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2 years ago

Kinetic energy increases as

Physics

2 answers:

HACTEHA [7]2 years ago

7 0

Answer is C
I did it in class

Snowcat [4.5K]2 years ago

3 0

C. Both mass and velocity increase
EXPLANATION:
Doubling the mass, doubles the kinectic energy. so it is higher.
Doubling the velocity, Quadruples the kinetic energy.
So Both mass and velocity increase will result in an increase in the kinetic energy

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A difference In density between gases creates buoyancy which is the ability of an object to float. 0°C helium has a density of .

rjkz [21]

Answer:

mass = 0.18 [kg]

Explanation:

This is a classic problem where we can apply the definition of density which is equal to mass over volume.

$density = \frac{mass}{volume} \\\\where:\\volume = 1 [m^3]\\density = 0.18[kg/m^3]$

mass = 0.18*1

mass = 0.18 [kg]

6 0

2 years ago

Which of the following metals require ultraviolet light to exhibit the photoelectric effect?The options available: a. Cs, work f

Eddi Din [679]

Answer:

b. AG, work function=4.74eV

Explanation:

Ultraviolet light starts at the end of the visible light spectrum, where violet light ends:

$\lambda=380 nm =3.8\cdot 10^{-7}m$ (wavelength of lowest-energy ultraviolet light)

So, the lowest energy of ultraviolet light can be found by using the formula

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

Substituting,

$E=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.8\cdot 10^{-7} m}=5.23\cdot 10^{-19}J$

And keeping in mind that

$1 eV = 1.6\cdot 10^{-19}J$

This energy converted into electronvolts is

$E=\frac{5.23\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=3.27 eV$

The work function of a metal is the minimum energy needed to extract a photoelectron from the surface of the metal. Therefore, the metals that exhibit photoelectric effect are the ones whose work function is larger than the energy we found previously, so:

b. AG, work function=4.74eV

Because for all the other metals, visible light will be enough to extract photoelectrons.

7 0

2 years ago

Conservation of Momentum Practice

Yakvenalex [24]

Answer:

Before: 0 m/s

After: -4 m/s

Explanation:

Before: Since you and your beau started at rest, your beau initial velocity is 0 m/s.

After: Since we have to conserve momentum,

momentum before push = momentum after push.

The momentum before push = 0 (since you and your beau are at rest)

momentum after push = m₁v₁ + m₂v₂ were m₁ = your mass = 60 kg, v₁ = your velocity after push = 3 m/s, m₂ = beau's mass = 45 kg and v₂ = beau's velocity.

So, m₁v₁ + m₂v₂ = 0

m₁v₁ = -m₂v₂

v₂ = -m₁v₁/m₂ = -60 kg × 3 m/s ÷ 45 kg = -4 m/s

So beau moves with a velocity of 4 m/s in the opposite direction

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2 years ago

The low-_____ areas in a sound wave are called rarefactions. A. period B. frequency C. energy D. density

Delicious77 [7]

You can picture a sound wave a lot like a Slinky wave . . . a
thicker, compressed blob moving along the path, with thinner,
stretched-out places before and after it.

The thicker parts of a sound wave, where the air is more dense,
are called compressions.

The thinner parts of a sound wave, where the air is less dense,
are called rarefactions.

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2 years ago

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which of the following is not a projectile? a.a satellite b.a throw ball c.a ball on the ground d.a soaring arrow

Delvig [45]

C a ball on the ground

7 0

3 years ago

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