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ch4aika [34]
2 years ago
8

A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of

35 m. If he completes the 200 m dash in 30.0 s and runs at constant speed throughout the race, what is the magnitude of his centripetal acceleration (in m/s2) as he runs the curved portion of the track? m/s2
Physics
1 answer:
Pie2 years ago
5 0

Answer

given,

distance = 200 m

radius of curvature = 35 m

time = 30 s

centripetal acceleration = ?

speed of the runner

v = \dfrac{d}{t}

v = \dfrac{200}{35}

v = 5.71 m/s

acceleration of the runner

a_c = \dfrac{v^2}{r}

a_c = \dfrac{5.71^2}{35}

a_c = 0.931 m/s^2

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For a two-level system, the weight of a given energy distribution can be expressed in terms of the number of systems, N, and the
Nikitich [7]

Answer:

W = N!/(n0! * n1!)

Explanation:

Let n0 = number of particles in the lowest energy state

n1 = number of particles in the excited energy state.

Using this, we can say that N = n0 + n1

From this we can then express the weight, W of the close system by finding the factorials of each particles

W = N!/(n0! * n1!)

Hence, the weight W is expressed as W = N!/(n0! * n1!)

7 0
2 years ago
The wavelength of violet light is about 425 nm (1 nanometer = 1 × 10−9 m). what are the frequency and period of the light waves?
BigorU [14]

1. Frequency: 7.06\cdot 10^{14} Hz

The frequency of a light wave is given by:

f=\frac{c}{\lambda}

where

c=3\cdot 10^{-8} m/s is the speed of light

\lambda is the wavelength of the wave

In this problem, we have light with wavelength

\lambda=425 nm=425\cdot 10^{-9} m

Substituting into the equation, we find the frequency:

f=\frac{c}{\lambda}=\frac{3\cdot 10^{-8} m/s}{425\cdot 10^{-9} m}=7.06\cdot 10^{14} Hz


2. Period: 1.42 \cdot 10^{-15}s

The period of a wave is equal to the reciprocal of the frequency:

T=\frac{1}{f}

The frequency of this light wave is 7.06\cdot 10^{14} Hz (found in the previous exercise), so the period is:

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4 0
3 years ago
Read 2 more answers
Two forces act on an object. The first force has a magnitude of 17.0 N and is oriented 48.0° counterclockwise from the x ‑axis,
Ivanshal [37]

Answer:

Fn: magnitude of the net force.

Fn=30.11N , oriented 75.3 ° clockwise from the -x axis

Explanation:

Components on the x-y axes of the 17 N force(F₁)

F₁x=17*cos48°= 11.38N

F₁y=17*sin48° = 12.63 N

Components on the x-y axes of the  the second force(F₂)

F₂x= −19.0 N

F₂y=   16.5 N

Components on the x-y axes of the net force (Fn)

Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N

Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N

Magnitude of the net force.

F_{n} =\sqrt{(F_{nx})^{2} +(F_{ny}) ^{2} }

F_{n} =\sqrt{(-7.62)^{2} +(29.13) ^{2} }

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Direction of the net force (β)

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In the attached graph we can observe the magnitude and direction of the net force

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3 years ago
What is the name for all the electromagnetic waves that exist
zheka24 [161]

The Electromagnetic spectrum.

8 0
3 years ago
Astronomers estimate that a 2
tia_tia [17]
M)³ / 6 = 4.2e9 m³ 
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Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

6 0
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