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3 years ago

B. Direction: In each of the following situations, identify the method of heat transfer that takes place

Physics

1 answer:

Ainat [17]3 years ago

4 0

Answer:

1) Conduction

2)Covection

3)Radiation

4)Convection (Land breeze one of the application of convection of heat)

5) Convection

6)Radiation

7) Radiation

8)Conduction

9) Conduction

10) Radiation

Hope it helps

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The sound intensity at a distance of 16 m from a noisy generator is measured to be 0.25 W/m2. What is the sound intensity at a d

Margarita [4]

Answer:

0.1111 W/m²

Explanation:

If all other parameters are constant, sound intensity is inversely proportional to the square of the distance of the sound. That is,

I ∝ (1/r²)

I = k/r²

Since k can be the constant of proportionality. k = Ir²

We can write this relation as

I₁ × r₁² = I₂ × r₂²

I₁ = 0.25 W/m²

r₁ = 16 m

I₂ = ?

r₂ = 24 m

0.25 × 16² = I₂ × 24²

I₂ = (0.25 × 16²)/24²

I₂ = 0.1111 W/m²

4 0

3 years ago

What is the time period of vibration?

Y_Kistochka [10]

Answer:

A time period is denoted by 'T' . It is the time to complete one cycle of vibration. As the frequency of a wave increases, the time period of the wave decreases. The unit for time period is 'seconds

HOPE IT HELPS :)

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4 0

3 years ago

Three persons wants to push a wheel cart in the direction marked x in Fig. The two person push with horizontal forces F1 and F2

Svetllana [295]

Answer:

I had to search the Figure on Google to solve this question.

a) The magnitude of the force F₃ is:

$F_{3} = 87.47 N$

And the direction of F₃:

$\alpha = 79.04 ^{\circ}$ (with respect to the y-direction, in the third quadrant)

b) P = 4.22 N

Explanation:

I had to search the Figure on Google to solve this question.

a) We can find the force of the third person as follows:

$\Sigma F_{x} = F_{1x} + F_{2x} + F_{3x} = 0$

$\Sigma F_{y} = F_{1y} - F_{2y} + F_{3y} = 0$

So, in x-direction we have:

$\Sigma F_{x} = 45 N*cos(70) + 75 N*cos(20) + F_{3x} = 0$

$F_{3x} = -85.87 N$

In y-direction we have:

$\Sigma F_{y} = 45 N*sin(70) - 75 N*sin(20) + F_{3y} = 0$

$F_{3y} = -16.63 N$

The magnitude of the force F₃ is:

$F_{3} = \sqrt{F_{3x}^{2} + F_{3y}^{2}} = \sqrt{(-85.87 N)^{2} + (-16.63 N)^{2}} = 87.47 N$

To find the direction of F₃ we need to calculate its angle with respect to the y-direction (in the third quadrant):

$tan(\alpha) = \frac{|F_{3x}|}{|F_{3y}|} = \frac{85.87 N}{16.63 N}$

$\alpha = 79.04 ^{\circ}$

b) If the third person exerts the force found in part (a) the car will stop, so the only way for the cart to accelerate at 200 m/s² is that the third person does not exert the force found in a.

To find the weight of the cart when it accelerates at 200 m/s², we need to consider: F₃ = 0.

First, we need to find the cart's mass. Since the car is moving in the x-direction we have:

$\Sigma F_{x} = F_{1x} + F_{2x} = ma$