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user100 [1]
3 years ago
11

An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s 1045 rad/s ). If a particular disk is

spun at 440.9 rad/s 440.9 rad/s while it is being read, and then is allowed to come to rest over 0.569 seconds 0.569 seconds , what is the magnitude of the average angular acceleration of the disk?
Physics
1 answer:
levacccp [35]3 years ago
6 0

Answer:

-775 rad/s^2

Explanation:

Knowing the initial and final angular speed is \omega_0 = 440.9 rad/s and \omega = 0 rad/s, respectively. We can use the following formula for equation of motion to calculate the average angular acceleration in t = 0.569 seconds

\Delta \omega = \alpha t

\omega - \omega_0 = \alpha t

0 - 440.9 = \alpha 0.569

\alpha = \frac{-440.9}{0.569} = -775 rad/s^2

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A punter wants to kick a football so that the football has a total flight time of 4.70s and lands 56.0m away (measured along the
Sindrei [870]

Answer:

Explanation:

How long does the football need to rise?

4.70/3 = 2.35 s

What height will the football reach?

h = ½(9.81)2.35² = 27.1 m

With what speed does the punter need to kick the football?

vy = g•t = 9.81(2.35) = 23.1 m/s

vx = d/t = 56.0/4.70 = 11.9 m/s

v = √(vx²+vy²) = 26.0 m/s

At what angle (θ), with the horizontal, does the punter need to kick the football?

θ = arctan(vy/vx) = 62.7°

3 0
2 years ago
A solution containing 10.0 g of an unknown liquid and 90.0 g water has a freezing point of -3.33 °C. Given Kf = 1.86 °C/m for wa
Fantom [35]

Answer:

62.06 g/mol

Explanation:

We are given that a solution containing 10 g of an unknown liquid and 90 g

Given mass of solute =W_B=10 g

Given mass of solvent=W_A=90 g

k_f=1.86^{\circ}C/m

Freezing point of solution =-3.33^{\circ}C

Freezing point of solvent =0^{\circ}C

Change in freezing point =Depression in freezing point

=Freezing point of solvent - freezing point of solution=0+3.33=3.33^{\circ}

\Delta T_f=\frac{W_B\times K_f\times 1000}{W_A\times M_B}

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M_B=62.06 g/mol

Hence, molar mass of unknown liquid is 62.06g/mol.

6 0
3 years ago
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A) Determine the x and y-components of the ball's velocity at t = 0.0s, 2.0, 3.0 secs.
malfutka [58]

The kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.

a) the position are

      time (s)  x (m)   y(m)

        0            0          0

        2.0         3.6        1.2

        3.0         5.4        0.9

b) The aceleration is  g = 0.6 m / s²

c) The launch angle      θ = 33.7º

given parameters

  • the initial velocity of the body vₓ = 1.8m / s and v_y = 1.2 m / s
  • the movement times t = 1.0s, 2.0s and 3.0 s

to find

    a) position

    b) acceleration

    c) launch angle

Projectile launch is an application of kinematics to the movement of the body in two dimensions where there is no acceleration on the x axis and the y axis has the planet's gravity acceleration

b) To calculate the acceleration of the plant acting on the y-axis, we use that the vertical velocity of the body at the highest point is zero.

         v_y = v_{oy} - g t

where v and v({oy}  are the velocities of the body, g the acceleration of the planet's gravity and t the time

          0 = v_{oy} - gt

           g = v_{oy} / t

from the graph we observe that the highest point occurs for t = 2.0 s

           g = 1.2 / 2.0

           g = 0.6 m / s²

 

a) The position is requested for several times

X axis

in this axis there is no acceleration so we can use the uniform motion relationships

          vₓ = x / t

          x = vₓ t

where x is the position, vx is the velocity and t is the time

we calculate for the time

t = 0.0 s

          x₀ = 0

           

t = 2.0 s

          x₂ = 1.8 2

          x₂ = 3.6 m

t = 3.0 s

          x₃ = 1.8 3

          x₃ = 5.4 m

Y axis

In this axis there is the acceleration of the planet, let us use for the position the relation

          y = v_{oy} t - ½ g t²

t = 0.0 s

          y₀ = 0

          y₀ = 0 m

t = 2.0 s

         y₂ = 1.2 2 - ½ 0.6 2²

         y₂ = 1.2 m

t = 3.0 s

        y₃ = 1.2  3 - ½  0.6  3²

        y₃ = 0.9 m

c) the launch angle use the trigonometry relation

        tan θ = \frac{v_y}{v_x}

        θ = tan⁻¹ \frac{v_y}{v_x}

        θ = tan⁻¹ \frac{1.2}{1.8}

        θ = 33.7º

measured counterclockwise from the positive side of the x-axis

With the kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.

a) the position are

      time (s)  x (m)   y(m)

        0            0          0

        2.0         3.6        1.2

        3.0         5.4        0.9

b) The aceleration is  g = 0.6 m / s²

c) The launch angle      θ = 33.7ºto)

learn more about projectile launch here:

brainly.com/question/10903823

4 0
2 years ago
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