PHYSICS COLLEGE

A solid cylindrical object has a mass of 2.0 kg, a diameterof0.10m, and a length of 0.18m. What is the moment of inertiaof the
body about an axis along the axis of thecylinder?

Answers

Answer 1
Answer:

Answer:

I = 0.0025 kg.m²

Explanation:

Given that

m= 2 kg

Diameter ,d= 0.1 m

Radius ,R=(d)/(2)

R=(0.1)/(2)

R=0.05 m

The moment of inertia of the cylinder about it's axis same as the disc and it is given as

I=(mR^2)/(2)

Now by putting the all values

I=(2* 0.05^2)/(2)

I = 0.0025 kg.m²

Therefore we can say that the moment of inertia of the cylinder will be  0.0025 kg.m².

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Consider the interactions involved when you use a TV remote control to change the channel. Classify each interaction as long range or short range.

Answers

Explanation:

Following are two interactions that are generally involved when we use a TV remote control to change the channel :

1. Figure touches remote buttons, and its a short range interaction.

2. Now remote sends signal to Television, this is a long range interaction.

Final answer:

The interactions of a TV remote and the TV involve short-range infrared communication, while the TV receives signals from long-range electromagnetic waves broadcasted for channels in frequency ranges for VHF and UHF.

Explanation:

When you use a TV remote control to change the channel, two main interactions are involved. The first interaction is the infrared communication between the remote and the TV, which is a form of electromagnetic radiation. Infrared signals require a direct line of sight, operating over a relatively short range. On the other hand, the TV itself receives broadcast signals through antennas that capture electromagnetic waves broadcasted over a long range - these signals can be VHF or UHF TV channels.

Additionally, the TV channels are broadcasted on frequencies ranging from 54 to 88 MHz and 174 to 222 MHz for VHF, while UHF channels utilize frequencies from 470 to 1000 MHz. These signals are sent over a significant distance to your TV’s antenna, showing that television broadcast interaction is long range. These broadcast signals are part of electromagnetic spectrum and carry a large range of frequencies due to the variety of content (audio and visual information) that needs to be transmitted.

Suppose that we are designing a cardiac pacemaker circuit. The circuit is required to deliver pulses of 1ms duration to the heart, which can be modeled as a 500 ohm resistance. The peak amplitude of the pulses is required to be 5 V. However, the battery delivers only 2.5 V. Therefore, we decide to charge two equal value capacitors in parallel from the 2.5V battery and then switch the capacitors in series with the heart during the 1ms pulse. What is the minimum value of the capacitances required so the output pulse amplitude remains between 4.9 V and 5.0 V throughout its 1ms duration

Answers

Answer:

Minimum capacitance = 200 μF

Explanation:

From image B attached, we can calculate the current flowing through the capacitors.

Thus;

Since V=IR; I = V/R = 5/500 = 0.01 A

Maximum charge in voltage is from 5V to 4.9V. Thus, each capacitor will have 2.5V. Hence, change in voltage(Δv) for each capacitor will be ; Δv = 0.05 V

So minimum capacitance will be determined from;

i(t) = C(dv/dt)

So, C = i(t)(Δt/Δv) = 0.01[0.001/0.05]

C = 0.01 x 0.0002 = 200 x 10^(-6) F = 200 μF

A Venturi tube may be used as the inlet to an automobile carburetor. If an inlet pipe with a diameter of 2.0 cm diameter narrows to diameter of 1.0 cm, determine the pressure drop in the constricted section for an initial airflow of 3.0 m/s in the 2-cm section? (Assume air density is 1.25 kg/m

Answers

The pressure drop is equal to 80.99 Pa

Given information:

d1 = 2 cm = 0.02 m

d2 = 1 cm = 0.01 m

v = 3 m/s

p = 1.25 kg/m^3

Here we use Bernoulli's principle for the Venturi Tube:

Calculation of pressure drop:

P1 - P2 = ((v^2* p)/ 2)* ((A1^2/ A2^2)-1)\n\nP1 - P2 = \Delta P = ((v1^2* p)/ 2)* ((A1^2/ A2^2)-1)

Now the following formula for area calculation should be used:

A1 = (\pi* d1^2)/ 4 = (\pi* (0.02 m)^2)/ 4 = 0.00031 m^2\n\nA2 = (\pi* (0.01 m)^2)/ 4 = 0.000079 m^2\n\n\Delta P = ((3 m/s)^2 *1.25 kg/m^3)/ 2) * ((0.00031 m^2)^2/(0.000079 m^2)^2)-1)

= 80.99

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Answer:

the pressure drop is equal to 80.99 Pa

Explanation:

we have the following data:

d1 = 2 cm = 0.02 m

d2 = 1 cm = 0.01 m

v = 3 m/s

p = 1.25 kg/m^3

ΔP = ?

For the calculation of the pressure drop we will use Bernoulli's principle for the Venturi Tube:

P1 - P2 = ((v^2*p)/2)*((A1^2/A2^2)-1)

where A = area

P1 - P2 = ΔP = ((v1^2*p)/2)*((A1^2/A2^2)-1)

for the calculation of the areas we will use the following formula:

A1 = (pi*d1^2)/4 = (pi*(0.02 m)^2)/4 = 0.00031 m^2

A2 = (pi*(0.01 m)^2)/4 = 0.000079 m^2

ΔP = ((3 m/s)^2*1.25 kg/m^3)/2)*((0.00031 m^2)^2/(0.000079 m^2)^2)-1) = 80.99 N/m^2 = Pa

An object of mass 10.0kg is released at point A, slidesto the bottom of the 30° incline, then collides with ahorizontal massless spring, compressing it a maximumdistance of 0.750m. (See below.) The spring constant is 500N/m, the height of the incline is 2.0 m, and the horizontalsurface is frictionless. (a) What is the speed of the object atthe bottom of the incline? (b) What is the work of frictionon the object while it is on the incline? (c) The springrecoils and sends the object back toward the incline. Whatis the speed of the object when it reaches the base of theincline? (d) What vertical distance does it move back up theincline?

Answers

Final answer:

The final speed at the bottom of the incline can be calculated using the conservation of energy principle. There is no work done against friction as the object is moving on a frictionless surface. The speed does not change when the spring pushes it back towards the base of the incline due to lack of friction and it moves to a certain height given the angle of the incline and the initial speed.

Explanation:

  1. "Speed at the bottom of the incline:" This can be calculated using conservation of energy. The potential energy at the top (m*g*h) will convert into kinetic energy at the bottom (1/2*m*v^2). Here, m is the mass, g is acceleration due to gravity, h is the height, and v is the velocity. Using this, we can solve for v.
  2. Work of friction on the incline: As per the question, the surface is frictionless. Therefore, the work done by friction is automatically 0 as there is no force of friction.
  3. Speed of the object when it reaches the base of the incline again: As the surface is frictionless, the object reaches the incline with the same speed with which it left as there are no opposing forces to reduce its momentum.
  4. Vertical distance it moves back up the incline: This can be calculated using the principles of conservation of energy and kinematic equations, taking into account the angle of the incline and the velocity of the object.

Learn more about Mechanics and Energy Conservation here:

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Which describes the motion of the box based on the resulting free-body diagram?1. It is moving up with a net force of 20 N.
2. It is moving to the right with a net force of 10 N.
3. It is in dynamic equilibrium with a net force of 0 N.
4. It is in static equilibrium with a net force of 0 N.

Answers

The statement "It is in dynamic equilibrium with a net force of 0 N" describes the motion of the box based on the resulting free-body diagram. (option 3)

What is a free-body diagram?

A free-body diagram is a diagram that shows all the forces acting on an object. If the net force on an object is zero, then the object is in equilibrium. This means that the object is not accelerating and is either at rest or moving with constant velocity.

In the case of the box in the free-body diagram, there are two forces acting on it: the force of gravity and the force of the table pushing up on the box. The force of gravity is pulling the box down, but the force of the table is pushing the box up.

These two forces are equal in magnitude and opposite in direction, so they cancel each other out. This means that the net force on the box is zero and the box is in dynamic equilibrium.

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Answer:

4. It is in static equilibrium with a net force of 0 N.

Explanation:

Just got it right :)

When water freezes, its volume increases by 9.05% (that is, ΔV / V0 = 9.05 × 10-2). What force per unit area is water capable of exerting on a container when it freezes? (It is acceptable to use the bulk modulus of water, B = 2.2 × 109 N/m2, for this problem.) Give your answer in N/cm2.

Answers

Answer:

P = 1.99 10⁸ Pa

Explanation:

The definition of the bulk module is

      B = - P / (ΔV / V)

The negative sign is included for which balk module is positive, P is the pressure and V that volume

They tell us that the variation in volume is 9.05%, that is

    ΔV / V = ​​9.0Δ5 / 100 = 0.0905

    P = - B DV / V

    P = 2.2 10⁹ (0.0905)

    P = 1.99 10⁸ Pa
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