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Kipish [7]
3 years ago
10

A5 kg frisbee is thrown from rest to a final speed of 12 m/s. What is the impulse of the frisbee?

Physics
1 answer:
lesya692 [45]3 years ago
3 0

Answer:

60kgm/s

Explanation:

Given parameters:

Mass of frisbee  = 5kg

Final speed  = 12m/s

Unknown:

Impulse of the frisbee  = ?

Solution:

The impulse of the frisbee is the same as the change in momentum.

It is given as:

 Impulse  = mass (Final velocity  -  Initial velocity)

 Impulse  = 5(12  - 0)  = 60kgm/s

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A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stati
Sergio [31]

Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

a)

  • If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:

       \tau = I * \alpha  (1)

  • Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
  • Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
  • τ = F*r (2)
  • For a solid uniform disk, the rotational inertia regarding an axle passing through its center  is just I = m*r²/2 (3).
  • Replacing (2) and (3) in (1), we can solve for α, as follows:

       \alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)

  • Since the angular acceleration is constant, we can use the following kinematic equation:

        \omega_{f}^{2}  - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)

  • Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:

       0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)

  • Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:

       \omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)

  • Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:

        v = \omega * r (8)

  • where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of  the disk, r becomes the radius of the disk, 0.200 m.
  • Replacing this value and (7) in (8), we get:

       v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)

b)    

  • There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:

       a_{t} = \alpha * r (9)

  • where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
  • Replacing this value and (4), in (9), we get:

       a_{t}  = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)

  • Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
  • The radial acceleration is just the centripetal acceleration, that can be expressed as follows:

       a_{c} = \omega^{2} * r  (11)

  • Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
  • Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
  • Replacing this value and (7) in (11) we get:

       a_{c} = \omega^{2} * r   = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)

  • The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
  • Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:

       a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)

6 0
2 years ago
The graph shows the amplitude of a passing wave over time in seconds (s). What is the approximate frequency of the wave shown?
belka [17]

Answer:

B) 0.3Hz

Explanation:

I just took the test i hope i helped and i hope you pass the test

7 0
3 years ago
Read 2 more answers
The sense of equilibrium responds to movements of the head. It is divided into two types--static and dynamic equilibrium. Descri
Sonja [21]

Answer:

nodding of head ,yes- static equilibrium

nodding of head, no- dynamic equilibrium.

Explanation:

static equilibrium monitors head position when body is not moving .

dynamic equilibrium monitors the angular or rotational movements of the head when body moves.

5 0
2 years ago
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Suppose an automobile has 2000-joules of kinetic energy. when it moves at twice the speed, what will be its kinetic energy? what
NeX [460]

Answer:

K.Eₓ = 4 K.E

K.Eₓ = 9 K.E

Explanation:

Th formula for the kinetic energy of a body is given as follows:

K.E = \frac{1}{2}mv^2\\   ---------------equation (1)

where,

K.E = Kinetic Energy of Automobile

m = mass of automobile

v = speed of automobile

For twice speed:

vₓ = 2v

then,

K.E_{x} = \frac{1}{2}mv_{x}^2\\K.E_{x} = \frac{1}{2}m(2v)^2\\K.E_{x} = 4\frac{1}{2}mv^2\\

using equation (1):

<u>K.Eₓ = 4 K.E</u>

For thrice speed:

vₓ = 3v

then,

K.E_{x} = \frac{1}{2}mv_{x}^2\\K.E_{x} = \frac{1}{2}m(3v)^2\\K.E_{x} = 9\frac{1}{2}mv^2\\

using equation (1):

<u>K.Eₓ = 9 K.E</u>

6 0
3 years ago
HELP PLEASE<br><br> answer choices: W X Y Z
Gala2k [10]
I believe it would be x because you are trying to find the width of the wave but i could be wrong 
6 0
3 years ago
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