Complete Question
The complete question is shown on the first uploaded image
Answer:
The expression for the change in the air temperature is [tex]\Delta T = \frac{Mv^2}{2 \rho_{air} c_{air}* V}[/tex]
Explanation:
From the question we are told that
The mass of the train is M
The speed of the train is v
The volume of the station is V
The density of air in the station is [tex]\rho_{air}[/tex]
The specific heat of air is [tex]c_{air}[/tex]
The workdone by the break can be mathematically represented as
[tex]W =\Delta KE = \frac{1}{2} Mv^2[/tex]
Now this is equivalent to the heat transferred to air in the station
Now the heat capacity of the air in the station is mathematically represented as
[tex]Q = \rho_{air} * m_{air} * c_{air} (\Delta T)[/tex]
Now Since this is equivalent to the workdone by the breaks we have that
[tex]\frac{1}{2} Mv^2 = m_{air} * c_{air} (\Delta T)[/tex]
=> [tex]\Delta T = \frac{Mv^2}{2 \rho_{air} c_{air}* V}[/tex]
Explain the force responsible for the formation of our solar system.
Answer:
gravity
Approximately 4.5 billion years ago, gravity pulled a cloud of dust and gas together to form our solar system.
Explanation:
Which multiplier does the word kilo represent? A. 100 B. 1,000 C. `(1)/(100)` D. `(1)/(10)`
Answer:
A. 1,000
Explanation:
Kilo is the prefix for thousands, thus, the multiplier that kilo represents is 1,000
A. 1,000
Answer:
B. 1,000
Explanation:
plato
Why will a change in velocity have a greater affect on KE than a change in mass?
Answer: the answer is corect of what everone else said
Explanation:
Which best explains why making a pancake from batter is an example of a chemical change?
-The pancake that forms is a different state of matter.
-The change from batter to pancake can be reversed.
-A new substance forms when the batter is cooked.
-The batter changes shape when it is cooked.
The correct answer is C. A new substance forms when the batter is cooked.
Explanation:
When a chemical change occurs the properties, and composition of substances change. This means atoms in the substance re-arrange to form a new substance. This only occurs when there is a chemical change, but not when physical changes occur, indeed a physical change only affects the state of the matter, shape, size, etc.
In the case of the pancake, this is an example of a chemical change because though the process of cooking the pancake changes its composition. Due to this, the properties of the cooked pancake, and the butter are not the same as a new substance forms. Also, in this and most chemical changes, reversibility is not possible, that is why you cannot reverse the process and make the cooked pancake batter once again.
Answer:
C. on edge
Explanation:
In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their components. (You must first specify a coordinate system in order to find the components of each arrow.) This problem gives you some practice with the components. Let vectors
A =(1,0,−3), B =(−2,5,1), and C =(3,1,1). Calculate the following, and express your answers as ordered triplets of values separated by commas.
a. A- B=______
b. B-C=_______
c. -A +B- C=______
d. 3A- 2C=______
e. -2A+3B- C=_____
f. 2A- 3(B-C)
Answer:
a)A-B =(1,0,-3) +(2,-5,-1) =(3, -5, -4);
b)B-C =(-2-3, 5-1, 1-1) = (-5,4,0);
c)-A +(B –C) = (-1,0,3) +(-5,4,0) =(-6,4,3);
d) =(3,0,-9) –(6,2,2) = (-3, -2, -11);
e) = (-2,0,6) +(-6,15,3) +(-3,-1,-1) =
d) = (2,0,-6) –(-5*3, 4*3, 0) =
Explanation:
1)C=k*A = (k*a1, k*a2, k*a3);
2)C=-A = -1*A;
3)C= A+B = (a1+b1, a2+b2, c2+c2);
4)C=A-B =A +(-B);
5)A+B=B+A;
6)A+B+C =(A+B)+C =A+(B+C
Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes the moon to orbit the earth. The force of gravitational attraction is represented by the equationF=Gm1m2r2where F is the magnitude of the gravitational attraction on either body, m1 and m2 are the masses of the bodies, r is the distance between them, and G is the gravitational constant. In SI units, the units of force are kgâ‹…m/s2, the units of mass are kg, and the units of distance are m. For this equation to have consistent units, the units of G must be which of the following?Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes the moon to orbit the earth. The force of gravitational attraction is represented by the equation,where is the magnitude of the gravitational attraction on either body, and are the masses of the bodies, is the distance between them, and is the gravitational constant. In SI units, the units of force are , the units of mass are , and the units of distance are . For this equation to have consistent units, the units of must be which of the following?A. kg3mâ‹…s2B. kgâ‹…s2m3C. m3kgâ‹…s2D. mkgâ‹…s2
Answer:
The S.I unit of G must be m³/kg.s² to keep the equation consistent.
Explanation:
We have the equation:
F = Gm₁m₂/r²
where,
F = Gravitational Force between Two Bodies
G = Gravitational Constant
m₁ = Mass of 1st Body
m₂ = Mass of 2nd body
r = Distance between the Bodies
The S.I Units of the quantities are:
F = Newton = kg.m/s²
m₁ = kg
m₂ = kg
r = meter = m
G = ?
Therefore, to find the units of Gravitational Constant (G), we substitute the known units in the formula:
kg.m/s² = G(kg)(kg)/m²
G = m³/kg.s²
What is the relationship Between frequency and sound?
Answer:
The sensation of a frequency is commonly referred to as the pitch of a sound. A high pitch sound corresponds to a high frequency sound wave and a low pitch sound corresponds to a low frequency sound wave.
Explanation:
Hope this helps : )
What an athlete do to increase his stability when he lands on the balance beam from a flip?
Answer:
A series of body movements in which a person leaps into the air and then rotates one or more times while airborne is referred to as a flip.
When an athlete flips and lands on the beam, to increase his or her balance, he or she must spread out their hands-on either sides.
In doing this, they redistribute their forces (centre of gravity) to gain equilibrium create inertia to dampen the kinetic energy during the flip.
Cheers!
Batman and Robin are attempting to escape that dastardly villain, the Joker, by hiding in a large pool of water (refractive index nwater = 1.333). The Joker stands gloating at the edge of the pool. (His makeup is watersoluble.) He holds a powerful laser weapon y1 = 1.49 m above the surface of the water and fires at an angle of θ1 = 27◦ to the horizontal. He hits the Boy Wonder squarely on the letter "R", which is located y2 = 3.77 m below the surface of the water. θ x y y 1 1 2 R J Batplastic surface Mirrored Surface water B How far (horizontal distance) is Robin from the edge of the pool? (Fear not, Batfans. The "R" is made of laser-reflective material.) Answer in units of m.
Answer:
x_total = 4.29m
Explanation:
To solve this exercise we must work in parts. Let's use the law of refraction to find the angle of the refracted ray and trigonometry to find the distances.
Let's start by looking for the angles that the laser refracts
n₁ sin θ₁ = n₂ sin θ₂
where n₁ is the air refraction compensation n₁ = 1, n₂ the water refractive index n₂ = 1,333
θ₂ = sin⁻¹ (n₁ sin θ₁/n₂)
θ₂ = sin⁻¹ (1 sin 27 / 1,333)
θ₂ = sin⁻¹ 0.34057
θ₂ = 19.9º
now let's find the distance from the edge of the pool to the point where the ₂lightning strikes the water
tan θ₁ = y₁ / x₁
x₁ = y₁ / tan θ₁
x₁ = 1.49 / tan 27
x₁ = 2,924 m
Now let's look for the waterfall in the water as far as Robin
tan θ₂₂ = y₂ / x₂
x₂ = y₂ / tan θ₂
x₂ = 3.77 / tan 19.9
x₂ = 1,364
the distance from the edge of the pool to Robin is
x_total = x₁ + x₂
x_total = 2,924 + 1,364
x_total = 4.29m
A car is travelling at 16.7m/s. If the car can slow at a rate of 21.5m/s^2, how much time does the driver need in order to stop the red light?
Answer:
About 0.7767 seconds
Explanation:
[tex]\dfrac{16.7m/s}{21.5m/s^2}\approx 0.7767s[/tex]
Hope this helps!
Explain why you cannot charge one end of a steel rod and leave the other end uncharged
Answer:
Electrons can be made to move from one object to another. However, protons do not move because they are tightly bound in the nuclei of atoms.Static charge occurs when electrons build up on an object. Static charge:
can only build up on objects which are insulators, eg plastic or wood
cannot build up on objects that act as conductors, eg metals
Conductors allow the electrons to flow away, forming an electric current.
When a static charge on an object is discharged, an electric current flows through the air. This can cause sparks. Lightning is an example of a large amount of static charge being discharged.
Explanation:
How does the periodic table organize atoms of elements with the same number of valence electrons?
1.in cells
2.in columns
3.in diagonals
4.in rows
Answer:
2.in columns
Explanation:
The elements of the periodic table is distribute as in 7 periods and 18 groups. The element is distribute on base on increasing atomic number. The period is called horizontal row and the group is called vertical column. in periodic number of sphere around the nucleus remain the same, but as we move from left to right, the valence electrons continue to increase.so correct option is 2.in columns
A lightbulb manufacturer makes bulbs with different "color temperatures," meaning that the spectrum of light they emit is similar to a blackbody with that temperature. Assuming the emitting areas of the filaments in two bulbs with color temperatures of 2,000 K and 4,000 K are the same, which of the two is the brighter?
Answer:
The bulb with higher temperature(4000 K) will be brighter
Explanation:
From the question we are told that
The color temperature for first bulb is [tex]T_1 = 2000K[/tex]
The color temperature for second bulb is [tex]T_2 = 4000K[/tex]
Generally the emission power of black body radiation is mathematically represented as
[tex]E = \sigma T^4[/tex]
Where [tex]\sigma[/tex] is the Stefan-Boltzmann constant with a value [tex]\sigma = 5.67 * 10^{-8} W m^{-2} K^{-4.}[/tex]
Now for [tex]T_1 = 2000K[/tex]
[tex]E_1 = 5.67*10^{-8} * (2000)^4[/tex]
[tex]E_1 = 907.2 \ W/m^2[/tex]
At [tex]T_2 = 4000K[/tex]
[tex]E_2 = 5.67*10^{-8} * 4000[/tex]
[tex]E_2 = 14515.2 \ KW/m^2[/tex]
Looking at the result we got we see that the emission power for the higher temperature bulb is higher, this means that its power to emit in the visible spectrum range would be higher
So the bulb with higher temperature will be brighter
A string, fixed at both ends and 2.14 m long, is vibrating in its second harmonic. It excites a 0.5-m long pipe closed on one end into its third harmonic (first overtone). Use a speed of sound in air of 345 m/s if needed. (a) Sketch the normal modes given above for both the string and the closed-open pipe. [8] (b) What is the wave speed on the string? [7] (c) Write down the wave function y(x,t) for the 2nd harmonic wave on the string. Assume an arbitrary amplitude. [10] (d) If the mass density of the string is 0.55 g/m, what is the tension in the string? [5]
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
Revlew Millikan's Photoelectric Experiment Robert A. Mlkan (1868 1953). although best known for his "oil-drop experiment," which measured the charge of an electron, also perfomed pioneering research on the photoelectric effect. In experiments on lithium, for example, Millikan observed a maximum kinetic energy of 0.550 eV when electrons were ejected with 433.9-nm light. When light of 253.5 m was used, he observed a maximum kinetic energy of 2.57 eV.
Part A What is the work function,W, for lithium, as determined from Milikan's results? Express your answer to three significant figures and include appropriate units.
Part B What maximum kinetic energy do you expet illikan found when he used light with a wavelength of 362.4 TIm? Express your answer to three significant figures and include appropriate units Value Units
Answer:
A.) Work function = 2.3 eV
B.) Max. K.E observed = 1.1 eV
Explanation:
A.) Millikan observed a maximum kinetic energy of 0.550 eV when electrons were ejected with 433.9-nm light. When light of 253.5 m was used, he observed a maximum kinetic energy of 2.57 eV.
work function (f) is the minimum energy required to remove an electron from the surface of the material.
hf = Ø + K.E (maximum)
Where
h = Plank constant 6.63 x 10-34 J s
Ø = work function
hc/λ = Ø + K.E (max)
(6.63×10^-34 × 3×10^8)/433.9×10^-9 = Ø + 0.550 × 1.6×10^-19
4.58×10^-19 = Ø + 8.8×10^-20
Ø = 4.58×10^-19 - 8.8×10^-20
Ø = 3.7 × 10^-19 J
Converting Joule to eV
Ø = 3.7 × 10^-19/1.6×10^-19
Ø = 2.3 eV
B.) When light of wavelength 362.4 m is used
The maximum K.E observed = incident light K.E - (the work function).
Incident K.E = hf = hc/λ
Incident K.E =
(6.63×10^-34 × 3×10^8)/362.4
Incident K.E = 5.5 × 10^-28J
Let's convert joule to eV
Incident K.E = 5.5×10^-28/1.6×10^-19
Incident K.E = 3.4 × 10^-9
Max. K.E observed = 3.4 - 2.3
Max. K.E observed = 1.1 eV
I Science Sem 1
1 7.1.2 Exam: Semester 1 Exam
Question 18 of 30
3 Points
How much force is required in Newtons) to accelerate a 12-kg bicycle, along
with its 50-kg rider, at 2 m/s2?
Give your answer as a number.
Answer here
Answer:
Force, F = 124 N
Explanation:
We have,
Mass of bicycle is 12 kg and mass of rider is 50 kg
Total mass of the system is 12 kg + 50 kg = 62 kg
Acceleration of the system is 2 m/s²
It is required to find the force required to accelerate the system. The force acting on an object is given by :
[tex]F=ma\\\\F=62\ kg\times 2\ m/s^2\\\\F=124\ N[/tex]
So, the force of 124 N is acting on the system.
Fill in the word that completes each statement.
Current cannot flow through a circuit when the switch is
Current flows through a circuit and all of the bulbs are lit only when the switch is
Answer:
Current cannot flow through a circuit when the switch is OPENCurrent flows through a circuit and all of the bulbs are lit only when the Switch is CLOSEExplanation:
In a series connected electrical circuit as also seen in all electrical Circuit the switch allows the flow of Electricity through out the circuit therefore causing the lighting of bulbs and also other electrical appliances to receive electricity.
When the switch is Open the flow of electricity in the circuit is disrupted while when the Switch is closed it allows the flow of electricity seamlessly through out the circuit. there are different types of switches but the most important is the MAIN SWITCH because it is used to shutoff electricity completely from the entire circuit during emergencies or repairs.
Answer:
Current cannot flow through a circuit when the switch is OPEN
Current flows through a circuit and all of the bulbs are lit only when the switch is CLOSED
Explanation:
Thinking about the winter we missed out on this year. Calvin and his tiger go sledding down a snowy hill. There is friction between the snow and the sled, and there is air resistance. Looking at the entire time that they spend on the hill (from when they start from rest at the top of the hill until they reach the bottom moving fast), indicate whether each of the following quantities is positive, negative, or zero, by writing either, -, or 0 for the questions below. We take Calvin, the tiger, and the sled as our system.
a. The work done by the normal force that the snow exerts on the system?
b. The work is done by the frictional force that the snow exerts on the system?
c.The change in the kinetic energy of the system?
d.The change in the gravitational potential energy of the system?
e. The total work is done on the system?
Answer:
a) W=0, b) Work is negative, c) work is positive and scientific energy variation is positive, d) the variation of the potential enrgy is negative,
e) total work is positive
Explanation:
Work in physics is defined by the scalar scalar product of force by displacement
W = F. dx
The bold are vectors; this can be written in the form of the mules of the quantities
W = F dx cos θ
where θ is the angle between force and displacement.
a) The normal force is perpendicular to the inclined plane which is perpendicular to the displacement, therefore the angle is
θ = 90 cos 90 = 0
W=0
In conclusion the work is zero
b) The friction force opposes the displacement whereby the angle is
θ = 180 cos 190 = -1
W = - fr d
Work is negative
c) To calculate the change in kinetic energy we use that the work is equal to the variation of the kinetic energy
m g sin θ L = ΔK
this magnitude is positive since the angle is zero cos 0 = 1
how the system starts from rest ΔK = Kf -K₀= + Kf -0
work is positive and scientific energy variation is positive
d) change in potential energy
The potential energy is is ΔU = Uf -U₀
we fix the reference system in the bases of the plane so Uf = 0
ΔU = -U₀
the variation of the potential enrgy is negative
e) The total work is formed by the work of the weight component, the work of the friction force
W_Total = W_weight - W_roce
as the body moves down
W_Total> 0
Therefore the total work is positive
A spring with a 6-kg mass and a damping constant 7 can be held stretched 0.5 meters beyond its natural length by a force of 1.5 newtons. Suppose the spring is stretched 1 meters beyond its natural length and then released with zero velocity, In the notation of the text, what is the value c2−4mk? m2kg2/sec2 Find the position of the mass, in meters, after t seconds. Your answer should be a function of the variable t with the general form c1eαtcos(βt)+c2eγtsin(δt)
Answer:
The value for [tex]c^2 - 4mk[/tex] is : [tex]\mathbf{-23 \ m^2kg^2/sec^2}[/tex]
The position of the mass (m) after t seconds is:
[tex]\mathbf{x(t) = e^{\frac{1}{2}t }( cos \frac{\sqrt{36}}{12}t + \frac{6}{\sqrt{36}} sin \frac{\sqrt{36}}{12}t)}[/tex]
Explanation:
The spring constant is :
[tex]F = kx \\ \\ k = \frac{F}{x} \\ \\ k = \frac{1.5}{0.5} \\ \\ k = 3 \ N/m[/tex]
The value for [tex]c^2 - 4mk[/tex] is :
= [tex]7^2 - 4(6)(3)[/tex]
= [tex]49 - 27[/tex]
= [tex]\mathbf{-23 \ m^2kg^2/sec^2}[/tex]
The differential equation for this system is :
[tex]m \frac{d^2x}{dt^2}+c \frac{dx}{dt}+kx = 0[/tex]
[tex]6 \frac{d^2x}{dt^2}+7\frac{dx}{dt}+3x = 0[/tex]
and the auxiliary equation for this differential equation is :
[tex]6m ^2+ 6m + 3 = 0[/tex]
using the quadratic formula :
[tex]\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}[/tex]
[tex]\frac{-6 \pm \sqrt{6^2 - 4(6)(3)} }{2(6)}[/tex]
= [tex]\frac{-6 \pm \sqrt{-36} }{12}[/tex]
= [tex]-\frac{1}{2} \pm \frac{\sqrt{36} }{12}i[/tex]
The general solution is :
[tex]x(t) = e^{-\frac{1}{2}t}}(c_1 cos {\frac{\sqrt{36} }{12}} t + c_2 sin \frac{\sqrt{36} }{12} t})[/tex]
Initial conditions : [tex]x(0) = 1 \ m , x' (0) = 0\\[/tex]
[tex]x(0) = ( c_1 cos 0 + c_2 sin 0)[/tex]
[tex]1 = c_1[/tex]
[tex]x't = e^{- \frac{1}{2}t}}}(- c_1 \frac{\sqrt{36} }{12} sin \frac{\sqrt{36} }{12}t + c_2 \frac{\sqrt{36} }{12} cos \frac{\sqrt{36} }{12}t) - e^{- \frac{1}{2}t}}} (\frac{1}{2}) (c_1cos \frac{\sqrt{36} }{12} t +c_2 sin \frac{\sqrt{36} }{12} t)[/tex]
[tex]x'(0)= e^{- \frac{1}{2}t}}}(- c_1 \frac{\sqrt{36} }{12} sin0 + c_2 \frac{\sqrt{36} }{12} cos 0) - e^{- \frac{1}{2}t}}} (\frac{1}{2}) (c_10+c_2 sin0)[/tex]
[tex]x'(0) = c_2 \frac{\sqrt{36} }{12}-c_1 \frac{1}{2}[/tex]
replacing 1 for [tex]c_1[/tex]
[tex]0 = c_2 \frac{\sqrt{36} }{12} -\frac{1}{2}[/tex]
[tex]c_2 \frac{\sqrt{36} }{12} = \frac{1}{2}[/tex]
[tex]c_2 = \frac{6}{\sqrt{36} }[/tex]
The position of the mass (m) after t seconds is:
[tex]\mathbf{x(t) = e^{\frac{1}{2}t }( cos \frac{\sqrt{36}}{12}t + \frac{6}{\sqrt{36}} sin \frac{\sqrt{36}}{12}t)}[/tex]
Throughout the reflection, make sure you have a copy of the Student Guide, PDF or Word, and your data tables. Use the drop-down menus to complete the statements.
This activity was a multistep laboratory. Think about each step and identify the variables.
In all three parts, the dependent variable was .
The independent variable is the one that the scientist intentionally manipulates. In Part I involving the golf ball and table tennis ball, the independent variable was .
In Part II involving the inflated and deflated footballs, the independent variable was .
In Part III involving the baseball, the independent variable was .
Answer:
-In all three parts, the dependent variable was density.
-The independent variable is the one that the scientist intentionally manipulates. In Part I involving the golf ball and table tennis ball, the independent variable was mass.
-In Part II involving the inflated and deflated footballs, the independent variable was volume.
-In Part III involving the baseball, the independent variable was material.
Explanation:
correct on edge! hope it helps :)
Answers:
in all three parts, the dependent variable was. density
The independent variable is the one that the scientist intentionally manipulates. In Part I involving the golf ball and table tennis ball, the independent variable was. mass
In Part II involving the inflated and deflated footballs, the independent variable was. volume
In Part III involving the baseball, the independent variable was .material
Explanation:
Some students conduct an experiment to find out how long different rubber balls would keep bouncing before they stop, when they dropped from a fixed height. which unit would be the most suitable for measuring the results of their experiment?
A. hours
B. kilograms
C. metres
D. seconds
What is the linear diameter (in meters) of an object that has an angular diameter of 110 arcseconds and a distance of 25,000 m?
Answer:
Explanation:
angle θ = 110 arcsec
= 110 /60 arcmin
= 1.8333 arcmin
= 1.8333 / 60 degree
= .030555 degree
= .030555 x π / 180 radian
= 5.33 x 10⁻⁴ radian .
linear diameter = angular diameter in radian x distance
= 5.33 x 10⁻⁴ x 25000
= 13.325 m
linear diameter of object = 13.325 m .
Sonia was experimenting with electric charges. She tied two inflated balloons together, held them next to each other, and rubbed both with a piece of wool.
What did Sonia observe, and why?
Answer: They will repel each other.
Explanation:
Two inflated balloons when rubbed with woolen cloth will lead to repeal each other because of the similar charges on both the balloons.
Rubbing both the balloons together by the woolen cloth will introduce negative charge in the balloons.
As, we know that the same charges repeal each other both of the balloons with be apart from each other.
This is due to the static electricity, the negatively charged particles jump to the positive one. When balloons are rubbed they become negatively charged.
Answer:
Sample Response: Sonia observed that the two balloons repelled each other. This is because both balloons acquired the same charge when she rubbed them with the piece of wool, and like charges repel each other.
Explanation:
Did it on Egde 2020
Consider a spinning plate is dropped onto a stationary plate (which is itself at rest on a frictionless surface). Both plates have a radius of 30cm and a mass of 1kg. The spinning plate is initially spinning at a rate of 0.7 revolutions per second. Hint: This is like a totally-inelastic collision.
Required:
a. After a sufficiently long time, what is the angular velocity of the initially-spinning plate? What about the initially-stationary plate?
b. Assume that the period of velocity matching happens over a course of 2 seconds. Further, assume that the torque exerted by each plate on the other is constant over time. In that case, what is the magnitude of the acceleration that each plate feels during those two seconds? Hint: Use the rotational impulse-momentum theorem.
Answer:
The final angular velocity is [tex]w_f = 2.1994 rad/sec[/tex]
The angular acceleration is [tex]\alpha = 1.099 \ rad/sec^2[/tex]
Explanation:
From the question we are told that
The radius of each plate is [tex]r = 30 \ cm = \frac{30}{100} = 0.3 \ m[/tex]
The mass of each plate is [tex]m_p = 1 \ kg[/tex]
The angular speed of the spinning plate is [tex]w = 0.7 \ rev \ per \ sec = 0.7 * 2 \pi = 4.3988 \ rad/sec[/tex]
From the law of conservation of momentum
[tex]L_i = L_f[/tex]
Where [tex]L_i[/tex] is the initial angular momentum of the system (The spinning and stationary plate ) which is mathematically represented as
[tex]L_i = I_1 w + 0[/tex]
here [tex]I_1[/tex] is the moment of inertia of the spinning plate which mathematically represented as
[tex]I_1 = \frac{m_pr^2}{2}[/tex]
and the zero signify that the stationary plate do not have an angular momentum as it is at rest at the initial state
[tex]L_f[/tex] is the final angular momentum of the system (The spinning and stationary plate) , which is mathematically represented as
[tex]L_f = (I_1 + I_2 ) w_f[/tex]
Where
[tex]I_2[/tex] is the moment of inertia of the second plate (This was stationary before but now it spinning due to the first pate ) and is equal to [tex]I_1[/tex]
and [tex]w_f[/tex] is the final angular speed
So we have
[tex]I_1 w = (I_1 + I_2)w_f[/tex]
[tex]\frac{m_p r^2}{2} * w = 2 * \frac{m_p r^2}{2} * w_f[/tex]
[tex]w = 2 * w_f[/tex]
substituting values
[tex]4.3988 = 2 * w_f[/tex]
[tex]w_f = \frac{4.3988 }{2}[/tex]
[tex]w_f = 2.1994 rad/sec[/tex]
The the rotational impulse-momentum theorem can be mathematially represented as
[tex]\tau * \Delta t = 0.09891[/tex]
Where [tex]\tau[/tex] is the torque and [tex]\Delta t[/tex] is the change in time
So at [tex]\Delta t = 2 \ sec[/tex]
[tex]\tau = \frac{0.09891}{2}[/tex]
[tex]\tau = 0.0995 \ Nm[/tex]
now the angular acceleation is mathematically represented as
[tex]\alpha = 2 * \frac{\tau}{m_p * r^2 }[/tex]
substittuting values
[tex]\alpha = 2 * \frac{0.0995}{1 * 0.3^2}[/tex]
[tex]\alpha = 1.099 \ rad/sec^2[/tex]
What do you think are the key consideration for machining parts and joints when designing self-assembly furniture like what you see in Ikea?
Answer:
then, that IKEA would be the authority on flat-pack ... Hardware for the Hopen Bed, via Swedish Furniture Parts.
Explanation:
Una furgoneta circula por una carretera a 55km/h. Diez km atrás , un coche circula en el mismo sentido a 85km/h ¿ En cuanto tiempo alcanzará el coche a la furgoneta? ¿A qué distancia se producirá el encuentro?
Answer:
t = 0.33h = 1200s
x = 18.33 km
Explanation:
If the origin of coordinates is at the second car, you can write the following equations for both cars:
car 1:
[tex]x=x_o+v_1t[/tex] (1)
xo = 10 km
v1 = 55km/h
car 2:
[tex]x'=v_2t[/tex] (2)
v2 = 85km/h
For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:
[tex]x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}[/tex]
[tex]t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s[/tex]
The position in which both cars coincides is:
[tex]x=(55km/h)(0.33h)=18.33km[/tex]
An object sits at rest at some position to the left of the origin. Draw what you expect the position vs time, velocity vs time, and acceleration vs time graphs to look like.
Answer:
see below
Explanation:
The position graph will be a horizontal line (of constant position) located at the negative value representing the position to the left of the origin.
The velocity graph will be a horizontal line at 0, since the object is at rest.
The acceleration graph will be a horizontal line at 0, since the object's velocity is not changing.
Let’s discuss some particle physics. In 1897, physicist J.J. Thomson conducted a seminalexperiment with a cathode ray tube. In the experiment, a beam of an unknown particle of chargeqandmassmis shot through the tube onto a fluorescent screen, at some speedvand in thex-direction. Uponimpact the beam leaves a bright spot on the screen. Sincevis very fast, effects of gravity are negligible andthe beam forms essentially a straight line normal to the screen.
To eliminate this deflection, an external uniform magnetic field of magnitudeBis turned on as well. What should be the direction and magnitudeBof this field such that ∆z= 0?
Answer:
The direction of the magnetic field should be in the x and y coordinates alone, and their should be no change towards the z coordinate. Also the magnitude of the magnetic field in the z direction would be zero if there is no magnetic field in that coordinate.
The coordinate should be:
B = bi + bj
When you turn down the volume on the television, you reduce the _______ carried by the sound waves, so you also reduce their ________
Answer:
when you turn down the volume on the television, you reduce the intensity carried by the sound waves, so you also reduce their amplitude.
Explanation:
When you turn down the volume of the television, you are actually reducing the intensity of the sound wave, which is directly proportional to the amplitude of the sound. Amplitude is height of the sound wave.
Therefore, when you turn down the volume on the television, you reduce the intensity carried by the sound waves, so you also reduce their amplitude.
It has been proposed that we could explore Mars using inflated balloons to hover just above the surface. The buoyancy of the atmosphere would keep the balloon aloft. The density of the Martian atmosphere is 0.0154 (although this varies with temperature). Suppose we construct these balloons of a thin but tough plastic having a density such that each square meter has a mass of 4.60 . We inflate them with a very light gas whose mass we can neglect. So far I found the following: What should be the radius of these balloons so they just hover above the surface of Mars? Radius of the balloon = /896 m What should be the mass of these balloons so they just hover above the surface of Mars? Mass of balloon = 4.64*10^-2 kg If we released one of the balloons from part A on earth, where the atmospheric density is 1.20 , what would be its initial acceleration assuming it was the same size as on Mars? If on Mars these balloons have five times the radius found in part A, how heavy an instrument package could they carry?
Answer:
the radius of the balloon r = 0.896 m and mass m = 4.64 × 10⁻² kg
the initial acceleration is a = 753.47 m/s²
the an instrument package could they carry a required mass of 4.64 kg
Explanation:
a) What should be the radius of these balloons so they just hover above the surface of Mars?
Given that :
The density of the Martian atmosphere, ρ = 0.0154 kg/m³
The volume of the sphere, V = (4/3)πr³
The area of the sphere, A = 4πr²
The mass of the balloon is m = (4.60 g/m²)A
m = (4.60×10⁻³ kg/m²)(4πr²)
The formula for the buoyant force is expressed as :
F = ρVg
m×g = ρ×V×g
m = ρ×V
Now;
(4.60×10⁻³ kg/m²)(4πr²)= ρ(4/3)πr³
r = 3(4.60×10⁻³ kg/m²)/ ρ
r = 3(4.60×10⁻³ kg/m²)/ 0.0154 kg/m³
r = 0.896 m
Thus; the radius of the balloon r = 0.896 m
The mass of the balloon is (4.60×10⁻³ kg/m²)(4πr²)
m = (4.60×10⁻³ kg/m²)(4π×0.896²)
m = 4.64 × 10⁻² kg
b) what would be its initial acceleration assuming it was the same size as on Mars?
The density of the air on earth, ρ = 1.20 kg/m³
The volume of the balloon is V = (4/3)(π)(0.896 m)³
V = 3.01156 m³
Considering the net force acting on the balloon ; we have
ΣF = ρVg - mg = ma
However; making the initial acceleration a of the balloon the subject ; we have:
a = (ρVg - mg)/m
a = (1.20 kg/m³)(3.01156 m³)(9.8 m/s²) - (4.64×10⁻² kg)(9.8 m/s²)]/(4.64×10⁻² kg)
a = 753.47 m/s²
c) If on Mars these balloons have five times the radius found in part A, how heavy an instrument package could they carry?
The volume of the total system is V' = (4/3)π(5r)³
V' = (4/3)π(5)³(0.896 m)³
V' =376.446 m³
The mass of the total system is m = (4.60×10⁻³ kg/m²) (4π(5r)²)
m = [4.60×10⁻³ kg/m²][4π][25](0.896 m)²
m = 1.159587 kg
We can then say that the buoyant force is equals to the weight of the total mass (balloon+load) and is expressed as:
F = (m + m')g
ρV'g = (m + m')g
ρV' = (m + m')
Thus; the required mass m' is = ρV' - m
m' = ρV' - m
m' = (0.0154 kg/m³)(376.446 m³) - (1.159587 kg)
m' = 4.64 kg
Thus; the an instrument package could they carry a required mass of 4.64 kg