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V125BC [204]
3 years ago
11

An elevator weighing 2.00 x 10 5 N is supported by a steel cable. What is the tension in the cable when the elevator is accelera

ted upward at a rate of 3.00 m/s2? (g = 9.81 m/s2 ) Show your work.
Physics
1 answer:
STALIN [3.7K]3 years ago
6 0
First, we will get the mass of the elevator:
Fw = mass * gravitational acceleration
2 * 10^5 = mass * 9.8
mass = 20.4 * 10^3 kg

Now, the total force acting on the elevator can be written as follows:
Fnet = Tension - Fw
mass * acceleration = Tension - 2*10^5
20.4 * 10^3 * 3 = Tension - 2*10^5
61200 = Tension - 2*10^5
Tension = (61200) + (2*10^5)
Tension = 2.61 * 10^5 newton
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A 730-keV gamma ray Compton-scatters from an electron. Find the energy of the photon scattered at 120°, the kinetic energy of th
romanna [79]

Answer:

Energy of scattered photon is 232.27 keV.

Kinetic energy of recoil electron is 497.73 keV.

The recoil angle of electron is 13.40°

Explanation:

The energy of scattered photon is given by the relation :

E_{2}=\frac{E_{1} }{1+(\frac{E_{1} }{m_{e}c^{2}  })(1-\cos\theta) }     .....(1)

Here E₁ is the energy of incident photon, E₂ is the energy of scattered photon, m_{e} is mass of electron and θ is scattered angle.

Substitute 730 keV for E₁, 511 keV for m_{e} and 120° for θ in equation (1).  

E_{2}=\frac{730 }{1+(\frac{730 }{511  })(1-\cos120) }

E₂ = 232.27 keV

Kinetic energy of recoil electron is given by the relation :

K.E. = E₁ - E₂ = (730 - 232.27 ) keV = 497.73 keV

The recoil angle of electron is given by :

\cot\phi=(1+\frac{E_{1} }{m_{e}c^{2}  })\tan\frac{\theta}{2}

Substitute the suitable values in above equation.

\cot\phi=(1+\frac{730 }{511  })\tan\frac{120}{2}

\cot\phi=4.20

\phi = 13.40°

8 0
3 years ago
Two carts (m1 = m2 = 0.400 kg) are placed on an aluminum track. The first cart is pushed with the initial velocity of 1.5 m/s to
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Answer:

That's just if it's a Flexible collision

5 0
3 years ago
8.92 A 45.0 kg woman stands up in a 60.0 kg canoe 5.00 m long. She walks from a point 1.00 m from one end to a point I .00 m fro
omeli [17]

The distance that the canoe moves in this process is 1.29 meters.

We first have to find the center of mass

X = \frac{MsXs+McXc}{Mw+Mc} \\\\

Where

Ms = Woman's mass = 45

Mc = Canoe's mass = 60kg

Xs = position from left= 1 cm

Xc = position from left end of canoe's mass = 2.5cm

When we put these values into the equation we have:

X=\frac{45*1+60*2.5}{45+60} \\\\= 1.857\\

The center of gravity lies at the center of this boat. Therefore,

Xc = \frac{L}{2} \\\\L = 5 m long\\\\\frac{5}{2} =2.5

5.00 - 1. 00 = 4 meters

\frac{45*4+60*2.5}{45+60} = 3.14meters\\\\

To get the distance that is moved by this canoe

distance = 3.143-1.857

= 1.286

≈ 1.29 meters

The distance that the canoe moves in this process is 1.29 meters.

Read more on brainly.com/question/13198009?referrer=searchResults

6 0
2 years ago
A 1000 kg rocket ship is traveling at 40 m/s. If the velocity changes to 70
nydimaria [60]

Answer: A

Explanation:

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3 years ago
Find the vector parallel to the resultant of the vector A=i +4j-2k and B=3i-5j+k​
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Answer:

2008

Explanation:

2000+3+5======2008

8 0
3 years ago
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