Answer:
Explanation:
Rotational kinetic energy of flywheel
= 1/2 Iω² where I is moment of inertia , ω is angular velocity
for wheel I = 1/2 m R² where m is mass and R is radius of flywheel
Putting the values
I = 1/2 x 370 x .5²
= 46.25 kg m²
ω = 2πn where n is frequency of revolution per second
ω = 2 x 3.14 x 540 = 3391.2
Rotational kinetic energy = .5 x 46.25 x 3391.2²
= 265.94 x 10⁶ J
If this energy is transferred to a car of mass 1600kg , velocity acquired by it be v , then
kinetic energy of car = rotational kinetic energy of flywheel
= 1/2 m v ² = 265.94 x 10⁶
.5 x 1600 v² = 265.94 x 10⁶
v² = 33.24 x 10⁴
v = 5.76 x 10²
= 576 m /s
A rigid tank A of volume 0.6 m3 contains 5 kg air at 320K and the rigid tank B is 0.4 m3 with air at 600 kPa, 360 K. They are connected to a piston cylinder initially empty with closed valves. The pressure in the cylinder should be 800 kPa to float the piston. Now the valves are slowly opened and the entire process is adiabatic. The internal energy of the mixture at final state is:_____.
a. 229 k/kg.
b. 238 kJ/kg
c. 257 kg
d. cannot be determined.
Answer:
the internal energy of the mixture at final state = 238kJ/kg
Explanation:
Given
V= 0.6m³
m=5kg
R=0.287kJ/kg.K
T=320 K
from ideal gas equation
PV = nRT
where P is pressure, V is volume, n is number of mole, R is ideal gas constant , T is the temperature.
Recall, mole = mass/molar mass
attached is calculation of the question.
A helicopter rotor blade is 3.40m long from the central shaft to the rotor tip. When rotating at 550rpm what is the radial acceleration of the blade tip expressed in multiples of g?
Answer:
a = 1.15 10³ g
Explanation:
For this exercise we will use the relations of the centripetal acceleration
a = v² / r
where is the linear speed of the rotor and r is the radius of the rotor
let's use the relationships between the angular and linear variables
v = w r
let's replace
a = w² r
let's reduce the angular velocity to the SI system
w = 550 rev / min (2pi rad / 1 rev) (1 min / 60 s)
w = 57.6 rad / s
let's calculate
a = 57.6² 3.4
a = 1.13 10⁴ m / s²
To calculate this value in relation to g, let's find the related
a / g = 1.13 10⁴ / 9.8
a = 1.15 10³ g
In a circus act, a uniform board (length 3.00 m, mass 25.0 kg ) is suspended from a bungie-type rope at one end, and the other end rests on a concrete pillar. When a clown (mass 79.0 kg ) steps out halfway onto the board, the board tilts so the rope end is 30∘ from the horizontal and the rope stays vertical. Calculate the force exerted by the rope on the board with the clown on it.
Answer:
Force of Rope = 122.5 N
Force of Rope = 480.2N
Explanation:
given data
length = 3.00 m
mass = 25.0 kg
clown mass = 79.0 kg
angle = 30°
solution
we get here Force of Rope on with and without Clown that is
case (1) Without Clown
pivot would be on the concrete pillar so Force of Rope will be
Force of Rope × 3m = (25kg)×(9.8ms²)×(1.5m)
solve it and we get
Force of Rope = 122.5 N
and
case (2) With Clown
so here pivot is still on concrete pillar and clown is standing on the board middle and above the centre of mass so Force of Rope will be
Force of Rope × 3m = (25kg+73kg)×(9.8ms²)×(1.5m)
solve it and we get
Force of Rope = 480.2N
A commuter train passes a passenger platform at a constant speed of 39.6 m/s. The train horn is sounded at its characteristic frequency of 350 Hz. (a) What overall change in frequency is detected by a person on the platform as the train moves from approaching to receding
Complete Question
A commuter train passes a passenger platform at a constant speed of 39.6 m/s. The train horn is sounded at its characteristic frequency of 350 Hz.
(a)
What overall change in frequency is detected by a person on the platform as the train moves from approaching to receding
(b) What wavelength is detected by a person on the platform as the train approaches?
Answer:
a
[tex]\Delta f = 81.93 \ Hz[/tex]
b
[tex]\lambda_1 = 0.867 \ m[/tex]
Explanation:
From the question we are told that
The speed of the train is [tex]v_t = 39.6 m/s[/tex]
The frequency of the train horn is [tex]f_t = 350 \ Hz[/tex]
Generally the speed of sound has a constant values of [tex]v_s = 343 m/s[/tex]
Now according to dopplers equation when the train(source) approaches a person on the platform(observe) then the frequency on the sound observed by the observer can be mathematically represented as
[tex]f_1 = f * \frac{v_s}{v_s - v_t}[/tex]
substituting values
[tex]f_1 = 350 * \frac{343 }{343-39.6}[/tex]
[tex]f_1 = 395.7 \ Hz[/tex]
Now according to dopplers equation when the train(source) moves away from the person on the platform(observe) then the frequency on the sound observed by the observer can be mathematically represented as
[tex]f_2 = f * \frac{v_s}{v_s +v_t}[/tex]
substituting values
[tex]f_2 = 350 * \frac{343}{343 + 39.6}[/tex]
[tex]f_2 = 313.77 \ Hz[/tex]
The overall change in frequency is detected by a person on the platform as the train moves from approaching to receding is mathematically evaluated as
[tex]\Delta f = f_1 - f_2[/tex]
[tex]\Delta f = 395.7 - 313.77[/tex]
[tex]\Delta f = 81.93 \ Hz[/tex]
Generally the wavelength detected by the person as the train approaches is mathematically represented as
[tex]\lambda_1 = \frac{v}{f_1 }[/tex]
[tex]\lambda_1 = \frac{343}{395.7 }[/tex]
[tex]\lambda_1 = 0.867 \ m[/tex]
If 62.9 cm of copper wire (diameter = 1.15 mm, resistivity = 1.69 × 10-8Ω·m) is formed into a circular loop and placed perpendicular to a uniform magnetic field that is increasing at the constant rate of 8.43 mT/s, at what rate is thermal energy generated in the loop?
Answer:
The answer is "[tex]\bold{7.30 \times 10^{-6}}[/tex]"
Explanation:
length of the copper wire:
L= 62.9 cm
r is the radius of the loop then:
[tex]r=\frac{L}{2 \pi}\\[/tex]
[tex]=\frac{62.9}{2\times 3.14}\\\\=\frac{62.9}{6.28}\\\\=10.01\\[/tex]
area of the loop Is:
[tex]A_L= \pi r^2[/tex]
[tex]=100.2001\times 3.14\\\\=314.628[/tex]
change in magnetic field is:
[tex]=\frac{dB}{dt} \\\\ = 0.01\ \frac{T}{s}[/tex]
then the induced emf is: [tex]e = A_L \times \frac{dB}{dt}[/tex]
[tex]=314.628 \times 0.01\\\\=3.14\times 10^{-5}V[/tex]
resistivity of the copper wire is: [tex]\rho =[/tex] 1.69 × 10-8Ω·m
diameter d = 1.15mm
radius (r) = 0.5mm
[tex]= 0.5 \times 10^{-3} \ m[/tex]
hence the resistance of the wire is:
[tex]R=\frac{\rho L}{\pi r^2}\\[/tex]
[tex]=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times (0.5 \times 10^{-3})^2}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.5 \times 0.5 \times 10^{-6}}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.25 \times 10^{-6}}\\\\=135.41 \times 10^{-2}\\=1.35\times 10^{-4}\\[/tex]
Power:
[tex]P=\frac{e^2}{R}[/tex]
[tex]=\frac{3.14\times 10^{-5}\times 3.14\times 10^{-5}}{1.35 \times 10^{-4}}\\\\=7.30 \times 10^{-6}[/tex]
The final answer is: [tex]\boxed{7.30 \times 10^{-6} \ W}[/tex]
In an oscillating LC circuit, the total stored energy is U and the maximum current in the inductor is I. When the current in the inductor is I/2, the energy stored in the capacitor is
Answer:
The definition of that same given problem is outlined in the following section on the clarification.
Explanation:
The Q seems to be endless (hardly any R on the circuit). So energy equations to describe and forth through the inducer as well as the condenser.
Presently take a gander at the energy stored in your condensers while charging is Q.
⇒ [tex]U =\frac{Qmax^2}{C}[/tex]
So conclude C doesn't change substantially as well as,
When,
⇒ [tex]Q=\frac{Qmax}{2}[/tex]
⇒ [tex]Q^2=\frac{Qmax^2}{4}[/tex]
And therefore only half of the population power generation remains in the condenser that tends to leave this same inductor energy at 3/4 U.
(Equation 17.6) Write the equation for the path-length difference at a bright fringe (constructive interference). Define all variables. What are the SI units of each variable
Answer:
d sin tea = m λ
Explanation:
When we have a two-slit system, the optical path difference determines whether the intensity reaching an observation screen is maximum or zero.
To find this difference in optical path, we assume that the screen is much farther than the gap is, we draw a perpendicular from ray 1 to the second ray
OP = d sin θ
now to have constructive interference and see a bright line this leg must be an integer number of wavelengths, ose
d sin tea = m λ
where
d is the distance between the two slits
θ complexion the angle sea the point hold it between the two slits
λ the wavelength of the coherent light used
m an integer, which counts the number of lines of interference
Units in the SI system
d, lam in meters
θ degrees
m an integer
A lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffraction standpoint is
The lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffraction standpoint is: in the near-ultraviolet.
What is diffraction?The act of bending light around corners such that it spreads out and illuminates regions where a shadow is anticipated is known as diffraction of light. In general, since both occur simultaneously, it is challenging to distinguish between diffraction and interference. The diffraction of light is what causes the silver lining we see in the sky. A silver lining appears in the sky when the sunlight penetrates or strikes the cloud.
Longer wavelengths of light are diffracted at a greater angle than shorter ones, with the amount of diffraction being dependent on the wavelength of the light. Hence, among the light waves of the visible, near-infrared, and near-ultraviolet range, near-ultraviolet waves have the shortest wavelengths. So, The best resolution of this lens from a diffraction standpoint is in the near-ultraviolet, where diffraction is minimum.
Learn more about diffraction here:
https://brainly.com/question/11176463
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Initially stationary, a train has a constant acceleration of 0.8 m/s2. (a) What is its speed after 27 s? m/s (b) What is the total time required for the train to reach a speed of 41 m/s?
Answer:
(a) v1 = 21.6 m/s
(b) t = 51.25 s
Explanation:
Use kinematics equation
v1 = v0 + at
Given
v0 = 0 = initial velocity
a = 0.8 m/s^2 = acceleration
(a) t = 27 seconds
v1 = v0 + at = 0 + 0.8*27 = 21.6 m/s
(b) v1 = 41 m/s
v1 = v0 + at
solve for t
t = (v1-v0)/a = (41-0)/0.8 = 51.25 s
Given:
Acceleration, a = 0.8 m/s²Time, t = 41 m/sSpeed, v = 41 m/sBy using Kinematics equation, we get
→ [tex]v_1 = v_0+at[/tex]
(a)
At t = 27 seconds,
→ [tex]v_1 = v_0 +at[/tex]
[tex]= 0+0.8\times 27[/tex]
[tex]= 21.6 \ m/s[/tex]
(b)
When [tex]v_1 = 41 \ m/s[/tex]
→ [tex]t = \frac{(v_1-v_0)}{a}[/tex]
[tex]= \frac{41.0}{0.8}[/tex]
[tex]= 51.25 \ s[/tex]
Thus the response above is right.
Learn more about acceleration here:
https://brainly.com/question/18401853
What is unique about the c-ray that is not about other rays? Note: Refer to the concave mirror video Select one: a. only ray whose angle of incidence = angle of reflection b. only ray that reflects back in the same direction it came from c. both the above statements are true d. none of the above
Answer:
b. only ray that reflects back in the same direction it came from
Explanation:
C-rays can be said to be a ray that comes from the center of the curvature. It is known that any ray that comes from the center of the curvature reflects back in the same direction it came from, this is because the line joining from the center of the curvature to any point in the mirror is perpendicular to the mirror.
Correct answer is option B.
C-ray is the only ray that reflects back in the same direction it came from.
Option A is incorrect because for other rays, angle of incidence = angle of reflection. This is not a property of c-ray.
A very bouncy ball is dropped from a height of 2.47 m to an asphalt playground surface and the height of its 4 th bounce is measured to be 1.71 m. Find the coefficient of restitution of the ball for a collision with asphalt.
Answer:
0.912
Explanation:
Given that
Height of bouncing of the ball, h = 1.71 m
Number of times the ball bounced, n = 4 times
Height from which the ball was dropped, H = 2.47
First, let's start by defining what coefficient of restitution means
Coefficient of Restitution, CoR is the "ratio of the final to initial relative velocity between two objects after they collide. It normally ranges from 0 to 1 where 1 would be a perfectly elastic collision."
It is mathematically represented as
CoR = (velocity after collision) / (velocity before collision)
1.71 = 2.47 * c^4, where c = CoR
1.71/2.47 = c^4
c^4 = 0.6923
c = 4th root of 0.6923
c = 0.912
3. Two spherical objects at the same altitude move with identical velocities and experience the same drag force at a time t. If Object 1 has twice (2x) the diameter of Object 2, which object has the larger drag coefficient? Explain your answer using the drag equation.
Answer:
Object 2 has the larger drag coefficient
Explanation:
The drag force, D, is given by the equation:
[tex]D = 0.5 c \rho A v^2[/tex]
Object 1 has twice the diameter of object 2.
If [tex]d_2 = d[/tex]
[tex]d_1 = 2d[/tex]
Area of object 2, [tex]A_2 = \frac{\pi d^2 }{4}[/tex]
Area of object 1:
[tex]A_1 = \frac{\pi (2d)^2 }{4}\\A_1 = \pi d^2[/tex]
Since all other parameters are still the same except the drag coefficient:
For object 1:
[tex]D = 0.5 c_1 \rho A_1 v^2\\D = 0.5 c_1 \rho (\pi d^2) v^2[/tex]
For object 2:
[tex]D = 0.5 c_2 \rho A_2 v^2\\D = 0.5 c_2 \rho (\pi d^2/4) v^2[/tex]
Since the drag force for the two objects are the same:
[tex]0.5 c_1 \rho (\pi d^2) v^2 = 0.5 c_2 \rho (\pi d^2/4) v^2\\4c_1 = c_2[/tex]
Obviously from the equation above, c₂ is larger than c₁, this means that object 2 has the larger drag coefficient
A box with an initial speed of 15 m/s slides along a surface where the coefficient of sliding friction is 0.45. How long does it take for the block to come to rest
Answer:
t = 3.4 s
The box will come to rest in 3.4 s
Explanation:
For the block to come to rest, the friction force must become equal to the unbalanced force. Therefore:
Unbalanced Force = Frictional Force
but,
Unbalanced Force = ma
Frictional Force = μR = μW = μmg
Therefore,
ma = μmg
a = μg
where,
a = acceleration of box = ?
μ = coefficient of sliding friction = 0.45
g = 9.8 m/s²
Therefore,
a = (0.45)(9.8 m/s²)
a = -4.41 m/s² (negative sign due to deceleration)
Now, for the time to stop, we use first equation of motion:
Vf = Vi + at
where,
Vf = Final Speed = 0 m/s (since box stops at last)
Vi = Initial Speed = 15 m/s
t = time to stop = ?
Therefore,
0 m/s = 15 m/s + (-4.41 m/s²)t
(-15 m/s)/(-4.41 m/s²) = t
t = 3.4 s
The box will come to rest in 3.4 s
The lower the value of the coefficient of friction, the____the resistance to sliding
Answer: lower
There are a number of factors that can affect the coefficient of friction, including surface conditions.
Values of the coefficient of sliding friction can be a good reference for specific combinations of materials. The frictional force and normal reaction are directly proportion but an increase or decrease in coefficient of friction will cause an increase or decrease in the resistance of sliding respectively
A 1500 kg car drives around a flat 200-m-diameter circular track at 25 m/s. What are the magnitude and direction of the net force on the car
Answer:
9,375
Explanation:
Data provided
The mass of the car m = 1500 Kg.
The diameter of the circular track D = 200 m.
For the computation of magnitude and direction of the net force on the car first we need to find out the radius of the circular path which is shown below:-
The radius of the circular path is
[tex]R = \frac{D}{2}[/tex]
[tex]= \frac{200}{2}[/tex]
= 100 m
after the radius of the circular path we can find the magnitude of the centripetal force with the help of below formula
[tex]Force F = \frac{mv^2}{R}[/tex]
[tex]= \frac{1500\times (25)^2}{100}[/tex]
= 9,375
Therefore for computing the magnitude of the centripetal force we simply applied the above formula.
A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction. At a particular instant, the velocity of the particle has components Vx=−1.68×10^4m/s, Vy=−2.61×10^4m/s, and Vz=5.85×10^4m/s. What is the z-component of the force on the particle at this time?
Answer:
The z-component of the force is [tex]\= F_z = 0.00141 \ N[/tex]
Explanation:
From the question we are told that
The charge on the particle is [tex]q = 7.76 *0^{-8} \ C[/tex]
The magnitude of the magnetic field is [tex]B = 0.700\r i \ T[/tex]
The velocity of the particle toward the x-direction is [tex]v_x = -1.68*10^{4}\r i \ m/s[/tex]
The velocity of the particle toward the y-direction is
[tex]v_y = -2.61*10^{4}\ \r j \ m/s[/tex]
The velocity of the particle toward the z-direction is
[tex]v_y = -5.85*10^{4}\ \r k \ m/s[/tex]
Generally the force on this particle is mathematically represented as
[tex]\= F = q (\= v X \= B )[/tex]
So we have
[tex]\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)[/tex]
[tex]\= F = q (v_y B(-\r k) + v_z B\r j)[/tex]
substituting values
[tex]\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)[/tex]
[tex]\= F= 0.00303\ \r j +0.00141\ \r k[/tex]
So the z-component of the force is [tex]\= F_z = 0.00141 \ N[/tex]
Note : The cross-multiplication template of unit vectors is shown on the first uploaded image ( From Wikibooks ).
A speed skater moving across frictionless ice at 9.2 m/s hits a 5.0 m wide patch of rough ice. She slows steadily, then continues on at 5.8 m/s. What is her acceleration on the rough ice?
Answer:
a = -5.10 m/s^2
her acceleration on the rough ice is -5.10 m/s^2
Explanation:
The distance travelled on the rough ice is equal to the width of the rough ice.
distance d = 5.0 m
Initial speed u = 9.2 m/s
Final speed v = 5.8 m/s
The time taken to move through the rough ice can be calculated using the equation of motion;
d = 0.5(u+v)t
time t = 2d/(u+v)
Substituting the given values;
t = 2(5)/(9.2+5.8)
t = 2/3 = 0.66667 second
The acceleration is the change in velocity per unit time;
acceleration a = ∆v/t
a = (v-u)/t
Substituting the values;
a = (5.8-9.2)/0.66667
a = -5.099974500127
a = -5.10 m/s^2
her acceleration on the rough ice is -5.10 m/s^2
Please help! Which statements correctly describe the effect of distance in determining the gravitational force and the electrical force? Check all that apply.
There are six statements on the list.
The first 2 are true, and the last 2 are true.
The 2 in the middle aren't true. They are false.
Can someone explain what is loss of seismic energy ?
Answer:
Seismic attenuation describes the energy loss experienced by seismic waves as they propagate. It is controlled by the temperature, composition, melt content, and volatile content of the rocks through which the waves travel.
Explanation:
Julie is playing with a toy car and is pushing it around on the floor. The little car has a mass of 6.3 g. The car has a velocity of 2.5 m/s. What is the car's momentum?
Answer:
Momentum of the car = [tex]1.575\times 10^{-2}[/tex] kg meter per second
Explanation:
Julie is playing with a car which has mass = 6.3 g = [tex]6.3\times 10^{-3}[/tex] kg
Velocity of the car is 2.5 meter per second
Since formula to calculate the momentum of an object is,
p = mv
Where, p = momentum of the object
m = mass of the object
v = velocity of the object
By substituting these values in the formula,
p = [tex](6.3\times 10^{-3})\times 2.5[/tex]
= [tex]1.575\times 10^{-2}[/tex] Kg meter per second
Therefore, momentum of the car will be [tex]1.575\times 10^{-2}[/tex] Kg meter per second.
What is the maximum height the rock will reach?
Answer:
Explanation:
initial vertical velocity = 17.5 m/s
using g=-9.81 m/s^2
apply kinematics equation
v1^2-v0^2=2gS
solve for S with v1=0, v0=+17.5
S = (v1^2-v0^2)/2g
=(0-17.5^2)/(2*(-9.81))
= 15.61 m
If a sample of 346 swimmers is taken from a population of 460 swimmers,
the population mean, w, is the mean of how many swimmers' times?
Answer:
It is the mean of 460 swimmers
Explanation:
In this question, we are concerned with knowing the mean of the population w
Now, according to the question at hand, we have a total population of 460 swimmers and we have taken out 346 swimmers for a study
The population mean in this case is simply the mean of the swimming times of the 460 swimmers
There is another related thing here called the sample mean. For the sample mean, we only make a reference to the mean of the 346 swimmers who were taken out from the population to conduct a separate study
So conclusively, the population mean w is simply the mean of the total 460 swimmers
A metal such as copper is a(n) _______________ because it provides a pathway for electric charges to move easily. A material such as rubber is a(n) _______________ because it _______________ the flow of electric charges. A material that partially conducts electric current is a(n) _______________. These materials include _______________ elements.
Explanation:
A metal such as copper is a conductor because it provides a pathway for electric charges to move easily. A material such as rubber is an insulator because it resists the flow of electric charges. A material that partially conducts electric current is a semiconductor. These materials include group 3 and group 5 elements.
Answer:
conductor
insulator
resists
semiconductor
group 3 and group 5
Explanation:
A swimmer standing near the edge of a lake notices a cork bobbing in the water. While watching for one minute, she notices the cork bob (from up to down to back up) 240 times. What is the frequency in Hz of the water wave going by?
Answer:
4Hz (240 cycles/60 seconds = 4 cycles/second)
Explanation:
hope this helped!
In a circus act a 64.3 kg magician lies on a bed of nail. The bed consists of a large number of evenly spaced, relatively sharp nails mounted in a board so that the points extend vertically outward from the board/while the magician lying down, approximately 1900 nails make contact with hisbody.
1. What is the average force exerted by each nail on the magician's body?
2. If the area of contact at the head of each nail is1.26x10-6 m2 , what is the average pressure at each contact?
Answer:
(a) Fn = 0.33 N
(b) Pn = 263.22 x 10³ Pa = 263.22 KPa
Explanation:
(a)
First, we need to calculate the total force exerted by all nails on the magician. This force must be equal to the weight of magician:
F = W
where,
F = Total Force exerted by all nails = ?
W = Weight of magician = mg = (64.3 kg)(9.8 m/s²) = 630.14 N
Therefore,
F = 630.14 N
Now, we calculate the force exerted by each nail:
Fn = F/n
where,
Fn = force exerted by each nail = ?
n = Total no. of nails = 1900
Therefore,
Fn = 630.14 N/1900
Fn = 0.33 N
(b)
The pressure exerted by each nail is given as:
Pn = Fn/An
where,
Pn = Pressure exerted by each nail = ?
An = Area of contact for each nail = 1.26 x 10⁻⁶ m²
Therefore,
Pn = 0.33 N/1.26 x 10⁻⁶ m²
Pn = 263.22 x 10³ Pa = 263.22 KPa
A uniform thin rod of mass ????=3.41 kg pivots about an axis through its center and perpendicular to its length. Two small bodies, each of mass m=0.249 kg , are attached to the ends of the rod. What must the length L of the rod be so that the moment of inertia of the three-body system with respect to the described axis is ????=0.929 kg·m2 ?
Answer:
The length of the rod for the condition on the question to be met is [tex]L = 1.5077 \ m[/tex]
Explanation:
The Diagram for this question is gotten from the first uploaded image
From the question we are told that
The mass of the rod is [tex]M = 3.41 \ kg[/tex]
The mass of each small bodies is [tex]m = 0.249 \ kg[/tex]
The moment of inertia of the three-body system with respect to the described axis is [tex]I = 0.929 \ kg \cdot m^2[/tex]
The length of the rod is L
Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as
[tex]I = I_r + 2 I_m[/tex]
Where [tex]I_r[/tex] is the moment of inertia of the rod about the describe axis which is mathematically represented as
[tex]I_r = \frac{ML^2 }{12}[/tex]
And [tex]I_m[/tex] the moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented as
[tex]I_m = m * [\frac{L} {2} ]^2 = m* \frac{L^2}{4}[/tex]
Thus [tex]2 * I_m = 2 * m \frac{L^2}{4} = m * \frac{L^2}{2}[/tex]
Hence
[tex]I = M * \frac{L^2}{12} + m * \frac{L^2}{2}[/tex]
=> [tex]I = [\frac{M}{12} + \frac{m}{2}] L^2[/tex]
substituting vales we have
[tex]0.929 = [\frac{3.41}{12} + \frac{0.249}{2}] L^2[/tex]
[tex]L = \sqrt{\frac{0.929}{0.40867} }[/tex]
[tex]L = 1.5077 \ m[/tex]
A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 40.0-gram mass is attached at the 14.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick?
Answer:
103 g
Explanation:
given that
Mass of the attached mass, m = 40 g
Initial distance of balance, x1 = 49.7 cm = 0.497 m
Final distance of balance, x2 = 39.2 cm = 0.392 m
Equilibrium point of attachment, x3 = 14 cm = 0.14 m
To get the mass of the meter stick, we use the relation
M = m.[(x2 - x3) / (x1 - x2)]
And on solving in full, we gave
M = 40.[(0.392 - 0.14) / [0.497 - 0.392)]
M = 40(0.258 / 0.105)
M = 40 * 2.457
M = 102.8 g
Therefore the needed work good for mass of the stick j
Is 103 g
In a sinusoidally driven series RLC circuit, the inductive resistance is XL = 100 Ω, the capacitive reactance is XC = 200 Ω, and the resistance is R = 50 Ω. The current and applied emf would be in phase if
Answer:
The current and the applied emf can be in phase if either of the two changes are made.
1) The inductance of the inductor is doubled, with everything else remaining constant.
2) The capacitance of the capacitor is doubled, with everything else remaining constant.
Explanation:
The current and applied emf for this type of circuit would be in phase when there is no phase difference between the two quantities. That is, Φ = 0°.
The phase difference between current and applied emf is given as
Φ = tan⁻¹ [(XL - Xc)/R]
XL = Impedance due to the inductor
Xc = Impedance due to the capacitor
R = Resistance of the resistor.
For Φ to be 0°, tan⁻¹ [(XL - Xc)/R] = 0
But only tan⁻¹ 0 = 0 rad
So, for the phase difference to be 0,
[(XL - Xc)/R] = 0
Meaning
XL = Xc
But for this question,
XL = 100 Ω, Xc = 200 Ω
For them to be equal, we have to find a way to increase the impedance of the inductor or reduce the impedance of the capacitor.
The impedance are given as
XL = 2πfL
Xc = (1/2πfC)
f = Frequency
L = Inductance of the inductor
C = capacitance of the capacitor
The impedance of the inductor can be increased from 100 Ω to 200 Ω by doubling the inductance of the inductor.
And the impedance of the capacitor can be reduced from 200 Ω to 100 Ω by also doubling the capacitance of the capacitor.
So, these are either of the two ways to make the current and applied emf to be in phase.
Hope this Helps!!!
Jerome places a bag of flour on a scale. The scale shows that the bag has a weight of 17 N. Which is the reaction force of the bag sitting on the scale?
The scale exerts a 17 N force up on the bag.
The scale exerts a 17 N force down on the counter.
Earth exerts a 17 N force down on the bag.
The bag exerts a 17 N force down on the scale.
Answer:
A. The scale exerts a 17 N force up on the bag.
Explanation:
I Just took the test
Reaction force of the bag sitting on the scale a)The scale exerts a 17 N force up on the bag.
What is newton's third law of motion ?
According to newton's third law of motion , every action have a equal and opposite reaction
When bag is being put on the scale to measure its weight , then bag must have exerted a force on the scale , in result of which an equal and opposite force must be exerted by the scale on the bag (according to newton's third law of motion) in order to keep an equilibrium state where both the forces are equal but opposite to each other .
correct option is a)The scale exerts a 17 N force up on the bag.
Learn more about newton's third law of motion :
https://brainly.com/question/974124?referrer=searchResults
# SPJ2
The aorta pumps blood away from the heart at about 40 cm/s and has a radius of about 1.0 cm. It then branches into many capillaries, each with a radius of about 5 x 10−4 cm carrying blood at a speed of 0.10 cm/s.
How many capillaries are there?
Answer:
n = 1.6*10^9 capillaries
Explanation:
In order to calculate the number of capillaries, you take into account that the following relation must be accomplished:
[tex]A_1v_1=nA_2v_2[/tex] (1)
A1: area of the aorta
v1: speed of the blood in the aorta = 40cm/s
n: number of capillaries = ?
A2: area of each capillary
v2: speed of the blood in each capillary
For the calculation of A1 and A2 you use the formula for the cross sectional area of a cylinder, that is, the area of a circle:
[tex]A=\pi r^2\\\\A_1=\pi r_1^2=\pi(1.0cm)^2=3.1415 cm^2\\\\A_2=\pi r_2^2=\pi (5*10^{-4}cm)^2=7.85*10^{-7}cm^2[/tex]
Where you have used the values of the radius for the aorta and the capillaries.
Next, you solve the equation (1) for n, and replace the values of all parameters:
[tex]n=\frac{A_1v_1}{A_2v_2}=\frac{(3.1415cm^2)(40cm/s)}{(7.85*10^{-7}cm^2)(0.10cm/s)}=1.6*10^9[/tex]
Then, the number of capillaries is 1.6*10^9