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AVprozaik [17]
3 years ago
15

The sun is 60° above the horizon. Rays from the sun strike the still surface of a pond and cast a shadow of a stick that is stuc

k in the sandy bottom of the pond. If the stick is 19 cm tall, how long is the shadow?

Physics
1 answer:
Kipish [7]3 years ago
5 0

Answer:

shadow length 7.67 cm

Explanation:

given data:

refractive index of water is 1.33

by snell's law we have

n_{air} sin30 =n_{water} sin\theta

1*0.5 = 1.33*sin\theta

solving for\theta

sin\theta = \frac{3}{8}

\theta = sin^{-1}\frac{3}{8}

\theta =  22 degree

from shadow- stick traingle

tan(90-\theta) = cot\theta = \frac{h}{s}

s = \frac{h}{cot\theta} = h tan\theta

s = 19tan22 = 7.67 cm

s = shadow length

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A 52 kg and a 95 kg skydiver jump from an airplane at an altitude of 4750 m, both falling in the pike position. Assume all value
Scilla [17]

Answer: 52 kg skydiver: 9.09 m/s and 522.55 s

              95 kg skydiver: 12.3 m/s and 386.2 s

Explanation: <u>Drag</u> <u>Force</u> is an opposite force when an object is moving in a fluid.

For skydivers, when falling through the air, the forces acting on it are gravitational and drag forces. At a certain point, drag force equals gravitational force, which is constant on any part of the planet, producing a net force that is zero. Since there is no net force, there is no acceleration and, consequently, velocity is constant. When that happens, the person reached the <u>Terminal</u> <u>Velocity</u>.

Drag Force and Velocity are proportional to the squared speed. So, terminal velocity is given by:

F_{G}=F_{D}

mg=\frac{1}{2}C \rho Av_{T}^{2}

v_{T}=\sqrt{\frac{2mg}{\rho CA} }

where

m is mass in kg

g is acceleration due to gravitational force in m/s²

ρ is density of the fluid in kg/m³

C is drag coefficient

A is area of the object in the fluid in m²

Calculating:

The 52kg skydiver has terminal velocity of:

v_{T}=\sqrt{\frac{2(52)(9.8)}{(1.21)(0.7)(0.14)} }

v_{T}= 9.09

The 95kg skydiver's terminal velocity is

v_{T}=\sqrt{\frac{2(95)(9.8)}{(1.21)(0.7)(0.14)} }

v_{T}= 12.3

The 52 kg and 95kg skydivers' terminal velocity are 9.09m/s and 12.3m/s, respectively.

The time each one will reach the floor will be:

52 kg at 9.09 m/s:

t=\frac{4750}{9.09}

t = 522.5

95 kg at 12.3 m/s:

t=\frac{4750}{12.3}

t = 386.2

The 52 kg and 95kg skydivers' time to reach the floor are 522.5 s and 386.2 s, respectively.

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Explanation:

6 0
2 years ago
An airplane is being pushed West by its engines with a force of 350 N as it starts to take off the runway at the Atlanta airport
Marizza181 [45]

Answer:

option C

Explanation:

given,

Force by the engine on plane in West direction = 350 N

Frictional force on the runway = 100 N in east

force exerted by the wind = 100 N in east

net force and direction = ?

consider west to be positive and east be negative.

when airplane will be moving there will be frictional as well as wind resistance will be acting in opposite direction of airplane

Net force = 350 N - 100 N - 100 N

                 = 150 N

as our answer comes out to be positive so the airplane will be moving in West

hence, the correct answer is option C

8 0
3 years ago
5. Two charged particles are separated by a distance of 12 meters. The Coulomb force between them is 20 N. What will the Coulomb
Leni [432]

Answer:

A) 80 N

Explanation:

The closer the particles get, the stronger the Coulomb force, which elongates choices C and D. The Coulomb force is inversely proportional to the distance squared. If the distance is cut in half, the force is multiplied by the reciprocal of (1/2)^2, which is 4. Multiplying it out, 20 times 4 is 80 N.

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Earths atmosphere heats up polars melt
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