble appears very bright, it means, there is a constructive interference takes place.

For the constructive interference of light through a thin film ( soap bubble), the condition of constructive interference is given as:

$2\mu t=\left ( m+\dfrac 12 \right )\lambda.$

where $m$ is the order of constructive interference.

Since the soap bubble is appearing very bright, the order should be 0, as $0^{th}$ order interference has maximum intensity.

Thus,

$2\mu t=\left (0+\dfrac 12\right )\lambda\\t=\dfrac{\lambda}{4\mu}\\\ \ = \dfrac{550\times 10^{-9}}{4\times 1.33}\\\ \ = 1.034\times 10^{-7}\ m.$

It is the possible thickness of the soap bubble.

6 0

3 years ago