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Phoenix [80]
3 years ago
13

As speed increases, how does the potential, kinetic, and total energy levels change?

Physics
1 answer:
Anastaziya [24]3 years ago
4 0
Potential equals kenecric at the bottom so potential would also increas
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Find the velocity in m/s of a swimmer who swims 110m toward the shore in 72 s
Sladkaya [172]
1.5 m/s toward shore 

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Water raises a boat once every 3.0 seconds. What is the frequency (f) of the waves passing the boat?
Goshia [24]

Answer:

I think its A

Explanation:

if its every 3 seconds wouldnt it be 3.0 Hz

6 0
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It takes a bus driver 30 minutes to pick up students from four stops. The last stop is at the corner of Green Street and Route 7
deff fn [24]

Answer:

No, the distance from the last stop to the school and the time it takes to travel that distance are required.

7 0
3 years ago
a car with a mass of 2000 kilograms is moving around a circular curve at a uniform velocity of 25 meters per second. The curve h
melisa1 [442]
In the given problem, we say various information's that are going to help us reach the ultimate answer to the question. Let us first write the information's that have been presented in front of us.
Mass of the car = 2000 kg
Velocity of the car = 25 m/s^2
Radius of the circle = 80 m
Now we already know the equation for calculating the centripetal force and that is
Centripetal Force = [mass * (velocity)^2]/Radius
                            = [2000 * (25)^2]/80
                            = (2000 * 625)/80
                            = 1250000/80
                            = 15625
So the centripetal force on the car is 15625 Newtons
  
4 0
3 years ago
Read 2 more answers
A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless, horizontal axles. A light cord wrapped around
Alex_Xolod [135]

Answer:

\alpha =\frac{m*g*R}{I-m*R^2}

a = \frac{m*g*R^2}{I-m*R^2}

T=\frac{I*m*g}{I-m*R^2}

Explanation:

By analyzing the torque on the wheel we get:

T*R=I*\alpha    Solving for T:   T=I/R*\alpha

On the object:

T-m*g = -m*a    Replacing our previous value for T:

I/R*\alpha-m*g = -m*a

The relation between angular and linear acceleration is:

a=\alpha*R

So,

I/R*\alpha-m*g = -m*\alpha*R

Solving for α:

\alpha =\frac{R*m*g}{I+m*R^2}

The linear acceleration will be:

a =\frac{R^2*m*g}{I+m*R^2}

And finally, the tension will be:

T =\frac{I*m*g}{I+m*R^2}

These are the values of all the variables: α, a, T

8 0
3 years ago
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