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kap26 [50]
3 years ago
7

What happens to the room temperature in ehich an air conditioner is lefy running on a table in the midle of the room

Physics
2 answers:
Katen [24]3 years ago
8 0
It will get cold only in that room. if you also have it in a room and close the door it will also get colder. sorry if this didnt help.
photoshop1234 [79]3 years ago
7 0
An air conditioner takes heat out of the air that runs through it. It has to put that heat somewhere. If it can't blow the heat out of the room then the heat stays in the room. But that's not all. While it's running the air conditioner also uses electrical energy to run. And that generates more heat. So if there's an air conditioner running in a closed room then the room will get warmer.
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Suppose a light source is emitting red light at a wavelength of 700 nm and another light source is emitting ultraviolet light at
klasskru [66]

Answer:

b) twice the energy of each photon of the red light.

Explanation:

\lambda = Wavelength

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

c = Speed of light = 3\times 10^8\ m/s

Energy of a photon is given by

E=h\nu\\\Rightarrow E=h\dfrac{c}{\lambda}

Let \lambda_1 = 700 nm

\lambda_2=350\\\Rightarrow \lambda_2=\dfrac{\lambda_1}{2}

For red light

E_1=\dfrac{hc}{\lambda_1}

For UV light

E_2=\dfrac{hc}{\dfrac{\lambda_1}{2}}

Dividing the equations

\dfrac{E_1}{E_2}=\dfrac{\dfrac{hc}{\lambda_1}}{\dfrac{hc}{\dfrac{\lambda_1}{2}}}\\\Rightarrow \dfrac{E_1}{E_2}=\dfrac{1}{2}\\\Rightarrow E_2=2E_1

Hence, the answer is  b) twice the energy of each photon of the red light.

7 0
3 years ago
Read 2 more answers
Recently, astronomers have observed stars and other objects that orbit the center of the Milky Way Galaxy farther out than our S
Lesechka [4]

Answer:

That scenario can be explained by the idea of the contribution of dark matter on that point.

Explanation:

It can be explained through the idea of dark matter, this one was born to explain why stars (or any object) that were farther for the supermassive black hole in the center of the Milky Way galaxy didn't decrease it rotational velocity as it was expected according to equation 1.

v = \sqrt{\frac{G M}{r}}  (1)

Where v is the rotational velocity, G is the gravitational constant, M is the mass of the supermassive black hole, and r is the orbital radius.

Notice, that If the distance increases the orbital speed decreases (inversely proportional).

7 0
3 years ago
One object (m1 = 0.220 kg) is moving to the right with a speed of 2.10 m/s when it is struck from behind by another object (m2 =
blagie [28]

Answer:

vf₁  = 6.86 m/s , to the right

vf₂ =  2.96 m/s, to the right

Explanation:

Theory of collisions  

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:  

p=m*v  

where  

p:Linear momentum  

m: mass  

v:velocity  

There are 3 cases of collisions : elastic, inelastic and plastic.  

For the three cases the total linear momentum quantity is conserved:  

P₀ = Pf Formula (1)  

P₀ :Initial linear momentum quantity  

Pf : Final linear momentum quantity  

Data

m₁= 0.220 kg : mass of  object₁

m₂= 0.345 kg : mass of  object₂

v₀₁ =  2.1 m/s ₁ , to the right : initial velocity of m₁

v₀₂=   6 m/s, to the right  i :initial velocity of m₂

Problem development

We appy the formula (1):

P₀ = Pf  

m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂  

We assume that the two objects move to the right at the end of the collision, so, the sign of the final speeds is positive:

(0.22)*(2.1) + (0.345)*(6) = (0.22)*vf₁ +(0.345)*vf₂

2.532 = (0.22)*vf₁ +(0.345)*vf₂ Equation (1)

Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.

e= \frac{v_{f2}-v_{f1} }{v_{o1} -v_{o2} }

1*(v₀₁ - v₀₂ )  = (vf₂ -vf₁)

(2.1 - 6 )  = (vf₂ -vf₁)

-3.9 =  (vf₂ -vf₁)

vf₂ = vf₁ - 3.9

vf₂ = vf₁ - 3.9 Equation (2)

We replace Equation (2) in the Equation (1)

2.532 = (0.22)*vf₁ +(0.345)*( vf₁ - 3.9)

2.532 = (0.22)*vf₁ +(0.345)* (vf₁) -(0.345)( 3.9)

2.532 + 1.3455 = (0.565)*vf₁

3.8775 = (0.565)*vf₁

vf₁  = (3.8775) / (0.565)

vf₁  = 6.86 m/s, to the right

We replace vf₁  = 6.86 m/s in the Equation (2)

vf₂ =  6.86 - 3.9

vf₂ =  2.96 m/s, to the right

8 0
3 years ago
In a particular metal, the mobility of the mobile electrons is 0.0033 (m/s)/(N/C). At a particular moment, the electric field ev
Lapatulllka [165]

Answer:

the average drift speed of the mobile electrons in the metal is 1.089 x 10⁻⁴ m/s.

Explanation:

Given;

mobility of the mobile electrons in the metal, μ = 0.0033 (m/s)/(N/C)

the electric field strength inside the cube of the metal, E = 0.033 N/C

The average drift speed of the mobile electrons in the metal is calculated as;

v = μE

v =  0.0033 (m/s)/(N/C) x 0.033 N/C

v = 1.089 x 10⁻⁴ m/s.

Therefore, the average drift speed of the mobile electrons in the metal is 1.089 x 10⁻⁴ m/s.

6 0
3 years ago
The scientific method can easily be misinterpreted as ___________. Question 1 options: a great magical trick, or slide of hands,
evablogger [386]

Answer:

I think the answer is  

a "cookbook" recipe for performing scientific investigations

Explanation:

5 0
3 years ago
Read 2 more answers
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