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Kitty [74]
3 years ago
6

A child and sled with a combined mass of 50.0 kg slide down a frictionless slope. if the sled starts from rest and has a speed o

f 4.00 m/s at the bottom, what is the height of the hill? m
Physics
2 answers:
hoa [83]3 years ago
8 0

Answer:

The height of the hill is 0.81 m

Explanation:

It is given that,

Mass of child and sled, m = 50 kg

The sled starts from rest and has a speed of v = 4 m/s

We have to find the height of the hill. It can be calculated using conservation of energy. The potential energy at the bottom of hill is 0 and at maximum height the kinetic energy is 0.

So, \dfrac{1}{2}mv^2=mgh

h=\dfrac{v^2}{2g}

h=\dfrac{(4\ m/s)^2}{2\times 9.8\ m/s^2}

h = 0.81 m

So, the height of the hill is 0.81 m. Hence, this is the required solution.                                        

Furkat [3]3 years ago
4 0
            <span> Using conservation of energy

Potential Energy (Before) = Kinetic Energy (After)

mgh = 0.5mv^2

divide both sides by m

gh = 0.5v^2

h = (0.5V^2)/g

h = (0.5*2.2^2)/9.81

h = 0.25m

</span>
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Elliot jumps up and down on a pogo stick. He weighs 600.N, and his pogo stick has a spring with spring constant 1100N/m. What is
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From conservation of energy, the height he will reach when he has gravitational potential energy 250J is 0.42 meters approximately

The given weight of Elliot is 600 N

From conservation of energy, the total mechanical energy of Elliot must have been converted to elastic potential energy. Then, the elastic potential energy from the spring was later converted to maximum potential energy P.E of Elliot.

P.E = mgh

where mg = Weight = 600

To find the height Elliot will reach, substitute all necessary parameters into the equation above.

250 = 600h

Make h the subject of the formula

h = 250/600

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Therefore, the height he will reach when he has gravitational potential energy 250J is 0.42 meters approximately

Learn more about energy here: brainly.com/question/24116470

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2 years ago
The force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of
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Explanation:

It is given that, the force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of the car and the square of the car's speed such that,

F\propto \dfrac{mgv^2}{r}

F=\dfrac{kmgv^2}{r}

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Case 1.

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A carnival Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute. What is the accelerat
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If a Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute then the acceleration of a passenger at his lowest point during the ride is 4.11 \text{m/s}^{2}.

Calculation:

Step-1:

It is given that the radius of the Ferris wheel is r=15 m, and the angular speed of the wheel is \omega=5rev/min.

It is required to find the angular acceleration of a passenger at his lowest point during the ride.

The formula required to calculate the angular acceleration is, a=\omega ^2 r.

Step-2:

Now substituting the given values into the equation to get the value of the angular acceleration.

\begin{aligned}a &=\omega^{2} r \\&=(5 \mathrm{rev} / \mathrm{min})^{2}(15 \mathrm{~m}) \\&=\left(5 \mathrm{rev} / \mathrm{min} \times \frac{\left(\frac{2 \pi}{60}\right) \mathrm{rad} / \mathrm{sec}}{1 \mathrm{rev} / \mathrm{min}}\right)^{2}(15 \mathrm{~m}) \\&=(0.2741 \times 15) \mathrm{m} / \mathrm{s}^{2} \\&=4.11 \mathrm{~m} / \mathrm{s}^{2}\end{aligned}

The acceleration is towards upwards that means towards the center of the wheel.

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