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Talja [164]
3 years ago
7

A ball is rolling along at speed v without slipping on a horizontal surface when it comes to a hill that rises at a constant ang

le above the horizontal. In which case will it go higher up the hill: if the hill has enough friction to prevent slipping, or if the hill is perfectly smooth. Justify your answer with a conservation of energy statement
Physics
1 answer:
MissTica3 years ago
6 0

Answer:

The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.

Explanation:

The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.

This is because:

If we consider the ball initially at rest on a frictionless surface and a force is exerted through the centre of mass of the ball, it will slide across the surface with no rotation, and thus, there will only be translational motion.

Now, if there is friction and force is again applied to the stationary ball, the frictional force will act in the opposite direction to the force but at the edge of the ball that rests on the ground. This friction generates a torque on the ball which starts the rotation.

Therefore, static friction is infact necessary for a ball to begin rolling.

Now, from the top of the ball, it will move at a speed 2v, while the centre of mass of the ball will move at a speed v and lastly, the bottom edge of the ball will instantaneously be at rest. So as the edge touching the ground is stationary, it experiences no friction.

So friction is necessary for a ball to start rolling but once the rolling condition has been met the ball experiences no friction.

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10. A loud motorcycle passes by the front of Bill's house. What medium
garik1379 [7]

Answer: Air

Explanation:

3 0
2 years ago
Read 2 more answers
(a) What is the fluid speed in a fire hose with a 9.00-cm diameter carrying 80.0 L of water per second? (b) What is the flow rat
son4ous [18]

Answer:

12.5752053801 m/s

80\times 10^{-3}\ m^3/s

No.

Explanation:

Q = Volume flow rate = 80\ L/s=80\times 10^{-3}\ m^3/s

d = Diameter of pipe = 9 cm

A = Area = \dfrac{\pi}{4}d^2

Volume flow rate is given by

Q=Av\\\Rightarrow v=\dfrac{Q}{A}\\\Rightarrow v=\dfrac{80\times 10^{-3}}{\dfrac{\pi}{4} (9\times 10^{-2})^2}\\\Rightarrow v=12.5752053801\ m/s

Velocity of fluid is 12.5752053801 m/s

The volume flow rate in m³/s is 80\times 10^{-3}\ m^3/s

The flow of fluid does not depend on the type of water used. Hence the answers would be same. If Q is constant v will be the same irrespective of the type of water used.

8 0
3 years ago
Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed o
kobusy [5.1K]

Answer:

\Delta \lambda=14.3\ nm

Explanation:

It is given that,

The number of lines per unit length, N = 900 slits per cm

Distance between the formed pattern and the grating, l = 2.3 m

n the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.98 mm, \Delta Y=2.98\ mm = 0.00298\ m

Let d is the slit width of the grating,

d=\dfrac{1}{N}

d=\dfrac{1}{900\ cm}

d=1.11\times 10^{-5}\ m

For the first wavelength, the position of maxima is given by :

y_1=\dfrac{L\lambda_1}{d}

For the other wavelength, the position of maxima is given by :

y_2=\dfrac{L\lambda_2}{d}

So,

\Delta \lambda=\dfrac{\Delta y d}{l}

\Delta \lambda=\dfrac{0.00298\times 1.11\times 10^{-5}}{2.3}

\Delta \lambda=1.43\times 10^{-8}\ m

or

\Delta \lambda=14.3\ nm

So, the difference between these wavelengths is 14.3 nm. Hence, this is the required solution.

3 0
2 years ago
PLEASE HELP ANSWER ASAP. 10 points. I will give brainliest.
alina1380 [7]

The answer is D. I know because I already answered the question.

3 0
3 years ago
A 2-kg wheel rolls down the road with a linear speed of 15m/s. Find its trwansitional and rotational kinetic energies.​
Elza [17]

Answer:

The translational kinetic energy is 225 J

The rotational kinetic energy is 225 J

Explanation:

Given;

mass of the wheel, m = 2-kg

linear speed of the wheel, v = 15 m/s

Transnational kinetic energy is calculated as;

E = ¹/₂MV²

where;

M is mass of the moving object

V is the velocity of the object

E =  ¹/₂ x 2 x (15)²

E = 225 J

Rotational kinetic energy is calculated as;

E = ¹/₂Iω²

where;

I is moment of inertia

ω is angular velocity

E = \frac{1}{2} I \omega^2\\\\E = \frac{1}{2} *mr^2*(\frac{v}{r})^2\\\\E =  \frac{1}{2} *mr^2*\frac{v^2}{r^2} \\\\E =  \frac{1}{2}mv^2

E =  ¹/₂ x 2 x (15)²

E = 225 J

Thus, the translational kinetic energy is equal to rotational kinetic energy

6 0
3 years ago
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