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Virty [35]
2 years ago
10

A car is traveling at 20 km in 15 minutes. What is its average velocity

Physics
1 answer:
yarga [219]2 years ago
7 0
The velocity is 4374.45 m/s.
I got the answer by using v=d/t.
20,000meters / 25min= 4,374.45 m/s.
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Explain what distinguishes acute and chronic sports injuries.
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An acute injury is sudden and severe such as a broken bone. A chronic injury develops and worsens over an extended period of time like shin splints

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Consider a double Atwood machine constructed as follows: A mass 4m is suspended from a string that passes over a massless pulley
kenny6666 [7]

Answer:

Hello your question is incomplete attached below is the complete question

Answer : x ( acceleration of mass 4m ) = \frac{g}{7}

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

Explanation:

Given data:

mass suspended = 4 meters

mass suspended at other end = 3 meters

first we have to express the kinetic and potential energy equations

The general kinetic energy of the system can be written as

T = \frac{4m}{2} x^2  + \frac{3m}{2} (-x+y)^2 + \frac{m}{2} (-x-y)^2

T = 4mx^2 + 2my^2 -2mxy  

also the general potential energy can be expressed as

U = -4mgx-3mg(-x+y)-mg(-x-y)+constant=-2mgy +constant

The Lagrangian of the problem can now be setup as

L =4mx^2 +2my^2 -2mxy +2mgy + constant

next we will take the Euler-Lagrange equation for the generalized equations :

Euler-Lagrange  equation = 4x-y =0\\-2y+x +g = 0

solving the equations simultaneously

x ( acceleration of mass 4m ) = \frac{g}{7}

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

8 0
3 years ago
Nora walks down a street and sees a ball dropped from a building
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It is gravity¿ what is the question?
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Absolute zero (K=0 or -273.15°C) is what temperature on the Farenheit scale? a) 459.4°F b) -301.43 °F c) 233.05°F d) -40.15°F e
ANEK [815]

Answer:

e) None of these is true

Explanation:

Given that

Temperature = 0 K

We know that relationship between kelvin and Farenheit scale

\dfrac{K-273}{100}=\dfrac{F-32}{180}

Now by putting the values

\dfrac{K-273}{100}=\dfrac{F-32}{180}

\dfrac{0-273}{100}=\dfrac{F-32}{180}

So F= - 459.67°F

So we can say that 0 K is equal to  - 459.67°F.

So the our option e is correct.

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