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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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17. A volleyball weighs about 300 grams.
atroni [7]

Answer:

PE = 44.1 J

Explanation:

Ok, to have the specific data, the first thing we must do is convert from grams to kilograms. Since mass must always be in kilograms (kg)

We have:

  • 1 kilograms = 1000 grams.

We convert it using a rule of 3, replacing, simplifying units and solving:

  • \boxed{\bold{x=\frac{gr*1\ kg}{1000\ gr}=\frac{300\ gr*1\ kg}{1000\ gr}=\frac{300\ kg}{1000}=\boxed{\bold{0.3\ kg}}}}

==================================================================

Earth's gravity is known to be 9.8 m/s², so we have:

Data:

  • m = 0.3 kg
  • g = 9.8 m/s²
  • h = 15 m
  • PE = ?

Use formula of potencial energy:

  • \boxed{\bold{PE=m*g*h}}

Replace and solve:

  • \boxed{\bold{PE=0.3\ kg*9.8\frac{m}{s^{2}}*15\ m}}
  • \boxed{\boxed{\bold{PE=44.1\ J}}}

Since the decimal number, that is, the number after the comma is less than 5, it cannot be rounded, then we have this result.

The potential energy of the volleyball is <u>44.1 Joules.</u>

Greetings.

8 0
3 years ago
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Leno4ka [110]

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7 0
3 years ago
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valina [46]

Answer:

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Explanation:

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3 years ago
A girl exerts a horizontal force of 109 N on a crate with a mass of 31.2 kg. HINT (a) If the crate doesn't move, what's the magn
vesna_86 [32]

Answer:

(a) Magnitude of static friction force is 109 N

(b) Minimum possible value of static friction is 0.356

Solution:

As per the question;

Horizontal force exerted  by the girl, F = 109 N

Mass of the crate, m = 31.2 kg

Now,

(a) To calculate the magnitude of static friction force:

Since, the crate is at rest, the forces on the crate are balanced and thus the horizontal force is equal to the frictional force, f:

F = f = 109 N

(b) The maximum possible force of friction between the floor and the crate is given by:

f_{m} = \mu_{s}N

where

N = Normal reaction = mg

Thus

f_{m} = \mu_{s}mg

For the crate to remain at rest, The force exerted on the crate must be less than or equal to the maximum force of friction.

f\leq f_{m}

f \leq \mu_{s}mg

109 \leq \mu_{s}\times 31.2\times 9.8

\mu_{s}\geq 0.356

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A shopper walks westward 5.4 meters and then eastward 7.8 meters
RoseWind [281]

Answer:

13.2 meters

Explanation:

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6 0
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