Our verbal and nonverbal _a______ provides clues to our attitude on a given topic.

At a given instant an object has an angular velocity. It also has an angular acceleration due to torques that are present. There

katen-ka-za [31]

a) Constant

b) Constant

Explanation:

a)

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

$\tau_{net} = I \alpha$ (1)

where

$\tau_{net}$ is the net torque acting on the object in rotation

I is the moment of inertia of the object

$\alpha$ is the angular acceleration

The angular acceleration is the rate of change of the angular velocity, so it can be written as

$\alpha = \frac{\Delta \omega}{\Delta t}$

where

$\Delta \omega$ is the change in angular velocity

$\Delta t$ is the time interval

So we can rewrite eq.(1) as

$\tau_{net}=I\frac{\Delta \omega}{\Delta t}$

In this problem, we are told that at a given instant, the object has an angular acceleration due to the presence of torques, so there is a non-zero change in angular velocity.

Then, additional torques are applied, so that the net torque suddenly equal to zero, so:

$\tau_{net}=0$

From the previous equation, this implies that

$\Delta \omega =0$

Which means that the angular velocity at that instant does not change anymore.

b)

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

$\tau_{net}=0$

Therefore, this means that

$\Delta \omega =0$

So also in this case, there is no change in angular velocity: this means that the angular velocity of the object will remain constant.

So cases (a) and (b) are basically the same situation, as the net torque is zero in both cases, so the object acts in the same way.

8 0

3 years ago

The labeled images each represent the wave patterns found in the electromagnetic wave spectrum. which of these images are correc

Phantasy [73]

Energy E of EM radiation is given by the equation E=hf, where h is Planck's constant and f is frequency. It means energy E and frequency f are proportional so as we increase the frequency, energy also increases. Also, the relationship between the wavelength and frequency is c=λ*f where λ is the wavelength and f is frequency and c is the speed of light. This tells us the wavelength and frequency are inversely proportional. So as we increase the frequency the wavelength is getting smaller. So as we go from left to right the frequency increases, energy also increases and the wavelength is decreasing. Or, on the left side we should have low frequency, low radiant energy, and long wavelength. On the right side we should have high frequency, high radiant energy and low wavelength. That is the third graph.

5 0

3 years ago

Two automobiles of equal mass approach an intersection. One vehicle is traveling with velocity 14.0 m/s toward the east and the

Sergeeva-Olga [200]

Answer:

kqwwwwj

Explanation:

4 0

3 years ago

The box leaves position x=0 with speed v0. The box is slowed by a constant frictional force until it comes to rest at position x

Diano4ka-milaya [45]

Answer:

$Ff=m\times \dfrac{V_o^2}{2X_1}$

Explanation:

Given that

At X=0 V=Vo

At X=X1 V=0

As we know that friction force is always try to oppose the motion of an object. It means that it provide acceleration in the negative direction.

We know that

$V^2=U^2-2aS$

$0=V_o^2-2a X_1$

$a=\dfrac{V_o^2}{2X_1}$

So the friction force on the box

Ff= m x a

$Ff=m\times \dfrac{V_o^2}{2X_1}$

Where m is the mass of the box.

4 0

3 years ago

A wheel of mass 4kg is pulled up a plane inclined at 30° to the horizontal by a force of 45N applied to the axle and parallel to

Len [333]

Answer:

v = 10 m/s

Explanation:

Let's assume the wheel does not slip as it accelerates.

Energy theory is more straightforward than kinematics in my opinion.

Work done on the wheel

W = Fd = 45(12) = 540 J

Some is converted to potential energy

PE = mgh = 4(9.8)12sin30 = 235.2 J

As there is no friction mentioned, the remainder is kinetic energy

KE = 540 - 235.2 = 304.8 J

KE = ½mv² + ½Iω²

ω = v/R

KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²

v = √(2KE / (m + I/R²))

v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6

v = 10.07968...

5 0

3 years ago