What number on this diagram indicates the trend line? A. 1 B. 2 C. 4 D. 7

The critical angle between a medium and air is 430. What is the speed of light in that medium?

vagabundo [1.1K]

Answer:

The speed of light is that medium is 281907786.2 m/s.

Explanation:

since the critical angle is Фc = 430, we know that the refractive index is given by:

n = 1/sin(Фc)

= 1/sin(430)

= 1.06

then if n is the refractive index of the medium and c is the speed of light, then the speed of light in the medium is given by:

v = c/n

= (3×10^8)/(1.06)

= 281907786.2 m/s

Therefore, the speed of light is that medium is 281907786.2 m/s.

8 0

3 years ago

A sailboat is traveling across the ocean with a speed of 4 m/s when large gusts of wind pick up that starts to accelerate the sa

Ivahew [28]

solution:

Using Cartesian co-ordinate system

Final velocity =v= -4m/s

Initial velocity = u

Acceleration a = m/s²

Time (t).= 60s

By the first kinematical equation

V= u +at

U = v – at

=(-4)-(-3)(60)

176m/s

So, initial velocity was 176m/s

8 0

4 years ago

Formula for calculating work done in machines

Archy [21]

Answer:

force×distance

Explanation:

i think that's the answer and it could also be expressed as m×g×d

7 0

3 years ago

A 70.0-kg person throws a 0.0430-kg snowball forward with a ground speed of 32.0 m/s. A second person, with a mass of 58.5 kg, c

Aleks04 [339]

Answer:

The velocities of the skaters are $v_{1} = 3.280\,\frac{m}{s}$ and $v_{2} = 0.024\,\frac{m}{s}$ , respectively.

Explanation:

Each skater is not under the influence of external forces during process, so that Principle of Momentum Conservation can be used on each skater:

First skater

$m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}$ (1)

Second skater

$m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}$ (2)

Where:

$m_{1}$ - Mass of the first skater, in kilograms.

$m_{2}$ - Mass of the second skater, in kilograms.

$v_{1,o}$ - Initial velocity of the first skater, in meters per second.

$v_{1}$ - Final velocity of the first skater, in meters per second.

$v_{b}$ - Launch velocity of the meter, in meters per second.

$v_{2}$ - Final velocity of the second skater, in meters per second.

If we know that $m_{1} = 70\,kg$ , $m_{b} = 0.043\,kg$ , $v_{b} = 32\,\frac{m}{s}$ , $m_{2} = 58.5\,kg$ and $v_{1,o} = 3.30\,\frac{m}{s}$ , then the velocities of the two people after the snowball is exchanged is:

By (1):

$m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}$

$m_{1}\cdot v_{1,o} - m_{b}\cdot v_{b} = m_{1}\cdot v_{1}$

$v_{1} = v_{1,o} - \left(\frac{m_{b}}{m_{1}} \right)\cdot v_{b}$

$v_{1} = 3.30\,\frac{m}{s} - \left(\frac{0.043\,kg}{70\,kg}\right)\cdot \left(32\,\frac{m}{s} \right)$

$v_{1} = 3.280\,\frac{m}{s}$

By (2):

$m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}$

$v_{2} = \frac{m_{b}\cdot v_{b}}{m_{2}+m_{b}}$

$v_{2} = \frac{(0.043\,kg)\cdot \left(32\,\frac{m}{s} \right)}{58.5\,kg + 0.043\,kg}$

$v_{2} = 0.024\,\frac{m}{s}$

5 0

3 years ago

A 1,250 W electric motor is connected to a 220 Vrms, 60 Hz source. The power factor is lagging by 0.65. To correct the pf to 0.9

Black_prince [1.1K]

Answer:

$C = 46.891 \mu F$

Explanation:

Given data:

v = 220 rms

power factor = 0.65

P = 1250 W

New power factor is 0.9 lag

we knwo that

$s = \frac{P}{P.F} < COS^{-1} 0.65$

$S = \frac{1250}{0.65} < 49.45$

s = 1923.09 < 49.65^o

s = [1250 + 1461 j] vA

$P.F new = cos [tan^{-1} \frac{Q_{new}}{P}]$

solving for $Q_{new}$

$Q_{new} = P tan [cos^{-1} P.F new]$

$Q_{new} = 1250 [tan[cos^{-1}0.9]]$

$Q_{new} = 605.40 VARS$

$Q_C = Q - Q_{new}$

$Q_C = 1461 - 605.4 = 855.6 vars$

$Q_C = \frac[v_{rms}^2}{xc} =v_{rms}^2 \omega C$

$C = \frac{Q_C}{ v_{rms}^2 \omega}$

$C = \frac{855.6}{220^2 \times 2\pi \times 60}$

$C = 4..689 \times 10^{-5}$ Faraday

$C = 46.891 \mu F$

6 0

4 years ago