A uniform steel rod of length 0.9 m and mass 3.8 kg has two point masses of 2.3 kg each at the two ends. Calculate the moment of inertia of the system about an axis perpendicular to the rod, and passing through its center.

Answers

Answer 1

Answer: [tex]2.4705\ kg.m^2[/tex]

Explanation:

Given

length of the rod is L=0.9 m

Mass of the rod m=3.8 kg

Point masses has mass of m=2.3 kg

Moment of Inertia of the rod about the center is

[tex]\Rightarrow I_o=\dfrac{1}{12}ML^2[/tex]

Moment of inertia of combined system is the sum of rod and two point masses.

[tex]\Rightarrow I=I_o+2mr^2[/tex]

[tex]\Rightarrow I=\dfrac{1}{12}3.8\times 0.9^2+2\times 2.3\times \left(\dfrac{0.9}{2}\right)^2\\\\\Rightarrow I=1.539+0.9315\\\Rightarrow I=2.4705\ kg-m^2[/tex]


Related Questions

how much amount of heat energy is required to convert 5 kg of ice at - 5° c into 100°c steam?​

Answers

Assuming no heat lost to the surrounding,

-5⁰C ice → 0⁰C ice

Specific heat capacity of ice = 2.0 x 10³ J/kg/⁰C

Q = mc∆θ

Q = 5(2.0 x 10³) x (0-(-5))

Q = 50000J

0⁰C ice → 0⁰C water

Specific latent heat of fusion of ice = 3.34 x 10⁵J/kg

Q = mLf

Q = 5(3.34 x 10⁵)

Q = 1670000J

0⁰C water → 100⁰C water

Specific heat capacity of water = 4.2 x 10³ J/kg/⁰C

Q = mc∆θ

Q = 5(4.2 x 10³) x (100-0)

Q = 2100000J

100⁰C water → 100⁰C steam

Specific latent heat of vaporization of water = 2.26 x 10⁶ J/kg

Q = mLv

Q = 5(2.26 x 10⁶)

Q = 11300000J

Total amount of heat required

= 50000 + 1670000 + 2100000 + 11300000

= 15120000J

A body of mass 2kg is released from from a point 100m above the ground level. calculate kinetic energy 80m from the point of released.​

Answers

Answer:

1568J

Explanation:

Since the problem states 80 m from the point of drop, the height relative to the ground will be 100-80=20m.

Use conservation of Energy

ΔUg+ΔKE=0

ΔUg= mgΔh=2*9.8*(20-100)=-1568J

ΔKE-1568J=0

ΔKE=1568J

since KEi= 0 since the object is at rest 100m up, the kinetic energy 20meters above the ground is 1568J

An electron has an initial speed of 8.06 x10^6 m/s in a uniform 5.60 x 10^5 N/C strength electic field.The field accelerates the electron in the direction opposite to its initial velocity.
(a) What is the direction of the electric field?
i. opposite
ii. direction to the electron's initial velocity
iii. same direction as the electron's initial velocity
iv. not enough information to decide
(b) How far does the electron travel before coming to rest? m
(c) How long does it take the electron to come to rest? s
(d) What is the electron's speed when it returns to its starting point?

Answers

Answer:

Explanation:

a)

The force on electron acts opposite to the velocity , and direction of force on electron is always opposite to direction of electric field .

Hence direction of electric field must be in the same  in which electrons travels.

Hence option iii is correct.

b )

deceleration a = force / mass

= qE / m

= 1.6 x 10⁻¹⁶ x 5.6 x 10⁵ / 9.1 x 10⁻³¹

= .98 x 10²⁰ m /s²

v² = u² - 2 a s

0 = (8.06 x 10⁶ )² - 2 x .98 x 10²⁰ s

s = 64.96 x 10¹² / 1.96 x 10²⁰

= 33.14 x 10⁻⁸ m

c ) time required

= 8.06 x 10⁶ / .98 x 10²⁰

= 8.22 x 10⁻¹² s .

d ) Its speed will be same as that in the beginning ie 8.06 x 10⁶ m/s .

Answer:

(a) Option (i)

(b) 6.6 x 10^-4 m  

(c) 8.2 x 10^-11 s

Explanation:

initial velocity, u = 8 .06 x 10^6 m/s

Electric field, E = 5.6 x 10^5 N/C

(a) The direction of field is opposite.

Option (i).

(b) Let the distance is s.  

Use third equation of motion

[tex]v^2 = u^2 + 2 a s \\\\0 = u^2 - 2 \times \frac{qE}{m}\times s\\\\8.06\times 10^6\times 8.06\times 10^6 = \frac {1.6\times 10^{-19}\times 5.6\times 10^5}{9.1\times 10^{-31}} s\\\\s = 6.6\times 10^{-4} m[/tex]

(c) Let the time is t.

Use first equation of motion.

[tex]v = u + a t \\\\0 = u - \times \frac{qE}{m}\times t\\\\8.06\times 10^6 = \frac {1.6\times 10^{-19}\times 5.6\times 10^5}{9.1\times 10^{-31}} t\\\\t = 8.2\times 10^{-11} s[/tex]

If you buy an amateur-sized reflecting telescope, say around 10 inches (25cm) aperture, it'll have something in it that sends the gathered starlight out the side of the telescope tube. What do we call this thing

Answers

Answer: objective lens

Explanation:

Light enters a refra

Light enters a telescope through a lens at the upper end, which focuses the light near the bottom of the telescope. An eyepiece then magnifies the image so that it can be viewed by the eye, or a detector like a photographic plate can be placed at the focus. The upper end of a reflecting telescope is open, and the light passes through to the mirror located at the bottom of the telescope. The mirror then focuses the light at the top end, where it can be detected. Alternatively, a second mirror may reflect the light to a position outside the telescope structure, where an observer can have easier access to it.

The velocity-time graph of a body is given. What quantities are represented by (a) slope of the graph and (b) area under the graph?​

Answers

Answer:

a) acceleration

b) displacement

Explanation:

The velocity-time graph is a graph of velocity versus time. The velocity (m/s) would be on the Y-axis while time (s) would be on the X-axis.

a) The slope of a graph is given by: change in Y-axis/change in X-axis = ΔY/ΔX

In a velocity-time graph, ΔY = change in velocity and ΔX = change in time.

Hence, the slope of a velocity-time graph becomes: change in velocity/change in time.

Also, acceleration = change in velocity/change in time.

Hence, the slope of a velocity-time graph = acceleration.

b) Assuming that the area under a velocity-time graph is a rectangle, the area is given as:

Area of a rectangle = length x breadth

                                  = velocity x time (m/s x s)

Also, displacement = velocity x time (m)

Hence, the area under a velocity-time graph of a body would give the displacement of the body.

Find the refractive index of a medium
having a velocity of 1.5 x 10^8*

Answers

Explanation:

refractive index ,is the ratio of velocity of light in vacuum to the velocity of light a medium

At which point is the kinetic energy of the pendulum the greatest?

C

A

D

B

Answers

Answer:

Point C

Explanation:

Greatest Kinetic Energy means lowest potential energy since energy is conserved. Lowest potential energy means lowest height which is at Point C.

A water-balloon launcher with mass 2 kg fires a 0.75 kg balloon with a
velocity of 14 m/s to the west. What is the recoil velocity of the launcher?
What is the answer

Answers

Answer:

5.25 m/s to the east

Explanation:

Applying,

MV = mv.............. Equation 1

Where M = mass of the launcher, V = recoil velocity of the launcher, m = mass of the balloon, v = velocity of the balloon

make V the subject of the equation

V = mv/M............ Equation 2

From the question,

M = 2 kg, m = 0.75 kg, v = 14 m/s

Substitute these values into equation 2

V = (0.75×14)/2

V = 5.25 m/s to the east

A uniform metre rule of mass 10g is balanced on a knife edge placed at 45cm mark. Calculate the distance of a mass 25g from the pivot​

Answers

Answer:

The distance of a mass 25g from the pivot​ is 18cm

Explanation:

Given

[tex]m_1 = 10g[/tex]

[tex]d_1 = 45cm[/tex]

[tex]m_2 = 25g[/tex]

Required

Distance of m2 from the pivot

To do this, we make use of:

[tex]m_1 * d_1 = m_2 * d_2[/tex] --- moments of the masses

So, we have:

[tex]10 * 45= 25* d_2[/tex]

[tex]450= 25* d_2[/tex]

Divide both sides by 25

[tex]18= d_2[/tex]

Hence:

[tex]d_2 = 18[/tex]

You are stranded in a stationary boat. Your friend is on a dock, but the boat is just beyond his reach. There is a 5 kg anchor in the boat. You'd like to get the boat to move closer to the dock so your friend can rescue you. Select from the following list what effect each change will have on the position of the boat relative to the dock. A. The boat will move closer to the dock. B. The boat will move away from the dock. C. The position of the boat relative to the dock will not change.

Answers

Answer:

running away and launching the anchor that will give a greater speed towards the dock v₄.

Explanation:

To try to bring the boat closer to the dock, several cases can be carried out.

* move inside the ship so that the center of mass changes and since moving away you have a speed v, the ship will approach the dock at a speed v₂,

* Throw the anchor in the opposite direction to the dock so that using the conservation of the moment the boat moves towards it, it moves at a speed v₃

* A combination of the two processes running away and launching the anchor that will give a greater speed towards the dock v₄.

In all cases, the friction must be zero.

All other movements move the ship away from the dock

Monique walks 360 meters to get to lanier on days when she is early and doesn't get caught by traffic it takes her 60 seconds to get to school how fast was she running

Answers

Answer:

6m/s

Explanation:

We are to calculate the speed of Monique

Speed = Distance/Time

Given

Distance = 360m

Time = 60secs

Substitute

Speed = 360m/60s

Soeed = 6m/s

Hence she was running at 6m/s

Please help,it is urgent!)

Answers

Answer:

answer is 18.58 because

Answer:

My answer is d.25.1 because

Differentiate between Scalar quantity and Vector quantity and give two examples each​

Answers

Answer:

A scalar quantity is a physical quantity which only has magnitude. Scalars can be added algebraically.A vector quantity is a physical quantity which has both magnitude and direction. Vectors should be added vectorially.

David is driving a steady 30 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.0 m/s2 at the instant when David passes. How far does Tina drive before passing David?

Answers

a. 441 m B: 46.0 m/s

Magnets produce _________ in the spaces surrounding them

Answers

Answer:

magnetic field

Explanation:

g As they reach higher temperatures, most semiconductors... Selected Answer: have an increased resistance. Answers: have a constant resistance. have an increased resistance. have a decreased resistance.

Answers

Answer:

have an increased resistance

Which image illustrates reflection?
A
B
с
D

Answers

Answer: I beleive A

Explanation:

Answer:

A

Explanation:

We can see the light being reflected off the mirror.

if a body covers 100m in 5 second from rest find the acceleration produced by a body in 10 second​

Answers

Answer:

a=10m/s^2

Explanation:

acceleration= final velocity+ initial velocity/time taken

v-u/t=a

100-0/5=a

100/5=a

a=20m/s^2

case2

100-0/10=a

100/10=a

a=10m/s^2

Don't forget to write the units.

Hope this helps

please mark me as brainliest.

convert 2.4 milimetres into metre​

Answers

0.0024

Milimetres are before centimetres and centimetres are before metres

Answer: 2.4 millimeters = 0.0024 meters

Explanation: A millimeter is 1/1000 of a meter. By diving 2.4 by 1000, you get 0.0024.

When an apple falls towards the earth,the earth moves up to meet the apple. Is this true?If yes, why is the earth's motion not noticeable?

Answers

Answer:

Yes, when an apple falls towards the earth, the apple gets accelerated and comes down due to the gravitational force of attraction used by the earth. The apple also exerts an equal and opposite force on the earth but the earth does not move because the mass of the apple is very small, due to which the gravitational force produces a large acceleration in it (a = F/m) but the mass of the earth is very large, the same gravitational force produces very small acceleration in the earth and we don't see the earth rising towards the apple.

Which of the following represents the velocity time relationship for a falling apple?

Answers

Answer "a" would be correct.

Answer:

d

Explanation:

There's an acceleration from gravity, thus the velocity is becoming faster and faster as it reaches the ground. Thus its D

Brainliest please~

The correct equation for the x component of a vector named A with an angle measured from the x axis would be which of the following?

Answers

Answer:

Acosθ

Explanation:

The x-component of a vector is defined as :

Magnitude * cosine of the angle

Maginitude * cosθ

The magnitude is represented as A

Hence, horizontal, x - component of the vector is :

Acosθ

Furthermore,

The y-component is taken as the sin of the of the angle multiplied by the magnitude

Vertical, y component : Asinθ

A wheel 30 cm in diameter accelerates uniformly from 245 rpm to 380 rpm in 6.1 s . Part A How far will a point on the edge of the wheel have traveled in this time

Answers

Answer:

A point on the edge of the wheel will travel 199.563 radians at the given time.

Explanation:

Given;

initial angular velocity of the wheel; [tex]\omega _i = 245 \ rev/\min = 245\ \frac{rev}{\min} \times \frac{2\pi}{1\ rev} \times \frac{1 \ \min}{60 \ s} = 25.66 \ rad/s[/tex]

final angular velocity of the wheel;

[tex]\omega _f = 380 \ rev/\min = 380 \ \frac{rev}{\min} \times \frac{2\pi}{1\ rev} \times \frac{1 \ \min}{60 \ s} = 39.80 \ rad/s[/tex]

radius of the wheel, d/2 = (30 cm ) / 2 = 15 cm = 0.15 m

time of motion, t = 6.1 s

The angular distance traveled by the edge of the wheel is calculated as;

[tex]\theta = (\frac{\omega_f + \omega_i}{2} )t\\\\\theta = (\frac{39.8 + 25.66}{2} )\times 6.1\\\\\theta = 199.653 \ radian[/tex]

Therefore, a point on the edge of the wheel will travel 199.563 radians at the given time.

A wheel rotates about a fixed axis with an initial angular velocity of 13 rad/s. During a 8-s interval the angular velocity increases to 57 rad/s. Assume that the angular acceleration was constant during this time interval. How many revolutions does the wheel turn through during this time interval

Answers

Answer:

The number of revolutions is 44.6.

Explanation:

We can find the revolutions of the wheel with the following equation:

[tex]\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}[/tex]

Where:

[tex]\omega_{0}[/tex]: is the initial angular velocity = 13 rad/s              

t: is the time = 8 s

α: is the angular acceleration

We can find the angular acceleration with the initial and final angular velocities:

[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]

Where:

[tex] \omega_{f} [/tex]: is the final angular velocity = 57 rad/s

[tex] \alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2} [/tex]

Hence, the number of revolutions is:

[tex] \theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev [/tex]

Therefore, the number of revolutions is 44.6.

       

I hope it helps you!

A cat's displacement is 15 meters to the right in 7.0 seconds. If, at the start of the 7.0 seconds, the cat was moving at a velocity of 2.0 m/s left what was its final velocity?

Answers

Answer:

6.3 m/s

Explanation:

From the given information:

The displacement (x) = 15 m

time (t) = 7.0 s

initial velocity = -2.0 m/s (since it is moving in the opposite direction)

We need to determine the acceleration then find the final velocity.

By applying the kinematics equation:

[tex]x = ut + \dfrac{1}{2}at^2[/tex]

[tex]15 = (-2.0)(7.0) + \dfrac{1}{2}a(7.0)^2\\ \\ 15 = -14.0 + \dfrac{49}{2}a \\ \\ 29= 24.5a \\ \\ a= \dfrac{29}{24.5} \\ \\ a = 1.184 \ m/s^2[/tex]

Now, to determine the final velocity by using the equation:

v = u + at

v = -2 + 1.184(7.0)

v = 6.288 m/s

v ≅ 6.3 m/s

A pilot flies her route in two straight-line segments. The displacement vector A for the first segment has a magnitude of 243 km and a direction 30.0o north of east. The displacement vector for the second segment has a magnitude of 178 km and a direction due west. The resultant displacement vector is R = A + B and makes an angle ? with the direction due east. Using the component method, find (a) the magnitude of R and (b) the directional angle ?.
(a) R = km
(b) ? = degrees

Answers

Answer:

a) [tex]R=126Km[/tex]

b) [tex]\theta=74.6\textdegree[/tex]

Explanation:

From the question we are told that:

1st segment

243km at Angle=30

2nd segment

178km West

Resolving to the X axis

[tex]F_x=243cos30+178[/tex]

[tex]F_x=33.44Km[/tex]

Resolving to the Y axis

[tex]F_y=243sin30+178sin0[/tex]

[tex]R=\sqrt{F_y^2+F_x^2}[/tex]

[tex]F_y=121.5Km[/tex]

Therefore

Generally the equation for Directional Angle is mathematically given by

[tex]\theta=tan^{-1}\frac{F_y}{F_x}[/tex]

[tex]\theta=tan^{-1}\frac{121.5}{33.44}[/tex]

[tex]\theta=74.6\textdegree[/tex]

Generally the equation for Magnitude is mathematically given by

[tex]R=\sqrt{F_y^2+F_x^2}[/tex]

[tex]R=\sqrt{33.44^2+121.5^2}[/tex]

[tex]R=126Km[/tex]

A 2090-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that its vertical velocity as a function of time is given by v(t) =At+Bt^2 , where A and B are constants and time is measured from the instant the fuel is ignited. The rocket has an upward acceleration of 1.50m/s 2 at the instant of ignition and, 1.00 s later, an upward velocity of 2.00 m/s. (a) Determine A and B , including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket’s weight. (d) What was the initial thrust due to the fuel?

Answers

Answer:

a) A = 1.50 m / s²,  B = 1.33 m/s³,  b) a = 12.1667 m / s²,

c)  I = M (1.5 t + 1.333 t²) ,  d)  ΔI = M 2.833   N

Explanation:

In this exercise give the expression for the speed of the rocket

         v (t) = A t + B t²

and the initial conditions

         a = 1.50 m / s² for t = 0 s

         v = 2.00 m / s for t = 1.00 s

a) it is asked to determine the constants.

Let's look for acceleration with its definition

         a = [tex]\frac{dv}{dt}[/tex]

         a = A + 2B t

we apply the first condition t = 0 s

         a = A

         A = 1.50 m / s²

we apply the second condition t = 1.00 s

          v = 1.5 1 + B 1²

          2 = 1.5 + B

          B = 2 / 1.5

          B = 1.33 m/s³

the equation remains

           v = 1.50 t + 1.333 t²

b) the acceleration for t = 4.00 s

           a = 1.50 + 1.333 2t

           a = 1.50 + 2.666 4

           a = 12.1667 m / s²

c) The thrust

           I = ∫ F dt = p_f - p₀

           

Newton's second law

          F = M a

          F = M (1.5 + 2 1.333 t) dt

           

we replace and integrate

         I = M ∫ (1.5 + 2.666 t) dt

         I = 1.5 t + 2.666 t²/2

         I = M (1.5 t + 1.333 t²) + cte

in general the initial rockets with velocity v = 0 for t = 0, where we can calculate the constant

         cte = 0

         I = M (1.5 t + 1.333 t²)

d) the initial push

For this we must assume some small time interval, for example between

t = 0 s and t = 1 s

        ΔI = I_f - I₀

        ΔI = M (1.5 1 + 1.333 1²)

        ΔI = M 2.833   N

3. You have a variable-voltage power supply and a capacitor in the form of two metal disks of radius 0.6 m, held a distance of 1 mm apart. What is the largest voltage you can apply to the capacitor without the air becoming highly conductive

Answers

Answer:

The breakdown of air occurs at a maximum voltage of 3kV/mm.

Explanation:

The breakdown of air occurs at a maximum voltage of 3kV/mm.

At this level of voltage the air between the plates become highly ionised and breakdown occurs. Since, the distance held between the plates is 1mm , it can withstand a maximum voltage of 3 kV.

After this voltage the air will become conductive in nature and will form ions in the air between the plates and ultimately breakdown will take place with a flash.

what does it mean to do science

Answers

Answer:

Doing science could be defined as carrying out scientific processes, like the scientific method, to add to science's body of knowledge.

Turning a corner at a typical large intersection in a city means driving your car through a circular arc with a radius of about 25 m. if the maximum advisable acceleration of your vehicle through a turn on wet pavement is 0.40 times the free-fall acceleration, what is the maximum speed at which you should drive through this turn?

Answers

Answer:

9.89 m/s.

Explanation:

Given that,

The radius of the circular arc, r = 25 m

The acceleration of the vehicle is 0.40 times the free-fall acceleration i.e.,a = 0.4(9.8) = 3.92 m/s²

Let v is the maximum speed at which you should drive through this turn. It can be solved as follows :

[tex]a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{3.92\times 25} \\\\=9.89 m/s[/tex]

So, the maximum speed of the car should be 9.89 m/s.

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