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blagie [28]
3 years ago
12

a car with a mass of 1200 kilograms is moving around a circular curve at a uniform velocity of 20 meters per second. the centrip

etal force on the car is 6000 newtons. what is the radius of the curve?
Physics
1 answer:
qaws [65]3 years ago
5 0
Well, first of all, a car moving around a circular curve is not moving
with uniform velocity.  The direction of motion is part of velocity, and
the direction is constantly changing on a curve.

The centripetal force that keeps an object moving in a circle is

         Force  =  (mass of the object) · (speed)² / (radius of the circle)

         F  =  m s² / r

We want to know the radius, to rearrange the formula to give us
the radius as a function of everything else.

                                          F     =  m s² / r

Multiply each side by 'r':       F· r  =  m · s²

Divide each side by 'F':            r  =  m · s² / F    

We know all the numbers on the right side,
so we can pluggum in:

                      r  =       m       ·        s²      /     F

                      r  =  (1200 kg) · (20 m/s)² / (6000 N) .

I'm pretty sure you can finish it up from here.

                                      


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Explanation:

The tangential speed of Andrea is given by :

v=r\omega

Where

r is radius of the circular path

ω is angular speed

The merry-go-round is rotating at a constant angular speed. Let the new distance from the center of the circular platform is r'

r' = 2r

New angular speed,

v'=r'\omega'\\\\v'=(2r)\omega\\\\v'=2r\omega\\\\v'=2v

New angular speed is twice that of the Chuck's speed.

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3 years ago
A 1,600 kg train car rolling freely on level track at 16 m/s bumps into a 1.0 × 103 kg train car moving at 10.0 m/s in the same
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This item is solved through the concept of the conservation of momentum which states that the momentum before and after collision should be equal. 
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3 years ago
The index of refraction of silicate flint glass for red light is 1.620 and for violet light is 1.660 . A beam of white light in
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Answer:

The angular separation equals 0.35^{o}

Explanation:

We have according to Snell's law

n_{1}sin(\theta _{1})=n_{2}sin(\theta _{2})

\therefore \theta _{2}=sin^{-1}(\frac{n_{1}sin(\theta _{1})}{n_{2}})

Using this equation for both the colors separately we have

\theta _{red}=sin^{-1}(\frac{sin(23.90^{o})}{1.62})\\\\\theta _{red}=14.48^{o}

Similarly for violet light we have

\theta _{violet}=sin^{-1}(\frac{sin(23.90^{o})}{1.660})\\\\\theta _{violet}=14.13^{o}

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A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/m. At t = 0 the spring is neither stretch
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Answer:

a) A = 0.98 m

b) Ф = 90°

c) x = -0.98sin(12.25t)

Explanation:

We know the value of the spring constant which is 300 N/m, the innitial apmplitude, that we will call it xo is 0, at t = 0, and the speed is 12 m/s

The expression for the amplitude under these conditions is:

A = √xo² + vx²/w² (1)

To calculate the angular speed w, we use the following expression:

w = √k/m  (2)

Calculating w:

w = √300/2 = 12.25 rad/s

Now, we replace this value into equation 1, along with the other known values and solve for A:

A = √0 + (12)²/(12.25)²

A = 0.98 m

b) In this part, is actually easy, the displacement of x in function of the time is given by:

x = A cos(wt - Ф) (4)

But at t = 0 we have then:

x = xo = A cosФ (5)

Solving for the angle Ф we have:

xo/A = cosФ

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Replacing the data in (6):

Ф = arccos(0/0.98)

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c) Equation (4) is the expression for the simple harmonic motion

x = A cos(wt - Ф)

And if we replace here the value of w and the previous angle, we can write an equation for x in function of t:

x = 0.98 cos (12.25t - 90)

x = 0.98 cos(12.25t - 90)

And we have an trigonometric expression for cos that is:

cos(α - π/2) = -sinα

in this case, α will be the value of w = 12.25 rad/s. and 90° is the same as rewritting as π/2, therefore:

cos(α - π/2) = cos(12.25t - 90)

x = -0.98sin(12.25t)

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