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valentinak56 [21]
3 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]3 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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2.(Ramp section) Suppose the height of the ramp is h1= 0.40m, and the foot of the ramp is horizontal, and is h2= 1.5m above the
frozen [14]

Answer:

a) the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b) the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

c) the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

 

Explanation:

Given that;

height of the ramp h1 = 0.40 m

foot of the ramp above the floor h2 = 1.50 m

assuming R = 15 mm = 0.015 m

density of steel = 7.8 g/cm³

density of aluminum =  2.7 g/cm³

a) distance that the solid steel sphere sliding down the ramp without friction;

we know that

distance = speed × time

d = vt --------let this be equ 1

according to the law of conservation of energy

mgh₁ = \frac{1}{2} mv²

v² = 2gh₁  

v = √(2gh₁)

from the second equation; s = ut +  \frac{1}{2} at²

that is; t = √(2h₂/g)

so we substitute for equations into equation 1

d = √(2gh₁) × √(2h₂/g)

d = √(2gh₁) × √(2h₂/g)

d = 2√( h₁h₂ )    

we plug in our values

d = 2√( 0.40 × 1.5 )

d = 1.55 m

Therefore, the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b)

distance that a solid steel sphere rolling down the ramp without slipping;

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{5}mR²) ω²

v = √( \frac{10}{7}gh₁  )

so we substitute √( \frac{10}{7}gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( \frac{10}{7}gh₁  ) × √(2h₂/g)  

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )  

d = 1.31 m

Therefore, the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

 

c)

distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping;

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{3}mR²) ω²

v = √( \frac{6}{5}gh₁ )

so we substitute √( \frac{6}{5}gh₁ ) for v and t = √(2h₂/g) in equation 1 again

d = vt

d = √( \frac{6}{5}gh₁ ) × √(2h₂/g)

d = 1.549√( h₁h₂ )

d = 1.549√( 0.4 × 1.5 )

d = 1.2 m

Therefore, the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) distance that a solid aluminum sphere rolling down the ramp without slipping.

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{5}mR²) ω²

v = √( \frac{10}{7}gh₁  )

so we substitute √( \frac{10}{7}gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( \frac{10}{7}gh₁  ) × √(2h₂/g)  

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )  

d = 1.31 m

Therefore, the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

8 0
3 years ago
A person drops a brick from the top of a building. The height of the building is 400 m and the mass of the brick is 2.00 kg. Wh
DaniilM [7]
-- It takes the brick 8.9 seconds to reach the ground. 

-- At the instant of the "splat", it's falling at 89 m/s.

-- The mass doesn't matter. If not for air resistance, every object
    would fall at the same rate.  The answer is the same for a feather,
    a rubber chicken, a brick, or a school bus.
5 0
3 years ago
A paleontologist estimates that when a particular rock formed, it contained 12 mg of the radioactive isotope potassium-40, which
leva [86]

Answer:

t = 2.52 billion \:years

Explanation:

As we know by radioactivity law

N = N_o e^{-\lambda t}

so here we will have

N = 3 mg

N_o = 12 mg

now we will have

3 = 12 e^{-\lambda t}

\lambda t = ln 4

now we also know that

\lambda = \frac{ln2}{1.26 \times 10^6 yrs}

t = 1.26\times 10^6\times \frac{ln4}{ln2}

t = 2.52 billion \:years

7 0
3 years ago
What is created by the flow of electric current?
grin007 [14]
C.) a magnetic field is the correct answer…
7 0
3 years ago
Read 2 more answers
Statistical time division multiplexing does not require the capacity of the circuit to be equal to the sum of the combined circu
aleksklad [387]

Answer:

The answer is True

Explanation:

Statistical Multiplexing is considered an example of communication link sharing which makes it comparable to DBA (Dynamic Bandwidth Allocation). Here, communication channels are broken down into data streams to optimize the communication process.

In Statistical Time-division Multiplexing, time slots are allocated to data streams for communication optimization. This method makes sure that no time slot or bandwidth is wasted.

Hence, the sum of combined circuits must not be equal to the capacity of the circuit to work effectively.

7 0
3 years ago
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