Answer:
Mass, M = 1000 kg
Speed, v = 90 km/h = 25 m/s
time, t = 6 sec.
Distance:
[tex]{ \tt{distance = speed \times time }} \\ { \tt{distance = 25 \times 6}} \\ { \tt{distance = 150 \: m}}[/tex]
Force:
[tex]{ \tt{force = mass \times acceleration}} \\ { \bf{but \: for \: acceleration : }} \\ from \: second \: equation \: of \: motion : \\ { \bf{s = ut + \frac{1}{2} {at}^{2} }} \\ \\ { \tt{150 = (0 \times 6) + ( \frac{1}{2} \times a \times {6}^{2} ) }} \\ \\ { \tt{acceleration = 8.33 \: {ms}^{ - 2} }} \\ \\ { \tt{force = 1000 \times 8.33}} \\ { \tt{force = 8333.3 \: newtons}}[/tex]
Do all substances conduct heat ?Why/ Why not ?
Answer:
no, all substances doesnot conduct heat
Answer:
No, all substances do not conduct heat easily because it depends on the nature of the substance. Some are good conductors of heat and some are bad. Therefore, it depends on their characteristics and their ability to conduct heat.
The bad conductors of heat are water, air, plastic, wood, etc.
Gold, Silver, Copper, Aluminium, Iron, etc. are good heat conductors as well as electrical conductors.
An object moves in a direction parallel to its length with a velocity that approaches the velocity of light. The length of this object, as measured by a stationary observer:________
a. approaches infinity.
b. approaches zero.
c. increases slightly.
d. does not change.
Answer:
b. approaches zero.
Explanation:
The phenomenon is known as length contraction.
Length contraction is a result of Einstein's special theory of relativity. This theory states that an observer in an inertial frame of reference will observe a decrease in the length of any moving object placed at another inertial frame of reference.
let the length of the train = L
Let the length observed when the train is in motion = L₀
Apply Einstein's special theory of relativity;
[tex]L_0 = L \times \sqrt{1 - \frac{v^2}{c^2} } \\\\where;\\\\v \ is \ the \ velocity \ of \ the \ train\\\\c \ is \ the \ speed \ of \ light\\\\[/tex]
from the equation above, when v = 0, the length observed is equal to the initial length of the train. (L₀ = L)
As the velocity of the train (v) approaches the speed of light (c), the length of the train observed (L₀) becomes smaller than the initial length of the train (L). (L₀ < L)
Eventually, when v equals c, we will have a square root of zero (0), and the length observed will become zero. (L₀ = 0)
Thus, the length of this object, as measured by a stationary observer approaches zero
you decide to work part time at a local supermarket. The job pays eight dollars and 60 per hour and you work 20 hours per week. Your employer withhold 10% of your gross pay federal taxes, 7.65% for FICA taxes, and 5% for state taxes
I guess that we want to find how much money you get each week.
We know that the job pays $8.60 per hour.
We know that you work 20 hours per week.
Then the gross pay (the total money that you earn) in a week is 20 times $8.60, or:
20*$8.60 = $172.
Now we know that your employer witholds:
10% + 7.65% + 5% = 22.65%
Then your employer withholds 22.65% of your gross pay.
if the 100% of your gross pay is $172
Then the 22.65% will be:
(22.65%/100%)*$172 = 0.2265*$172 = $38.96
This means that your employer withholds $38.96 of your weekly gross pay.
Then each week you get:
$172 - $38.96 = $133.04
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A spherical conductor of radius = 1.5 cm with a charge of 3.9 pC is within a concentric hollow spherical conductor of inner radius = 3 cm, and outer radius = 4 cm, which has a total charge of 0 pC. What is the magnitude of the electric field 2.3 cm from the center of these conductors?
Answer:
The answer is "66.351 N/C"
Explanation:
Given:
[tex]a=1.5\ cm= 1.5 \times 10^{-2}\ m\\\\q_1=3.9\ pc\\\\b=3\ cm\\\\c= 4\ cm\\\\q_2=0 \ pc\\\\[/tex]
Using Gauss Law:
[tex]\oint \vec{E} \cdot \vex{dA}= \frac{Q_{enc}}{\varepsilon_0 }[/tex]
[tex]E \times 4 \pi\ r^2=\frac{Q_{enc}}{\varepsilon_0}\\\\E= \frac{Q_{enc}}{4 \pi\ r^2 \varepsilon_0}= \frac{1}{4 \pi \varepsilon_0} \frac{Q_{enc}}{r^2}= \frac{k_e\ Q_{enc}}{r^2}\\\\[/tex]
[tex]=\frac{9\times 10^{9} \times 3.9 \times 10^{-12}}{(2.3\times 10^{-2})^2}\\\\=\frac{35.1\times 10^{-3}\ }{(2.3\times 10^{-2})^2}\\\\=\frac{35.1\times 10^{-3}\ }{5.29 \times 10^{-4}}\\\\=\frac{35.1\times 10 }{5.29 }\\\\=\frac{351}{5.29 }\\\\=66.351\ \frac{N}{C}[/tex]
A falcon is hovering above the ground, then suddenly pulls in its wings and begins to fall toward the ground. Air resistance is not negligible.
Identify the forces on the falcon.
a. Kinetic friction
b. Weight w
c. Static friction
d. Drag D
e. Normal force n
f. Thrust
g. Tension T
Answer:
Explanation:
When a falcon is hovering, the force of up thrust is balanced by the weight.
When it begins to fall towards the ground, the weight acts downwards, kinetic friction is upwards, drag is upwards, normal force is upwards, thrust is upwards.
A baseball pitcher brings his arm forward during a pitch, rotating the forearm about the elbow. If the velocity of the ball in the pitcher's hand is 34.0 m/s and the ball is 0.310 m from the elbow joint, what is the angular velocity (in rad/s) of the forearm
Answer:
[tex]\omega=109.67\ rad/s[/tex]
Explanation:
Given that,
The speed of the ball, u = 34 m/s
The ball is 0.310 m from the elbow joint.
We need to find the angular velocity (in rad/s) of the forearm.
We know that,
[tex]v=r\omega\\\\\omega=\dfrac{v}{r}\\\\\omega=\dfrac{34}{0.31}\\\\\omega=109.67\ rad/s[/tex]
So, the required angular velocity of the forearm is 109.67 rad/s.
A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The ball moves so fast that the string is always taut and perpendicular to the velocity of the ball. As the ball swings from its lowest point to its highest point Group of answer choices the work done on it by gravity is -118 J and the work done on it by the tension in the string is zero. the work done on it by gravity is -118 J and the work done on it by the tension in the string is 118 J. the work done on it by gravity and the work done on it by the tension in the string are both equal to -118 J. the work done on it by gravity is 118 J and the work done on it by the tension in the string is -118 J. the work done on it by gravity and the work done on it by the tension in the string are both equal to zero.
Answer:
The ball moves from lowest to highest point:
W = M g h = 3 * 9.8 * 4 = 118 J
This is work done "against" gravity so work done by gravity is -118 J
The tension of the string does no work because the tension does not
move thru any distance W = T * x = 0 because the length of the string is fixed.
True or False: The forces applied by our muscles on our bones are usually several times larger than the forces we exert on the outside world with our limbs.
Answer:
True
Explanation:
This is because of the point where the forces are applied by our muscles and
the angle they have about the bones. Take for example the diagram I uploaded.
If we do a free body diagram and a sum of torques, we would get that:
[tex]F_{muscle}sin \theta r1 - mg r2 = 0[/tex]
In this case, mg is the same in magnitude as the force made by the hand to hold the ball, so:
[tex]F_{muscle}sin \theta r_{1} - F_{hand} r_{2} = 0[/tex]
If we solve the equation for the force of the muscle we would get that:
[tex]F_{muscle}=\frac{F_{hand}r_{2}}{r_{1}sin \theta}[/tex]
Since r2 is greater than r1 and the sin function can only return values that are less than 1, this means that the force of the muscle is much greater than the force used by the hand to hold the weight.
Let's use some standard values to prove this, let's say that r1=10cm, r2=35cm and theta=60 degrees. When inputing the values into the equation we get:
[tex]F_{muscle}=\frac{F_{hand}(35cm)}{(10cm)sin (60^{o})}[/tex]
which yields:
[tex]F_{muscle}=4.04 F_{hand}[/tex]
so in this example, the force made by the muscle is 4 times as big as the force exerted by the hand.
A negative point charge q1 = 25 nC is located on the y axis at y = 0 and a positive point charge q2 = 10 nC is located at y =14 cm. Find the y coordinate of the points where the net electric potential due to these two charges is zero.
Answer:
y = 0.1 m
Explanation:
The electrical power for point loads is
V = [tex]k \sum \frac{q_i}{r_i}[/tex]k Sum qi / ri
in this case
V = k ([tex]- \frac{q_1}{r_1 } + \frac{q_2}{r_2}[/tex])
indicate that V = 0
[tex]\frac{q_1}{r_1} = \frac{q_2}{r_2}[/tex]
r₂ = [tex]\frac{q_2}{q_1} r_1[/tex]
the distance r1 is
r₁ = y -0
the distance r2
r₂ = 0.14 -y
we substitute
0.14 - y = [tex]\frac{10}{25}[/tex] y
y ( [tex]\frac{10}{25} + 1[/tex]) = 0.14
y 1.4 = 0.14
y = 0.14 / 1.4
y = 0.1 m
Question 1 of 10
Which nucleus completes the following equation?
239UHe+?
A. 228 Th
B. 2220
c. 23. Pu
D. 78Th
SUBMIT
Answer:
Option D. ²²²₉₀Th
Explanation:
Let the unknown be ⁿₘZ. Thus, the equation becomes:
²²⁶₉₂U —> ⁴₂He + ⁿₘZ
Next, we shall determine n, m and Z. This can be obtained as follow:
For n:
226 = 4 + n
Collect like terms
226 – 4 = n
222 = n
n = 222
For m:
92 = 2 + m
Collect like terms
92 – 2 = m
90 = m
m = 90
For Z:
ⁿₘZ => ²²²₉₀Z => ²²²₉₀Th
Therefore, the complete equation becomes:
²²⁶₉₂U —> ⁴₂He + ⁿₘZ
²²⁶₉₂U —> ⁴₂He + ²²²₉₀Th
Thus, the unknown is ²²²₉₀Th
The acceleration vector of a particle in uniform circular motion:___________
a) points outward from the center of the circle.
b) points toward the center of the circle.
c) is zero.
d) points along the circular path of the particle and opposite the direction of motion.
e) points along the circular path of the particle and in the direction of motion.
(B)
Explanation:
Centripetal means "towards the center" so the acceleration vector of an object undergoing UCM is always pointed towards the center.
The acceleration vector of a particle in a uniform circular motion points toward the center of the circle, The correct option is option (b).
Centripetal force is the force acting on an object in curvilinear motion directed towards the axis of rotation or center of curvature. The unit of centripetal force is Newton.
Centripetal means "towards the center" so the acceleration vector of an object undergoing circular motion is always pointed towards the center.
Therefore, The acceleration vector of a particle in a uniform circular motion points toward the center of the circle, The correct option is option (b).
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Use the following information to answer the next question.
Environmental Concerns
1. release of greenhouse gases
2. release of gases that cause acid rain
3. release of excess heat
4. depletion of solar energy
5. depletion of geothermal energy
6. flooding of land
Which of the above environmental concerns are associated with the production of electricity?
Select one:
O A. 2, 3, and 4
O B. 1, 2, 3, and 6
O C. 1, 2, 3, 5 and 6
O D. 1, 3, and 5
Answer:
1.Emitted primarily through the burning of fossil fuels (oil, natural gas, and coal), solid waste, and trees and wood products. Changes in land use also play a role. Deforestation and soil degradation add carbon dioxide to the atmosphere, while forest regrowth takes it out of the atmosphere.
2.Acid rain is caused by a chemical reaction that begins when compounds like sulfur dioxide and nitrogen oxides are released into the air. These substances can rise very high into the atmosphere, where they mix and react with water, oxygen, and other chemicals to form more acidic pollutants, known as acid rain.
3.Untreated, heat exhaustion can lead to heatstroke, a life-threatening condition that occurs when your core body temperature reaches 104 F (40 C) or higher. Heatstroke requires immediate medical attention to prevent permanent damage to your brain and other vital organs that can result in death.
4.The loss of solar energy in passing through the atmospheric layers is called the atmospheric deflection. ... The longer the path traversed, the greater the amount of radiant energy depleted. Various processes whereby heat energy is lost through the atmosphere are known as scattering, diffusion, absorption, and reflection.
5.Geothermal energy is renewable because the Earth has retained a huge amount of the heat energy that was generated during formation of the planet. In addition, heat is continuously produced by decay of radioactive elements within the Earth. The amount of heat within the Earth, and the amount that is lost though natural processes (e.g. volcanic activity, conduction/radiation to the atmosphere), are much, much more than the amount of heat lost through geothermal energy production. At any one geothermal field, however, the temperature of the geothermal reservoir or the fluid levels/fluid pressure in the reservoir may decrease over time as fluids are produced and energy is extracted. Produced fluids can be re-injected to maintain pressures, although this may further cool down the reservoir if care is not taken. Over time, it is commonly necessary to drill additional wells in order to maintain energy production as temperatures and/or reservoir fluid pressures decline.
6.Floods, Floodplains, and Flood-Prone Areas. ... Flooding is a result of heavy or continuous rainfall exceeding the absorptive capacity of soil and the flow capacity of rivers, streams, and coastal areas. This causes a watercourse to overflow its banks onto adjacent lands.
The environmental concerns associated with the production of electricity are the release of greenhouse gases, the release of gases that cause acid rain, the release of excess heat, the flooding of land, and the depletion of geothermal energy so, option C is correct.
What is electricity?The presence or movement of charged particles is electricity. The movement of electrons through a circuit is known as an electric current. The accumulation of electrons on an insulator causes static electricity.
Mostly released when solid trash, trees, and wood products are burned, along with fossil fuels (coal, gas, and oil).
Land use changes also have an impact. Carbon dioxide is released into the atmosphere by deforestation and soil erosion, while it is removed from it by forest regeneration.
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the product 17.10 ✕
Explanation:
pls write the full question
A motorcycle daredevil jumps off a 33.0 ramp at 20.3 m/s. The landing ramp is at the same height, 28.0 m away. What is the height of the motorcycle when it reaches the landing ramp? (Unit = m)
The height of the motorcycle daredevil when it reaches the landing ramp is 4.93 m.
Since the ramp is a 33.0° ramp and the motorcycle daredevil jumps off with a speed of 20.3 m/s, the motorcycle dare devil has a horizontal component of speed u = 20.3cos33.0° m/s and a vertical component of speed v = 20.3sin33.0° m/s.
Now, since the other ramp is d = 28.0 m away, it takes the time it takes the motorcycle dare devil to reach it is t.
Considering motion in the horizontal direction, d = ut.
Thus, t = d/u
= 28.0 m/20.3cos33.0° m/s
= 28.0 m/(20.3 × 0.8387) m/s
= 28.0 m/17.025 m/s
= 1.645 s
Let h be the height of the motorcycle daredevil when it reaches the landing ramp in time, t.
Considering the vertical motion and using h = vt - 1/2gt² where v = vertical velocity of motorcycle daredevil = 20.3sin33.0°, t = time taken to reach landing ramp = 1.645 s and g = acceleration due to gravity = 9.8 m/s² (Note that there is a negative in front of g since it is directed downwards)
So, substituting the values of the variables into the equation, we have
h = vt - 1/2gt²
h = 20.3sin33.0° m/s × 1.645 s - 1/2 × 9.8 m/s² × (1.645 s)²
h = 20.3 × 0.5446 m/s × 1.645 s - 1/2 × 9.8 m/s² × 2.706025 s²
h = 18.187 m - 1/2 × 26.519 m
h = 18.187 m - 13.26 m
h = 4.927
h ≅ 4.93 m
So, the height of the motorcycle daredevil when it reaches the landing ramp is 4.93 m.
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B.F.Skinner emphesized the importance of-----?
Answer:
BFSkinner enfatizó la importancia de creía en la importancia de desarrollar la psicología experimental y dejar atrás el psicoanálisis y las teorías acerca de la mente basadas en el simple sentido común.
Explanation:
friction between two flat surfaces can be divided into two categories. what are the two most common kinds of friction?
Answer:
kinetic and static
Explanation:
hope it helps! ^w^
A spinning wheel having a mass of 20 kg and a diameter of 0.5 m is positioned to rotate about its vertical axis with a constant angular acceleration, a of 6 rad/s If the initial angular velocity is 1.5 rad/s, determine The maximum angular velocity and linear velocity of the wheel after 1 complete revolution.
Answer:
ωf = 8.8 rad/s
v = 2.2 m/s
Explanation:
We will use the third equation of motion to find the maximum angular velocity of the wheel:
[tex]2\alpha \theta = \omega_f^2 -\omega_I^2[/tex]
where,
α = angular acceleration = 6 rad/s²
θ = angular displacemnt = 1 rev = 2π rad
ωf = max. final angular velocity = ?
ωi = initial angular velocity = 1.5 rad/s
Therefore,
[tex]2(6\ rad/s^2)(2\pi\ rad)=\omega_f^2-(1.5\ rad/s)^2\\\omega_f^2=75.4\ rad/s^2+2.25\ rad/s^2\\\omega_f = \sqrt{77.65\ rad/s^2}[/tex]
ωf = 8.8 rad/s
Now, for linear velocity:
v = rω = (0.25 m)(8.8 rad/s)
v = 2.2 m/s
A Man has 5o kg mass man in the earth and find his weight
Answer:
49 N
Explanation:
Given,
Mass ( m ) = 50 kg
To find : Weight ( W ) = ?
Take the value of acceleration due to gravity as 9.8 m/s^2
Formula : -
W = mg
W = 50 x 9.8
W = 49 N
The elastic extensibility of a piece of string is .08. If the string is 100 cm long, how long will the string be when it is stretched to the point where it becomes plastic?
Answer:
The elastic extensibility of a piece of string is .08. If the string is 100 cm long, how long will the string be when it is stretched to the point where it becomes plastic? is your ansewer dont take tension
The string will be 108 cm long when it is stretched to the point where it becomes plastic.
What is elasticity?Elasticity in physics and materials science refers to a body's capacity to withstand a force that causes distortion and to recover its original dimensions once the force has been withdrawn.
When sufficient loads are applied, solid objects will deform; if the material is elastic, the object will return to its original size and shape after the weights have been removed. Unlike plasticity, which prevents this from happening and causes the item to stay deformed,
Given parameters:
The elastic extensibility of a piece of string is 0.08.
The string is 100 cm long.
Hence, it becomes plastic, after it is stretched up to = 100 × 0.08 cm = 8 cm. The string will be 108 cm long.
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A wire carrying a 30.0-A current passes between the poles of a strong magnet that is perpendicular to its field and experiences a 2.15-N force on the 4.00 cm of wire in the field. What is the average field strenth?
Answer:
1.79 T
Explanation:
Applying,
F = BILsin∅................ Equation 1
Where F = Force, B = magnetic field, I = current flowing through the wire, L = length of the wire, ∅ = angle between the magntic field and the force
make B the subject of the equation
B = F/ILsin∅............. Equation 2
From the question,
Given: F = 2.15 N, I = 30 A, L = 4.00 cm = 0.04 m, ∅ = 90° (perpendicular to the field)
Substitute these values into equation 2
B = 2.15/(30×0.04×sin90°)
B = 2.15/1.2
B = 1.79 T
Hence the average field strength is 1.79 T
A positive statement is:________. a. reflects oneâs opinions. b. can be shown to be correct or incorrect. c. a value judgment. d. based upon an optimistic judgment.
Answer:
b
Explanation:
The patellar tendon attaches to the tibia at a 20 deg angle 3 cm from the axis of rotation at the knee. If the force generated in the patellar tendon is 400 N, what is the resulting angular acceleration, in rad/s2), if the lower leg and foot have a combined mass of 4.2kg and a given radius of gyration of 25 cm
Answer:
the resulting angular acceleration is 15.65 rad/s²
Explanation:
Given the data in the question;
force generated in the patellar tendon F = 400 N
patellar tendon attaches to the tibia at a 20° angle 3 cm( 0.03 m ) from the axis of rotation at the knee.
so Torque produced by the knee will be;
T = F × d⊥
T = 400 N × 0.03 m × sin( 20° )
T = 400 N × 0.03 m × 0.342
T = 4.104 N.m
Now, we determine the moment of inertia of the knee
I = mk²
given that; the lower leg and foot have a combined mass of 4.2kg and a given radius of gyration of 25 cm ( 0.25 m )
we substitute
I = 4.2 kg × ( 0.25 m )²
I = 4.2 kg × 0.0626 m²
I = 0.2625 kg.m²
So from the relation of Moment of inertia, Torque and angular acceleration;
T = I∝
we make angular acceleration ∝, subject of the formula
∝ = T / I
we substitute
∝ = 4.104 / 0.2625
∝ = 15.65 rad/s²
Therefore, the resulting angular acceleration is 15.65 rad/s²
A free undamped spring/mass system oscillates with a period of 4 seconds. When 10 pounds are removed from the spring, the system then has a period of 2 seconds. What was the weight of the original mass on the spring? (Round your answer to one decimal place.)
Answer:
13.3 pounds.
Explanation:
For a spring of constant K, with an attached object of mass M, the period can be written as:
T = 2*π*√(M/K)
Where π = 3.14
First, we know that the period is 4 seconds, then we have:
4s = (2*π)*√(M/K)
We know that if the mass is reduced by 10lb, the period becomes 2s.
Then the new mass of the object will be: (M - 10lb)
Then the period equation becomes:
2s = (2*π)*√((M-10lb)/K)
So we have two equations:
4s = (2*π)*√(M/K)
2s = (2*π)*√((M-10lb)/K)
We want to solve this for M.
First, we need to isolate K in one of the equations.
Let's isolate K in the first one:
4s = (2*π)*√(M/K)
(4s/2*π) = √(M/K)
(2s/π)^2 = M/K
K = M/(2s/π)^2 = M*(π/2s)^2
Now we can replace it in the other equation.
2s = (2*π)*√((M-10lb)/K)
First, let's simplify the equation:
2s/(2*π) = √((M-10lb)/K)
1s/π = √((M-10lb)/K)
(1s/π)^2 = ((M-10lb)/K
K*(1s/π)^2 = M - 10lb
Now we can use the equation: K = M*(π/2s)^2
then we get:
K*(1s/π)^2 = M - 10lb
(M*(π/2s)^2)*(1s/π)^2 = M - 10lb
M/4 = M - 10lb
10lb = M - M/4
10lb = (3/4)*M
10lb*(4/3) = M
13.3 lb = M
What is the approximate radius of an equipotential spherical surface of 30 V about a point charge of +15 nC if the potential at an infinite distance from the surface is zero?
Answer:
V = k Q / R potential at distance R for a charge Q
R = k Q / V
R = 9 * 10E9 * 15 * 10E-9 / 30 = 9 * 15 / 30 = 4.5 m
Note: Our equation says that if R if infinite then V must be zero.
A boy of mass 50 kg on a motor bike is moveny coith 20m/see what is hio k.E
What bet force is required to stop a 2250 kg car if the decelerates at a rate of -4.3 m/s^2 please answer fast
Answer:
Force = Mass × Acceleration
[tex]{ \tt{force = 2250 \times 4.3}} \\ = { \tt{9675 \: newtons}}[/tex]
4. Which of the following statements best describes the relationship
between mechanical, kinetic and potential energies of an object of mass
m kg that is thrown vertically upwards with in initial velocity of v. m/s.
A. Kinetic energy increases while potential energy decrease and mechanical
energy remains constant.
B. Kinetic energy decreases, while potential energy increases and mechanical
energy remains constant.
C. Both kinetic and potential energies decrease while mechanical energy
increases.
D. Both kinetic and potential energies increase while mechanical energy
remains constant.
(1)
21
Answer:
D
Explanation:
increase while mechanical energy remains constant
What is the result of (305.120 + 267.443) x 0.50? How many answers can be written based on the principle of significant digits?
Answer:
The answer is 286.2815.
the current through a wire is measured as the potential difference is varied what is the wire resistance
Answer:
Resistance, R = 0.02 Ohms
Explanation:
Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.
Mathematically, Ohm's law is given by the formula;
V = IR
Where;
V is the voltage or potential difference.
I is the current.
R is the resistance.
From the attachment, we would pick the following values on the graph of current against voltage;
Voltage, V = 0.5 V
Current = 25 A
To find resistance;
R = V/I
R = 0.5/2.5
Resistance, R = 0.02 Ohms
Note:
Resistance (R) is the inverse of slope i.e change in current with respect to change in voltage.
A fan that is rotating at 960 rev/s is turned off. It makes 1500 revolutions before it comes to a stop. a) What was its angular acceleration(assuming it was constant)
Answer:
α = 1930.2 rad/s²
Explanation:
The angular acceleration can be found by using the third equation of motion:
[tex]2\alpha \theta=\omega_f^2-\omega_i^2[/tex]
where,
α = angular acceleration = ?
θ = angular displacement = (1500 rev)(2π rad/1 rev) = 9424.78 rad
ωf = final angular speed = 0 rad/s
ωi = initial angular speed = (960 rev/s)(2π rad/1 rev) = 6031.87 rad/s
Therefore,
[tex]2\alpha(9424.78\ rad) = (0\ rad/s)^2-(6031.87\ rad/s)^2\\\\\alpha = -\frac{(6031.87\ rad/s)^2}{(2)(9424.78\ rad)}[/tex]
α = - 1930.2 rad/s²
negative sign shows deceleration
