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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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A steel rod with a length of l = 1.55 m and a cross section of A = 4.45 cm2 is held fixed at the end points of the rod. What is
Blababa [14]

To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.

F = AY\alpha \Delta T

Where

A = Cross-sectional Area

Y = Young's modulus

\alpha= Coefficient of linear expansion for steel

\Delta T= Temperature Raise

Our values are given as,

A = 4.45cm^2

T = 37K

\alpha = 1.17*10^{-5}K^{-1}

Y = 200*10^9Gpa

Replacing we have,

F = (4.45*10^{-4})(200*10^9)(1.17*10^{-5})(37)

F = 38526.1N

Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N

7 0
2 years ago
What is the thickness required of a masonry wall having thermal conductivity 0.75 W/m K if the heat rate is to be 80% of the hea
Leya [2.2K]

Answer:

the thickness required of a masonry wall L = 375mm

Explanation:

The detailed steps and appropriate use of fourier's law of heat conduction is as shown in the attached file.

5 0
2 years ago
The suspension cable of a 1,000 kg elevator snaps, sending the elevator moving downward through its shaft. The emergency brakes
tester [92]

Answer:

option (E) 1,000,000 J

Explanation:

Given:

Mass of the suspension cable, m = 1,000 kg

Distance, h = 100 m

Now,

from the work energy theorem

Work done by the gravity = Work done by brake

or

mgh = Work done by brake

where, g is the acceleration due to the gravity = 10 m/s²

or

Work done by brake  = 1000 × 10 × 100

or

Work done by brake = 1,000,000 J

this work done is the release of heat in the brakes

Hence, the correct answer is option (E) 1,000,000 J

4 0
2 years ago
In coordinates with the origin at the barn door, the cow walks from x 0 to x 6.9 m as you apply a force with x component Fx 320.
Stella [2.4K]

Answer:

-209.42J

Explanation:

Here is the complete question.

A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component Fx=−[20.0N+(3.0N/m)x]. How much work does the force you apply do on the cow during this displacement?

Solution

The work done by a force W = ∫Fdx since our force is variable.

Since the cow moves from x₁ = 0 m to x₂ = 6.9 m and F = Fx =−[20.0N+(3.0N/m)x] the force applied on the cow.

So, the workdone by the force on the cow is  

W = ∫₀⁶°⁹Fx dx = ∫₀⁶°⁹−[20.0N+(3.0N/m)x] dx

= ∫₀⁶°⁹−[20.0Ndx - ∫₀⁶°⁹(3.0N/m)x] dx

= −[20.0x]₀⁶°⁹ - [3.0x²/2]₀⁶°⁹

= -[20 × 6.9 - 20 × 0] - [3.0 × 6.9²/2 - 3.0 × 0²/2]

= -[138 - 0] - [71.415 - 0] J = (-138 - 71.415) J

= -209.415 J ≅ -209.42J

5 0
2 years ago
A projectile has an initial horizontal velocity of 15 meters per second and an initial vertical velocity of 25 meters per second
Artyom0805 [142]

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

v_x = 15 m/s

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

d= v_x t =(15 m/s)(5 s)=75 m

5 0
2 years ago
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