Answer:
1) I_ pendulum = 2.3159 kg m² , 2) I_pendulum = 24.683 kg m²
Explanation:
In this exercise we are asked to calculate the moment of inertia of a physical pendulum, let's start by calculating the center of mass of each elements of the pendulum and then the center of mass of the pendulum
Sphere
They indicate the density of the sphere roh = 37800 kg / m³ and its radius
r = 5 cm = 0.05 m
we use the definition of density
ρ = M / V
M = ρ V
the volume of a sphere is
V = 4/3 π r³
we substitute
M = ρ 4/3 π r³
we calculate
M = 37800 4/3 π 0.05³
M = 19,792 kg
Bar
the density is ρ = 32800 kg / m³ and its dimensions are 1 m,
0.8 cm = 0.0008 m and 4cm = 0.04 m
The volume of the bar is
V = l w h
m = ρ l w h
we calculate
m = 32800 (1 0.008 0.04)
m = 10.496 kg
Now we can calculate the center of mass of the pendulum, we use that the center of mass of the sphere is its geometric center, that is, its center and the center of mass of the bar is where the diagonals intersect, in this case it is a very bar. long and narrow, whereby the center of mass is about half the length. It's mass scepter of the pendulum is
r_cm = 1 / M (M r_sphere + m r_bar)
M = 19,792 + 10,496 = 30,288 kg
r_cm = 1 / 30,288 (10,496 0.5 + 19.792 (1 + 0.05))
r_cm = 1 / 30,288 (5,248 + 20,7816)
r_cm = 0.859 m
This is the center of mass of the pendulum.
1) Now we can calculate the moment of inertia with respect to this center of mass, for this we can use the theorem of parallel axes and that the moments of inertia of the bodies are:
Sphere I = 2/5 M r2
Bar I = 1/12 m L2
parallel axes theorem
I = I_cm + m D²
where m is the mass of the body and D is the distance from the body to the axis of rotation
Sphere
m = 19,792 ka
the distance D is
D = 1.05 -0.85
D = 0.2 m
we calculate
I_sphere = 2/5 19.792 0.05 2 + 19.792 0.2 2 = 0.019792 +0.79168
I_sphere = 0.811472 kg m²
Bar
m = 10.496 kg
distance D
D = 0.85 - 0.5
D = 0.35 m
I_bar = 1/12 10.496 0.5 2 + 10.496 0.35 2 = 0.2186 + 1.28576
I_bar = 1.5044 kg m²
The moment of inertia is a scalar quantity whereby the moment of inertia of the body is the sum of the moment of the parts
I_pendulum = I_sphere + I_bar
I_pendulum = 0.811472 +1.5044
I_ pendulum = 2.3159 kg m²
this is the moment of inertia of the pendulum with respect to its center of mass located at r = 0.85 m
2) The moment is requested with respect to the pivot point at r = 0 m
Sphere
D = 1.05 m
I_sphere = 2/5 M r2 + M D2
I_sphere = 2/5 19.792 0.05 2 + 19.792 1.05 2 = 0.019792 +21.82
I_sphere = 21.84 kg m²
Bar
D = 0.5 m
I_bar = 1/12 10.496 0.5 2 + 10.496 0.5 2 = 0.21866 + 2.624
I_bar = 2,84266 kg m 2
The pendulum moment of inertia is
I_pendulum = 21.84 +2.843
I_pendulum = 24.683 kg m²
This moment of inertia is about the turning point at r = 0 m
Tony completed a 720 km journey with an average speed of
90 km/h. For the first 285 km, he travelled at an average speed
of 95 km/h. What was the average speed for the remaining
journey?
Answer:
87 km/h
Explanation:
If a person has a velocity of 5 meters per second, how much time will they take to travel 1.5 kilometers
Answer:
300 secsExplanation:
Velocity is the change in displacement of a body with respect to time.
Velocity = Displacement/Time
Given velocity = 5m/s
Displacement = 1.5 km
Displacement = 1.5km * 1000 = 1500m
From the formula, Time = Displacement/Velocity
Time = 1500/5
Time = 300 secs
Hence the time it will take to travel 1.5km is 300secs
A plastic wire is wind on a wheel on a radius 3 cm. The whole system is initially at rest. A force is applied on the plastic wire and the wire unwinds from the wheel from its axis of rotation. The wheel is given an angular acceleration of 92.00 rad/s2 for:______.
Answer:
The torque of the wheel is 0.92 N-m.
Explanation:
Given that,
Radius of wire = 3 cm
Angular acceleration = 92.00 rad/s²
Suppose, Mass of wheel is 4 kg and radius is 5 cm. find the torque of the wheel.
We need to calculate the torque of the wheel
Using formula of torque
[tex]\tau=I\alpha[/tex]
[tex]\tau=MR^2\times\alpha[/tex]
Where, M = mass of wheel
R = radius of wheel
[tex]\alpha[/tex] = angular acceleration
Put the value into the formula
[tex]\tau=4\times(5\times10^{-2})^2\times92[/tex]
[tex]\tau=0.92\ N-m[/tex]
Hence, The torque of the wheel is 0.92 N-m.
A flea jumps straight up to a maximum height of 0.500 m . What is its initial velocity v0 as it leaves the ground? Express your answer in meters per second to three significant figures.
Answer:
v0 = 3.13m/sExplanation:
Using the equation of motion to get the initial velocity as it leaves the ground. According to the equation of motion;
v² = v0²+2as where;
v is the final velocity
v0 is the initial velocity
a is the acceleration due to gravity
s is the height.
At the maximum height, the final velocity v = 0m/s
Given s = 0.500m and a = g = -9.81m/s² (acceleration due to gravity is negative due t the upward movement of the flea)
0² = v0² - 2(9.81)(0.500)
0 = v0² - 9.81
- v0² = -9.81
v0 = √9.81
v0 = 3.13m/s
Hence the initial velocity v0 of the flea as it leaves the ground is 3.13m/s (to three significant figures)
.A full length image of a distance tall building can definitely be seen by using
Answer:
a full length of image of a distance tall building can definitely be seen by using telescope
A train is moving at a constant speed on a surface inclined upward at 15.0° with the horizontal and travels 300 meters in 5 seconds. Calculate the horizontal velocity of the train at the end of 3 seconds.
Answer:
57,9 m/s
Explanation:
A train is moving at a constant speed on a surface inclined upward at 15.0° with the horizontal and travels 300 meters in 5 seconds. The horizontal velocity of the train at the end of 3 seconds is 100m/s.
The first formula we will use is:
v = d/t
300m ÷ 5s = 60m/s
The last formula we will use is:
vh = v·cosx
60m/s·cos15.0°
60m/s · 0,965 = 57,9 m/s
Therefore, the answer is 57,9 m/s.
Hope this helped!
57,9 m/s
Explanation:velocity formula
v = d/t
= 300m/5s
= 60m/s
horizontal velocity formula
vh = v·cosx
= 60m/s×cos15°
= 60m/s×0,965
= 57,9 m/s
Question 4 of 10
Which option is part of designing a set of experimental procedures?
O A. Developing a hypothesis
B. Determining how data will be gathered
C. Submitting conclusions to the peer-review process
O D. Beginning to take measurements
Answer:
Determining how data will be gathered
Explanation:
As a scientist considers the use of a particular set of experimental procedures to carry out his research, an important consideration is the method of data collection.
The experimental procedures chosen must be those that make the process of data collection much easier, efficient and lead to collection of reliable data.
Hence determining the method of data collection is very important when designing a set of experimental procedures.
A student dropped a textbook from the top floor of his dorm and it fell according to the formula s(t) = -16t2 + 8 underroot t, where t is the time in seconds and s(t) is the distance in feet from the top of the building. A) Write a formula for the average velocity of the ball for t near 4. B) Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds. C) What is your estimate for the instantaneous velocity of the ball at t = 4.
Answer:
Explanation:
s(t) = -16t2 + 8√t
A )
Average velocity
s(t) / t = (-16t2 + 8√t)/t
A(t)= -16t + 8 / √t
average velocity of the ball for t near 4.
A(t) = -16t + 8 / √t
Lt t⇒4
B )
Distance covered in 4 s
-16t2 + 8√t
= - 16 x 16 + 8 x 2
= - 240
Distance covered in 5 s
= - 16 x 25 + 8 √5
= -400 + 17.88
= -382.12
distance covered in duration from 4 to 5 sec
= -142.12
velocity = - 142.12 / 1 = - 142.12 m /s
Distance covered in 4.5 s
= -16 x 4.5² + 8√4.5
= -324 + 16.97
= -307
distance covered during 4 to 4.5
= 67
velocity during 4 to 4.5
= -67 / .5
- 134 m /s
distance covered in 4.05 s
-16 x 4.05² + 8√4.05
-262.44 + 16.1
-246.34
distance covered during 4 to 4.05
= -6.34
velocity during 4 to 4.05
= -6.34 / .05
- 126.8 m /s
C )
Instantaneous velocity at t = 4
= - 120 m /s
(A) As 't' approaches to 4s, the formula of average velocity is
[tex]v_{avg} _{|(t\rightarrow4)}=Lim_{|(t\rightarrow4)}[-16t+ \frac{8}{\sqrt{t}} ][/tex]
(B) The average velocity for the time interval starting at t = 4 with a duration of 1 second is [tex]142.11\,m/s[/tex].
The average velocity for the time interval starting at t = 4 with a duration of 0.5 seconds is [tex]-134.058\,m/s[/tex]
The average velocity for the time interval starting at t = 4 with a duration of 0.05 seconds is [tex]-126.8\,m/s[/tex].
(C) The instantaneous velocity of the ball at 4s is [tex]-126\,m/s[/tex].
The answers are explained as follows.
Given that the distance is a function of time.
[tex]s(t)=-16\,t^2\,+\,8\sqrt{t}[/tex]
(A) The average velocity can be given by,
[tex]v_{avg}=\frac{-16\,t^2\,+\,8\sqrt{t} }{t} =-16t+ \frac{8}{\sqrt{t}}[/tex]
As 't' approaches to 4s, the formula becomes;
[tex]v_{avg} _{|(t\rightarrow4)}=Lim_{|(t\rightarrow4)}[-16t+ \frac{8}{\sqrt{t}} ][/tex]
(B) We know that the average velocity or in this case speed is the total distance by the total time taken.
Distance covered in 4 s can be found by putting [tex]t=4s[/tex] in the distance formula.
[tex]s(4)=(-16\times16)+(8\times2)=-240\,m[/tex]Distance covered in 5 s can also be found by the same method
[tex]s(5)=(-16\times25)+(8\times\sqrt{5} )=-382.11\,m[/tex]Therefore, the distance covered in from 4 to 5 seconds is;
[tex]s(5) -s(4)=-382.11\,m-(-240\,m)=-142.11\,m[/tex]So, the average velocity here = [tex]\frac{-142.11\,m}{5\,s-4\,s}=-142.11\,m/s[/tex]Distance covered in 4.5 s is given by,
[tex]s(4.5)=(-16\times4.5^2)+(8\times\sqrt{4.5} )=-307.029\,m[/tex]Therefore, the distance covered in 4 to 4.5 seconds is;
[tex]s(4.5)-s(4)=-307.029\,m-(-240\,m)=67.029\,m[/tex]So, the average velocity here = [tex]\frac{-67.029\,m}{4.5\,s-4\,s}=-134.058\,m/s[/tex]Distance covered in 4.05 s is given by,
[tex]s(4.05)=(-16\times4.05^2)+(8\times\sqrt{4.05} )=-246.34\,m[/tex]Therefore, the distance covered in 4 to 4.05 seconds is;
[tex]s(4.05)-s(4)=-246.34\,m-(-240\,m)=-6.34\,m[/tex]So, the average velocity here = [tex]\frac{-6.34\,m}{4.05\,s-4\,s}=-126.8\,m/s[/tex](C) The instantaneous velocity of the ball can be found by differentiating the function [tex]s(t)[/tex].
[tex]v(t)=\frac{ds(t)}{dt} =\frac{d}{dt}(-16\,t^2\,+\,8\sqrt{t})=-32t+\frac{4}{\sqrt{t} }[/tex]For the instantaneous velocity of the ball at 4s, substitute [tex]t=4\,s[/tex] in the above equation.[tex]v(t)=(-32\times4)+\frac{4}{2}=-126\,m/s[/tex]Learn more about finding the velocity at a given time here:
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4. Describe how the velocity of an object changes if it undergoes uniformly acceleration motion. Can its direction change?
Answer:
n the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.
In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.
Explanation:
Velocity is a vector therefore it has magnitude and direction, a change in either of the two is the consequence of an acceleration on the system.
In the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.
[tex]a_{t}[/tex] = (v₂-v₁)/Δt
In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.
[tex]a_{c}[/tex] = v2/R
In the general case, both the module and the address change
a = Ra ( a_{t}^2 + a_{c}^2)
The velocity of an object changes if it undergoes uniformly acceleration
motion by considering if it is:
Positive constant accelerationNegative constant accelerationWhat is Velocity?This is the rate of change of an object's position with time. If the object has
positive constant acceleration, the slope goes upward while if it is a
negative constant acceleration, the slope goes downward.
The direction doesn't change as a result of the uniform speed and
direction that is involved.
Read more about Velocity here https://brainly.com/question/6504879
Daniel has a bill of 2750 on his credit card. If interest is charged at a rate of 15% p.a., calculate the amount of interest that Daniel must pay for the month?
Answer:
412.5
Explanation:
all you have to do is multiply .15
Sympathy, Unproductive, Perspectives
Empathetic people try to look at things from different _______
People with low empathy often are unable to understand others perspective and involve in _____arguments.
_________ is when you feel sorry for the other person and there is no urge, no impelling reason to be feel others feelings.
Answer:
Empathetic people try to look at things from different perspectives
People with low empathy often are unable to understand others perspective and involve in unproductive arguments.
sympathy is when you feel sorry for the other person and there is no urge, no impelling reason to be feel others feelings.
Explanation:
A machine, modeled as a simple spring-mass system, oscillates in simple harmonic motion. Its acceleration is measured to have an amplitude of 10,000 mm/s2 with a frequency of 8 Hz. Compute the maximum displacement the machine undergoes during this oscillation.
Answer:
3.96*10^-3 m
Explanation:
Given that
Amplitude, A = 10000 mm/s² = 10 m/s²
Frequency, f = 8 hz
the maximum acceleration of a simple harmonic motion is given as
|a|(max) = Aw², where
A = amplitude of the simple harmonic motion
w = angular frequency of the simple harmonic motion
Remember, w = 2πf, so
w = 2 * 3.142 * 8
w = 50.272 rad/s
Rearranging the formula, and making a the subject, we have
A = |a|(max) / w², so that
A = 10 / 50.272²
A = 10 / 2527.27
A = 0.00396 m
A = 3.96*10^-3 m
What is true about all uranium atoms
Answer:
erm they all should have same numbers of protons
Answer:
They each have the same number of nuclear particles.
Explanation:
What is true about all uranium atoms? They each have the same number of nuclear particles. They each have the same number of neutral particles. They each have the same number of neutrons . Uranium is an element that is often used in nuclear power plants. I hope this helped :)
What do you think is the
difference between
speed and acceleration?
What do you think is the
difference between mass
and weight?
A soccer ball is at rest. A player kicks the ball from the ground. What impact will the energy transfer have on the ball?
A. The ball will stay at rest
B. The ball will move forward
C. The ball will move backward.
D. The ball will stay in the same position.
Answer:
Its B
Explanation:
the force of the ball is being kicked from the back side of the ball will causes a pushing motion and the ball will move forward
A shopper walks eastward 3.2 meters and then westward
7.2 meters.
Pro
For this motion, what is the distance moved?
i NED HELP i’m so confused
The element found in Period 2, Group 8 is called Argon. True False
can someone define Newton's 1st 2nd and 3rd law
Answer:
HEY THERE! HERE'S YOUR ANSWER!
Explanation:
In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.
HOPE THIS HELPS YOU BUDDY!
A particle moves along the x-axis according to the equation
S = 4+6t-2t^2
where S is in meters and t is in seconds. At t = 3.0 s, calculate
(a) the position of the particle
(b) its instantaneous velocity
(c) its instantaneous acceleration.
◈ A particle is moving along the x-axis according to the equation
S = 4 + 6t - 2t²____________________________
✪ Position at t = 3s :
➝ S = 4 + 6t - 2t²
➝ S = 4 + 6(3) - 2(3)²
➝ S = 4 + 18 - 18
➝ S = 4m
✪ Instantaneous velocity at t = 3s :
➝ v = dx/dt = d(4 + 6t - 2t²)/dt
➝ v = 6 - 4t
➝ v = 6 - 4(3)
➝ v = 6 - [tex]\sf{12}[/tex]
➝ v = -12m/s
✪ Instantaneous acc. at t = 3s :
➝ a = dv/dt = d(6 - 4t)/dt
➝ a = -4m/s²
[Acceleration does not depend on time]
A car is moving at 6 m/s and then accelerates at 1.7 m/s2 for 4.2 seconds. What is the final velocity of the car?
Explanation:
Hey there!!
Here,
Initial velocity (u) = 6m/s.
Acceleration (a) = 1.7m/s^2.
Time (t) = 4.2s.
final velocity (v) = ?
We have,
[tex]a = \frac{v - u}{t} [/tex]
Putting their values,
[tex]1.7 = \frac{v - 6}{4.2} [/tex]
7.14 = v - 6
v = 7.14 + 6
Therefore, the final velocity is 13.14 m/s.
Hope it helps....
"Find the total weight of an 18-ft3 tank of oxygen if the oxygen is pressurized to 184 psia, the tank itself weighs 150 lbf, and the temperature is 95°F."
Answer:
Explanation:
1 psi = 6894.76 Pa
P = 184 psi = 12.686 x 10⁵ Pa .
Temperature T = 95⁰F = 35⁰C= 308 K
volume V = 18 ft³ = 18 x 0.0283168 m³
= .51 m³
From the gas law
PV/RT = n where n is mole of gas
= 12.686 x 10⁵ x .51 / 8.31 x 308
= 252.78 gm mole
= 252.78 x 32 gm
= 8.08896 kg
= 2.20462 x 8.08896 lb
= 17.833 lb
= 17.833 / 32 lbf
= .5573 lbf
weight of tank = 150 lbf
Total weight = 150 + .5573
= 150.5573 lbf .
What feature of the sun is a section that is cooler than its surroundings
Answer:
sunspots looking dark
a graph compares the weight and height of several different breeds of cat. What type of model is the graph?
Answer: Mathematical Model
Explanation:
Took the test
what is a hypothisis
Answer:
a precise, testable statement of what the reseatcher predict will be the outcome of the study
Explanation:
Answer:
supposition or proposed explanation
Explanation:
its made on the basis of limited evidence as a starting point for further investigation.
Convert the following to SI units: a) 75 in. b) 3.45 x 106 yr.c) 62 ft/day.d) 2.2 x 104 mi2.
Answer:
A. 0.025inches= 1m
So 75in= 0.025*75= 1.88m
B 3.45*10^6yrs= 3.45E6* 365x 86900s
= 1.09*10^8s
C. 62ft/day= 62* 0.3048/86900= 2.1*10^-4
D. 2.2*10^4mi² x ( 1609.3)²
= 6.1*10^7m²
A sailboat took 25 hours to cover 1/4 of a journey. Then, it
covered the remaining 144 miles in 3.5 hours. What was the
average speed for the whole journey?
mph
Answer:
32
Explanation:
The average speed of the car for the remaining is 32 miles/h.
What is average speed?The average speed of any moving object is the ratio of the total distance covered and the total time taken to cover that distance.
Average Speed S= distance /time
Given is a sailboat took 2.5 hours to cover 1/4 of a journey. Then, it covered the remaining 144 miles in 3.5 hours.
If x be the total distance covered during journey, then
3/4 x = 144
x = 192 miles
The total time taken for journey is 2.5 +3.5 = 6 h
The average speed for the whole journey is
Avg speed S= 192 / 6
S = 32 miles/h
Thus, the average speed of the car for the remaining is 32 miles/h.
Learn more about average speed.
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Which word describes a statement that has been accepted, tested, and supported by multiple sets of evidence? A. Hypothesis B. Theory C. Procedure D. Data
Answer:
Data
Explanation:
A softball player is testing her physics knowledge and tosses a softball upward from the top of a building. The building is 100 m tall and the ball starts with a velocity of 10 m/s.a. What is the maximum height the ball reaches and at what time? b. How much time does it take for the ball to go from h=50 m to h=0 m on the way down? c. What is the velocity when h=50 m?
Answer:
Explanation:
We shall apply newton's laws formula
a )
initial velocity in upward direction u = 10 m/s
acceleration due to gravity g = 9.8/ m .s²
Let h be the maximum height where v = o
v² = u² - 2gh
0 = 10² - 2 gh
h = 10² / 2g
= 10² / 2 x 9.8
= 5.10 m
Since the ball was thrown from height of 100 m , total maximum height of ball
= 100 + 5.10
= 105.10 m
Let t be the time taken
v = u - gt
0 = 10 - gt
t = 10 / 9.8
= 1.02 s
b )
when h = 50 on its way downwards , velocity
v² = u² + 2 g s
v² = 0 + 2 x 9.8 x ( 105.10 - 50 )
[ distance travelled by ball at this point from top = 105.1 - 50 = 55.10 ]
v = 32.86 m / s
Let us find out final velocity of touching the ground . For it distance travelled = 105.10
v² = u² + 2gh
v² = 0 + 2 x 9.8 x 105.1
v = 45.39 m /s
Now velocity at h = 50 is 32.86
velocity at h = 0 is 45.39
time taken to travel fro h = 50 to h = 0
v = u + gt
45 .39 = 32.86 + 9.8 x t
t = 1.28 s .
a quarterback throws a football in a high, arching spiral to a receiver running down the field. where is the football traveling the fastest? where is it traveling the slowest?
Answer:
Travelling fastest: when the football leaves the hands of the quarterback and when the receiver gets it
Travelling slowest: at the very top of then arched trajectory
Explanation:
Notice that this example can be analyzed with a two dimensional pattern as in the launching of a cannonball, where we study separately the velocities and acceleration acting vertically and horizontally.
The football is thrown at an agle with the horizontal with an initial velocity that is decomposed in the vertical and in the horizontal axes.
The Horizontal movement of the football is that of an object with constant velocity (that of the horizontal component of the initial velocity imparted by the quarterback) .
The vertical movement of the football is that of an object moving in an accelerated fashion, with constant acceleration due to the gravitational field, and with an initial velocity opposite to that constant acceleration. The initial velocity is that of the vertical component of the initial velocity imparted by the quarterback.
AT every point on the path of the ball, its velocity can be calculated by the vector addition of the horizontal component (with constant velocity as we discussed above) and the vertical component (this velocity is changing since its is under accelerated motion) of the object's velocity at any given time through the path.
The maximum velocity of the football will be at the point where the two components are at their maximum (that is when the football leaves the hand of the thrower, and when it gets to the hands of the receiver.
While the minimum is going to be when the vertical component of the velocity is at is minimum (zero) at the top of the arched trajectory when the football changes vertical direction and starts heading downwards.