Ruth is 1.23m tall lee is 6 cm shorter than ruth. how tall is lee

A ray of light (f = 5.09 x 1014 Hz) is incident on the boundary between air and an unknown material X at an angle of incidence o

11Alexandr11 [23.1K]

Answer:

A) Material X is flint glass

B) Speed of light in the X material is 1.807 x $10^{8}$ m/s

C) Angle of refraction of light in medium X is $29.57^{o}$

Explanation:

Given data:

frequency of the light f = 5.09 x $10^{14}$ Hz

angle of incidence Θ $_{i}$ = $55^{o}$

index of refraction of material $n_{2}$ = 1.66

A) To find material X

Given the index of refraction is 1.66 and hence the material is the flint glass

B) To calculate the speed of light in the material.

We know that the relation between index of refraction (n), velocity of light (c = 3 x $10^{8}$ m/s) and velocity of light is given by the equation:

n = c/v

Hence,

Speed of light in the X material v = c/n

= $\frac{3 x 10^{8} }{1.66}$

= 1.807 x $10^{8}$ m/s

C) To calculate angle of refraction of light in medium X

We know that the Snell's law states that

$n_{1} sin$ Θ $_{i}$ = $n_{2} sin$ Θ $_{r}$

$n_{1}$ = incident index

$n_{2}$ = refracted index

Θ $_{i}$ = incident angle

Θ $_{r}$ = refracted angle

In given problem, $n_{1}$ = 1 since medium is air

Substituting the known values, we get

1 x $sin$ $55^{o}$ = 1.66 x $sin$ Θ $_{r}$

$sin$ Θ $_{r}$ = $sin$ $55^{o}$ /1.66

= 0.4935

Hence, angle of refraction of light in medium X Θ $_{r}$ = $sin^{-1}$ (0.4935)

= $29.57^{o}$

4 0

3 years ago

A ball is being thrown straight upward and rises to a maximum height of 12.0m above its launch point. At what height above it’s

AlexFokin [52]

Answer:

9.0 m

Explanation:

Let the initial velocity be 'u'.

Given:

Final velocity is half of initial velocity.

Maximum height reached by ball (H) = 12.0 m

Acceleration of the ball is due to gravity (g) = -9.8 m/s²(Downward)

Now, first, we will find initial velocity of the ball using equation of motion given as:

$v^2=u^2+2a(\Delta y)$

For maximum height, final velocity is 0 as the ball stops at the maximum height temporarily. So, $v=0\ m/s$

Also, $\Delta y=H=12\ m$

Now, plug in all the values and solve for 'u'.

$0^2=u^2+2(-9.8)(12)\\\\u^2=235.2\\\\u=\sqrt {235.2} =15.34\ m/s$

Now, consider the motion of the ball till the velocity reaches half of initial velocity.

So, final velocity (v) = $\frac{u}{2}=\frac{15.34}{2}=7.67\ m/s$

Now, again using the same equation and finding the new height now. Let the new height be 'h'.

So, equation of motion is given as:

$v^2=u^2+2ah\\7.67^2=15.34^2+2\times -9.8\times h\\58.83=235.2-19.6h\\\\19.6h=235.2-58.83\\\\19.6h=176.37\\\\h=\frac{176.37}{19.6}\approx9.0\ m$

Therefore, the height reached by the ball when velocity is decreased to one-half of the initial velocity is 9.0 m.

8 0

3 years ago

What is the temperature increase of 4.0 kg of water when heated by an 800-W immersion heater for 10 min? (cw = 4 186 J/kg⋅°C)

Llana [10]

Answer:

option C

Explanation:

given,

mass of water = 4 Kg

Water is heated to = 800 W

time of immersion = 10 min

= 10 x 60 = 600 s

using equation of specific heat

Q = m S ΔT

S is the specific heat capacity of water which is equal to 4182 J/kg°C.

and another formula of heat

Q = Pt

now,

P t = m S ΔT

800 x 600 = 4 x 4182 x ΔT

ΔT = 29° C

temperature increased is equal to ΔT = 29° C

Hence, the correct answer is option C

3 0

3 years ago

By how many newtons does the weight of a 85.9-kg person lose when he goes from sea level to an altitude of 6.33 km if we neglect

sergejj [24]

Answer:

$Weight\ loss=1.6321N$

Explanation:

From the question we are told that:

Weight $W=85.9kg$

Altitude $h= 6.33 km$

Let

Radius of Earth $r=6380km$

Gravity $g=9.8m/s^2$

Generally the equation for Gravity at altitude is mathematically given by

$g_s=9.8(\frac{6380}{6380+6.33})^2$

$g_s=9.781m/s^2$

Therefore

Weight at sea level

$W_s=9.8*85.9$

$W_s=841.82N$

Weight at 6.33 altitude

$W_a=9.781*85.9$

$W_a=840.2N$

Therefore

$Weight loss=W_s-W_b$

$Weight loss=841.82-840.2$

$Weight loss=1.6321N$

3 0

3 years ago

A couple of astronauts agree to rendezvous in space after hours. Their plan is to let gravity bring them together. She has a mas

Oduvanchick [21]

Answer:

(a) Acceleration of female astronaut = 9.33*10^-12 m/s^2; Acceleration of male astronaut = 7.56*10^-12 m/s^2

(b) 27.013 days

(c) No, their accelerations would not be constant.

Explanation:

In the given question, we have:

Mass of female astronaut $M_{1}$ = 60.0 kg

Mass of male astronaut $M_{2}$ = 74.0 kg

Distance between them (S) = 23.0 m

(a) The free-body diagram is shown in the attached figure.

Using the equations below, the initial accelerations of the two astronauts can be calculated:

Force of gravity (F) = $\frac{G*M_{1}*M_{2}}{S^{2} }$

G = 6.67*10^-11 $\frac{m^{3} }{kg*s^{2} }$

F = (6.67*10^-11 *60*74)/23^2 = 5.598*10^-10 N

For the female astronaut, her initial acceleration = F/ $M_{1}$ = 5.598*10^-10/60 = 9.33*10^-12 m/s^2

For the male astronaut, his acceleration = F/ $M_{2}$ = 5.598*10^-10/74 = 7.56*10^-12 m/s^2

(b) Since the different between their mass is not much, we can deduce that:

$a_{average} = \frac{a_{1}+a_{2}}{2}$ = (9.33*10^-12 + 7.56*10^-12)/2 = 8.445*10^-12 m/s^2

Using the equation below, we can calculate the the time:

S = ut + 1/2 (at^2) where u = 0

23 = 1/2 (8.445*10^-12)*t^2

t^2 = 5.447*10^12

t = 2333883.044 s = 27.013 days

(c) No, their accelerations will not be constant. It will increase because their radii would be decreasing.

8 0

3 years ago