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vredina [299]
2 years ago
6

What is the Net Force of the applied forces? *

Physics
1 answer:
Oduvanchick [21]2 years ago
3 0
All of anove
Bc it make sense
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The Michelson-Morley experiment was designed to measure Group of answer choices the velocity of the Earth relative to the ether.
NISA [10]

Answer:

The Michelson-Morley was designed to detect the motion of the earth through the ether.

No such relation was found and the speed of light is assumed to be the same in all reference frames.

4 0
3 years ago
“Is it correct to say that a radio wave can be considered a low-frequency light wave?
klio [65]

It's weird but technically correct to say that a radio wave can be considered a low-frequency light wave.  Radio and light are both electromagnetic waves.  The only difference is that radio waves have much much much longer wavelengths, and much much much lower frequencies, than light waves have.  But they're both the same physical phenomenon.

However, a radio wave CAN'T also be considered to be a sound wave.  These two things are as different as two waves can be.

-- Radio is an electromagnetic wave. Sound is a mechanical wave.

-- Radio waves travel more than 800 thousand times faster than sound waves do.

-- Radio waves are transverse waves. Sound waves are longitudinal waves.

-- Radio waves can travel through empty space. Sound waves need material stuff to travel through.

-- Radio waves can be detected by radio, TV, and microwave receivers. Sound waves can't.

-- Sound waves can be detected by our ears. Radio waves can't.

-- Sound waves can be generated by talking, or by hitting a frying pan with a spoon. Radio waves can't.

-- Radio waves can be generated by an alternating current flowing through an isolated wire. Sound waves can't.

4 0
3 years ago
A honeybee with a mass of 0.150 g lands on one end of a floating 4.75-g popsicle stick, as shown in (Figure 1) . After sitting a
Alekssandra [29.7K]

Answer:

0.0443 m/s

Explanation:

m_1 = Mass of honeybee = 0.15 g

m_2 = Mass of popsicle stick = 4.75 g

v_1 = Velocity of honeybee

v_2 = Velocity of stick = 0.14 cm/s

In this system the linear momentum is conserved

m_1v_1=m_2v_2\\\Rightarrow v_1=\dfrac{m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{4.75\times 0.14}{0.15}\\\Rightarrow v_1=4.43\ m/s

The velocity of the bee is 4.43 cm/s or 0.0443 m/s

5 0
2 years ago
If the system is operated on mars, through what distance would the 18.0-kg mass have to fall to give the same amount of kinetic
-Dominant- [34]
The previous part of the exercise says:
"<span>Engineers are designing a system by which a falling mass m imparts kinetic energy to a rotating uniform drum to which it is attached by thin, very light wire wrapped around the rim of the drum. There is no appreciable friction in the axle of the drum, and everything starts from rest. This system is being tested on Earth, but it is to be used on Mars, where the acceleration due to gravity is 3.71 m/s². In the Earth tests, when m is set to 18.0 kg and allowed to fall through 5.50 m, it gives 300.0 J of kinetic energy to the drum."

Since Kearth = Kmars, we have, for conservation of energy, that also the potential energies must be equal:
Uearth = Umars

which means:
m </span>· gearth · hearth = m · gmars <span>· hmars

we can solve for hmars:
hmars = (gearth / gmars) </span>· hearth
           = (9.8 / 3.71) · 5.50
           = 14.53m

Therefore, the correct answer will be: the mass would have to fall from an height of 14.53m.

5 0
3 years ago
Help me please I have other ones like this too on my page please help!
Dvinal [7]
Balanced equation is:
2Mn+4CuCl——->4Cu+2MnCl2
a=2
b=4
c=4
d=2
5 0
2 years ago
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