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kap26 [50]
3 years ago
12

an 80 kg cart goes around the inside of a vertical loop of a roller coaster. the radius of the loop is 5m and the cart moves at

a speed of 8m/s at the top. what was the force exerted by the track on the cart at the top of the loop?

Physics
1 answer:
GarryVolchara [31]3 years ago
3 0

Answer:

F=  224 N

Explanation:

Given that

mass ,m = 80 kg

Radius ,r= 5 m

speed at the top v= 8 m/s

The force at the top = F

Now by using the Second law of Newton's

F+mg=\dfrac{mv^2}{r}

Now by putting the values

F+80\times 10 = \dfrac{80\times 8^}{5}

Take  g = 10 m/s²

F= \dfrac{80\times 8^2}{5}-800

F=  224 N

Therefore the force exerted by the track at the top position will be 224 N.

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Westkost [7]
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Torque = moment of inertia * angular acceleration

For this, we require the angular acceleration, α. We may calculate this using:
α = Δω/Δt
The time taken to achieve rotational speed may be calculated using:
time = 1 revolution * 2π radians per revolution / 3.5 radians per second
time = 1.80 seconds

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The moment of inertia of a thin disc is given by:
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4 0
3 years ago
The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have st
irina [24]

Complete Question:

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm . The spring constant for the tendon is the same for both groups, 31 {\rm {N}/{mm}}. What is the difference in maximum stored energy between the sprinters and the nonathlethes?

Answer:

\triangle E = 12.79 J

Explanation:

Sprinters' tendons stretch, x_s = 43 mm = 0.043 m

Non athletes' stretch, x_n = 32 mm = 0.032 m

Spring constant for the two groups, k = 31 N/mm = 3100 N/m

Maximum Energy stored in the sprinter, E_s = 0.5kx_s^2

Maximum energy stored in the non athletes, E_m = 0.5kx_n^2

Difference in maximum stored energy between the sprinters and the non-athlethes:

\triangle E = E_s - E_n = 0.5k(x_s^2 - x_n^2)\\\triangle E = 0.5*3100* (0.043^2 - 0.032^2)\\\triangle E = 0.5*31000*0.000825\\\triangle E = 12.79 J

4 0
3 years ago
What factors affect the speed of water waves
Aneli [31]
Hey there,

Your question states: What factors affect the speed of water waves
Let's get one thing out the way, (wavelength) does NOT affect the the speed of water. If anything, it would be how high the wavelength's are. The higher the wavelengths are, the more that it would affect the speed, because there very high, but if it were to go longer on the width side, that would increase the speed, but that's not the case. Your correct answer would be (higher wavelength).

Hope this really helps you.
6 0
3 years ago
Suppose you design a new thermometer called the "x" thermometer. on the x scale, the boiling point of water is 130.0 ox and the
Hoochie [10]

You've told us:

-- 130°x  =  212°F

and

-- 10°x  =  32°F

Thank you.  Those are two points on a graph of °x vs °F .  With those, we can figure out the equation of the graph, and easily convert ANY temperature on one scale to the equivalent temperature on the other scale.

-- If our graph is going to have °x on the horizontal axis and °F on the vertical axis, then the two points we know are  (130, 212)  and  (10, 32) .

-- The slope of the line through these two points is

Slope = (32 - 212) / (10 - 130)

Slope = (-180) / (-120)

Slope = 1.5

So far, the equation of the graph is

F = 1.5 x + (F-intercept)

Plug one of the points into this equation.  I'll use the second point  (10, 32) just because the numbers are smaller:

32 = 1.5 (10) + F-intercept

32 = 15 + (F-intercept)

F-intercept = 17

So the equation of the conversion graph is

F = 1.5 x + 17

There you are !  Now you can plug ANY x temperature in there, and the F temperature jumps out at you.

The question is asking what temperature is the same on both scales. This seems tricky, but it's not too bad.  Whatever that temperature is, since it's the same on both scales, you can take the conversion equation, and write the same variable in BOTH places.

We can write [ x = 1.5x + 17 ], solve it for  x, and the solution will be the same temperature in  F  too.

or

We can write [ F = 1.5F + 17 ], solve it for  F, and the solution will be the same temperature in  x  too.

F = 1.5F + 17

Subtract  F  from each side:  0.5F + 17 = 0

Subtract 17 from each side:   0.5F = -17

Multiply each side by 2 :  F = -34

That should be the temperature that's the same number on both scales.

Let's check it out, using our handy-dandy conversion formula (the equation of our graph):

F = 1.5x + 17

Plug in -34 for  x:  

F = 1.5(-34) + 17

F = -51 + 17

<em>F = -34</em>

It works !  -34 on either scale converts to -34 on the other one too. If the temperature ever gets down to -34, and you take both thermometers outside, they'll both read the same number.

<em>yay !</em>

6 0
3 years ago
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Alika [10]

Answer:

Please help on any part you can. I know it is a lot but any help I’d greatly appreciate. I attempted the problem but still do not understand. Thank you so much!

Explanation:

Please help on any part you can. I know it is a lot but any help I’d greatly appreciate. I attempted the problem but still do not understand. Thank you so much!

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7 0
2 years ago
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