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tigry1 [53]
4 years ago
8

Bill and Julie observe the size of the opening of a laser and compare it with the size of a bright spot on the wall toward which

it points. They discover that the spot on the wall is larger than the opening of the laser. Julie says that the spot on the wall is bigger than the laser opening because some light from the laser goes in all directions while most light travels straight to the wall. She suggests the following picture to illustrate her reasoning: Bill disagrees and says that light travels in a straight line and then "splashes" like water from a water hose when it hits the wall. He suggests the following picture to illustrate his reasoning: Describe the design of an experiment that you could perform to decide if one or both models should be rejected. Be sure to make a prediction of the outcome of your experiment based on both models (i.e., 2 predicted outcomes, one for each model).

Physics
1 answer:
belka [17]4 years ago
4 0

Answer:

attached below

Explanation:

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If 61.4 cm of copper wire (diameter = 1.08 mm, resistivity = 1.69 × 10-8Ω·m) is formed into a circular loop and placed perpendic
harkovskaia [24]

Answer:

the thermal energy generated in the loop = 6.64*10^{-4}  \ W

Explanation:

Given that;

The length of the copper wire L = 0.614 m

Radius of the loop  r = \frac{L}{2 \pi}

r = \frac{0.614}{2 \pi}

r = 0.0977 m

However , the area of the loop is :

A_L = \pi r^2

A_L = \pi (0.0977)^2

A_L = 0.02999 \ m^2

Change in the magnetic field is \frac{dB}{dt}= 0.0914 \ T/s

Then the induced emf e = A_L \frac{dB}{dt}

e = 0.02999 * 0.0914

e = 2.74 × 10⁻³ V

resistivity of the copper wire \rho = 1.69* 10^{-8} Ω m

diameter of the wire = 1.08 mm

radius of the wire = 0.54 mm = 0.54 × 10⁻³  m

Thus, the resistance of the wire  R = \frac {\rho L}{\pi r^2}

R =  \frac{(1.69*10^{-8})(0.614)}{   \pi (0.54*10^{-3})^2}

R = 1.13× 10⁻² Ω

Finally,  the thermal energy generated in the loop (i.e the power) = \frac{e^2}{R}

= \frac{(2.74*10^{-3})^2}{1.13*10^{-2}}

= 6.64*10^{-4}  \ W

8 0
3 years ago
Weber theorized that we would eventually get stuck in ______, where we move from one formalized structure to the next, unable to
dsp73

Answer:

The correct answer is

d. An iron cage

Max Weber is well known for developing the theoretical concept of the iron cage

Explanation:

The theory of the iron was first presented by the sociologist Max Weber in the work  The Protestant Ethic and the Spirit of Capitalism.

Here Weber clarified that with the diminishing effect of Protestantism within our social life the capitalist system remained as well as the bureaucratic principles and social structures that came about along with came along with it, such the worldviews and values becomes major controlling factors in socil life which is tantamount to living in a strong house maade of steel

8 0
3 years ago
Determine the energy in joules of a photon whose frequency is 3.55 x10^17 hz
asambeis [7]
By using the Plancks-Einstein equation, we can find the energy;
E = hf
where h is the plancks constant = 6.63 x 10⁻³⁴
f = frequency = 3.55 x 10¹⁷hz
E = (6.63 x 10⁻³⁴) x (3.55 x 10¹⁷)
E = 2.354 x 10⁻¹⁶J
3 0
3 years ago
A ski jumper travels down a slope and leaves
Serhud [2]

Question seems to be missing. Found it on google:

a) How long is the ski jumper airborne?

b) Where does the ski jumper land on the incline?

a) 4.15 s

We start by noticing that:

- The horizontal motion of the skier is a uniform motion, with constant velocity

v_x = 28 m/s

and the distance covered along the horizontal direction in a time t is

d_x = v_x t

- The vertical motion of the skier is a uniformly accelerated motion, with initial velocity u_y = 0 and constant acceleration g=9.8 m/s^2 (where we take the downward direction as positive direction). Therefore, the vertical distance covered in a time t is

d_y = \frac{1}{2}gt^2

The time t at which the skier lands is the time at which the skier reaches the incline, whose slope is

\theta = 36^{\circ} below the horizontal

This happens when:

tan \theta = \frac{d_y}{d_x}

Substituting and solving for t, we find:

tan \theta = \frac{\frac{1}{2}gt^2}{v_x t}= \frac{gt}{2v_x}\\t = \frac{2v_x}{g}tan \theta = \frac{2(28)}{9.8} tan 36^{\circ} =4.15 s

b) 143.6 m

Here we want to find the distance covered along the slope of the incline, so we need to find the horizontal and vertical components of the displacement first:

d_x = v_x t = (28)(4.15)=116.2 m

d_y = \frac{1}{2}gt^2 = \frac{1}{2}(9.8)(4.15)^2=84.4 m

The distance covered along the slope is just the magnitude of the resultant displacement, so we can use Pythagorean's theorem:

d=\sqrt{d_x^2+d_y^2}=\sqrt{(116.2)^+(84.4)^2}=143.6 m

7 0
3 years ago
How will you determine the direction<br>of a torque? Explain.​
asambeis [7]
You use the right hand rule. With your thumb out and the rest of your fingers curved ( like a thumbs up) curve your fingers to the direction of the torque. The direction your thumbs points at is the direction of the torque
7 0
3 years ago
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