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Express the following speeds as a function of the speed of light, c: (a) an automobile speed (93 km/h) (b) the speed of sound (329 m/s) (c) the escape velocity of a rocket from the Earth's surface (12.1 km/s) (d) the orbital speed of the Earth about the Sun (Sun-Earth distance 1.5×108 km).

Answers

Answer 1

Answer:

Answer:

(a). An automobile speed as a function of speed of light is $8.61*10^(-8)\ m/s$

(b). The speed of sound as a function of speed of light is $10.9*10^(-8)\ m/s$

(c). The escape velocity of a rocket as a function of speed of light is $4.03*10^(-8)\ m/s$

(d). The orbital speed of the Earth about the Sun is $3*10^(8)\ m/s$

Explanation:

Given that,

Express the following speeds as a function of the speed of light.

Automobile speed = 93 km/h

We know that,

A function of speed of light is

$c=3*10^(8)\ m/s$

(a). Automobile speed = 93 km/h

Speed $v_(a)=93*(5)/(18)$

$v_(a)=25.83\ m/s$

We need to express the speed of automobile speed as a function of speed of light

Using formula of speed

$v=(v_(a))/(v_(l))$

Put the value into the formula

$v=(25.83)/(3*10^(8))$

$v=8.61*10^(-8)\ m/s$

(b). The speed of sound is 329 m/s.

We need to express the speed of sound as a function of speed of light

Using formula of speed

$v=(v_(s))/(v_(l))$

Put the value into the formula

$v=(329)/(3*10^(8))$

$v=10.9*10^(-8)\ m/s$

(c). The escape velocity of a rocket from the Earth's surface is 12.1 m/s

We need to express the speed of sound as a function of speed of light

Using formula of speed

$v=(v_(e))/(v_(l))$

Put the value into the formula

$v=(12.1)/(3*10^(8))$

$v=4.03*10^(-8)\ m/s$

(d). The orbital speed of the Earth about the Sun

Distance = 1.5\times10^{8}[/tex]

We know that,

The sun rays reached on the earth in 8 min 20 sec.

We need to express the speed of sound as a function of speed of light

Using formula of speed

$v=(d)/(t)$

Put the value into the formula

$v=(1.5*10^(8)*1000)/(500)$

$v=3*10^(8)\ m/s$

Hence, (a). An automobile speed as a function of speed of light is $8.61*10^(-8)\ m/s$

(b). The speed of sound as a function of speed of light is $10.9*10^(-8)\ m/s$

(c). The escape velocity of a rocket as a function of speed of light is $4.03*10^(-8)\ m/s$

(d). The orbital speed of the Earth about the Sun is $3*10^(8)\ m/s$

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Two particles, one with charge -6.29 × 10^-6 C and one with charge 5.23 × 10^-6 C, are 0.0359 meters apart. What is the magnitude of the force that one particle exerts on the other?

Answers

Answer:

Force, F = −229.72 N

Explanation:

Given that,

First charge particle, $q_1=-6.29* 10^(-6)\ C$

Second charged particle, $q_2=5.23* 10^(-6)\ C$

Distance between charges, d = 0.0359 m

The electric force between the two charged particles is given by :

$F=k(q_1q_2)/(d^2)$

$F=9* 10^9* (-6.29* 10^(-6)* 5.23* 10^(-6))/((0.0359)^2)$

F = −229.72 N

So, the magnitude of force that one particle exerts on the other is 229.72 N. Hence, this is the required solution.

A uniform electric field, with a magnitude of 370 N/C, is directed parallel to the positive x-axis. If the electric potential at x = 2.00 m is 1 000 V, what is the change in potential energy of a particle with a charge of + 2.80 x 10-3 C as it moves from x = 1.9 m to x = 2.1 m?

Answers

Answer:

$\Delta U = 0.2072 J$

Explanation:

Potential difference between two points in constant electric field is given by the formula

$\Delta V = E.\Delta x$

here we know that

$E = 370 N/C$

also we know that

$\Delta x = 2.1 - 1.9 = 0.2 m$

now we have

$\Delta V = 370 (0.2) = 74 V$

now change in potential energy is given as

$\Delta U = Q\Delta V$

$\Delta U = (2.80 * 10^(-3))(74)$

$\Delta U = 0.2072 J$

How long would it take a 500. W electric motor to do 15010 J of work?

Answers

time = energy / power = 15010 / 500 = .... seconds

URGENT!!!!!! Assume that a wire has 1.5 ohms of resistance. If the wire is connected to two batteries with a total voltage of 3.0 V, how much current will flow through the wire? 3.0 amps 2.3 amps 2.0 amps 1.0 amps

Answers

Answer:

2.0 amps

Explanation:

Current is the ratio of voltage to resistance:

I = V/R = (3.0)/(1.5) = 2.0

The current in the wire is 2.0 amps.

Complete the calculations for total magnification produced by various combinations of the eyepiece and objective lenses. You may assume that the magnification for the eyepiece is 10X for each question. 1. When the scanning (4X) objective is used the total magnification will be:________
2. When the low power (10X) objective is used the total magnification will be:________
3. When the high power (40X) objective is used the total magnification will be:________
4. When the oil immersion (100X) objective is used the total magnification will be:_________

Answers

Answer:

a) m_ttoal = 40x, b) m_total = 100X, c) m_total = 400X,

d) m_total = 1000 X

Explanation:

La magnificación o aumentos es el incremento de del tamaño de la imagen con respecto al tamaño original del objeto, en la mayoria del os sistema optico la magnificacion total es el producoto de la magnificación del objetivo por la magnificación del ocular

m_total = m_ objetivo * m=ocular

apliquemos esto a nuestro caso

1) m_total = 4 x * 10 x

m_ttoal = 40x

2) m_total = 10X * 10X

m_total = 100X

3)mtotal = 40X * 10X

m_total = 400X

4) m _totla = 100x * 10 X

m_total = 1000 X

en este ultimo caso para magnificación grandes es decalcificar el objeto

The total magnification produced by different combinations of eyepiece and objective lenses in a microscope.

1. When the scanning (4X) objective is used, the total magnification will be 40X because the eyepiece magnification is 10X and the objective magnification is 4X.

2. When the low power (10X) objective is used, the total magnification will be 100X because the eyepiece magnification is 10X and the objective magnification is 10X.

3. When the high power (40X) objective is used, the total magnification will be 400X because the eyepiece magnification is 10X and the objective magnification is 40X.

4. When the oil immersion (100X) objective is used, the total magnification will be 1000X because the eyepiece magnification is 10X and the objective magnification is 100X.

Learn more about Microscope magnifications here:

brainly.com/question/36268796

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The time taken by a mass projected verticallyupwards to reach the maximum height (with air
resistance not neglected) is 10 sec. The time of
descent of the mass from the same height will be

Answers

Answer:

10s

Explanation:

The time to get to the maximum would be the same as the time to get down to the maximum unless somehow gravity’s changes during the duration it goes up to and from maximum height.

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