A constant net torque is applied to an object. Which one of the following will NOT be constant?

Physics

1 answer:

love history [14]3 years ago

7 0

Answer:

B) Angular velocity

Explanation:

The equivalent of Newton's second law for the rotational motions can be written as:

$\tau = I \alpha$

where

$\tau$ is the net torque applied to the object

I is the moment of inertia

$\alpha$ is the angular acceleration

From the formula we see that when a constant net torque $\tau$ is applied, then the object also has a constant angular acceleration, $\alpha$ .

But we also know that

$\alpha = \frac{d\omega}{dt}$

where $\omega$ is the angular velocity: so, a constant angular acceleration means that the angular velocity of the object is changing, so the correct answer is

B) Angular velocity

(moment of inertia and center of gravity do not change since they only depend on the mass and the geometry/shape of the object, which do not change)

You might be interested in

What happens to the saturation for when adding salt to water at room temperature

saw5 [17]

Saturated and Unsaturated Solutions
Table salt (NaCl) readily dissolves in water. Suppose that you have a beaker of water to which you add some salt, stirring until it dissolves. So you add more and that dissolves. You keep adding more and more salt, eventually reaching a point that no more of the salt will dissolve no matter how long or how vigorously you stir it. Why? On the molecular level, we know that action of the water causes the individual ions to break apart from the salt crystal and enter the solution, where they remain hydrated by water molecules. What also happens is that some of the dissolved ions collide back again with the crystal and remain there. Recrystallization is the process of dissolved solute returning to the solid state. At some point the rate at which the solid salt is dissolving becomes equal to the rate at which the dissolved solute is recrystallizing. When that point is reached, the total amount of dissolved salt remains unchanged. Solution equilibrium is the physical state described by the opposing processes of dissolution and recrystallization occurring at the same rate.
While this shows the change of state back and forth between solid and aqueous solution, the preferred equation also shows the dissociation that occurs as an ionic solid dissolves.

When the solution equilibrium point is reached and no more solute will dissolve, the solution is said to be saturated. A saturated solution is a solution that contains the maximum amount of solute that is capable of being dissolved. At 20°C, the maximum amount of NaCl that will dissolve in 100. g of water is 36.0 g. If any more NaCl is added past that point, it will not dissolve because the solution is saturated. What if more water is added to the solution instead? Now more NaCl would be capable of dissolving in the additional solvent. An unsaturated solution is a solution that contains less than the maximum amount of solute that is capable of being dissolved.

When 30.0 g of NaCl is added to 100 ml of water, it all dissolves, forming an unsaturated solution. When 40.0 g is added, 36.0 g dissolves and 4.0 g remains undissolved, forming a saturated solution.

How can you tell if a solution is saturated or unsaturated? If more solute is added and it does not dissolve, then the original solution was saturated. If the added solute dissolves, then the original solution was unsaturated. A solution that has been allowed to reach equilibrium but which has extra undissolved solute at the bottom of the container must be saturated.

5 0

3 years ago

An insulated pipe carries steam at 300°C. The pipe is made of stainless steel (with k = 15 W/mK), has an inner diameter is 4 cm,

insens350 [35]

Answer:

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

$Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}$