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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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Answer:

Explanation:

Given:

Steam Mass rate, ms = 1.5 kg/min

= 1.5 kg/min × 1 min/60 sec

= 0.025 kg/s

Air Mass rate, ma = 100 kg/min

= 100 kg/min × 1 min/60 sec

= 1.67 kg/s

A.

Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.

xf, quality = 0.9.

Tsat = 89.9°C

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hfg = 2283.38 kJ/kg

Using the equation for specific enthalpy,

hi = hf + (hfg × xf)

= 376.57 + (2283.38 × 0.9)

= 2431.552 kJ/kg

The specific enthalpy of the outlet, h2 = hf

= 376.57 kJ/kg

B.

Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy

= ms × (hi - h2)

= 0.025 × (2431.552 - 376.57)

= 0.025 × 2055.042

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5 0
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How do you compare the energies of the different parts of the electromagnetic spectrum?
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The energy associated with electromagnetic radiation is proportional to frequency and inversely proportional to wavelength. So, electromagnetic waves with shorter wavelengths have more energy.

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The total momentum of the system is preserved through the collision.

Note that momentum is
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where m = mass
v = velocity.

Initial momentum:
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Final momentum:
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Answer: 1.0 m/s
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