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Vera_Pavlovna [14]
2 years ago
5

Wave energy can only be transmitted through a material medium. wave energy can only be transmitted through a material medium.

Physics
1 answer:
S_A_V [24]2 years ago
3 0
The correct answer is False.
In fact, electromagnetic waves (which carry energy), do not need a material medium to travel, because they can travel in vacuum as well. So, wave energy can be transmitted also through vacuum.
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The acceleration due to gravity on the moon is about 5.4ft/s2. if your weight is 150lbf on earth
Ivenika [448]

... then your weight is <em>25.2 lbf</em> on the moon.

6 0
3 years ago
What is the work done by the friction when the body slides against a rough horizontal surfaces?
katovenus [111]

Negative work

Hope this helps :)

4 0
2 years ago
The latent heat of fusion of alcohol is 25 kcal/kg and its melting point is -114 o C. It has a specific heat of 0.60 in its liqu
UNO [17]

Answer:

=170kcal

Explanation:

We first calculate the amount of energy required to melt the alcohol using the formula: MLf, where Lf is the latent heat of fussion

We then calculate amount of heat required to raise the temperature of liquid alcohol to -14° C using MC∅.  We then add the two.

Thus ΔH=MLf+MC∅

ΔH=2kg×25kcal/kg+ 2kg×(0.6kcal/kg.K×(-14-⁻114)

=50kcal+120kcal

=170kcal

3 0
2 years ago
There are four charges, each with a magnitude of 4.25 C. Two are positive and two are negative. The charges are fixed to the cor
VMariaS [17]

Answer:

 F = 7.68 10¹¹ N,  θ = 45º

Explanation:

In this exercise we ask for the net electric force. Let's start by writing the configuration of the charges, the charges of the same sign must be on the diagonal of the cube so that the net force is directed towards the interior of the cube, see in the attached numbering and sign of the charges

The net force is

          F_ {net} = F₂₁ + F₂₃ + F₂₄

bold letters indicate vectors. The easiest method to solve this exercise is by using the components of each force.

let's use trigonometry

          cos 45 = F₂₄ₓ / F₂₄

          sin 45 = F_{24y) / F₂₄

          F₂₄ₓ = F₂₄ cos 45

          F_{24y} = F₂₄ sin 45

let's do the sum on each axis

X axis

          Fₓ = -F₂₁ + F₂₄ₓ

          Fₓ = -F₂₁₁ + F₂₄ cos 45

Y axis  

         F_y = - F₂₃ + F_{24y}

         F_y = -F₂₃ + F₂₄ sin 45

They indicate that the magnitude of all charges is the same, therefore

         F₂₁ = F₂₃

Let's use Coulomb's law

         F₂₁ = k q₁ q₂ / r₁₂²

       

the distance between the two charges is

         r = a

         F₂₁ = k q² / a²

we calculate F₂₄

           F₂₄ = k q₂ q₄ / r₂₄²

the distance is

           r² = a² + a²

           r² = 2 a²

         

we substitute

           F₂₄ = k  q² / 2 a²

we substitute in the components of the forces

          Fx = - k \frac{q^2}{a^2} +  k \frac{q^2}{2 a^2}  \ cos 45

          Fx = k \frac{q^2}{a^2}  ( -1 + ½ cos 45)

          F_y = k \frac{q^2}{a^2} ( -1 +  ½ sin 45)    

         

We calculate

            F₀ = 9 10⁹ 4.25² / 0.440²

            F₀ = 8.40 10¹¹ N

       

            Fₓ = 8.40 10¹¹ (½ 0.707 - 1)

            Fₓ = -5.43 10¹¹ N

         

remember cos 45 = sin 45

             F_y = - 5.43 10¹¹  N

We can give the resultant force in two ways

a) F = Fₓ î + F_y ^j

          F = -5.43 10¹¹ (i + j)   N

b) In the form of module and angle.

For the module we use the Pythagorean theorem

          F = \sqrt{F_x^2 + F_y^2}

          F = 5.43 10¹¹  √2

          F = 7.68 10¹¹ N

in angle is

           θ = 45º

7 0
2 years ago
If you wanted to find the area of the hot filament in a light bulb, you would have to know the temperature (determinable from th
Sladkaya [172]

Answer:

To find out the area of the hot filament of a light bulb, you would need to know the temperature, the power input, the Stefan-Boltzmann constant and <u>Emissivity of the Filament</u>.

Explanation:

The emissive power of a light bulb can be given by the following formula:

E = σεAT⁴

where,

E = Power Input or Emissive Power

σ = Stefan-Boltzmann constant

ε = Emissivity

A = Area

T = Absolute Temperature

Therefore,

A = E/σεT⁴

So, to find out the area of the hot filament of a light bulb, you would need to know the temperature, the power input, the Stefan-Boltzmann constant and <u>Emissivity of the Filament</u>.

3 0
3 years ago
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