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Bad White [126]
2 years ago
6

In an experiment, a torque of a known magnitude is exerted along the edge of a rotating disk. The disk rotates about its center.

All frictional forces are considered to be negligible. Which of the following quantities should a student collect in order to determine the change in angular momentum of the disk for a specific time interval? Justify your selection.
A) The initial angular velocity of the disk, because this quantity is related to the angular impulse of the disk.
B) The amount of time the torque is applied to the disk, because the time interval is related to the angular impulse of the disk.
C) The mass of the disk, because this quantity is related to the rotational inertia of the disk. The radius of the disk, because this quantity is related to the rotational inertia of the disk.
Physics
1 answer:
lisabon 2012 [21]2 years ago
3 0

Answer:

B) The amount of time the torque is applied to the disk, because the time interval is related to the angular impulse of the disk.

Explanation:

Angular impulse = Torque x time

= change in angular momentum

So,

Torque x time  = change in angular momentum

change in angular momentum = Torque x time

Torque is already known .

Hence to know the change in angular momentum what is needed to know is time duration of torque acting on the body .

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Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 4.15 m/s . Her husband Bruce suddenly reali
aleksandr82 [10.1K]

Answer:

a 15.22 m/s

b 45.65 m

Explanation:

Using the same formula,

x = vt, where

x is now 45.65, and

t is 3 s, then

45.65 = 3v

v = 45.65/3

v = 15.22 m/s

See the attachment for the part b. We used the distance gotten in part B, to find question A

5 0
1 year ago
A student has made the statement that the electric flux through one half of a Gaussian surface is always equal and opposite to t
nata0808 [166]

Answer:

E.true only when no charge is enclosed within the Gaussian surface.

Explanation:

Because Gauss’s law states that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge.

6 0
1 year ago
A 0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa
gregori [183]

The magnitude of the change in momentum of the stone is about 18.4 kg.m/s

\texttt{ }

<h3>Further explanation</h3>

Let's recall Impulse formula as follows:

\boxed {I = \Sigma F \times t}

<em>where:</em>

<em>I = impulse on the object ( kg m/s )</em>

<em>∑F = net force acting on object ( kg m /s² = Newton )</em>

<em>t = elapsed time ( s )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of ball = m = 0.500 kg

initial speed of ball = vo = 20.0 m/s

final kinetic energy = Ek = 70% Eko

<u>Asked:</u>

magnitude of the change of momentum of the stone = Δp = ?

<u>Solution:</u>

<em>Firstly, we will calculate the final speed of the ball as follows:</em>

Ek = 70\% \ Ek_o

\frac{1}{2} m v^2 = 70\% \ ( \frac{1}{2} m (v_o)^2 )

v^2 = 70 \% \ (v_o)^2

v = - v_o \sqrt{70 \%} → <em>negative sign due to ball rebounds</em>

v = - v_o \sqrt{0.7} \texttt{ m/s}

\texttt{ }

<em>Next, we could find the magnitude of the change of momentum of the stone as follows:</em>

\Delta p_{stone} = - \Delta p_{ball}

\Delta p_{stone} = - [ mv - mv_o ]

\Delta p_{stone} = m[ v_o - v ]

\Delta p_{stone} = m[ v_o + v_o\sqrt{0.7} ]

\Delta p_{stone} = mv_o [ 1 + \sqrt{0.7} ]

\Delta p_{stone} = 0.500 ( 20.0 ) [ 1 + \sqrt{0.7} ]

\Delta p_{stone} \approx 18.4 \texttt{ kg.m/s}

\texttt{ }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Average Speed of Plane : brainly.com/question/12826372
  • Impulse : brainly.com/question/12855855
  • Gravity : brainly.com/question/1724648

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

8 0
2 years ago
Why does the closed top of a convertible bulge when the car is riding along a highway? Why does the closed top of a convertible
iren2701 [21]

Answer:

Option D

The air pressure inside the car is greater than the pressure outside.

Explanation:

When considering airflow over and around a surface, from Bernoulli's equation, air flow regions with higher velocity have a lower pressure, and regions with lower velocity have a higher pressure.

The air outside the convertible is moving faster than the air inside the convertible. This leads to a higher pressure zone just below the surface of the roof (inside the car) causing the roof of the convertible to bulge upwards

4 0
2 years ago
Modern wind turbines generate electricity from wind power. The large, massive blades have a large moment of inertia and carry a
horsena [70]

Answer:

a)106.48 x 10⁵ kg.m²

b)144.97 x 10⁵  kgm² s⁻¹  

Explanation:

a)Given

m = 5500 kg

l = 44 m

Moment of inertia of one blade

I= 1/3 x m l²

where m is mass of the blade

l is length of each blade.

Putting all the required values, moment of inertia of one blade will be

I= 1/3 x 5500 x 44²  

I= 35.49 x 10⁵ kg.m²

Moment of inertia of 3 blades

I= 3 x 35.49 x 10⁵ kg.m²

I= 106.48 x 10⁵ kg.m²

b) Angular momentum 'L' is given by

L =I x ω

where,

I= moment of inertia of turbine i.e  106.48 x 10⁵ kg.m²

ω=angular velocity =2π f

f is frequency of rotation of blade i.e  13 rpm

f = 13 rpm=>= 13 / 60 revolution per second

ω = 2π f =>  2π  x  13 / 60 rad / s

L=I x ω =>106.48 x 10⁵ x   2π  x  13 / 60

  = 144.97 x 10⁵  kgm² s⁻¹    

7 0
1 year ago
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