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valentinak56 [21]
2 years ago
8

An early method of measuring the speed of light makes use of a rotating slotted wheel. A beam of light passes through one of the

slots at the outside edge of the wheel, travels to a distant mirror, and returns to the wheel just in time to pass through the next slot in the wheel. One such slotted wheel has a radius of 6.0 cm and 400 slots at its edge. Measurements taken when the mirror was L = 500 m from the wheel indicated a speed of light of 3.0 ✕ 105 km/s. (a) What was the (constant) angular speed of the wheel?
Physics
1 answer:
Readme [11.4K]2 years ago
5 0

Answer:

4.7 x 10³ rad / s

Explanation:

During the time light goes and comes back , one slot is replaced by next slot while rotating  before the light source

Time taken by light to travel a distance of 2 x 500 m is

= (2 x 500) / 3 x 10⁸

= 3.333 x 10⁻⁶ s .

In this time period, two consecutive slots come before the source of light one after another by rotation. There are 400 slots so time taken to make one rotation

= 3.333 x 10⁻⁶ x 400

= 13.33 x 10⁻⁴ s

This is the time period so

T = 13.33 X 10⁻⁴

Angular speed

= 2π / T

= \frac{2\times3.14}{13.33\times10^{-4}}

4.7 x 10³ rad / s

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At a distance of 11 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per seco
Alborosie

To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.

The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,

A_1*m=M*A_2

A_i =Area

M,m = Counts per second

Our radios are given by

r_1 = 11cm

R_2 = 20cm

m = 65cps

Therefore replacing we have that,

A_1*m=M*A_2

4\pi r_1^2*m = M * 4\pi R_2^2 M

r^2*m=MR^2

M = \frac{m*r^2}{R^2}

M = \frac{65*11^2}{20^2}

M = 19.6625cps

Therefore the number of counts expect at a distance of 20 cm is 19.66cps

7 0
3 years ago
Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe
AlexFokin [52]

Answer:

0.56 atm

Explanation:

First of all, we need to find the number of moles of the gas.

We know that

m = 1.00 g is the mass of the gas

Mm=44.0 g/mol is the molar mass of the carbon dioxide

So, the number of moles of the gas is

n=\frac{m}{M_m}=\frac{1.00 g}{44.0 g/mol}=0.023 mol

Now we can find the pressure of the gas by using the ideal gas equation:

pV=nRT

where

p is the pressure

V=1.00 L = 0.001 m^3 is the volume

n = 0.023 mol is the number of moles

R=8.314 J/mol K is the gas constant

T=25.0^{\circ}+273=298 K is the temperature of the gas

Solving the equation for p, we find

p=\frac{nRT}{V}=\frac{(0.023 mol)(8.314 J/mol K)(298 K)}{0.001 m^3}=5.7 \cdot 10^4 Pa

And since we have

1 atm = 1.01\cdot 10^5 Pa

the pressure in atmospheres is

p=\frac{5.7\cdot 10^4 Pa}{1.01\cdot 10^5 Pa/atm}=0.56 atm

5 0
3 years ago
An arrow is moving at 35 m/s and travels for 5 seconds. how far did the arrow travel?
pashok25 [27]
The answer is 175 meters
8 0
3 years ago
1. A car is stopped at a red light. When the light turns green, the car starts
Lubov Fominskaja [6]

Answer:

Because of inertia (Newton's First Law of motion)

Explanation:

According to Newton's First Law of motion:

"An object at rest (or in motion at constant velocity) will tend to stay at rest (or tend to keep moving with same velocity) unless acted upon an unbalanced force"

In this problem, the object we are analyzing is the coffee cup.

At the beginning, the cup is at rest, together with the car.

Later, the car starts moving when the light turns green.

If we apply Newton's First Law of motion to the cup, we see that the coffee cup tends to keep its state of rest: for this reason, as the car moves forward, the coffee in the cup will spill backward, into the rear seat. This property of an object to mantain its state of motion is also called as inertia.

8 0
3 years ago
An object attached to one end of a spring makes 20 vibrations in 10 seconds. Its angular frequency is: 0. 79 rad/s 1. 57 rad/s 2
morpeh [17]

Angular frequency in radian per second for 20 vibrations in 10 seconds is 12.6 rad/s

<h3>What is Angular frequency?</h3>

Angular frequency is the number of vibrations in radian per second.

The total number of vibrations n is 20 and the time taken for these vibrations is 10 s

The frequency of the vibrations will be

f = 20 / 10 = 2 Hz

Angular frequency is related to the frequency as

ω = 2πf

ω=2π × 2

ω = 12.6 rad/s

Thus, the angular frequency is 12.6 rad/s.

Learn more about Angular frequency.

brainly.com/question/14244057

#SPJ

5 0
2 years ago
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