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svetlana [45]
2 years ago
8

A machine is applying a torque to rotationally accelerate a metal disk during a manufacturing process. An engineer is using a gr

aph of torque as a function of time to determine how much the disk’s angular speed increases during the process.
The graph of torque as a function of time starts at an initial torque value and is a straight line with positive slope. What aspect of the graph and possibly other quantities must be used to calculate how much the disk’s angular speed increases during the process?
Physics
2 answers:
Viefleur [7K]2 years ago
8 0

Explanation:

Question 9 A machine is applying a torque to rotationally accelerate a metal disk during a manufacturing process. An engineer is using a graph of torque as a function of time to determine how much the disk's angular speed increases during the process The graph of torque as a function of time starts at an initial torque value and is a straight line with positive slope. What aspect of the graph and possibly other quantities must be used to calculate how much the disk's angular speed increases during the process? The slope of the graph multiplied by the disk's radius will equal the change in angular speed The area under the graph multiplied by the disk's radius will equal the change in angular speed. The slope of the graph divided by the disk's rotational inertia will equal the change in angular speed. The area under the graph divided by the disk's rotational inertia will equal the change in angular speed. The area under the graph multiplied by the disk's rotational inertia will equal the change in angular speed E

KATRIN_1 [288]2 years ago
3 0

Answer:

(D) The area under the graph divided by the disk's rotational inertia will equal the change in angular speed.

Explanation:

Angular momentum is equal to the product of rotational inertia and angular speed. Therefore, angular speed would be angular momentum divided by rotational inertia.

Torque is the slope of angular momentum, so the area under the torque curve would be equal to the angular momentum.

In conclusion, the angular speed would be the area under the torque curve divided by the rotational inertia.

Hope this helps!

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a mass of 1.00 kg of water at temperature T is poured from a height of 0.100 km into a vessel containing water of the same tempe
Mariana [72]

Answer:

1.34352 kg

Explanation:

m_w = Mass of water falling = 1 kg

h = Height of fall = 0.1 km

\Delta T = Change in temperature = 0.1

c = Specific heat of water = 4186 J/kg K

g = Acceleration due to gravity = 9.81 m/s²

m_v = Mass of water in the vessel

Here the potential energy will balance the internal energy

m_wgh=m_wc\Delta T+m_vc\Delta T\\\Rightarrow m_v=\dfrac{m_wgh-m_wc\Delta T}{c\Delta T}\\\Rightarrow m_v=\dfrac{m_wgh}{c\Delta T}-m_w\\\Rightarrow m_v=\dfrac{1\times 9.81\times 100}{4186\times 0.1}-1\\\Rightarrow m_v=1.34352\ kg

Mass of the water in the vessel is 1.34352 kg

6 0
3 years ago
The magnitude of the weight of a 3.0 kg object on the surface of the earth is 29 N. True False
madreJ [45]
True

In fact, the weight of an object on the surface of the Earth is given by:
F=mg
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12. A rocket, initially at rest on the ground, accelerates vertically. It accelerates uniformly until it
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Answer:

We kindly invite you to read carefully the explanation and check the image attached below.

Explanation:

According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:

v_{f} = v_{o}+a\cdot (t-t_{o}) (1)

Where:

v_{o} - Initial velocity, measured in meters per second.

v_{f} - Final velocity, measured in meters per second.

a - Acceleration, measured in meters per square second.

t_{o} - Initial time, measured in seconds.

t - Final time, measured in seconds.

Now we obtain the kinematic equations for thrust and free fall stages:

Thrust (v_{o} = 0\,\frac{m}{s}, a = 30\,\frac{m}{s^{2}}, t_{o} = 0\,s, 0\,s\le t< 30\,s)

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v = 900-9.81\cdot (t-30) (3)

Now we created the graph speed-time, which can be seen below.

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