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3 years ago

Which of the following values represents an index of refraction of an actual material?

1 answer:

3 0

Sadly, we're forced to answer the question without the benefit of the
list of choices, which, for some reason, you decided not to let us see.

Index of refraction of a substance =

(speed of light in vacuum) / (speed of light in the substance).

Any number greater than ' 1 ' can be an index of refraction. A number
less than ' 1 ' can't be . . . that would be saying that the speed of light
in this substance is greater than the speed of light in vacuum.

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One of your summer lunar space camp activities is to launch a 1090 kg rocket from the surface of the Moon. You are a serious spa

Ludmilka [50]

Answer:

ΔP.E = 6.48 x 10⁸ J

Explanation:

First we need to calculate the acceleration due to gravity on the surface of moon:

g = GM/R²

where,

g = acceleration due to gravity on the surface of moon = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of moon = 7.36 x 10²² kg

R = Radius of Moon = 1740 km = 1.74 x 10⁶ m

Therefore,

g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²

g = 2.82 m/s²

now the change in gravitational potential energy of rocket is calculated by:

ΔP.E = mgΔh

where,

ΔP.E = Change in Gravitational Potential Energy = ?

m = mass of rocket = 1090 kg

Δh = altitude = 211 km = 2.11 x 10⁵ m

Therefore,

ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)

ΔP.E = 6.48 x 10⁸ J

7 0

3 years ago

Where does energy originate ina food web

ValentinkaMS [17]

$\huge{\underline{\underline{\red{Answer}}}}$

In a food web the energy is originated from the SUN.

4 0

2 years ago

Read 2 more answers

A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car

kogti [31]

Answer:

$t_1 = \frac{v_i}{a_i}$

$t_2 = \frac{v_i}{a_i}$

Δd = $v_it_1 = v_i^2/a_i$

Explanation:

As $v(t) = v_i + at$ , when the car is making full stop, $v(t_1) = 0$ . $a = -a_i$ . Therefore,

$0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}$

Apply the same formula above, with $v(t_2) = v_i$ and $a = a_i$ , and the car is starting from 0 speed, we have

$v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}$

As $s(t) = vt + \frac{at^2}{2}$ . After $t = t_1 + t_2$ , the car would have traveled a distance of

$s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}$

Hence $s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}$

As $t_1 = t_2$ we can simplify $s(t) = v_it_1$

After t time, the train would have traveled a distance of $s(t) = v_i(t_1 + t_2) = 2v_it_1$

Therefore, Δd would be $2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i$

8 0

2 years ago

Eric has a mass of 19.0 kg on the earth. What is Eric's weight on earth? What is Eric's weight on Mars? where the acceleration o

Tom [10]

Answer:

Weight on Earth = We = 186.2 N

Weight on Mars = Wm = 70.94 N

Explanation:

The weight of an object is defined as the force applied on the object by the gravitational field. The magnitude of weight is given by the following formula:

W = mg

were,

W= Weight of Eric

m = mass of Eric

g = acceleration due to gravity

ON EARTH:

W = We = Eric's Weight on Earth = ?

m = Eric's Mass on Earth = 19 kg

ge = acceleration due to gravity on Earth = 9.8 m/s²

Therefore,

We = (19 kg)(9.8 m/s²)

We = 186.2 N

ON MARS:

W = Wm = Eric's Weight on Mars = ?

m = Eric's Mass on Mars = 19 kg

gm = acceleration due to gravity on Mars = 0.381(ge) = (0.381)9.8 m/s² = 3.733 m/s²

Therefore,

Wm = (19 kg)(3.733 m/s²)

Wm = 70.94 N

3 0

2 years ago

An LR circuit contains an ideal 60-V battery, a 42-H inductor having no resistance, a 24-ΩΩ resistor, and a switch S, all in ser

s2008m [1.1K]

Answer:

1.6 s

Explanation:

To find the time in which the potential difference of the inductor reaches 24V you use the following formula:

$V_L=V_oe^{-\frac{Rt}{L}}$

V_o: initial voltage = 60V

R: resistance = 24-Ω

L: inductance = 42H

V_L: final voltage = 24 V

You first use properties of the logarithms to get time t, next, replace the values of the parameter:

$\frac{V_L}{V_o}=e^{-\frac{Rt}{L}}\\\\ln(\frac{V_L}{V_o})=-\frac{Rt}{L}\\\\t=-\frac{L}{R}ln(\frac{V_L}{V_o})\\\\t=-\frac{42H}{24\Omega}ln(\frac{24V}{60V})=1.6s$

hence, after 1.6s the inductor will have a potential difference of 24V

3 0

2 years ago