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V125BC [204]
2 years ago
14

A uniform rod of mass M and length L is free to swing back and forth by pivoting a distance x from its center. It undergoes harm

onic oscillations by swinging back and forth under the influence of gravity. Randomized Variables M=2.2 kg x= 0.49 m 33% Part (a) In terms of M, L, and x, what is the rod' s moment of inertia / about the pivot point?
Physics
1 answer:
Alisiya [41]2 years ago
5 0

Answer:

The moment of inertia is 0.7500 kg-m².

Explanation:

Given that,

Mass = 2.2 kg

Distance = 0.49 m

If the length is 1.1 m

We need to calculate the moment of inertia

Using formula of moment of inertia

I=\dfrac{1}{12}ml^2+mx^2

Where, m = mass of rod

l = length of rod

x = distance from its center

Put the value into the formula

I=\dfrac{1}{12}\times2.2\times(1.1)^2+2.2\times(0.49)^2

I=0.7500\ kg-m^2

Hence, The moment of inertia is 0.7500 kg-m².

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The air in a pipe resonates at 150 Hz and 750 Hz, one of these resonances being the fundamental. If the pipe is open at both end
Xelga [282]

Answer:

Explanation:

Two frequencies with magnitude 150 Hz and 750 Hz are given

For Pipe open at both sides

fundamental frequency is 150 Hz as it is smaller

frequency  of pipe is given by

f=\frac{nv}{2L}

where L=length of Pipe

v=velocity of sound

f=150\ Hz for n=1

and f=750 is for n=5

thus there are three resonance frequencies for n=2,3 and 4

For Pipe closed at one end

frequency is given by

f=\frac{(2n+1)}{4L}\cdot v

for n=0

f_1=\frac{v}{4L}

f_1=150\ Hz

for n=2

f_2=\frac{5v}{4L}

Thus there is one additional resonance corresponding to n=1 , between f_1 and f_2

8 0
2 years ago
A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a
Lapatulllka [165]

Answer:

Amplitude, A = 0.049 meters

Explanation:

Given that,

A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a function of time according to the equation :

y = 0.049 \cos(7t) .......(1)

The general equation of a wave is given by :

y=A\cos(\omega t) .......(2)

A is amplitude of wave

On comparing equation (1) and (2) we get :

A = 0.049 meters

So, the amplitude of the wave is 0.049 meters.

3 0
2 years ago
A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direc
Alborosie

The given question is incomplete. The complete question is as follows.

A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direction, and a magnetic field of magnitude 1.25 T points in the positive z direction.

A) If the net force acting on the particle is 6.21 \times 10^{-3} N in the positive x direction, find the components of the particle's velocity. Assume the particle's velocity is in the x-y plane.

Enter your answers numerically separated by commas.

Explanation:

The given data is as follows.

           Q = 6.50 \times 10^{-6} C

           E = 1300 N/C in the +x direction

           B = 1.02 T in the +z direction

and,    F_{net} = 6.25 \times 10^{-3} N in the +x direction

Also,       F_{net} = F_{E} - F_{b}

                         = qE - qvB

Now, we will calculate the value of v as follows.

             v = (\frac{1}{B}) \times (E - \frac{F_{net}}{q})

                 = (\frac{1}{1.02 T}) \times (1300 - \frac{6.25 \times 10^{-3}}{6.50 \times 10^{-6}})

                v = 458.507 m/s

Using the value for velocity, we need to know which direction it's going.

You know +x direction for E, +z direction for B and +x for F_{net}.

Using the right hand rule where:

your right thumb goes toward the F_{net}, then your index finger points to B (z direction) Then curl your middle, ring, and pink 90 angle. This shows where v is going which is -y direction.

Thus, we can conclude that v_{x}, v_{y}, v_{z} = 0, -(458.507), 0.

8 0
2 years ago
Corey, whose mass is 95 kg, stands on a bathroom scale in an elevator. The scale reads 850 N for the first 3.0 s after the eleva
lyudmila [28]

Answer:

v₂ = 2.568 m/s

Explanation:

given,

mass of Corey = 95 Kg

reading of sale for first 3 s when elevator start to move = 850 N

scale reading for the next 3.0 s = 930 N

Gravitation force acting =

  F = m g

  F = 95 x 9.8

  F = 931 N

using newtons second law, due to movement of elevator

    F_{net} = m a

 W - N = m a₁

931- 850 = 95 x a₁

    a₁ = 0.852 m/s²

now,

velocity calculation

v₁ = a₁t

v₁ = 0.852 x 3 = 2.557 m/s

now, For second case

931 - 930 = 95 x a₂

a₂ = 0.011 m/s²

now, velocity after 4 s

v₂ = v₁ + a₂ t

v₂ = 2.557+ 0.011 x (4 - 3)

(4-3) because velocity after 3 second is calculate we need to calculate velocity after 4 s from beginning.

v₂ = 2.557 + 0.011

v₂ = 2.568 m/s

velocity of the elevator is equal to v₂ = 2.568 m/s

7 0
2 years ago
During a hurricane, the atmospheric pressure inside a house may blow off the roof because of the reduced pressure outside. If ai
m_a_m_a [10]

Answer:

817.5 Pa

Explanation:

From Bernoulli's equation, considering thst there is no height difference then

P1+½d(v1)²=P2+½d(v2)²

P1-P2=½d(v2²-v1²)

∆P=½d(v2²-v1²)

Where P represent pressure, d is density and v is velocity. Subscripts 1 and 2 represent inside and outside. ∆P is tge change in pressure

Given the speed at roof top as 128 km/h, we convert it to m/s as follows

128*1000/3600=35.555555555555=35.56 m/s

Velocity at the bottom of roof is 0 m/s

Density is given as 1.293 kg/m³

∆P=½*1.293*(35.56²-0)=817.5 Pa

5 0
2 years ago
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