Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore frictional losses. (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?

Answers

Answer 1
Answer:

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_(f) = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_(cm )^(2) + ½ I_(cm)

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s


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45
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Answers

Answer:

I want to say a because you want to subtract and simplify

A glass plate 2.95 mmmm thick, with an index of refraction of 1.60, is placed between a point source of light with wavelength 600 nmnm (in vacuum) and a screen. The distance from source to screen is 1.25 cm. How many wavelengths are there between the source and the screen?

Answers

Answer:

N_T=2086285.67

Explanation:

Given;

Thickness of the glass plate, x=2.95* 10^(-3)\ m

refractive index of the glass plate, n=1.6

wavelength of light source in vacuum, \lambda=600* 10^(-9)\ m

distance between the source and the screen, d=1.25\ m

Distance travelled by the light from source to screen in vacuum:

d_v=d-x

d_v=1.25-0.00295

d_v=1.24705\ m

So the no. of wavelengths in the vacuum:

N=(d_v)/(\lambda)

N=(1.24705)/(6* 10^(-7))

N\approx2.0784* 10^(6)  .......................(1)

Now we find the wavelength of the light wave in the glass:

n=(\lambda)/(\lambda')

where:

\lambda'= wavelength of light in the medium of glass.

1.6=(600* 10^(-9))/(\lambda')

\lambda'=375* 10^(-9)\ m=375\ nm

Now the no. of wavelengths in the glass:

N'=(x)/(\lambda')

N'=(2.95* 10^(-3))/(375* 10^(-9))

N'=7.8667* 10^(3) ............................(2)

From (1) & (2):

  • total no. of wavelengths are there between the source and the screen:

N_T=N+N'

N_T=2086285.67

If a photon has a frequency of 5.20 x 10^14 hertz, what is the energy of the photon ? Given : Planck's constant is 6.63 x 10^-34 joule-seconds.

Answers

For this you would use planck's equation.

E = hv, where v = the frequency and h = planck's constant.

So E = 5.20 x10^14  x  6.63 x 10^-34
= 3.45 x 10^-19 Joules

Which lists mixtures, in order, from the smallest particles to the largest particles?O solution, colloid, suspension
solution, suspension, colloid
O suspension, colloid, solution
O suspension, solution, colloid

Answers

The order of mixtures according to their particle size from smallest to the largest is: solution< colloid< suspension. So, the first option is correct.

What is meant by solvent and solute in a mixture ?

Solvent is defined as the substance in which the solute particles dissolve and forms a mixture. Solute is defined as the particles that get dissolved in the solvent to form the mixture.

Here,

The particles of the solutions, colloids and suspension varies according to their size and the type of mixture they comes under.

Colloid is a heterogenous mixture, that means it has a non-uniform composition of particles. In colloid, the particles have an intermediate size between 1 nm to 1000 nm which is the size range between that of solution and suspension.

Solution is a homogenous mixture in which the particles have a diameter of less than 1 nm. The size of the particles in solution is the smallest when compared to that of colloid and suspension.

Suspension is a homogenous mixture in which the size of the particles is larger than 10000 nm. The particles of suspension are the largest among the particles of colloid and solution.

Hence,

The order of mixtures according to their particle size from smallest to the largest is: solution< colloid< suspension.

To learn more about solute and solvent, click:

brainly.com/question/7932885

#SPJ7

Answer:

A or 1

Explanation:

solution, colloid, suspension

A train station bell gives off a fundamental tone of 505 Hz as the train approaches the station at a speed of 27.6 m/s. If the speed of sound in air on that day is 339 m/s, what will be the apparent frequency of the bell to an observer riding the train

Answers

Answer:

Apparent frequency of the bell to the observer is 546.12 Hz

Explanation:

The frequency of train bell (frequency of source) = 505 Hz

The speed of train (observer) = 27.6 m/s

The speed of sound in the air is (velocity of sound) = 339 m/s

The apparent frequency of the bell to the observer is calculated as follows:

Apparent frequency of bell to the observer.

= \text{frequency of source} * (Observer + velocity \ of \ sound )/( velocity \ of \ sound ) \n= 505 * (27.6 + 339)/(339) \n= 546.12 Hz

A uniform, 4.5 kg, square, solid wooden gate 2.0 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.2 kg raven flying horizontally at 4.5 m/s flies into this door at its center and bounces back at 1.5 m/s in the opposite direction. What is the angular speed of the gate just after it is struck by the unfortunate raven?

Answers

Answer:

Explanation:

Mass of the gate, m_1 = 4.5 kg

Mass of the raven, m_2 = 1.2 kg

Initial speed of raven, v_1 = 4.5 m/s

Final speed of raven, v_2 = - 1.5 m/s

Moment of Inertia of the gate about the axis passing through one end:

I = (1)/(3) m_1 a^2\nI = (1)/(3) *4.5 * 2^2\nI = 6 kg m^2

Angular momentum of the gate, L = I \omega

L = 5.33 \omega

Using the law of conservation of angular momentum:

m_2 v_f (a/2) + I\omega = m_2v_i (a/2)\nI\omega = m_2 (a/2)(v_i - v_f)\n