Based on the path of a light ray through air, water, and glass, which medium has the greatest index of refraction?

3 A 100 g steel ball falls from a height of 1.8 m on to a metal plate and rebounds to a height of 1.25 m.

BigorU [14]

Given values:

Mass of the steel ball, m = 100 g = 0.1 kg

Height of the steel ball, h1 = 1.8 m

Rebound height, h2 = 1.25 m

a. PE= mgh

0.1 x 9.8 x 1.8 =

1.764 Joules

b. KE = PE ->

1.764 Joules

c. KE= 1/2 mv square

so v = square root 2ke/m

square root 2 x 1.764/ 0.1

= 5.93 m/s

d. KE=PE=mgh square

0.1 x 9.8 x 1.21 =

1.186 joules

velocity of rebond is square root 2x 1.186/ 0.1 = 4.87 m/s

6 0

3 years ago

Read 2 more answers

The focal length of a concave lens 13.3 cm.Calculate its power.

cricket20 [7]

Explanation:

Formula for finding power : 1/f

Substitution of values:

1 /13.3 = 0.07518 cm

Please mark it as Brainliest!

4 0

3 years ago

While on stage introducing the iPhone, Steve jobs walked 3 meters to the right, 9 meters to the left, 7 meters to the right, and

Y_Kistochka [10]

Answer:

Hey

The distance is easy, just add 3+9+7+9.

Answer <em>28m</em>

Distance is not to be confused with displacement (the total distance from your starting point).

<em>Wbob1314</em>

7 0

3 years ago

The critical angle in air for a particular type of glass is 39.0°. What is the speed of light in this class glass? (c = 3.00 × 1

hodyreva [135]

Answer:

speed of light in glass ≈ 1.89* $10^{8}m/s$

Explanation:

refractive index (n) = $\frac{1}{sinC}$ .......................equation 1

refractive index (n) = velocity of light in air/ velocity of light in medium ....equ2

equate equation 1 and two;

$\frac{1}{sinC} = \frac{c}{speed of light in glass}$

speed of light in glass = c*sinC

= $3*10^{8}*sin39$ °

speed of light in glass = 188,796,117.314 m/s

speed of light in glass ≈ 1.89* $10^{8}m/s$

7 0

3 years ago

Hi If the coefficient of kinetic friction between the 5.0 kg mass and the table is 0.305, what is the tension in the string?

statuscvo [17]

Answer:

18 N

Explanation:

Draw a free body diagram for each block.

There are four forces acting on block I:

Weight force Mg pulling down

Normal force N pushing up

Tension force T pulling right

Friction force Nμ

There are two forces acting on block II:

Weight force mg pulling down

Tension force T pulling up

Sum of forces on block I in the +y direction:

∑F = ma

N − Mg = 0

N = Mg

Sum of forces on block I in the +x direction:

∑F = ma

T − Nμ = Ma

T − Mgμ = Ma

Sum of forces on block II in the -y direction:

∑F = ma

mg − T = ma

Solve for a in the first equation, then substitute into the second.

a = (T − Mgμ) / M

mg − T = m (T − Mgμ) / M

mMg − MT = mT − mMgμ

mMg + mMgμ = mT + MT

mMg (1 + μ) = (m + M) T

T = mMg (1 + μ) / (m + M)

T = (2) (5) (9.8) (1 + 0.305) / (2 + 5)

T = 18.27

Rounding to two significant figures, the tension is 18 N.

4 0

3 years ago