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Nana76 [90]
3 years ago
12

A shunting engine in a rail yard is working on a straight east-west section of line. First it travels west for 150 s with an ave

rage velocity of 4 m/s, to hook up to 10 empty freight cars. Then it pulls the cars east with an average velocity of 3 m/s for 200 s, to a loading area. Calculate the engine's final displacement.
Physics
2 answers:
Ymorist [56]3 years ago
7 0

Displacement-1

\\ \rm\longmapsto Velocity(Time)

\\ \rm\longmapsto 150(4)

\\ \rm\longmapsto 600m

Displcaement-2

\\ \rm\longmapsto 200(3)

\\ \rm\longmapsto 600m

Total displacement=600+600=1200m

irinina [24]3 years ago
7 0

<u>Displacement-1</u>

⟼Velocity(Time)

⟼150(4)

⟼600m

<u>Displcaement-2</u>

200(3)

⟼200(3)

⟼600m

Total displacement=600+600=1200m.

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Kepler’s
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We’re trying to find M, so:

M=4pi^2*r^3/(G*T^2)

M=4pi^2*(1.496
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M=1.48× 10^40(m^3)/((N*m^2/kg^2)*days^2))

Let’s work
with the units:

(m^3)/((N*m^2/kg^2)*days^2))=

=(m^3*kg^2)/(N*m^2*days^2)

=(m*kg^2)/(N*days^2)

=(m*kg^2)/((kg*m/s^2)*days^2)

=(kg)/(days^2/s^2)

=(kg*s^2)/(days^2)

So:

M=1.48× 10^40(kg*s^2)/(days^2)

Now we need to convert days to seconds in order to cancel
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1 day=24 hours=24*60minutes=24*60*60s=86400s

M=1.48× 10^40(kg*s^2)/((86400s)^2)

M=1.48× 10^40(kg*s^2)/(
86400^2*s^2)

M=1.48× 10^40kg/86400^2

M=1.98x10^30kg

The
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× 10^30

(it may vary
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White raven [17]

Answer:

kinetic energy

Explanation:

hop it is helpful

mark me brainlist

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3 years ago
You wiggle a string,that is fixed to a wall at the other end, creating a sinusoidalwave with a frequency of 2.00 Hz and an ampli
FinnZ [79.3K]

Answer:

Explanation:

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f(x,t)=Acos(kx-\omega t)

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k: wave number

w: angular frequency

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\omega = 2\pi f=2\pi (2.00Hz)=12.56\frac{rad}{s}

k=\frac{\omega}{v}=\frac{12.56\frac{rad}{s}}{12.0\frac{m}{s}}=1.047\ m^{-1}

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d) the speed of the medium:

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Two identical sticky masses m are moving in the xy-plane, with their momenta at an angle of φ with one another. They are each mo
postnew [5]

Answer:

ucosφ=-v2cosθ2\\\\φ=cos^{-1} (\frac{-v2cosθ2}{cosφ} )

Explanation:

Two identical sticky masses m are moving in the xy-plane, with their momenta at an angle of φ with one another. They are each moving at the same speed v when they collide at the origin of the coordinates and stick together. After the collision, the masses move at an angle −θ2 with respect to the +x axis at speed v2 .1. What was the angle φ?

from the principle of momentum

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Take note that momentum is the product of mass and velocity

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Note that the collision is inelastic , since they both moved with the same velocity

umcosφ+umcosφ=(m+m)v2cos−θ2

2mucosφ=2mv2cos−θ2

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3 years ago
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