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trapecia [35]
3 years ago
13

wo people on skates, one with a mass of 61 kg and the other with a mass of 40 kg, stand on ice (frictionless surface) holding a

10 m pole (negligible mass). The skaters pull themselves from the end of the pole until they meet. How far does the 40 kg skater move. Answer:
Physics
1 answer:
poizon [28]3 years ago
5 0

Answer:

x=6.19m

Explanation:

Let m1, m2 be the two masses

and x the distance moved by mass 65kg

Therefore:- distance two=10-x

\\m_{1}x_{1}=m_{2}x_{2}\\(65*x)=40(10-x)\\=>65x=400-40x\\x=3.81\\Therefore:- \\(10-x)=10-3.81=6.19

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Hi! Can somebody please help?
dezoksy [38]

Answer:

Diagram A will reach the top first.

Explanation:

If it is going straight, it will go slower. The higher the movement speed the faster it is. Hope this helps!

7 0
3 years ago
Will the velocity of the book change as it moves across the surface with NO friction? Explain your answer.
MakcuM [25]

No velocity will not be changed

Why?

According to Newtons 1st law the velocity of a moving object remains unchanged unless a external force affect that.

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3 years ago
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I need help for a assignment NOT A QUIZ
Debora [2.8K]

Answer:

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3 years ago
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The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and
nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}

k = 1.4

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K

Work done is given as;

W = \frac{1}{2} *m*(v_i^2 - v_e^2)

inlet velocity is negligible;

v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41  \ m/s

Therefore, the exit velocity is 629.41 m/s

6 0
3 years ago
05. The time required to complete one lap around a perfectly circular track having a radius of 1,835 meters is 86
likoan [24]

Answer:

v = 134.06 m/s

Explanation:

Given that,

Radius of a circular track is 1,835 m

Time required to complete one lap around a perfectly circular track is 86 seconds

We need to find the car's velocity. Velocity is equal to,

v=d/t

On circular path,

v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 1835}{86}\\\\v=134.06\ m/s

So, car's velocity is 134.06 m/s.

7 0
3 years ago
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