Find refractive index first
[tex]\\ \rm\Rrightarrow \mu=\dfrac{c}{v}[/tex]
[tex]\\ \rm\Rrightarrow \mu=\dfrac{1.0003}{1.33}[/tex]
[tex]\\ \rm\Rrightarrow \mu =0.75[/tex]
Now
[tex]\\ \rm\Rrightarrow \dfrac{sini}{sinr}=\mu[/tex]
[tex]\\ \rm\Rrightarrow \dfrac{sin45}{sinr}=0.75[/tex]
[tex]\\ \rm\Rrightarrow \dfrac{sin45}{0.75}=sinr[/tex]
[tex]\\ \rm\Rrightarrow sinr=0.94[/tex]
[tex]\\ \rm\Rrightarrow r=sin^{-1}(0.94)[/tex]
[tex]\\ \rm\Rrightarrow r=70^{\circ}[/tex]
According to the __________perspective, people who suffer from PTSD are classically conditioned.
According to the pavlovian perspective, people who suffer from PTSD are classically conditioned.
What is PSTD?
This is referred to as post traumatic stress disorder which is a mental disorder from a terrifying incident.
The pavlovian perspective however view this disorder as being a type of classical conditioned one.
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A 3.42 kg steel ball strikes a massive wall at 14.3 m/s at an angle of α = 61.5° with respect to the plane of the wall. It bounces off with the same speed and angle as seen in the figure.
If the ball is in contact with the wall for 0.207 s, what is the average force exerted on the ball by the wall?
The magnitude of the average force exerted on the ball by the wall is 225.469 N.
What is impulse?The change in momentum is equal to the product of impact force applied while colliding and time for that impact.
Impulse F. t = m (Vf -Vi)
where, Vf is the final velocity and Vi is the initial velocity.
Given is a 3.42 kg steel ball strikes a massive wall at 14.3 m/s at an angle of α = 61.5° with respect to the plane of the wall. It bounces off with the same speed and angle. The ball is in contact with the wall for 0.207 s
Substitute the values into the expression, we get
Impulse = 2 x mvcosθ
Impulse= 2 x 3.42 x 14.3 x cos 61.5°
Impulse = 46.672 kg.m/s
The impact force can be written as
F.t = I
Put the given values, we have
F = 46.672 / 0.207
F = 225.469 N
Thus, the magnitude of force exerted by the wall is 225.469 N
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A spring is compressed by 0.0880 m and is used to launch an object horizontally with a speed of 2.76 m/s. If the object were attached
to the spring, at what angular frequency (in rad/s) would it oscillate?
Answer:
Approximately [tex]3.14\; {\rm rad \cdot s^{-1}}[/tex].
Explanation:
Fact: the angular velocity [tex]\omega[/tex] of a simple harmonic oscillator is the ratio between the maximum velocity [tex]v_{\text{max}}[/tex] and the maximum displacement [tex]x_\text{max}[/tex] of this oscillator. In other words:
[tex]\begin{aligned} \omega &= \frac{v_{\text{max}}}{x_{\text{max}}}\end{aligned}[/tex].
Derivation of the previous equation:
Let [tex]A[/tex] denote the amplitude of this oscillation, and let [tex]\omega[/tex] denote the angular velocity.
The displacement of the oscillator at time [tex]t[/tex] would be:
[tex]x(t) = A\, \sin(\omega\, t)[/tex].
The maximum displacement of this oscillator would be [tex]x_\text{max} = A[/tex].
The velocity of this oscillator at time [tex]t[/tex] is the derivative of displacement with respect to time:
[tex]\begin{aligned} v(t) &= \frac{d}{d t}\, [x(t)] \\ &= \frac{d}{d t} [A\, \sin(\omega\, t)] \\ &= A\, \omega\, \cos(\omega\, t)\end{aligned}[/tex].
The maximum velocity of this oscillator would be [tex]v_\text{max} = A\, \omega[/tex].
Notice that dividing [tex]v_\text{max} = A\, \omega[/tex] by [tex]x_\text{max} = A[/tex] would give:
[tex]\displaystyle \frac{v_\text{max}}{x_\text{max}} = \frac{A\, \omega}{A} = \omega[/tex].
It is given that [tex]v_\text{max} = 2.76\; {\rm m\cdot s^{-1}}[/tex] while [tex]x_\text{max} = 0.0880\; {\rm m}[/tex]. Therefore:
[tex]\begin{aligned} \omega &= \frac{v_{\text{max}}}{x_{\text{max}}} \\ &= \frac{2.76\; {\rm m\cdot s^{-1}}}{0.0880\; {\rm m}} \\ &\approx 3.14\; {\rm s^{-1}}\end{aligned}[/tex].
(Radians per second.)
A small sphere of mass 10 kg
is released from rest at a height of
15.0 m above the ground level.
The sphere experiences a constant
resistive force (due to air
resistance) of magnitude R = 10.0
N.
a) Calculate the speed of the
sphere after it has fallen
through a distance of 5.00 m
bCalculate the speed of the ball just before a it hits the gound.
Answer:
Approximately [tex]9.39 \; {\rm m\cdot s^{-1}}[/tex] after the sphere has travelled a distance of [tex]5\; {\rm m}[/tex].
Approximately [tex]16.3\; {\rm m\cdot s^{-1}}[/tex] right before touching the ground (a distance of [tex]15\; {\rm m}[/tex].)
Assumption: [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex].
Explanation:
Weight of the sphere: [tex]m\, g = 9.81\; {\rm N \cdot kg^{-1}} \times 10\; {\rm kg} = 98.1\; {\rm N}[/tex], downwards.
Drag on the sphere: [tex]10.0\; {\rm N}[/tex] upwards.
Net force on the sphere: [tex]98.1\; {\rm N} - 10\; {\rm N} = 88.1\; {\rm N}[/tex] downwards.
Acceleration of the sphere: [tex]a = F_\text{net} / m = 88.1\; {\rm N} / (10\; {\rm kg}) = 8.81\; {\rm m\cdot s^{-2}}[/tex].
Apply the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex], where [tex]v[/tex] is the final velocity, [tex]u[/tex] is the initial velocity ([tex]0[/tex] in this case, as the sphere was released from rest,) and [tex]x[/tex] is the distance (displacement) that the sphere has travelled so far.
Rearrange this equation to obtain an expression for [tex]v[/tex]:
[tex]\displaystyle v = \sqrt{2\, a\, x + u^{2}}[/tex].
For example, after the ball travelled a distance of [tex]5.00\; {\rm m}[/tex], [tex]x = 5.00 \; {\rm m}[/tex]:
[tex]\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 5.0\; {\rm m} + 0} \\ &\approx 9.39\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Similarly, [tex]x = 15.0\; {\rm m}[/tex] right before landing, such that:
[tex]\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 15.0\; {\rm m} + 0} \\ &\approx 16.3\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
6. A metal block increases in temperature from 15 °C to 60 °C when supplied with 13 500 J of heat energy.
a) Calculate the heat capacity of the metal.
b) Calculate the specific heat capacity of the metal, if this sample has a mass of 0.75 kg.
Answer:
Since 1 cal = 4.19 J the heat Q received by the metal is
Q = 13,500 J * 1 C / 4.19 J = 3,222 calories
S = ΔQ / (ΔT * ΔM) = 3222 cal / (45 deg C / 750 gm)
S = .095 cal / gm deg C
Note that specific heat capacity for Cu is .093 cal / gm deg C
what's the difference between coplanar forces and resultant forces?
Answer:
coplanar When all forces are acting in the same
resultunt the single force and associated torque obtained by combining a system of forces and torques acting on a rigid body via vector addition
Resolve the vector shown below into its components.
Answer:
D. s=3x+4y
Explanation:
The line is at the 3rd column in the 4th row.
Using Hooke's law, what happens when the displacement is tripled?
If the displacement is tripled then the spring force also is tripled.
What is Hooke's law?The amount of force required to stretch or compress the spring is known as the spring force. Its unit is Newton(N).
Froom the hooks law;
F = kx
Where,
F is the applied force
x is the displacement in case 1
Case 2;
x'=3x
F'=kx'
F'=3KX
F'=3x'
Where,
x' is the displacement when the force is tripled
F' is the force in the case2
Hence, if the displacement is tripled then the spring force also be tripled.
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During World War II, it was found that r, the radius of the shockwave produced during an atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density, .
The atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.
What is an atomic bomb?An atomic bomb is type of nuclear weapon that releases a vast amount of energy upon explosion in the form of fission reaction.
During an explosion of atomic bomb, a cloud of mushroom fire is formed which vaporises anything found within it. The extent of destruction depends on:
energy released, W, the elapsed time, t, and the air density.Therefore, the atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.
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A local municipality placed an advert in the local newspaper for a competition of local high schools to participate to build a security system in the community hall to avoid vandalism and break ins
We will use the design process to meet the requirements of the competition by the given explained in the main answer.
What is the competition?The initiative was taken by two or more parties operating independently to win the business of a third party by competing to hire contractors on the best terms possible.
The complete question is
"A local businessperson placed an advert in the local newspaper for a competition for girls only from different schools to participate in the competition. The competition is about drawing ideas for a shopping complex for the local community. Girls from the different schools are excited about the competition. you are a group of boys in the local schools who saw the advert and feel it is unfair for you to be excluded from the competition.
consider yourself as one of the learners in the local schools.
Explain why the project should be given to boys also who have better skills for this project than the girls do.
1.2 think about the design process skills, investigates, design, make, evaluate and communicate."
Explain how you will use the design process to meet the requirements of the competition.
[ 1.1 ]Boys should also be given the project because some of them have more artistic talent than girls do when it comes to sketching designs for local shopping centers.
Boys can also present a clear picture of how local communities can expand their shopping options.
For instance, they can encourage local merchants to offer seasonal items because doing so will benefit their businesses.
They can also suggest that retailers encourage customers to bring a friend to participate in competitions, which will be highly regarded.
[1.2] In order to fulfill the needs of the competition, I would make sure to explore how a shopping center for the neighborhood functions, for instance, by making sure I am aware of the strategies one may use to enhance his or her firm in the neighborhood.
In order to win the competition, I would make sure that I was confident in what I was saying and that I was audible enough to be heard during the competition.
Finally, I would make sure I met the competition's conditions by staying on-topic and solely addressing the issues raised in the advertisement
Hence will use the design process to meet the requirements of the competition.
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the question in photo
Answer:
a
Explanation:
What is the energy of a photon of blue light with a frequency of 6.53 × 10¹4 Hz?
Answer:
V = 6.53 × 10^14
h = 6.626 × 10^-34
E = hV
E = (6.626 × 10^-34) (6.53 × 10^14)
E = 4.33 × 10^-19
Erbium-165 has a half-life of 10.4 hours. If you start with 1,000 grams of
erbium-165, how much time will it take to have 125 grams of erbium-165 left
in the sample?
A. 41.6 hours
OB. 20.8 hours
C. 31.2 hours
D. 10.4 hours
The amount of time it will take to have 125 grams of erbium-165 left
in the sample is; C: 31.2 hours
How to calculate decay time?The formula for amount of substance after decay is;
N(t) = N₀(¹/₂)^(t/t_¹/₂)
We are given;
Half life; t_¹/₂ = 10.4 hours
Initial amount; N₀ = 1000 g
Amount left; N(t) = 125 g
Thus;
125 = 1000 * (¹/₂)^(t/10.4)
125/1000 = (¹/₂)^(t/10.4)
In 0.125 = (t/10.4) In 0.5
-2.08 = -0.693(t/10.4)
t/10.4 = -2.08/-0.693
t = 10.4 * 3
t = 31.2 hours
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a car is traveling along a straight road at a velocity of 30m/s when its engine cuts off. For the next ten seconds, the car slows down, and its average acceleration is a1. For the next five seconds, the car slows down further at a velocity of 24m/s, and its average acceleration is a2. The ratio of the average acceleration values is a1/a2=1.5. Find the velocity of the car at the end of the initial ten-second interval.
Answer:
See below....the question is unclear....pick the answer you think they want
Explanation:
Car goes from 30 m/s to 24 m/s in 10 seconds
acceleration = change in velocity/change in time
= -6 m/s / 10 s = - 3/5 m/s^2
The wording of the question is not very good.....perhaps they meant to say that after the 5 seconds the velocity is now 24 m/s ?
a1 *10 + a2 *5 = -6 m/s
( 6 m/s is the change in velocity from 30 to 24 m/s)
a1/a2 = 1.5 so a2 = a1/1.5 (substitute this in)
10 a1 + 5 ( a1/1.5) =-6
15 a1 + 5 a1 = -9
a1 = - .45 m/s^2
The velocity of the car at the end of the initial ten-second interval, given that the car further slows down for the next five seconds, is 25.5 m/s
How to determine the velocity of the car at the end of the 10 s interval?From the first statement, we have:
Initial velocity (u₁) = 30 m/sTime (t₁) = 10 secondsVelocity at the end of 10 s (v₁) = ?v₁ = u₁ + a₁t₁
v₁ = 30 + (a₁ × 10)
v₁ = 30 + 10a₁ ......(1)
From the second statement, we have:
Time (t₂) = 5 sFinal velocity (v₂) = 24 m/sRation of acceleration (a₁/a₂) = 1.5Initial velocity (u₂) = Velocity at the end of 10 sv₂ = u₂ + a₂t₂
24 = u₂ + (a₂ × 5)
24 = u₂ + 5a₂ .......(2)
But,
u₂ = v₁ = 30 + 10a₁
Thus, we have:
24 = 30 + 10a₁ + 5a₂
But,
a₁/a₂ = 1.5
a₁ = 1.5a₂ ..... (3)
Thus, we have:
24 = 30 + 10a₁ + 5a₂
24 = 30 + 10(1.5a₂) + 5a₂
24 = 30 + 15a₂ + 5a₂
24 = 30 + 20a₂
Collect like terms
24 - 30 = 20a₂
-6 = 20a₂
Divide both sides by 20
a₂ = -6 / 20
= -0.3 m/s²
Substituting the value of a₂ into equation 3, we have
a₁ = 1.5a₂
= 1.5 × -0.3
= -0.45 m/s²
Substitute the value of a₁ into equation 1 to obtain the velocity, v₁ at the end of the initial to 10 s
v₁ = 30 + 10a₁
= 30 + (10 × -0.45)
= 30 - 4.5
= 25.5 m/s
Thus, velocity of the car at the end of the initial ten-second interval is 25.5 m/s
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differentiate e^x - 8x +7
please answer the questions asap
Answer:
eˣ-8.
Explanation:
(eˣ-8x+7)'=eˣ-8.
Which graph shows constant acceleration with correctly placed independent and dependent variables?
Answer:
a = (V2 - V1) / t equation for constant acceleration
if t the independent variable is placed on the abcissa or x-axis and V the dependent variable is placed on the ordinate or y-axis then the acceleration would be the slope at the point considered.
The action force is the ballon pushing the air out. What is the magnitude of the reaction force of the air pushing the ballon
The magnitude of the reaction force of the air pushing the balloon would be equal and opposite.
How is the reaction force equal and opposite?A push or a pull that an object experiences as a result of interacting with another item is known as a reaction force.
Newton's third law is officially expressed as follows: There is an equal and opposite reaction to every action. The implication of the statement is that there are always two forces acting on the two interacting objects. The force acting on the first object is equal in size to the force acting on the second. The force acting on the first object is acting in the opposite direction to the force acting on the second object. Force pairs—equal and opposing action-reaction force pairs—always exist in pairs.
Knowing that everything has an equal and opposite reaction according to Newton's second law. The preasured air that a balloon had to push out into the free air acts as the reaction force.
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A force of looN acts upwards on a mass of 50 kg, another of
80N acts on the Same Mass downwards. Calculate the acceleration?
of the body and in which direction will be th acceleration?
Answer:
a force of acts on a body of mass 50 kg for 10 sec when the first of acting on the body the body covers 80 and in the next 10 second
Answer:
a = 0.4 ms⁻²
Explanation:
• First calculate the resultant force on the body:
If we consider upwards to be the positive direction and downwards to be the negative direction, then the resultant force will be:
F = +100 + (-80)
= +20 N
The resultant force is positive, so it will act upwards.
• Next calculate the acceleration:
F = ma
a = F/m
a = 20/ 50
a = 0.4 ms⁻²
As the force is acting upwards, the acceleration will also be upwards.
Exam Which is a true statement about the energy level of gas? A. Gases have very little energy. B. Gases have more energy than liquids and solids. C. Gases have no energy. D. Gases have less energy than solids. Copyright © 2003-2022 International Academy of Science. All Rights Reserved. Exam Which is a true statement about the energy level of gas ? A. Gases have very little energy . B. Gases have more energy than liquids and solids . C. Gases have no energy . D. Gases have less energy than solids . Copyright © 2003-2022 International Academy of Science . All Rights Reserved .
Statement B is true about the energy level of gas, Gases have more energy than liquids and solids.
What is gas?A gas is a sample of matter that adopts the shape of the container in which it is housed and develops a uniform density inside the container.
Liquid molecules are more energetic than solid molecules. Further heating will cause the molecules to move so quickly that they won't stick together at all. The gas molecules have the highest energy content.
Gases have more energy than liquids and solids.
Hence statement B is true about the energy level of gas.
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Which of the following statements is false?
A Heat is a form of energy while temperature is a measure of degree of hotness
B Heart and temperature are linearly related
C Heat can be measured by physical changes due to change in temperature
D Temperature can be measured by physical changes due to change in heat
d. is wrong...the temperature cannot be measured due to physical changes
A bike accelerates uniformly from rest to a speed of 10 m/s over a distance of 50 m. (a) Determine the acceleration of the bike. (b) how long will take to do that?
Answer:
a) 0.2m/s
b) 5m/s
Explanation:
a) acceleration=∆v/∆t
v=10
t=50
10/50=1/5
=0.2m/s
b) time= d/s
d=50m
s=10
50/10
=5m/s
Answer:
See below
Explanation:
Average speed = (0+ 10)/2 = 5 m/s
then to cover 50 m will take 50 m / 5 m/s = 10 seconds
change in velocity/ change in time = acceleration = 10/10 = 1 m/s^2
Lester plays middle linebacker for dices football team. During one play, he made the following movements. First, he back-pedaled in the southern direction for 2.6 meters
Answer:
Explanation:
missing info .....
Determine the speed at which the medicine leaves the needle
The speed at which the medicine leaves the needle is 2.462 m/s
What is Bernoulli's theorem?When an incompressible, ideal fluid is flowing through a tube or pipe, the total energy remains constant.
p₁ /ρg + v₁²/2g +z₁ = p₂ /ρg + v₂²/2g +z₂
Where, p/ρg = pressure energy
v²/2g = kinetic energy
z = potential energy
Given is during an injection, pressure in the barrel of syringe is 1.03 atm while pressure in the needle section is 1.00 atm. Assuming the syringe lays horizontally and mass density of the liquid medicine ρ =1000 kg/m³.
The fluid is initially at rest.
Using the Bernoulli's equation, we have
v₁² - v₂²= 2 x (p₂ -p₁) / ρ
Substituting the values, we get
0 - v₂² = 2 x (1.00 -1.03) x 1.01 x 10⁵ /1000
v₂² = 6.06
v₂ = 2.462 m/s
Thus, the speed at which the medicine leaves the needle is 2.462 m/s
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(i). A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long. The ball moves in a horizontal circle. If the cord can withstand a maximum tension of 64.0 N, a. What is the maximum speed at which the ball can move before the cord breaks? Assume the string remains horizontal during the motion. (5 marks) b. Suppose the ball moves in a circle of larger radius at the same speed v. Is the cord more likely or less likely to break? ( i ) . A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long . The ball moves in a horizontal circle . If the cord can withstand a maximum tension of 64.0 N , a . What is the maximum speed at which the ball can move before the cord breaks ? Assume the string remains horizontal during the motion . ( 5 marks ) b . Suppose the ball moves in a circle of larger radius at the same speed v . Is the cord more likely or less likely to break ?
(a) Let [tex]v[/tex] be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude
[tex]a_{\rm rad} = \dfrac{v^2}R[/tex]
where [tex]R[/tex] is the radius of the circle.
The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is
[tex]F = (1.500\,\mathrm{kg}) a_{\rm rad}[/tex]
so that
[tex](1.500\,\mathrm{kg}) \dfrac{v^2}{1.50\,\rm m} = 64.0\,\mathrm N[/tex]
Solve for [tex]v[/tex] :
[tex]v^2 = \dfrac{(64.0\,\mathrm N)(1.50\,\mathrm m)}{1.500\,\rm kg} \\\\ \implies \boxed{v = 8.00 \dfrac{\rm m}{\rm s}}[/tex]
(b) The net force equation in part (a) leads us to the relation
[tex]F = \dfrac{mv^2}R \implies v = \sqrt{\dfrac{FR}m}[/tex]
so that [tex]v[/tex] is directly proportional to the square root of [tex]R[/tex]. As the radius [tex]R[/tex] increases, the maximum linear speed [tex]v[/tex] will also increase, so the cord is less likely to break if we keep up the same speed.
What is the wavelength of a compression wave with a speed of 500 m/s and a frequency of 250 Hz?
Use this formula: λ = s/f
Select one:
a. 2 m
b. 1.25 m
c. 250 m
d. 750 m
Answer:
a (2m)
Explanation:
according to wavelenght = speed / frequency,
the wavelength is 2 m
11. Trait theory claims that
O A. people from the same locations
share the same personality type.
B. you always behave the way your
personality type says you will.
O C. your personality is made up of a
number of traits.
O D. you have one characteristic that
defines your entire personality.
Calculate the magnitude of the linear momentum for the following cases.
(a) a proton with mass 1.67 x 10-27 kg, moving with a speed of 5.05 x 106 m/s
(d) the Earth (mass - 5.98 x 1024 kg) moving with an orbital speed equal to 2.98 x 10 m/s.
Answer:
8.4335 x 10^-²¹, 1.78204 x 10²⁶
Explanation:
p = mv
p = (5.05 x 10⁶)(1.67 x 10^-²⁷)
p = 8.4335 x 10^-²¹
p = mv
p = (5.98 x 10²⁴)(2.98 x 10)
p = 1.78204 x 10²⁶
A force of 20N is applied to the end of a wire of length 5m to produce an extension of 0.20mm
calculate: the stress on the wire.
The stress on the wire is determined as 6.37 x 10⁶ N/m².
Stress on the wireThe stress on the wire is calculated as follows;
σ = F/A
where;
F is the force applied on the wireA is area of the wireLet the diameter of the wire = 2 mm
Area = πr²
Area = π(1 x 10⁻³)²
Area = 3.142 x 10⁻⁶ m²
Stress = 20/(3.142 x 10⁻⁶)
Stress = 6.37 x 10⁶ N/m²
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What must the charge (sign and magnitude) of a particle of mass 1.45 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 600 N/C ? Use 9.80 m/s2 for the magnitude of the free-fall acceleration.
[tex]q = -21 * 10^{-6} C[/tex]
What is Free-fall acceleration?The acceleration is constant and equal to the gravitational acceleration g which is 9.8 m/s at sea level on the Earth.
As we know that charge and its sign that remains in equilibrium is under gravity must be such that it will balance the gravitational force by electric force,
mg =qE
[tex]1.45 * 10^{-3}(9.80)=q(600)\\\\q = 21 *10^{6} C\\\\[/tex]
and its sign must be negative so that it will have upward electric force
so it is
[tex]q = -21 * 10^{-6} C[/tex]
The charge of a particle of mass is [tex]-21 * 10^{-6} C[/tex]
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5. A uniform solid ring of mass M = 10 kg and radius R = 2 m is initially rotating in mid-air with angular speed 20 rad/s [the Before state in figure (I) below]. It is then placed in contact with a surface (assume that the surface immediately feels the full weight of the ring). (1) M Before Wi (II) wf V Ө ə Ө After Ө wf H ? (a) What kind of frictional force acts on the ring upon contact with the surface? (b) If the ring begins to roll smoothly to the right with speed v = 10 m/s after 5 seconds of contact (the After state), find the coefficient of friction corresponding to the frictional force you mentioned in (a). (c) What is the increase in the thermal energy of the ring/surface system from right before the ring contacts the surface, to when the ring starts to roll smoothly? The ring then rolls smoothly up a ramp of 0 = π/6 rad and H = 5 m [see figure (II)] (d) What is the horizontal distance, from the end of the ramp, at which the ring lands?
(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.
(b) If the ring begins to roll smoothly to the right with speed v = 10 m/s after 5 seconds of contact (the After state), the coefficient of friction corresponding to the frictional force is 0.612
(c) The increase in the thermal energy of the ring/surface system from right before the ring contacts the surface, to when the ring starts to roll smoothly is 7500 J.
(d) The horizontal distance, from the end of the ramp, at which the ring lands is 7.78 m
What is frictional force?When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction force.
Given is a uniform solid ring of mass M = 10 kg and radius R = 2 m is initially rotating in mid-air with angular speed 20 rad/s [the Before state in figure (I) below]. It is then placed in contact with a surface (assume that the surface immediately feels the full weight of the ring).
(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.
(b) angular frequency = velocity / radius
ωf = 10/2 = 5 rad/s and ωi = 20 rad/s
Angular acceleration, α = ωf - ωi /t
Put the values, we get
α = -15/5 = -3 rad/s²
Coefficient of friction, μ = a/g = rα/g
Plug the values, we get
μ = 0.612
Thus, the coefficient of friction corresponding to the frictional force is 0.612.
(c) The energy lost = heat generated
energy lost = 1/2 Iω² + 1/2 Mv²
energy lost = 1/2 MR²ω² + 1/2 Mv²
Plug the values, we get
energy lost = 7500 J
Thus, the increase in the thermal energy is 7500 J.
(d) The horizontal distance, from the end of the ramp, at which the ring lands
s= (v- 2gH) sin2θ /g
s = (10 - 2x (-9.81)x5 ) sin (2x π/6) / 9.81
s = 7.78m
Thus, the horizontal distance is 7.78 m
Learn more about friction force.
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