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Elanso [62]
3 years ago
12

Is The force of friction always opposite to the motion? ​

Physics
1 answer:
Taya2010 [7]3 years ago
8 0

Answer:

Friction force always acts tangent to the surface at points of contact. Friction force acts opposite to the direction of motion.

Explanation:

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PLEASE HELP!!!<br>How are middle latitude cyclones and tropical cyclones different from each other?
defon
In contrast, extratropical cyclones have their strongest winds near the tropopause, which is about 8 miles above the surface. These differences are due to the tropical cyclone being “warm-core” in the troposphere, whereas extra-tropical cyclones are “warm-core” in the stratosphere and “cold-core” in the troposphere.
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3 years ago
Consider the following waves representing electromagnetic radiation: An illustration shows two waves representing electromagneti
Murljashka [212]

Answer:

a) red wave hs a longer wavelength than the green wave

b)f = 1.875 10¹¹ Hz ,  f_green = 5.45 10¹⁴Hz

c)   E = 1.24 10⁻²² J , E_green = 3.6 10⁻¹⁹ J

d) The red wave is in the infrared range, heat waves

The green wave is in the visible wavelength

Explanation:

a) The green wave are on the left in the electromagnetic spectrum so the red wave has a longer wavelength than the green wave

The green wavelength is in the range of 550 10⁻⁹ m

The speed of the wave is

            c = λ f

            f = c /λ

b) The frequency of the red wave is

            f = 3 10⁸ / 1.6 10⁻³

            f = 1.875 10¹¹ Hz

For the green wave

           f_green = 3 10⁸/550 10⁻⁹

           f_green = 5.45 10¹⁴Hz

c) The photon energy is given by the Planck equation

             E = h f

             E = 6.63 10⁻³⁴ 1.875 10¹¹

             E = 1.24 10⁻²² J

For the green wave

              E_green = 6.63 10⁻³⁴ 5.45 10¹⁴

              E_green = 3.6 10⁻¹⁹ J

d) The speed of electromagnetic waves is constant and has a value of 3 108 m / s

e)  

The red wave is in the infrared range, heat waves

The green wave is in the visible wavelength

6 0
3 years ago
What is the final velocity of a car that starts at 22 m/s and accelerates at 3.78 m/s for distance of 45 m
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v^2 = v0^2 +2ad v^2 = 22^2 + 2*3.78*45 = 824.2 v= √824.2 = 28.7 m/s

5 0
3 years ago
Hanna tosses a ball straight up with enough speed to remain in the air for several seconds?
olga_2 [115]
A) the velocity is 0 m/s
3 0
3 years ago
Read 2 more answers
The rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between
erma4kov [3.2K]

Answer:

Considering first question

    Generally the coefficient of performance of the air condition  is mathematically represented as

   COP  =  \frac{T_i}{T_o - T_i}

Here T_i is the inside temperature

while  T_o is the outside temperature

What this coefficient of performance represent is the amount of heat the air condition can remove with 1 unit of electricity

So it implies that the air condition removes   \frac{T_i}{T_o - T_i} heat with 1 unit of electricity

Now from the question we are told that the rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside. This can be mathematically represented as

         Q \ \alpha \ (T_o - T_i)

=>        Q= k (T_o - T_i)

Here k is the constant of proportionality

So  

    since  1 unit of electricity  removes   \frac{T_i}{T_o - T_i}  amount of heat

   E  unit of electricity will remove  Q= k (T_o - T_i)

So

      E =  \frac{k(T_o - T_i)}{\frac{T_i}{ T_h - T_i} }

=>   E = \frac{k}{T_i} (T_o - T_i)^2

given that  \frac{k}{T_i} is constant

    =>  E \  \alpha  \  (T_o - T_i)^2

From this above equation we see that the  electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square of the temperature difference.

 Considering the  second question

Assuming that  T_i   =  30 ^oC

 and      T_o  =  40 ^oC

Hence  

     E = K (T_o - T_i)^2

Here K stand for a constant

So  

        E = K (40 -  30)^2

=>      E = 100K

Now if  the  T_i   =  20 ^oC

Then

       E = K (40 -  20)^2

=>      E = 400 \ K

So  from this see that the electricity require (cost of powering and operating the air conditioner)when the inside temperature is low  is  much higher than the electricity required when the inside temperature is higher

Considering the  third question

Now in the case where the  heat that enters the building is at a rate proportional to the square-root of the temperature difference between inside and outside

We have that

       Q = k (T_o - T_i )^{\frac{1}{2} }

So

       E =  \frac{k (T_o - T_i )^{\frac{1}{2} }}{\frac{T_i}{T_o - T_i} }

=>   E =  \frac{k}{T_i} * (T_o - T_i) ^{\frac{3}{2} }

Assuming \frac{k}{T_i} is a constant

Then  

     E \ \alpha \ (T_o - T_i)^{\frac{3}{2} }

From this above equation we see that the  electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square root  of the cube of the  temperature difference.

   

4 0
3 years ago
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