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Molodets [167]
3 years ago
7

The radius of the earth's very nearly circular orbit around the sun is 1.5 1011 m. find the magnitude of the earth's velocity, a

ngular velocity, and centripetal acceleration as it travels around the sun. assume a year of 365 days. (a) the earth's velocity
Physics
1 answer:
kotykmax [81]3 years ago
5 0
The radius of Earth's circular trajectory is
R=1.5*10^{11} m
The time for the Earth to travel around the Sun is
t=365(days)*24(hours)*3600(seconds) =3.1536*10^7 (s)
Thus the velocity is velocity
v=2\pi R/t =2\pi *1.5*10^{11}/(3.1536*10^{7})=2.99*10^4 (m/s)
From this we can deduce the centripetal acceleration
a=v^2/R =(2.99*10^4)^2/(1.5*10^{11}) =5.96*10^{-3}  (m/s^2)
Angular velocity is
\omega =v/R =(2.99*10^4)/(1.5*10^{11})=2*10^{-7} (rad/s)

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Calculate the energy of a photon of electromagnetic radiation whose frequency is 2.94 x 10 to the 14th xbox one
DIA [1.3K]

Answer:

1.95\cdot 10^{-19} J

Explanation:

The energy of a photon is given by:

E=hf

where

h=6.63\cdot 10^{-34}Js is the Planck constant

f=2.94\cdot 10^{14} Hz is the frequency of the photon

Substituting the numbers into the equation, we find:

E=(6.63\cdot 10^{-34} Js)(2.94\cdot 10^{14} Hz)=1.95\cdot 10^{-19} J

3 0
2 years ago
The starship Enterprise approaches the planet Risa at a speed of 0.8c relative to the planet. On the way, it overtakes the inter
Alchen [17]

Answer:

<em>0.3c</em>

<em></em>

Explanation:

The speed of Enterprise relative to Risa is 0.8c

Relative speed of both ships as measured from Enterprise is 0.5c

therefore, relative speed of Astra to Enterprise is 0.8c - 0.5c = <em>0.3c</em>

<em>this is also the relative speed with which Astra approaches the planet Risa since Enterprise's speed was calculated relative to Risa.</em>

8 0
2 years ago
The highest freefall jump on record is from a height of almost 38,000 m. At this height, the acceleration of gravity is slightly
lina2011 [118]

Increasing his acceleration will impact his velocity and rate of displacement covered in that as the speed increases due to the increased rate of acceleration, the rate of air resistance also increases.  

<h3>What is air resistance?</h3>

Air resistance is a force created by air. When an item moves through the air, the force operates in the opposite direction.

When a diver descends, the force of air resistance acts to counteract the force of gravity. As the skydiver falls faster and faster, the quantity of air resistance grows until it equals the magnitude of gravity's force.

A balance of forces is achieved when the force of gravity equals the force of air resistance, and the skydiver no longer accelerates. The skydiver reaches what is known as terminal velocity.

Learn more about air resistance:

brainly.com/question/16859536
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4 0
1 year ago
A subway train is traveling at 22.2 m/s when it approaches a slower train 50m ahead traveling in the same direction at 6.94 m/s.
Amiraneli [1.4K]

Answer:

Time that they collide = 4.99s

Relative speed of the trains when they collide: The relative speed of The first train relative to the second, slower train at collision = 4.781 m/s

Explanation:

We will use the equations of motion to obtain the solution required

At time t = 0

speed of first train = 22.2 m/s

Initial space between the two trains = 50 m

Speed of second train = 6.94 m/s

For the first car, distance covered by the first train = y

y = distance covered between the beginning of the deceleration and the point where the the two trains hit one another.

u = initial velocity = 22.2 m/s

t = time taken for all this to happen

a = deceleration = - 2.1 m/s²

y = ut + (1/2)at²

y = 22.2t - 1.05t² (eqn 1)

For the second train,

At t = 0, y = 50 m

Let the new distance moved by the second train before collision = (y - 50)

u = initial velocity = 6.94 m/s

t = time taken = t

a = acceleration of the second train = 0 m/s² (constant velocity)

(y - 50) = ut + (1/2)at²

y - 50 = 6.94t

y = 6.94t + 50 (eqn 2)

substituting for y in eqn 2 using the expression obtained in eqn 1

y = 22.2t - 1.05t²

y = 6.94t + 50

22.2t - 1.05t² = 6.94t + 50

1.05t² - 15.26t + 50 = 0

Solving this quadratic equation

t = 4.99 s or 9.54 s

The position of the two trains are the same at those two times, but the first time is when they hit each other.

t = 4.99 s

At 4.99 s, the the velocity of the first train

v = u + at

v = 22.2 + (-2.1×4.99) = 11.721 m/s in the same direction as the second train.

Relative velocity at this point will be

= 11.721 - 6.94 = 4.781 m/s

Relative speed of the trains when they collide: The relative speed of The first train relative to the second, slower train at collision = 4.781 m/s

Hope this Helps!!!

4 0
2 years ago
The isobars in the conventional series that will be needed to complete the pressure analysis between the lowest and highest valu
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The isobars in the conventional series that will be needed to complete the pressure analysis between the lowest and highest values on this map are: 1008, 1012, 1016, 1020.

 

To add, an isobar is <span>a line on a map connecting points having the same atmospheric pressure at a given time or on average over a given period.</span>

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3 years ago
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