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V125BC [204]
1 year 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]1 year 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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A frog leaps up from the ground and lands on a step 0.1 m above the ground 2 s later. We want to find the vertical velocity of t
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Answer:

<em>You would use the kinematic formula:</em>

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Explanation:

The upwards vertical motion is ruled by the equation:

        y=y_0+V_{0y}\times t-g\times t^2/2

Where:

       y \text{ is the position at the time }t:y=0.1m

       y_0\text{ is the initial position: }y_0=0

       t=2s

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Naming Δy = y - y₀, the equation becomes:

      \Delta y=V_{0y}\times t-g\times t^2/2

Then, you just need to substitute with Δy = 0.1m, t = 2s, and g = 9.8m/s², ans solve for the intital vertical velocity.

7 0
1 year ago
Read 2 more answers
A bar of silicon is 4 cm long with a circular cross section. If the resistance of the bar is 270 ω at room temperature, what is
vova2212 [387]

solution:

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7 0
2 years ago
Two window washers, Bob and Joe, are on a 3.00 m long, 395 N scaffold supported by two cables attached to its ends. Bob weighs 8
WINSTONCH [101]

Answer:

- the forces on the left hand side is 1.038 kN

- the forces on the right hand side is 1.483 kN

Explanation:

Given the data in the question, as illustrated in the image below;

Length of the scaffold = 3 m

weight of the scaffold = 395 N

Weight of Bob = 805 N and stands 1 m from the left end

weight of washing equipment = 500N and on sits 2 m from the left end

Weight of Joe = 820 N and stand 0.500 m from the right end

so the force on the left cable will be;

T_{left = \frac{1}{3m}[ (805 N)( (3-1) m) + ( 395 N )( \frac{3}{2} m) + ( 500 N )(1m ) + ( 820 N)( 0.500m ) ]

T_{left =  \frac{1}{3m}[ 1610 + 592.5 + 500 + 410 ]

T_{left =  \frac{1}{3m}[ 3112.5 ]

T_{left =  1037.5 N

T_{left =  1.038 kN

Therefore, the forces on the left hand side is 1.038 kN

On the right hand side;

T_{Right =  ( 805 N + 395 N + 500 N + 820 N ) - 1037.5 N

T_{Right =  2520 N - 1037.5 N

T_{Right =  1482.5 N

T_{Right =  1.483 kN

Therefore, the forces on the right hand side is 1.483 kN

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1 year ago
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Leto [7]

Answer:

Explanation:

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initial total momentum = final angular momentum

I₁ X ωi  = ( I₁ + I₂ )ωf

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ωf =  ωi   / ( 1 + m r²/MR² )

6 0
1 year ago
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exis [7]
<span>Acceleration is the change in velocity divided by time taken. It has both magnitude and direction. In this problem, the change in velocity would first have to be calculated. Velocity is distance divided by time. Therefore, the velocity here would be 300 m divided by 22.4 seconds. This gives a velocity of 13.3928 m/s. Since acceleration is velocity divided by time, it would be 13.3928 divided by 22.4, giving a final solution of 0.598 m/s^2.</span>
7 0
2 years ago
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