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3 years ago

Which best describes what happens when light traveling through air enters water at an angle?

Physics

2 answers:

vovangra [49]3 years ago

3 0

I think it is "A'' 'It moves along straight lines in air and changes direction when it enters water.

galben [10]3 years ago

3 0

The correct choice is

A. It moves along straight lines in air and changes direction when it enters water.

This is because of phenomenon of refraction which a ray of light experience when it moves from one medium to another medium. The light ray bends away from the normal to the surface if it travels from denser to rarer medium. The light ray bends towards from the normal to the surface if it travels from rarer to denser medium.

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A car is traveling at a speed of 20m/s and has a mass of 1200kg how much kinetic energy does it have

Oxana [17]

Answer:

Explanation:

The formula for Kinetic Energy is

$KE=\frac{1}{2}mv^2$ . Filling in:

$KE=\frac{1}{2}(1200)(20)^2$ It looks like we only need 1 significant digit here but I'll give you 2 and you can round how you want.

KE = 2.4 × 10⁵ J

4 0

3 years ago

Suppose Tom Harmon17 is standing at the exact center of the Ohio State football field18 on the 50 yard line. The field is 300 fe

nignag [31]

Answer:

a) x = 40 t , y = 39 t , z = 6 + 32 t - 16 t ², b) x = 80 feet , y = 78 feet , the ball came into the field

Explanation:

a) This is a projectile launch exercise, where in the x and y axes there is no acceleration and in the z axis the acceleration of the acceleration of gravity, let's write the equations of motion in each axis

Since the cast is in the center of the field, let's place the coordinate system

x₀ = 0

y₀ = 0

z₀ = 6 feet

x-axis (towards end zone, GOAL zone)

x = xo + v₀ₓ t

x = 40 t

y-axis (field width)

y = y₀ + $v_{oy}$ t

y = 39 t

z axis (vertical)

z = z₀ + v_{oz} t - ½ g t²

z = 6 + 32 t - ½ 32 t²

z = 6 + 32 t - 16 t ²

b) The player catches the ball at the same height as it came out, so we can find the time it takes to arrive

z = 6

6 = 6 + 32 t - 16 t²

(t - 2)t = 0

t=0 s

t= 2 s

The ball position

x = 40 2

x = 80 feet

y = 39 2

y = 78 feet

the dimensions of the field from the coordinate system (center of the field) are

x_total = 150 feet

y _total = 80 feet

so we can see that the ball came into the field

6 0

3 years ago

two masses are kept 2 metre apart there is gravitational force of 2 Newton what is the gravitational force when they are kept at

sukhopar [10]

Answer: 0.5N

Explanation:

Gravitational force is calculated using the formula :

F = Gm1m2/r^2

Where G is the gravitational constant (6.67 × 10^-11)

At a distance 'r' of 2metres apart:

Mass of objects are m1 and m2

Gravitational force 'F1' = 2N

Inputting values into the formula :

2 = Gm1m2 / 2^2 - - - - - (1)

At a distance 'r' of 4meters apart:

Mass of objects are m1 and m2

Gravitational force 'F2' = y

Inputting values

F2 = Gm1m2 / 4^2 - - - - - (2)

Dividing equations 1 and 2

2 = Gm1m2 / 2^2 ÷ F2 = Gm1m2 / 4^2

2 / F2 = (Gm1m2 / 4) / (Gm1m2 / 16)

2 / F2 = (Gm1m2 / 4) × (16 / Gm1m2)

2/F2 = 16 / 4

Cross multiply

2 × 4 = 16 × F2

8 = 16F2

F2 = 8/16

F2 = 0.5N

7 0

3 years ago

A 900 kg car is traveling at 20 m/s along the road. What force must be applied to the car to stop it in a distance of 30 m2 Assu

serg [7]

Answer:

A. 6000 N

Explanation:

v²=u²+2as

0²=20²+2x30xa

-400=60a

a=-400/60

a =-6.667m/s²

f =ma

f = 900 x 6.667 = 6003N

F = 6000N

5 0

3 years ago

66. Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, and (d) 2.00 s for a ball thrown st

kozerog [31]

Answer:

a) t=1s

y = 10.1m

v=5.2m/s

b) t=1.5s

y =11.475 m

v=0.3m/s

c) t=2s

y =10.4 m

v=-4.6m/s (The minus sign (-) indicates that the ball is already going down)

Explanation:

Conceptual analysis

We apply the free fall formula for position (y) and speed (v) at any time (t).

As gravity opposes movement the sign in the equations is negative.:

y = vi*t - ½ g*t2 Equation 1

v=vit-g*t Equation 2

y: The vertical distance the ball moves at time t

vi: Initial speed

g= acceleration due to gravity

v= Speed the ball moves at time t

Known information

We know the following data:

Vi=15 m / s

$g =9.8 \frac{m}{s^{2} }$

t=1s ,1.5s,2s

Development of problem

We replace t in the equations (1) and (2)

a) t=1s

$y = 15*1 - ½ 9.8*1^{2}$ =15-4.9=10.1m

v=15-9.8*1 =15-9.8 =5.2m/s

b) t=1.5s

$y = 15*1.5 - ½ 9.8*1.5^{2}$ =22.5-11.025=11.475 m

v=15-9.8*1.5 =15-14.7=0.3m/s

c) t=2s

$y = 15*2 - ½ 9.8*2^{2}$ = 30-19.6=10.4 m

v=15-9.8*2 =15-19.6=-4.6m/s (The minus sign (-) indicates that the ball is already going down)

3 0

3 years ago