When the disks collide and stick together, their temperature rises. Calculate the increase in internal energy of the disks, assuming that the process is so fast that there is insufficient time for there to be much transfer of energy to the ice due to a temperature difference. (Also ignore the small amount of energy radiated away as sound produced in the collisions between the disks.)
Answer:
ΔT = [tex]\frac{\Delta K}{(m_1+m_2) c_e }[/tex]
Explanation:
This is an interesting problem, no data is given, so the result is a general expression.
Suppose that the disks are initially rotating with angular velocity w₁ and w₂, as well as that they have radii r₁ and r₂ and masses m₁ and m₂
we start the problem finding odl final angular velocity of the discs together, for this we define a system formed by the two discs, in this case the torques during the collision are internal and the angular momentum is conserved
initial instant. Just before the crash
L₀ = L₁ + L₂
with
L₁ = I₁ w₁
the moment of inertia of a disc with an axis passing through its center is
I₁ = ½ m₁ r₁²
we substitute
I₀ = ½ m₁ r₁² w₁ + ½ m₂ r₂² w₂
final instant. Right after the crash
L_f = I w
in angular momentum it is a scalar quantity, so it is additive
I = I₁ + I₂
angular momentum is conserved
L₀ = L_f
I₁ w₁ + I₂ w₂ = I w
w = [tex]\frac{ I_1 w_1 + I_2 w_2 }{I}[/tex] (1)
We already have the angular velocities of the system, let's find the kinetic energy of it
initial
K₀ = K₁ + K₂ = ½ I₁ w₁² + ½ I₂ w₂²
final
K_f = K = ½ I w²
the variation of the kinetic energy is the loss in the increase of the temperature of the system, they indicate us that we neglect the other possible losses
ΔK = K_f -K₀
ΔK = ½ I w² - (½ I₁ w₁² + ½ I₂ w₂²) (2)
In this chaos we know all the values for which the numerical value of ΔK can be calculated, the symbolic substitution gives expressions with complicated
Now if all this variation of energy turns into heat
Q = ΔK
m_{total} c_e ΔT = ΔK
where the specific heat of the bear discs must be known, suppose they are of the same material
ΔT = [tex]\frac{\Delta K}{(m_1+m_2) c_e }[/tex] (3)
to make a special case, we suppose some data
the discs have the same mass and radius, disc 2 is initially at rest and the discs are made of bronze that has c_e = 380 J / kg ºC
we look for the angular velocity
I₁ = I₂ = I₀
I = 2 I₀
we substitute in 1
w = [tex]\frac{I_o w_1 + I_o 0 }{2I_o}[/tex] I₀ w₁ + I₀ 0 / 2Io
w = w₁ /2
we look for the variation of the kinetic energy with 2
ΔK = ½ (2I₀) (w₁ /2)² - (½ I₀ w₁² + ½ I₀ 0)
ΔK = ¼ I₀ w₁² -½ I₀ w₁²
ΔK = - ¼ I₀ w₁²
the negative sign indicates that the kinetic energy decreases
We look for the change in Temperature with the expression 3
ΔT = [tex]\frac{ \Delta K}{(m_1 +m_2) c_e}[/tex]ΔK / (m1 + m2) ce
ΔT = [tex]\frac{ \frac{1}{4} I_o w_1^2 }{ 2m c_e}[/tex]
ΔT = [tex]\frac{1}{8} \frac{ (\frac{1}{2} m r_1^2 ) w_1^2 }{ m c_e}[/tex]
ΔT = [tex]\frac{1}{16} r_1^2 w_1^2 / c_e[/tex]
in this expression all the terms are contained
The increase in internal energy of the disks will be [tex]\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}[/tex].
What is internal energy?The energy contained within a thermodynamic system is known as its internal energy. It's the amount of energy required to build or prepare a system in any given internal state.
The given data in the problem is;
[tex]\rm \omega_1[/tex] is the angular velocity of disk 1
[tex]\rm \omega_2[/tex] is the angular velocity of disk 2
r₁ is the radius of disk 1
r₂ is the radius of disk 2
m₁ is the mass of disk 1
m₂ is the mass of disk 2
Momentum before the collision;
[tex]\rm L_1 = I_1 \omega_1[/tex]
The moment of inertia of disc 1
[tex]\rm i_1 = \frac{1}{2} m_1r_1^2[/tex]
The momentum gets conserved;
[tex]\rm L_0 = L_f \\\\ I_1 \omega_1 + I_2\omega_2 = I \omega \\\\ \rm \omega= \frac{I_1 \omega_1 + I_2\omega_2}{I}[/tex]
The change in the kinetic energy is;
[tex]\traingle KE= K_f - K_0 \\\\ \traingle KE= \frac{1}{2} I \omega^2-(\frac{1}{2} I_1\omega_1^2 + (\frac{1}{2} I_2\omega_2^2 )[/tex]
The change in the energy gets converted into heat;
[tex]\rm Q= \triangle k \\\\\ m_{total } c_e dt = \triangle k[/tex]
The change in the temperature is
[tex]\triangle T= \frac{\triangle k }{(m_1+m_2)c_e}[/tex]
The internal energy change is found by;
[tex]\rm \triangle E = mc_v dt[/tex]
[tex]\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}[/tex]
Hence the increase in internal energy of the disks will be [tex]\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}[/tex].
To learn more about the internal energy refer to the link;
https://brainly.com/question/11278589
A vertical straight wire 35.0 cmcm in length carries a current. You do not know either the magnitude of the current or whether the current is moving upward or downward. If there is a uniform horizontal magnetic field of 0.0300 TT that points due north, the wire experiences a horizontal magnetic force to the west of 0.0180 NN. Find the magnitude of the current.
Answer:
[tex]1.714\ \text{A}[/tex]
Explanation:
F = Magnetic force = 0.018 N
B = Magnetic field = 0.03 T
L = Length of wire = 35 cm
[tex]\theta[/tex] = Angle between current and magnetic field = [tex]90^{\circ}[/tex]
Magnetic force is given by
[tex]F=IBL\sin\theta\\\Rightarrow I=\dfrac{F}{BL\sin\theta}\\\Rightarrow I=\dfrac{0.018}{0.03\times 35\times 10^{-2}\times \sin90^{\circ}}\\\Rightarrow I=1.714\ \text{A}[/tex]
The magnitude of the current is [tex]1.714\ \text{A}[/tex].
pls helpone phyics question lots of points!
Answer:
PE = 0.73J
Explanation:
Remember that in conservative spring systems,
Total energy = potential + kinetic energy.
On the y-axis lies the kinetic energy and the question asks for the potential energy.
PE + KE must always equal the same result.
In this case, KE + PE = 1
So rearranging the equation,
PE = 1 - KE
KE = 0.27 (as we can see from the graph)
Therefore,
PE = 1 - 0.27 = 0.73J
Bonus tip: The graphs of potential and kinetic energy will look the exact opposite in this case. When the KE graph is at 0J, the PE graph is at 1J and vice versa. And they always cross over at 0.5J
Define the average acceleration of a particle
between two given instants.
ASAP 20 POINTS!!
The air also contained a small amount of argon
As the temperature of the air decreased from 20C to -190 C the argon changed
Explain the changes in arrangement and movement of the particles of the argon as the temperature of the air decreased
Answer:
See explanation
Explanation:
Let us recall that temperature is a measure of the average kinetic energy of the molecules of a body.The higher the temperature, the higher the kinetic energy of the molecules of the body.
As temperature decreases, the kinetic energy of the molecules of a substance also decreases rapidly and the magnitude of intermolecular interaction between molecules of the substance increases.
Hence, as argon gas is cooled from 20°C to -190°C the kinetic energy of the gas molecules decreases an the magnitude of intermolecular interaction increases hence the gas changes into liquid and subsequently changes into a solid at -190°C.
A twin-sized air mattress used for camping has dimensions of 100 cm by 194 cm by 14 cm when blown up. The weight of the mattress is 4 kg. How heavy a person (in N) could the air mattress hold if it is placed in freshwater
Answer:
[tex]2625.156\ \text{N}[/tex]
Explanation:
Dimensions of mattress 100 cm by 194 cm by 14 cm
[tex]m_m[/tex] = Mass of mattress = 4 kg
[tex]\rho[/tex] = Density of water = [tex]1000\ \text{kg/m}^3[/tex]
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
Volume of mattress
[tex]V=100\times 194\times 14=271600\ \text{cm}^3=0.2716\ \text{m}^3[/tex]
Weight of water displaced is equal to the buoyant force
Mass of water
[tex]m=\rho V\\\Rightarrow m=1000\times 0.2716\\\Rightarrow m=271.6\ \text{kg}[/tex]
Mass of person would be
[tex]m_p=m-m_m=271.6-4\\\Rightarrow m_p=267.6\ \text{kg}[/tex]
Weight of the person would be
[tex]w=m_pg\\\Rightarrow w=267.6\times 9.81\\\Rightarrow w=2625.156\ \text{N}[/tex]
The air mattress could hold a person that weighs up to [tex]2625.156\ \text{N}[/tex].
A small mirror is attached to a vertical wall, and it hangs a distance of 1.87 m above the floor. The mirror is facing due east, and a ray of sunlight strikes the mirror early in the morning and then again later in the morning. The incident and reflected rays lie in a plane that is perpendicular to both the wall and the floor. Early in the morning, the reflected ray strikes the floor at a distance of 3.56 m from the base of the wall. Later on in the morning, the ray is observed to strike the floor at a distance of 1.46 m from the wall. The earth rotates at a rate of 15.0o per hour. How much time (in hours) has elapsed between the two observations
Answer:
t = 1.62 h
Explanation:
A flat mirror fulfills the law of reflection where the incident angle is equal to the reflected angle.
θ_i = θ_r
If we use trigonometry to find the angles, the mirror is at a height of L = 1.87 m, and the reflected rays reach a distance x1 = 3.56 m
tan θ₁ = x₁ / L
tan θ₁ = [tex]\frac{3.56}{1.87}[/tex]
θ₁ = tan⁻¹ 1.90
θ₁ = 62.29º
for the second case x₂ = 1.46 m
tan θ₂ = x₂ / L
θ₂ = tan⁻¹ [tex]\frac{1.46}{1.87}[/tex]
θ₂ = 37.98º
the difference in degree traveled is
Δθ = θ₁- θ₂
Δθ = 62.29 - 37.98
Δθ = 24.31º
as in the exercise they indicate that every 15º there is an hour
t = 24.31º (1h / 15º)
t = 1.62 h
Which measures the amount of work you did?
I don’t get it
Answer:
Power is often measured in joules of work per second. The unit of measurement for power is the (W). One watt is equal to one joule of work done in one second. If an object does a large amount of work, its power is usually measured in units of 1000 watts, or kilowatts.
Explanation:
An ideal gas undergoes an adiabatic expansion, a process in which no heat flows into or out of the gas. As a result, (a) the temperature of the gas remains constant and the pressure decreases. (b) both the temperature and pressure of the gas decrease. (c) the temperature of the gas decreases and the pressure increases. (d) both the temperature and volume of the gas increase. (e) both the temperature and pressure of the gas increase. Group of answer choices a b c d e
Answer:
(b) both the temperature and pressure of the gas decrease.
Explanation:
An ideal gas undergoes an adiabatic expansion, a process in which no heat flows into or out of the gas. As a result, both the temperature and pressure of the gas decrease.
Gay Lussac states that when the volume of an ideal gas is kept constant, the pressure of the gas is directly proportional to the absolute temperature of the gas.
Mathematically, Gay Lussac's law is given by;
[tex] PT = K[/tex]
Also, according to the first law of thermodynamics which states that energy cannot be created or destroyed but can only be transformed from one form to another. Thus, the ideal gas does work on the environment with respect to the volume and temperature.
2.4 What is the radiation error of a temperature measurement?
I
Answer:
diameter of the wire = 0.05 in =0.05 /12 =4.167 *10 ^-3 ft
area of cross section of the wire = A = 22/7 * ( d /2 ) ^2 =0.786 * ( 4.167 *10 ^-3 ) ^2 =1.365 *10 ^-5 ft2
E =...
Explanation:
Help me, 100 points to answer right, answer without context will be reported
1. In the situation below, a tractor pulls a 850 sledge along a ramp of height ℎ = 1 and large = 30 °. If the tractor applies a constant force to the sledge = 6750 , at an angle = 36.9 °, determine the total work performed by all forces on the sledge to move it along the ramp. The coefficient of kinetic friction between the sledge and the plane is = 0.3. Tip: for the calculation, remember that only the components of the windows that are parallel to the direction of travel contribute to the work. Disregard the dimensions of the sled.
2) When firing a 2 projectile at a 1.4 bloco block, initially at rest, it is observed that the projectile is stuck in the block and the system moves together for a distance = 0.1 before stop. If the coefficient of kinetic friction between the block and the surface is = 0.25, determine what was the velocity of the projectile in the instant before impact. Tip: here you must use the conservation of linear momentum and also energy, considering the work done by the frictional force
Consider a long, thin rod with a length of 3 m rotating about it's end. This rod has a moment of inertia of 12 kg·m2 about this pivot.
What is the mass of the rod? Give your answer in kilograms to two decimal places.
Answer:
The mass of the rod is 16 kg.
Explanation:
Given that,
The length of a rod, L = 3 m
The moment of inertia of the rod, I = 12 kg-m²
We need to find the mass of the rod. The moment of inertia of the rod of length L is given by :
[tex]I=\dfrac{ML^2}{12}[/tex]
Where
M is mass of the rod
[tex]M=\dfrac{12I}{L^2}\\\\M=\dfrac{12\times 12}{(3)^2}\\\\M=16\ kg[/tex]
So, the mass of the rod is 16 kg.
One thing that animals have not demonstrated the ability to do with language (that humans can) is use language to describe itself. This use is called
a. reflexiveness.
b. specialization.
C. prevarication.
d. duality of patterning sounds, which is used to produce an infinite number of unique utterances.
Answer:
The answer is letter B hope it helps
If a car is traveling 103 miles per hour and passes another vehicle which is at a complete stop, how many miles would it take the stopped vehicle to catch the traveling vehicle?
Answer:
206 mph i think
Explanation:
A 120-gram toy airplane flies in a straight line at a speed of 1.3 m/s. How much kinetic energy does the airplane have?
Answer:
0.1014 J
Explanation:
This is the answer if the toy airplane was 120-GRAMS
if you meant KG then it's 101.4 J
which two options describes behaviors of particles that are related to the chemical properties of the materials
a- forming hydrogen bonds between them
b- reacting quickly with water
c- having a high mass
d- forming bonds with other atoms
Answer:
The two correct answers are B.) reacting quickly with water, and D.) forming bonds with other atoms.
Explanation:
I took the quiz on a.pex and these were correct.
A car travels at a constant speed of 25 m/s. Find the power supplied by the engine if it can supply a maximum force of 18,000 N
Answer:
720
Explanation:
helllllpppppppppppp.
pleaseeeeeeee
Calculate the amount of torque of an object being pushed by 6 N force along a circular path of a radius of 1x10^-2 mat 30 degree angle
Answer:
[tex]\tau=0.03\ N-m[/tex]
Explanation:
Given that,
Force acting, F = 6N
The radius of the path, [tex]r=10^{-2}\ m[/tex]
Angle, [tex]\theta=30^{\circ}[/tex]
We need to find the amount of torque acting on the object. The formula for torque is given by :
[tex]\tau=Fr\sin\theta\\\\\tau=6\times 10^{-2}\times \sin(30)\\\\\tau=0.03\ N-m[/tex]
So, the required torque is equal to 0.03 N-m.
ocean currents are always cold true or false
Energy is transferred between the ocean and the air to make sure that the temperature in the air is higher than the temperature on the surface False True
When a wave enters a new medium from an angle, both the speed and the ________ change
a
The frequency
b
The amplitude
c
The energy
d
The angle
Answer:
B: Amplitude
Explanation:
When a wave travels from one medium to the other from an angle, the things that change are amplitude, wavelength, intensity and velocity.
The frequency doesn't change because the frequency depends upon the source of the wave and not the medium by which the wave is propagated.
Answer:
The angle
Explanation:
One day, Peter put a leaf in hot water. After a while, he noted that a lot of air bubbles appeared on the lower surface of the leaf, but only a few air bubbles were found on the upper surface. Where did the air bubbles come from (name the part)?
Answer:
The air bubbles are oxygen coming out of the leaf
Explanation:
Though the leaf is in hot water, it is still using light to continue on the path of photosynthesis. Now, this path of photosynthesis involves letting oxygen out of the leaf. Thus, this oxygen is the one that we see as bubbles coming out of the water.
The reason for having more air bubbles at the lower surface than at the upper surface is because the leaf cells at the lower surface of the leaf possess faster photosynthetic rate in releasing
oxygen.
How much work is done in pushing an object 7.0 m across a floor with a force of 50 N and then
pushing it back to its original position? How much power is used if this work is done in 20 sec?
Answer:
35/2 J/s
Explanation:
Just use the 2 formulas
Work done = Force * distance moved
Power = Work done/time
WD = 7 * 50 = 350
Power = 350 / 20
= 35/2 J/s
A concave lens cannot produce a real image.
A. True
B. False
Answer:
B. False
A concave mirror and a converging lens will only produce a real image if the object is located beyond the focal point.
~Hoped this helped~
~Brainiliest?~
A sports car accelerates at a constant rate from rest to a speed of 90 km/hr in 8 s. What is its acceleration?
3.13 m/s2
4.22 m/s2
5.31 m/s2
6.67 m/s2
none of the above
A 4768-kg roller coaster train full of riders approaches
Answer: ?
Explanation:
True or false? A system must contain more than one object.
Answer:
true
Explanation:
normally -No system has ever performed well with one object.
A system must contain more than one object is a true statement.
What is system?A system is a group of interacting or interrelated objects that act according to a set of rules to form a unified whole.
Normally, no system has ever performed well with one object.
To learn more about System here
https://brainly.com/question/24893867
#SPJ2
30. Easy Guided Online Tutorial One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other words, they stick together after the collision and move off with a common velocity. Momentum is conserved. The speed of the object that is moving initially is 25 m/s. The masses of the two objects are 3.0 and 8.0 kg. Determine the final speed of the two-object system after the collision for the case when the large-mass object is the one moving initially and the case when the small-mass object is the one moving initially.
Answer:
[tex]18.18\ \text{m/s}[/tex]
[tex]6.82\ \text{m/s}[/tex]
Explanation:
[tex]m_1[/tex] = Mass of large object = 8 kg
[tex]m_2[/tex] = Mass of smaller object = 3 kg
When large mass is moving
[tex]u_1[/tex] = 25 m/s
[tex]u_2[/tex] = 0
For completely inelastic collision we have the relation
[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{8\times 25+3\times 0}{8+3}\\\Rightarrow v=18.18\ \text{m/s}[/tex]
Speed of the combined mass when the larger object is moving is [tex]18.18\ \text{m/s}[/tex]
When smaller mass is moving
[tex]u_1[/tex] = 0
[tex]u_2[/tex] = 25 m/s
[tex]v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{8\times 0+3\times 25}{8+3}\\\Rightarrow v=6.82\ \text{m/s}[/tex]
Speed of the combined mass when the smaller object is moving is [tex]6.82\ \text{m/s}[/tex]
At the same instant that a 0.50-kg ball is dropped from 25 m above Earth, a second ball, with a mass of 0.25 kg, is thrown straight upward from Earth's surface with an initial velocity of 19.6 m/s. They move along nearby lines and pass each other without colliding. When the second ball is at its highest point, what is the velocity of the center of mass of the two-ball system
Answer:
The velocity of the center of mass of the two-ball system is 13.1 m/s.
Explanation:
Given;
mass of the first ball, m₁ = 0.5 kg
mass of the second ball, m₂ = 0.25 kg
initial velocity of the second ball, u₂ = 19.6 m/s
At the highest point the velocity of the second ball, v₂ = 0
The highest point reached by the second ball is calculated as;
v₂² = u₂² - 2gh
0 = u₂² - 2gh
2gh = u₂²
h = u₂² / 2g
h = (19.6²) / (2 x 9.8)
h = 19.6 m
The final velocity of the first ball when it had traveled 19.6 m down;
v₁² = u₁² + 2gh
v₁² = 0 + 2gh
v₁ = √2gh
v₁ = √(2 x 9.8 x 19.6)
v₁ = 19.6 m/s
The velocity of the center of mass of the two-ball system is calculated as;
[tex]v = \frac{m_1v_1 \ + \ m_2v_2}{m_1 \ + \ m_2} \\\\v = \frac{0.5\times 19.6 \ + \ 0.25\times 0}{0.5 \ + \ 0.25} \\\\v = 13.1 \ m/s[/tex]
Suppose that you changed the area of the bottom surface of the friction cart without changing its mass, by replacing the Teflon slab with one that was smaller but thicker. The contact area would shrink, but the normal force would be the same as before. Would this change the friction force on the sliding cart
Answer:
in this case the weight of the vehicle does not change , consequently the friction force should not change
Explanation:
The friction force is a macroscopic manifestation of the interactions of the molecules between the two surfaces, this force in the case of solid is expressed by the relation
fr = μ N
W-N= 0
N = W
as in this case the weight of the vehicle does not change nor does the Normal one, consequently the friction force should not change