Answer:
θ = 57.4°
Explanation:
The complete formula to find out the work done by the plane is as follows:
[tex]W = FdCos\theta[/tex]
where,
W = Work = 200000 J
F = Force = Tension = 2560 N
d = distance = 145 m
θ = angle between rope and horizontal = ?
Therefore,
[tex]200000\ J = (2560\ N)(145\ m)Cos\theta\\\\Cos\theta = \frac{200000\ J}{371200\ J}\\\\\theta = Cos^{-1}(0.539)[/tex]
θ = 57.4°
Which of the following is a noncontact force?
O A. Friction between your hands
O B. A man pushing on a wall
O C. Air resistance on a car
D. Gravity between you and the Sun
Answer:
Gravity between you and the sun
Electrical resistance is a measure of resistance to the flow of _?____
Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms, symbolized by the Greek letter omega (Ω). Ohms are named after Georg Simon Ohm (1784-1854), a German physicist who studied the relationship between voltage, current and resistance.
Hope this helps!!!!
Answer:
electric current
Explanation:
The answer is electric current
When you hammer a nail into wood, the nail heats up. 30 Joules of energy was absorbed by a 5-g nail as it was hammered into place. How much does the nail's temperature increase (in °C) during this process? (The specific heat capacity of the nail is 450 J/kg-°C, and round to 3 significant digits.
Answer:
13.33 K
Explanation:
Given that,
Heat absorbed, Q = 30 J
Mass of nail, m = 5 g = 0.005 kg
The specific heat capacity of the nail is 450 J/kg-°C.
We need to find the increase in the temperature during the process. The heat absorbed in a process is as follows:
[tex]Q=mc\Delta T\\\\\Delta T=\dfrac{Q}{mc}\\\\\Delta T=\dfrac{30}{0.005\times 450}\\\\=13.33\ K[/tex]
So, the increase in temperature is 13.33 K.
g A mass of 2.0 kg traveling at 3.0 m/s along a smooth, horizontal plane hits a relaxed spring. The mass is slowed to zero velocity when the spring has been compressed by 0.15 m. What is the spring constant of the spring
By the work-energy theorem, the total work done on the mass by the spring is equal to the change in the mass's kinetic energy:
W = ∆K
and the work done by a spring with constant k as it gets compressed a distance x is -1/2 kx ²; the work it does is negative because the restoring force of the spring points opposite the direction in which it's getting compressed.
So we have
-1/2 k (0.15 m)² = 0 - 1/2 (2.0 kg) (3.0 m/s)²
Solve for k to get k = 800 N/m.
The weight of a hydraulic barber's chair with a client is 2100 N. When the barber steps on the input piston with a force of 44 N, the output plunger of a hydraulic system begins to lift the chair. Determine the ratio of the radius of the output plunger to the radius of the input piston.
Answer:
[tex]\frac{r_1}{r_2}=6.9[/tex]
Explanation:
According to Pascal's Law, the pressure transmitted from input pedal to the output plunger must be same:
[tex]P_1 = P_2\\\\\frac{F_1}{A_1}=\frac{F_2}{A_2}\\\\\frac{F_1}{F_2}=\frac{A_1}{A_2}\\\\\frac{F_1}{F_2}=\frac{\pi r_1^2}{\pi r_2^2}\\\\\frac{F_1}{F_2}=\frac{r_1^2}{r_2^2}[/tex]
where,
F₁ = Load lifted by output plunger = 2100 N
F₂ = Force applied on input piston = 44 N
r₁ = radius of output plunger
r₂ = radius of input piston
Therefore,
[tex]\frac{r_1^2}{r_2^2}=\frac{2100\ N}{44\ N}\\\\\frac{r_1}{r_2}=\sqrt{\frac{2100\ N}{44\ N}} \\\\\frac{r_1}{r_2}=6.9[/tex]
a) Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy source is to store antimatter charged particles in a vacuum chamber, circulating in a magnetic field, and then extract them as needed. Antimatter annihilates with normal matter, producing pure energy. What strength magnetic field is needed to hold antiprotons, moving at 5.00 x10^7 m/s in a circular path 2.00m in radius? Antiprotons have the same mass as protons but the opposite (negative) charge.b) Is this field strength obtainable with today's technology or is it a futuristic possibility?
Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 2 m/s, exactly how fast (in m2/s) is the area of the spill increasing when the radius is 39 m?
Explanation:
The area of a circle of radius r is given by
[tex]A = \pi r^2[/tex]
Taking the derivative of A with respect to time t, we get
[tex]\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}[/tex]
We also know that
[tex]\dfrac{dr}{dt} = 2\:\text{m/s}\:\text{at}\:r = 39\:\text{m}[/tex]
[tex]\dfrac{dA}{dt} = 2\pi (39\:\text{m})(2\:\text{m/s})= 490\:\text{m}^2\text{/s}[/tex]
Calculate the change in length of a 90.5 mm aluminum bar that has increased in temperature by from -14.4 oC to 154.6 oC
Take the coefficient of expansion to be 25 x 10-6 (oC)-1 . Write the answer in meters with three significant figures
Answer:
ΔL = 3.82 10⁻⁴ m
Explanation:
This is a thermal expansion exercise
ΔL = α L₀ ΔT
ΔT = T_f - T₀
where ΔL is the change in length and ΔT is the change in temperature
Let's reduce the length to SI units
L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m
let's calculate
ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))
ΔL = 3.8236 10⁻⁴ m
using the criterion of three significant figures
ΔL = 3.82 10⁻⁴ m
A proton enters a region of constant magnetic field, perpendicular to the field and after being accelerated from rest by an electric field through an electrical potential difference of 330 V. Determine the magnitude of the magnetic field, if the proton travels in a circular path with a radius of 23 cm.
Answer:
B = 1.1413 10⁻² T
Explanation:
We use energy concepts to calculate the proton velocity
starting point. When entering the electric field
Em₀ = U = q V
final point. Right out of the electric field
em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
q V = ½ m v²
v = [tex]\sqrt{2qV/m}[/tex]
we calculate
v = [tex]\sqrt{\frac{ 2 \ 1.6 \ 10^{-19} \ 300}{1.67 \ 0^{-27}} }[/tex]
v = [tex]\sqrt{632.3353 \ 10^8}[/tex]
v = 25.15 10⁴ m / s
now enters the region with magnetic field, so it is subjected to a magnetic force
F = m a
the force is
F = q v x B
as the velocity is perpendicular to the magnetic field
F = q v B
acceleration is centripetal
a = v² / r
we substitute
qvB =1/2 m v² / r
B = v[tex]\frac{m v}{2 q r}[/tex]
we calculate
B = [tex]\frac{1.67 \ 10^{-27} 25.15 \ 10^4 }{1.6 \ 10^{-19} 0.23}[/tex]
B = 1.1413 10⁻² T
Which simple machine is shown in the diagram?
a wedge
a screw
an inclined plane
a wheel and axle
Answer:
Wheel and axle
Explanation:
Which simple machine is shown in the diagram?
a wheel and axle
From the given diagram, the machine shown is actually a wheel and axle
Description of wheel and axle
The wheel and axle is a machine consisting of a wheel attached to a smaller axle so that these two parts rotate together in which a force is transferred from one to the other.
Answer:
Wheel and axle
Explanation:
A planet of mass m moves around the Sun of mass M in an elliptical orbit. The maximum and minimum distance of the planet from the Sun are r1 and r2, respectively. Find the relation between the time period of the planet in terms of r1 and r2.
Answer:
the relation between the time period of the planet is
T = 2π √[( r1 + r2 )³ / 8GM ]
Explanation:
Given the data i the question;
mass of sun = M
minimum and maximum distance = r1 and r2 respectively
Now, using Kepler's third law,
" the square of period T of any planet is proportional to the cube of average distance "
T² ∝ R³
average distance a = ( r1 + r2 ) / 2
we know that
T² = 4π²a³ / GM
T² = 4π² [( ( r1 + r2 ) / 2 )³ / GM ]
T² = 4π² [( ( r1 + r2 )³ / 8 ) / GM ]
T² = 4π² [( r1 + r2 )³ / 8GM ]
T = √[ 4π² [( r1 + r2 )³ / 8GM ] ]
T = 2π √[( r1 + r2 )³ / 8GM ]
Therefore, the relation between the time period of the planet is
T = 2π √[( r1 + r2 )³ / 8GM ]
A point charge of -3.0 x 10-5C is placed at the origin of coordinates. Find the electric field at the point 3. r= 50 m on the x-axis
Answer: -5×10-3
Explanation:
E=kq/r
The gravitational field strength due to its planet is 5N/kg What does it mean?
Answer:
The weight of an object is the force on it caused by the gravity due to the planet. The weight of an object and the gravitational field strength are directly proportional. For a given mass, the greater the gravitational field strength of the planet, the greater its weight.
Weight can be calculated using the equation:
weight = mass × gravitational field strength
This is when:
weight (W) is measured in newtons (N)
mass (m) is measured in kilograms (kg)
gravitational field strength (g) is measured in newtons per kilogram (N/kg)
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?
NEED HELP ASAP- Please show work
The angular position of an object is given by θ = 4t3 +10t −40 , where θ is in radians and t is in seconds what is:
(a) (5 points) The angular velocity at t = 2 s?
(b) (5 points) The angular acceleration at t = 2 s?
Answer:
Look at work
Explanation:
Θ= 4t^3+10t-40
a) In order to find ω, we need to find displacement so plug in t=2 to find Θ.
Θ= 4*8+20-40=12
use ω=Θ/t
Plug in values
ω=6 rad/s
b) In order to find α we use ω/t.
Plug in values
α=6/2= 3 rad/s^2
Which physical phenomenon is illustrated by the fact that the prism has different refractive indices for different colors
Answer:
The incoming white light is composed of light of different colors,
Since these different colors have different refractive indices they are refracted at different angles from one another.
The output light is then separated by color creating a color spectrum.
Since n is greater for shorter wavelengths (violet colors) these wavelengths are refracted thru the larger angles.
Cell phone conversations are transmitted by high-frequency radio waves. Suppose the signal has wavelength 35 cm while traveling through air. What are the
(a) frequency and
(b) wavelength as the signal travels through 3-mm-thick window glass into your room?
Answer:
(a) 8.57 x 10^8 Hz
(b) 23.3 cm
Explanation:
Wavelength = 35 cm = 0.35 m
speed =3 x10^8 m/s
Let the frequency is f.
(a) The relation is
speed = frequency x wavelength
3 x 10^8 = 0.35 x f
f = 8.57 x 10^8 Hz
(b) refractive index of glass is 1.5
The relation for the refractive index and the wavelength is
wavelength in glass= wavelength in air/ refractive index.
Wavelength in glass= 35/1.5 = 23.3 cm
Which of the following represents the velocity time relationship for a falling apple?
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~
A mass-spring system oscillates with an amplitude of 4.20 cm. If the spring constant is 262 N/m and the mass is 560 g, determine the mechanical energy of the system.
Answer:
[tex]M.E=41J[/tex]
Explanation:
From the question we are told that:
Amplitude [tex]a=4.20cm[/tex]
Spring Constant [tex]K=262N/m[/tex]
Mass [tex]m=560g[/tex]
Generally the equation for mechanical energy is mathematically given by
[tex]M.E=\frac{1}{2}km^2[/tex]
[tex]M.E=0.5*262*0.56^2[/tex]
[tex]M.E=41J[/tex]
A 10.0 L tank contains 0.329 kg of helium at 28.0 ∘C. The molar mass of helium is 4.00 g/mol . Part A How many moles of helium are in the tank? Express your answer in moles.
Answer:
82.25 moles of He
Explanation:
From the question given above, the following data were obtained:
Volume (V) = 10 L
Mass of He = 0.329 Kg
Temperature (T) = 28.0 °C
Molar mass of He = 4 g/mol
Mole of He =?
Next, we shall convert 0.329 Kg of He to g. This can be obtained as follow:
1 Kg = 1000 g
Therefore,
0.329 Kg = 0.329 Kg × 1000 g / 1 Kg
0.329 Kg = 329 g
Thus, 0.329 Kg is equivalent to 329 g.
Finally, we shall determine the number of mole of He in the tank. This can be obtained as illustrated below:
Mass of He = 329 g
Molar mass of He = 4 g/mol
Mole of He =?
Mole = mass / molar mass
Mole of He = 329 / 4
Mole of He = 82.25 moles
Therefore, there are 82.25 moles of He in the tank.
The following two waves are sent in opposite directions on a horizontal string so as to create a standing wave in a vertical plane: y1(x, t) = (8.20 mm) sin(4.00πx - 430πt) y2(x, t) = (8.20 mm) sin(4.00πx + 430πt), with x in meters and t in seconds. An antinode is located at point A. In the time interval that point takes to move from maximum upward displacement to maximum downward displacement, how far does each wave move along the string?
Answer:
Explanation:
From the information given:
The angular frequency ω = 430 π rad/s
The wavenumber k = 4.00π which can be expressed by the equation:
k = ω/v
∴
4.00 = 430 /v
v = 430/4.00
v = 107.5 m/s
Similarly: k = ω/v = 2πf/fλ
We can say that:
k = 2π/λ
4.00 π = 2π/λ
wavelength λ = 2π/4.00 π
wavelength λ = 0.5 m
frequency of the wave can now be calculated by using the formula:
f = v/λ
f = 107.5/0.5
f = 215 Hz
Also, the Period(T) = 1/215 secs
The time at which particle proceeds from point A to its maximum upward displacement and to its maximum downward displacement can be computed as t = T/2;
Thus, the distance(x) covered by each wave during this time interval(T/2) will be:
x = v * t
x = v * T/2
x = λ/2
x = 0.5/2
x = 0.25 m
Why don’t you see tides ( like those of the ocean ) in your swimming pool ?
A body starts from rest and accelerates uniformly at 5m/s. Calculate the time taken by the body to cover a distance of 1km
Answer:
20 seconds
Explanation:
We are given 2 givens in the first statement
v0=0 and a=5
And we are trying to find time needed to cover 1km or 1000m.
So we use
x-x0=v0t+1/2at²
Plug in givens
1000=0+2.5t²
solve for t
t²=400
t=20s
Two blocks in contact with each other are pushed to the right across a rough horizontal surface by the two forces shown. If the coefficient of kinetic friction between each of the blocks and the surface is 0.30, determine the magnitude of the force exerted on the 2.0-kg block by the 3.0-kg block.
I assume the blocks are pushed together at constant speed, and it's not so important but I'll also assume it's the smaller block being pushed up against the larger one. (The opposite arrangement works out much the same way.)
Consider the forces acting on either block. Let the direction in which the blocks are being pushed by the positive direction.
The 2.0-kg block feels
• the downward pull of its own weight, (2.0 kg) g
• the upward normal force of the surface, magnitude n₁
• kinetic friction, mag. f₁ = 0.30n₁, pointing in the negative horizontal direction
• the contact force of the larger block, mag. c₁, also pointing in the negative horizontal direction
• the applied force, mag. F, pointing in the positive horizontal direction
Meanwhile the 3.0-kg block feels
• its own weight, (3.0 kg) g, pointing downward
• normal force, mag. n₂, pointing upward
• kinetic friction, mag. f₂ = 0.30n₂, pointing in the negative horizontal direction
• contact force from the smaller block, mag. c₂, pointing in the positive horizontal direction (this is the force that is causing the larger block to move)
Notice the contact forces form an action-reaction pair, so that c₁ = c₂, so we only need to find one of these, and we can get it right away from the net forces acting on the 3.0-kg block in the vertical and horizontal directions:
• net vertical force:
n₂ - (3.0 kg) g = 0 ==> n₂ = (3.0 kg) g ==> f₂ = 0.30 (3.0 kg) g
• net horizontal force:
c₂ - f₂ = 0 ==> c₂ = 0.30 (3.0 kg) g ≈ 8.8 N
1. A block of mass m = 10.0 kg is released with a speed v from a frictionless incline at height 7.00 m. The
block reaches the horizontal ground and then slides up another frictionless incline as shown in Fig. 1.1. If the
horizontal surface is also frictionless and the maximum height that the block can slide up to is 26.0 m, (a) what
is the speed v of the block equal to when it is released and (b) what is the speed of the block when it reaches
the horizontal ground? If a portion of length 1 2.00 m on the horizontal surface is frictional with coefficient
of kinetic friction uk = 0.500 (Fig. 1.2) and the block is released at the same height 7.00 m with the same
speed v determined in (a), (c) what is the maximum height that the block can reach, (d) what is the speed of the
block at half of the maximum height, and (e) how many times will the block cross the frictional region before
it stops completely?
1 = 2.00 m (frictional region)
Let A be the position of the block at the top of the first incline; B its position at the bottom of the first incline; C its position at the bottom of the second incline; and D its position at the top of the second incline. I'll denote the energy of the block at a given point by E (point).
At point A, the block has total energy
E (A) = (10.0 kg) (9.80 m/s²) (7.00 m) + 1/2 (10.0 kg) v₀²
E (A) = 686 J + 1/2 (10.0 kg) v₀²
At point B, the block's potential energy is converted into kinetic energy, so that its total energy is
E (B) = 1/2 (10.0 kg) v₁²
The block then slides over the horizontal surface with constant speed v₁ until it reaches point C and slides up a maximum height of 26.0 m to point D. Its total energy at D is purely potential energy,
E (D) = (10.0 kg) (9.80 m/s²) (26.0 m) = 2548 J
Throughout this whole process, energy is conserved, so
E (A) = E (B) = E (C) = E (D)
(a) Solve for v₀ :
686 J + 1/2 (10.0 kg) v₀² = 2548 J
==> v₀ ≈ 19.3 m/s
(b) Solve for v₁ :
1/2 (10.0 kg) v₁² = 2548 J
==> v₁ ≈ 22.6 m/s
Now if the horizontal surface is not frictionless, kinetic friction will contribute some negative work to slow down the block between points C and D. Check the net forces acting on the block over this region:
• net horizontal force:
∑ F = -f = ma
• net vertical force:
∑ F = n - mg = 0
where f is the magnitude of kinetic friction, a is the block's acceleration, n is the mag. of the normal force, and mg is the block's weight. Solve for a :
n = mg = (10.0 kg) (9.80 m/s²) = 98.0 N
f = µn = 0.500 (98.0 N) = 49.0 N
==> - (49.0 N) = (10.0 kg) a
==> a = - 4.90 m/s²
The block decelerates uniformly over a distance 2.00 m and slows down to a speed v₂ such that
v₂² - v₁² = 2 (-4.90 m/s²) (2.00 m)
==> v₂² = 490 m²/s²
and thus the block has total/kinetic energy
E (C) = 1/2 (10.0 kg) v₂² = 2450 J
(c) The block then slides a height h up the frictionless incline to D, where its kinetic energy is again converted to potential energy. With no friction, E (C) = E (D), so
2450 J = (10.0 kg) (9.80 m/s²) h
==> h = 25.0 m
(d) At half the maximum height, the block has speed v₃ such that
2450 J = (10.0 kg) (9.80 m/s²) (h/2) + 1/2 (10.0 kg) v₃²
==> v₃ ≈ 15.7 m/s
The block loses speed and thus energy as it moves between B and C, but its energy is conserved elsewhere. If we ignore the inclines and pretend that the block is sliding over a long horizontal surface, then its velocity v at time t is given by
v = v₁ + at = 22.6 m/s - (4.90 m/s²) t
The block comes to a rest when v = 0 :
0 = 22.6 m/s - (4.90 m/s²) t
==> t ≈ 4.61 s
It covers a distance x after time t of
x = v₁t + 1/2 at ²
so when it comes to a complete stop, it will have moved a distance of
x = (22.6 m/s) (4.61 s) + 1/2 (-4.90 m/s²) (4.61 s)² = 52.0 m
(e) The block crosses the rough region
(52.0 m) / (2.00 m) = 26 times
What is the total surface charge qint on the interior surface of the conductor (i.e., on the wall of the cavity)
Answer: hello your question is incomplete below is the missing part
A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q.
answer:
- q
Explanation:
Since the spherical cavity was carved out of a neutral conducting sphere hence the electric field inside this conductor = zero
given that there is a point charge +q at the center of the spherical cavity hence for the electric field inside the conductor to be = zero the total surface charge qint on the wall of the cavity will be -q
need help pleaseee,question is in the pic
Explanation:
For engine 1,
Energy removed = 239 J
Energy added = 567 J
[tex]\eta_1=\dfrac{239}{567}\cdot100=42.15\%[/tex]
For engine 2,
Energy removed = 457 J
Energy added = 789 J
[tex]\eta_2=\dfrac{457}{789}\cdot100=57.92\%[/tex]
For engine 3,
Energy removed = 422 J
Energy added = 1038 J
[tex]\eta_3=\dfrac{422}{1038}\cdot100=40.65\%[/tex]
So, the engine 2 has the highest thermal efficiency.
A cannon and a supply of cannonballs are inside a sealed railroad car of length L, as in Fig. 7-33. The cannon fires to the right; the car recoils to the left. The cannonballs remain in the car after hitting the far wall. (a) After all the cannonballs have been fired, what is the greatest distance the car can have moved from its original position
Answer:
Initially let n cannonballs with a total mass of m be to the left of the center of mass at L /2 and the mass of the car at L/2
x1 = [-m / (m + M)] * L / 2 is the original position of the CM
x2 = (m (x + L/2) + M x) / (m + M) * L/2 final position of CM with all cannon balls to the right
[-m x - m L / 2 + m x - M x] / (M + m) * L/2
= - ( m L / 2 + M x) / (m + M) * L/2 = Xcm
Check the math, but maximum distance occurs when the cannonballs of mass m move from -L/2 to L/2 and the car of mass M moves from zero to -x
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?
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]
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.
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]
