ANALOGY: Mars bar: planet as Milky Way: _________

The speed of an electromagnetic wave traveling in a transparent nonmagnetic substance is v = 1/ κμ0ε0 , where κ is the dielectri

vladimir1956 [14]

Answer:

$v=2.24\times 10^{8}\ m/s$

Explanation:

Given that,

The speed of an electromagnetic wave traveling in a transparent nonmagnetic substance is given by :

$v=\dfrac{1}{\sqrt{k\mu_o\epsilon_o}}$

Where

k is the dielectric constant of the substance.

v is the speed of light in water

$c=\dfrac{1}{\sqrt{1.78\times 4\pi \times 10^{-7}\times 8.85\times 10^{-12}}}$

$v=2.24\times 10^{8}\ m/s$

So, the speed of light in water is $2.24\times 10^{8}\ m/s$

7 0

3 years ago

A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi

nadezda [96]

To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

$W = F * d$

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

$d=h*N$ , for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

$V_f^2-V_i^2 = 2a\Delta X$

Where,

$V_f =$ Final velocity

$V_i =$ Initial Velocity

a = Acceleration

$\Delta X =$ Displacement

PART A) For the particular case of work we know then that,

$W = F*d$

$W = m*g*(h*N)$

$W = 50*9.8*(0.3*30)$

$W = 4.41kJ$

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

$V_f^2-V_i^2 = 2a\Delta X$

Here,

$V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow$ 3 steps in one second

$v_f = 0$

Replacing,

$V_f^2-V_i^2 = 2a\Delta X$

$0-0.9^2=2a(30*0.3)$

Re-arrange for a,

$a = -\frac{0.9^2}{2*30*0.3}$

$a = -45*10^{-3}m/s^2$

At this point we can calculate the time, which is,

$t = \frac{\Delta V}{a}$

$t = \frac{0-0.9}{-45*10^{-3}}$

$t = 20s$

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

6 0

3 years ago

Different ____ of light through two separate mediums causes the bending of waves fronts associated with light rays.

Fittoniya [83]

Different speeds of light through two separate media ... and the difference in wavelength caused by the difference ... causes the bending of waves fronts associated with light rays.

AFTER the bend, since the light rays then travel in a different direction, we may also say that the 'velocity' has changed.

8 0

2 years ago

Give a specific example of a machine,and describe how its mechanical efficiency might be calculated

natta225 [31]

A machine that is interesting will be the homework machine. The way it works is you put your homework inside a slot, and you have to write any three letter word so the machine knows what handwriting to use. In 2 minutes your homework will be complete in your writing.

3 0

2 years ago

The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experi

melomori [17]

Complete Question:

The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experiences the smallest drop in temperature, and which one experiences the largest drop? Sample A: 4.0 kg of water [c = 4186 J/(kg·C°)] Sample B: 2.0 kg of oil [c = 2700 J/(kg·C°)] Sample C: 9.0 kg of dirt [c = 1050 J/(kg·C°)]

Answer:

A. Smallest B. Largest.

Explanation:

Assuming no heat exchange except for the heat removed from any sample (which we know is the same for the three ones), and that the process is done using only conduction, we can use the equation that relates the heat lost or gained by one object, with the mass of the object and the consequent change in temperature, as follows:

Q = c*m*ΔT, where c, is a proportionality constant called specific heat, which is different for each material.

As we know that the heat removed is the same for the three samples, we can equate the right sides of the equation for each sample, as follows:

cw*mw*ΔTw = co*mo*ΔTo = cd*md*ΔTd

Replacing by the givens, we have:

4.0 kg. 4,186 J/kgºC*ΔT(ºC) = 2.0 kg*2,700 J/kgºC*ΔT(ºC) =9.0kg*1,050J/kgºC*ΔT(ºC)

As the three expressions must be equal each other, it's clear that the unknown term (the drop in temperature) must compensate the product of the mass times the specific heat.

This product is the following for the three samples:

Water: 4.0 kg*4,186 J/kgºC = 16,744 J/ºC

Oil : 2.0 kg*2,700 J/kgºC = 5,400 J/ºC

Dirt: 9.0 * 1,050 J/kgºC = 9,450 J/ºC

Clearly, we see that in order to keep the heat exchange equations equal each other, the water must suffer the smallest drop in temperature, and the oil must experience the largest one.

So, the sample A experiencies the smallest drop in temperature, and sample B does the largest one.

5 0

2 years ago

ANALOGY: Mars bar: planet as Milky Way: __________