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ohaa [14]
3 years ago
5

It's nighttime, and you've dropped your goggles into a 3.2-mm-deep swimming pool. If you hold a laser pointer 1.1 mm above the e

dge of the pool, you can illuminate the goggles if the laser beam enters the water 2.2 mm from the edge. How far are the goggles from the edge of the pool?
Physics
1 answer:
Savatey [412]3 years ago
5 0

Answer: 5.30m

Explanation:

depth of pool = 3.2 m

i = 67.75°

Using snell's law, we have,

n₁ × sin(i) = n₂ × 2 × sin(r)

n₁ = 1, n₂ =1.33, r= 44.09°

Hence,

Distance of Google from edge if pool is:

2.2 + d×tan(r) = 2.2 + (3.2 × tan(44.09°) =5.30m

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8 0
3 years ago
A gun shoots a bullet with a velocity of 500 m/s. The gun is aimed horizontally and fired from a height of 1.5 m. How far does t
MrMuchimi

The bullet travels a horizontal distance of 276.5 m

The bullet is shot forward with a horizontal velocity u_x. It takes a time <em>t</em> to fall a vertical distance <em>y</em> and at the same time travels a horizontal distance <em>x. </em>

The bullet's horizontal velocity remains constant since no force acts on the bullet in the horizontal direction.

The initial velocity of the bullet has no component in the vertical direction. As it falls through the vertical distance, it is accelerated due to the force of gravity.

Calculate the time taken for the bullet to fall through a vertical distance <em>y </em>using the equation,

y=u_yt+\frac{1}{2} gt^2

Substitute 0 m/s for u_y, 9.81 m/s²for <em>g</em> and 1.5 m for <em>y</em>.

y=u_yt+\frac{1}{2} gt^2\\ 1.5 m=(0m/s)t+\frac{1}{2} (9.81m/s^2)t^2\\ t=\sqrt{\frac{2(1.5m)}{9.81m/s^2} } =0.5530s

The horizontal distance traveled by the bullet is given by,

x=u_xt

Substitute 500 m/s for u_x and 0.5530s for t.

x=u_xt\\ =(500m/s)(0.5530s)\\ =276.5m

The bullet travels a distance of 276.5 m.


5 0
3 years ago
While walking past a construction site, a person notices a pipe sticking out of a second floor window with water pouring out. As
tia_tia [17]

Answer:

Its diameter increases as it flows down from the pipe. Assuming laminar flow for the water, then Bernoulli's equation can be applied.

P1-P2 + (rho)g(h1 - h2) + 1/2(rho)(v1² - v2²) = 0

Explanation:

P1 = P2 = atmospheric pressure so, P1 - P2 = 0

h1 is greater than h2 so h1-h2 is positive. Rearranging the equation above 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho = v2²

From the continuity equation for fluids

A1v1 = A2v2

v2 = A1v1/A2

Substituting into the equation above

(A1v1/A2)² = 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho

Making A2² the subject of the formula,

A2² = (A1v1)²× rho/(2{ (rho)g(h1-h2) + 1/2(rho)v1²}

The denominator will be greater than the numerator and as a result the diameter of the flowing stream decreases.

Thank you for reading.

4 0
3 years ago
As the diver jumps, the board bends down. What kind of the energy does the board now have?
Vlada [557]
Hi there!

The energy that is about to or can be released from an object after energy has been transferred to it is called potential energy. In this case, as the diver jumps, potential energy is stored in the board as it bends. The potential energy is then released as kinetic energy when the board springs back up and the diver actually jumps. 

Hope this helps!
8 0
3 years ago
When an electron falls from a higher energy level to a lower energy level how is the energy released?
Illusion [34]
When a valence electron absorbs energy in the form of heat or light it uses that energy to jump to an excited state (outer level)
5 0
3 years ago
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