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r-ruslan [8.4K]
2 years ago
6

Which scientists supported the idea that light is made of streams of particles? Check all that apply.

Physics
1 answer:
otez555 [7]2 years ago
3 0

Answer:

hi, this is the answer

<em>Albert Einstein</em>

<em></em>

<em>Explanation:</em>

<em>Albert Einstein explained this "photoelectric" effect by proposing that light – thought to only be a wave – is also a stream of particles. Even though a variety of experiments have successfully observed both the particle- and wave-like behaviors of light, they have never been able to observe both at the same time</em>

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An air-track cart with mass m1=0.28kg and initial speed v0=0.75m/s collides with and sticks to a second cart that is at rest ini
arsen [322]
Kinetic energy is calculated through the equation,

   KE = 0.5mv²

At initial conditions,

  m₁:  KE = 0.5(0.28 kg)(0.75 m/s)² = 0.07875 J

  m₂ : KE = 0.5(0.45 kg)(0 m/s)² = 0 J

Due to the momentum balance,

   m₁v₁ + m₂v₂ = (m₁ + m₂)(V)

Substituting the known values,

   (0.29 kg)(0.75 m/s) + (0.43 kg)(0 m/s) = (0.28 kg + 0.43 kg)(V)

   V = 0.2977 m/s

The kinetic energy is,
   KE = (0.5)(0.28 kg + 0.43 kg)(0.2977 m/s)²
   KE = 0.03146 J

The difference between the kinetic energies is 0.0473 J. 
7 0
2 years ago
The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio
Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

\gamma = 5/3

T1 = 525 K

T2 = 275 K

We know that

P_1 = \frac{nRT_1}{v}

P_2 = \frac{nrT_2}{v}

n and v remain same at both side. so we have

\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}

P_1 = \frac{21}{11} P_2 ..............1

let final pressure is P and temp  T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3} ..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3} .............3

divide 2 equation by 3rd equation

\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

T_1 {f} = 1.48 T_2 {f}

thus, temperature on left side is 1.48 times the temperature on right

6 0
2 years ago
A falling skydiver opens his parachute. A short time later, the weight of the skydiver-parachute system and the drag force exert
Dimas [21]

A falling skydiver opens his parachute. A short time later, the weight of the skydiver-parachute system and the drag force exerted on the system are equal in magnitude. The following statements predicts the motion of the skydiver at this time

<u>The skydiver is moving downward with constant speed.</u>

Explanation:

Immediately on leaving the aircraft, the skydiver accelerates downwards due to the force of gravity. There is no air resistance acting in the upwards direction, and there is a resultant force acting downwards. The skydiver accelerates towards the ground.

The forces acting on a falling leaf are : gravity and air resistance.

The net force and the acceleration on the falling skydiver is upward.

An upward net force on a downward falling object would cause that object to slow down. The skydiver thus slows down.

As the speed decreases, the amount of air resistance also decreases until once more the skydiver reaches a terminal velocity.

<u>A skydiver falling at a constant speed opens his parachute. When the skydiver is falling, the forces are unbalanced.</u>

6 0
2 years ago
A 5kg bucket hangs from a ceiling on a rope. A student attaches a spring scale to the buckets handle and pulls horizontally on t
7nadin3 [17]
I don’t know what the angle is in your diagram so I used the angle from the vertical.

6 0
2 years ago
A real heat engine operates between temperatures tc and th. during a certain time, an amount qc of heat is released to the cold
tino4ka555 [31]

q_{c} = Heat released to cold reservoir

q_{h} = Heat released to hot reservoir

W_{max} = maximum amount of work

t_{c} = temperature of cold reservoir

t_{h} = temperature of hot reservoir

we know that

\frac{q_{c}}{q_{h}}=\frac{t_{c}}{t_{h}}

q_{h} = (\frac{t_{h}}{t_{c}})q_{c}                                eq-1

maximum work is given as

W_{max} = q_{h} - q_{c}

using eq-1

W_{max} =  (\frac{t_{h}}{t_{c}})q_{c} - q_{c}



6 0
2 years ago
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