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krek1111 [17]
3 years ago
13

In your physics lab you are given a 10.1-kg uniform rectangular plate with edge lengths 68.7 cm by 47.5 cm. Your lab instructor

requires you to rotate it about an axis perpendicular to its plane and passing through one of its corners and then prepare a report on the project. For your report you will need the plate's moment of inertia with respect to this axis. Calculate it here first.
Physics
1 answer:
VladimirAG [237]3 years ago
4 0

Answer:

2.35 kgm^2

Explanation:

we take length 68.7 cm as x-axis and 47.5 cm as y-axis then the axis about which we have to find out moment of inertia will be z-axis.

moment of inertia about x-axis

I_x = ML^2 /3 = 10.1\times 0.4752 /3 = 0.7596 kg-m2

I_y = 10.1\times 0.6872 / 3 = 1.5889 kgm^2

by perpendicular axis theorem

I_z = I_x + I_y = 0.7596 + 1.5889 = 2.35 kgm^2

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7 0
3 years ago
Read 2 more answers
A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a
SOVA2 [1]

Answer:

a)    x = v₀² sin 2θ / g

b)    t_total = 2 v₀ sin θ / g

c)    x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

         v₀ₓ = v₀ cos θ

         v_{oy} = 13.5 sin 32 = 7.15 m / s

         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

          v_{y} = v_{oy} - gt

          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

           t_total = 2 t

           t_total = 2 v₀ sin θ / g

c) we calculate

          x = 13.5 2 sin (2 32) / 9.8

          x = 16.7 m

5 0
3 years ago
brainly a child on a swing set swings back and forth with a period of 3.3 s and an amplitude of 25°. what is the maximum speed o
beks73 [17]

Maximum speed of the child as she swings is  2.23 m/s.

<h3>Step by Step Calculation:</h3>

T=3.3 s is the oscillation's time period.

The swing's greatest angle is 25° (max).

The swing's bottom will have the following kinetic energy:

k=12mv2...........(1)

The mass in this situation is m, and the speed is v.

The potential energy change is expressed as,

∆U=mgL1-cosθmax...............(2)

Here, L is a string's length and g is the acceleration caused by gravity. L is given as,

L=gT24π2

Combine equation (1) with (2)

12mv2=mgL1-cos, maxv=2g, maxv=gT24, maxv=g2T22, maxv=9.8 m/s

22.33 m/s, 22.31 s22.31 cos25°

Therefore, the child's top speed is 2.23 m/s.

<h3>What is Oscillation ?</h3>
  • The process of any quantity or measure fluctuating repeatedly about its equilibrium value in time is known as oscillation.
  • A periodic change in a substance's value between two values or around its central value is another way to define oscillation.

To learn more about Oscillation refer to:

brainly.com/question/28312746

#SPJ1

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1 year ago
The momentum of a 0.1 kg object traveling at 2000 m/s is 20,000 kg·m/s. True or False
Alina [70]

That's false.  

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"20,000 kg-m/s" has the correct units resulting from multiplication, but the number could only be the result of division.

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frosja888 [35]

Answer:

2 seconds

Explanation:

The frequency of a wave is related to its wavelength and  speed by the equation

f=\frac{v}{\lambda}

where

f is the frequency

v is the speed of the wave

\lambda is the wavelength

For the wave in this problem,

v = 2 m/s

\lambda=8 m

So the frequency is

f=\frac{2}{8}=0.25 Hz

The period of a wave is equal to the reciprocal of the frequency, so for this wave:

T=\frac{1}{f}=\frac{1}{0.25}=4 s

This means that the wave takes 4 seconds to complete one full cycle.

Therefore, the time taken for the wave to go from a point with displacement +A to a point with displacement -A is half the period, therefore for this wave:

t=\frac{T}{2}=\frac{4}{2}=2 s

4 0
2 years ago
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