Answer:
Kinetic Energy
Explanation:
I can't really explain it, but that's the correct answer.
Explain the differences between getting burned by fire and getting burned by dry ice?
Need it ASAP
ufl.edu A wire of length 27.7 cm carrying a current of 4.11 mA is to be formed into a circular coil and placed in a uniform magnetic field of magnitude 5.85 mT. The torque on the coil from the field is maximized. (a) What is the angle between and the coil's magnetic dipole moment? (b) What is the number of turns in the coil? (c) What is the magnitude of that maximum torque?
Answer:
A)90°
Explanation:
ufl.edu A wire of length 27.7 cm carrying a current of 4.11 mA is to be formed into a circular coil and placed in a uniform magnetic field of magnitude 5.85 mT. The torque on the coil from the field is maximized. (a) What is the angle between and the coil's magnetic dipole moment? (b) What is the number of turns in the coil? (c) What is the magnitude of that maximum torque?
(a) What is the angle between and the coil's magnetic dipole moment.
The angle between and the coil's magnetic dipole moment is 90°, because it is perpendicular to the field since there is maximum torgue.
(b) What is the number of turns in the coil?
Which of The following statements best describes the objects motion between zero and four seconds?
Answer:
A OR C
Nearly A.
Explanation:
If the ball is released from rest at a height of 0.63 m above the bottom of the track on the no-slip side, what is its angular speed when it is on the frictionless side of the track
Answer:
When the ball is on the frictionless side of the track , the angular speed is 89.7 rad/s.
Explanation:
Consider the ball is a solid sphere of radius 3.8 cm and mass 0.14 kg .
Given , mass, m=0.14 kg
Ball is released from rest at a height of, h= 0.83 m
Solid sphere of radius, R = 3.8 cm
=0.038 m
From the conservation of energy
ΔK = ΔU
[tex]\frac{1}{2}mv^2 +\frac{1}{2} I\omega^2=mgh[/tex]
Here , [tex]I=\frac{2}{5} MR^2 , v= R \omega[/tex]
[tex]\frac{1}{2}mv^2\frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R^2} )=mgh[/tex]
[tex]\frac{1}{2} [v^2+\frac{2}{5}v^2]= gh[/tex]
[tex]\frac{7}{10} v^2=gh[/tex]
[tex]0.7v^2=gh[/tex]
v=[tex]\sqrt{[gh/(0.7)][/tex]
=[tex]\sqrt{ [(9.8 m/s^2)(0.83 m) / (0.7) ][/tex]
= 3.408 m/s
Hence, angular speed when it is on the frictionless side of the track,
[tex]\omega=\frac{v}{R}[/tex]
= (3.408 m/s)/(0.038 m)
[tex]\omega[/tex] = 89.7 rad/s
Hence , the angular speed is 89.7 rad/s
Jack drops a stone from rest off of the top of a bridge that is 23.4 m above the ground. After the stone falls 7.2 m, Jill throws a second stone straight down. Both rocks hit the water at the exact same time. What was the initial velocity of Jill's rock
Explanation:dropped stone hits water:
24.4-4.9t^2=0
t=2.23
stone has fallen 6.6m:
4.9t^2 = 6.6
t = 1.16
So, Jill's thrown stone only has 2.23-1.16=1.07 seconds to hit the water:
24+1.07v-4.9*1.07^2 = 0
v = -17.19 m/s
Both the rocks hit the water at the exact same time, and the initial velocity of Jill's rock will be -19.28 m/s.
What is velocity?A vector measurement of the rate and direction of motion is what is meant by the term "velocity." Simply said, velocity is the rate of movement in a single direction. Velocity may be used to gauge both the speed of a rocket launching into space and the speed of a car travelling north on a busy freeway.
According to the question, initial velocity will be 0.
By using equation of motion :
Let Jill is x and Jack is y.
s=ut+1/2at² where,
a is acceleration,
u is initial velocity and,
t is the time period.
23.4=(-1/2)(9.8)×t²
t(y)=[tex]\sqrt{(23.4)/(4.9)}[/tex]
=2.185 seconds.
Time taken by Jack's stone that it travel the first 7.2 meters.
t=[tex]\sqrt{(2*7.2)/9.8}[/tex] =1.21 seconds.
So, time taken by stone thrown by jack,
t(Jill)=t(jack)-t
=23.4=-ut(x)-1/2(gt²)(y)
-23.4=(-u×0.975)-(1/2×9.8×(0.975)²)
u= -(23.4-4.6/0.975)
u=-19.28 m/s
Hence, the initial velocity of the Jill's stone will be -19.28 m/s.
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The speed of a 4.0-N hockey puck, sliding across a level ice surface, decreases at the rate of 2 m/s 2. The coefficient of kinetic friction between the puck and ice is:
Answer:
μ = 0.2041
Explanation:
Given
[tex]a = -2m/s^2[/tex]
Required
Determine the coefficient of friction (μ)
The acceleration is negative because it is decreasing.
Solving further, we have that:
[tex]F_f =[/tex] μmg
Where
μ = Coefficient of kinetic friction
[tex]F_f = ma[/tex]
So, we have:
ma = μmg
Divide both sides by m
a = μg
Substitute values for a and g
2 = μ * 9.8
Solve for μ
μ = 2/9.8
μ = 0.2041
A particle with charge q = +5.00 C initially moves at v = (1.00 î + 7.00 ĵ ) m/s. If it encounters a magnetic field B = 10 k, find the magnetic force vector on the particle.
Given :
Force on object moving with constant velocity due to magnetic field is given by :
[tex]\vec{F}=q\vec{v}\times \vec{B}\\\\\vec{F}=-5[ ( \hat{i} + 7\hat{j}) \times ( 10\hat{k})]\\\\\vec{F}=-5[ 10(-\hat{j})+70(\hat{i})]\\\\\vec{F}=50\hat{j}-350\hat{i}\ N[/tex]
Now,
[tex]F = \sqrt{50^2+350^2}\\\\F=353.55\ N[/tex]
Hence, this is the required solution.
A particle with charge q = +5.00 C that is moving with some velocity, then the magnetic force vector on the particle will be 353.55 N.
What is a vector?The phrase "vector" in physics and mathematics is used popularly to refer to some values that cannot be stated by a simple integer (a scalar) or to certain elements of high - dimensional space.
For things like displacements, forces, and velocity can have both a magnitude and direction, vectors were initially introduced in geometry and physics. Similar to how lengths, masses, and time are represented by real numbers, these quantities were represented by geometric vectors.
In some contexts, tuples—finite sequences of numbers with a definite length—are sometimes referred to as vectors.
From the data given in the question,
F = qv × B
F = -5 [(î +7 ĵ) ×(10 k)]
F = 50 ĵ - 350 î N
Now, calculate the force,
F = √50² + 350²
F = 353.55 N.
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differentiate between speed and velocity
Explanation:
Speed - The rate at which something moves
Velocity - The speed of something in a specific direction
Velocity is kind of a specific type of speed.
When an unknown resistance RxRx is placed in a Wheatstone bridge, it is possible to balance the bridge by adjusting R3R3 to be 2500ΩΩ. What is RxRx if R2R1=0.625R2R1=0.625?
Answer:
Rₓ = 1562.5 Ω
Explanation:
The formula for the wheat stone bridge in balanced condition is given as follows:
R₁/R₂ = R₃/Rₓ
where,
Rₓ = Unknown Resistance = ?
R₃ = 2500 Ω
R₂/R₁ = 0.625
R₁/R₂ = 1/0.625 = 1.6
Therefore,
1.6 = 2500 Ω/Rₓ
Rₓ = 2500 Ω/1.6
Rₓ = 1562.5 Ω
The microwave background radiation is observed at a wavelength of 1.9 millimeters. The red shift of these photons is observed to be raoughly 1100. What would be the emitted wavelength?
Answer:
1730nm
Explanation:
Observed wavelength λ= 1.9 millimeters
Red shift of photons = 1100
Emitted wavelength = λo
Red shift = λ - λo/ λo = 1100
= [(1.9x10^-3 - λo)/ λo] = 1100
= 1.9x10^-3 - λo = 1100 λo
= 1.9x10^-3 = 1100 λo + λo
= 1.9x10^-3 = 1101 λo
We divide through by 1101 to get the value of lambda
1.9x10^-3/1101 = λo
λo = 0.000001726
λo = 1726nm
This is approximately
λo = 1730nm
Thank you!
1.A large beach ball weighs 4.0 N. One person pushes it with a force of 7.0 N due South while another person pushes it 5.0 N due East. Find the acceleration on the beach ball.
2. What is the weight of a 70 kg astronaut on the earth, on the moon, (g=1.6 m/s2), on Venus (g = 18.7 m/s2) and in outer space traveling at a constant velocity
Answer:
1. force applied southward = -4 j
force applied eastward = 5 i
total force applied = 5i - 4j
magnitude of total force applied = √(5)²+(-4)²
magnitude of total force applied = √25 + 16 = √41
magnitude of total force applied = 6.4N
But the beach ball also weighs 4 N,
which means that a force of 4N is required to overcome the inertia of the ball
Net force applied on the ball = total force applied - force applied by inertia
Net force applied on the ball = 6.4 - 4
Net force applied on the ball = 2.4 N
Mass of the ball:
Mass of the ball = weight of the ball / gravitational constant
Mass of the ball = 4 / 9.8 = 0.4 kg
Acceleration of Ball:
from newton's second law of motion:
F = ma
replacing the variables
2.4 = 0.4 * a (where a is the acceleration of the ball)
a = 2.4/0.4
a = 6 m/s²
2. Mass of astronaut = 70 kg
Weight of Earth:
Weight = Mass * acceleration due to gravity
Weight = 70 * 9.8
Weight = 686 N
Weight on Moon:
Weight = Mass * acceleration due to gravity
Weight = 70 * 1.6 (we are given that g = 1.6 on moon)
Weight = 112 N
Weight on Venus:
Weight = Mass * acceleration due to gravity
Weight = 70 * 18.7 (we are given that g = 18.7 on Venus)
Weight = 1309 N
A long thin rod of length 2L rotates with a constant angular acceleration of 8.0 rad/s2 about an axis that is perpendicular to the rod and passes through its center. What is the ratio of the tangential acceleration of a point on the end of the rod to that of a point a distance L/2 from the end of the rod
Answer:
Explanation:
angular acceleration ω = 8 rad /s²
tangential acceleration a = angular acceleration x radius of circle of rotation
For a point on the end of the rod , radius of circle = L .
tangential acceleration = 8 L .
For a point a distance L/2 from the end of the rod, radius of circle = L /2 .
tangential acceleration = 8 L/2 = 4 L .
It is so because angular acceleration will be same for this point , only radius changes .
required ratio = 8L / 4L
= 2.
s this statement true or false?
Hurricanes are classified by three stages in which updrafts of air billow up, updrafts and downdrafts swirl the wind up and down, then downdrafts cause clouds to come apart.
true
false
Answer:
True
Explanation:
State two function of the mass M placed on the metre rule
correct answer is mass media is correct answer OK
Help pleaseeeee!
If a force of 12 N is applied to a 4.0 kg object, what acceleration is produced?
Answer:
The answer is 3 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]a = \frac{f}{m} \\ [/tex]
where
f is the force
m is the mass
From the question we have
[tex]a = \frac{12}{4} \\ [/tex]
We have the final answer as
3 m/s²Hope this helps you
A 3.00-kg block starts from rest at the top of a 35.5° incline and slides 2.00 m down the incline in 1.45 s. (a) Find the acceleration of the block.
Answer:
0.344 m/s^2
Explanation:
The first step is to calculate V
V= distance /time
= 2/1.45
= 1.379 m/s
Therefore the acceleration of the block can be calculated as follows
a= V - U/2d
= 1.379 - 0/2×2
= 1.379/4
= 0.344 m/s^2
A 1500 kg car hits a haystack at 8m/s and comes to a stop after 0.6 seconds
A)what is the change in momentum of the car ?
B) what is the impulse that acts on the car ?
C) what is the average force that acts on the car during the collision?
D) what is the work done by the haystack in stopping the car ?
Answer:
Kindly check explanation
Explanation:
Change in momentum:
m(v - u)
M = mass = 1500kg
v = final velocity = 0 m/s
u = initial velocity = 8m/s
Momentum = 1500(8 - 0)
Momentum = 1500 * 8
Momentum = 12000 kgm/s
B.) impulse that acts on the car
Impulse = Force * time
Impulse = change in momentum with time
Impulse = m(v - u)
Hence, impulse = 12000Ns
C.) Average force action on the car during collision
Impulse = Force * time
12000 = force * 0.6
12000/ 0.6 = force
Force = 20000 N
D.) Workdone by haystack in stopping the car :
Workdone = Force * Distance
Distance = speed * time
Distance = 8 * 0.6
Distance = 4.8m
Hence,
Workdone = 20,000 * 4.8 = 96000 J
A Ferris wheel rotates at an angular velocity of 0.25 rad/s. Starting from rest, it reaches its operating speed with an average angular acceleration of 0.027 rad/s2. How long does it take the wheel to come up to operating speed?
Answer:
t = 9.25 s
Explanation:
Given that,
Initial angular velocity, [tex]\omega_o=0[/tex] (at rest)
Final angular velocity, [tex]\omega_o=0.25\ rad/s[/tex]
Angular acceleration, [tex]\alpha =0.027\ rad/s^2[/tex]
We need to find the time it take the wheel to come up to operating speed. We know that the angular acceleration in terms of angular speed is given by :
[tex]\alpha =\dfrac{\omega_f-\omega_o}{t}\\\\t=\dfrac{\omega_f-\omega_o}{\alpha }\\\\t=\dfrac{0.25-0}{0.027}\\\\t=9.25\ s[/tex]
So, it will reach up to the operating speed in 9.25 s.
If the rock is kicked at initial velocity of 12 m/s from a height of 106 m above the ground what horizontal distance does the rock travel before striking the ground?
Answer:
74 m.
Explanation:
From the question given above, the following data were obtained:
Height (h) = 186 m
Initial velocity (u) = 12 m/s
Horizontal distance (d) =?
Next, we shall determine the time taken for the rock to get to the ground. This can be obtained as follow:
Height (h) = 186 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
H = ½gt²
186 = ½ × 9.8 × t²
186 = 4.9 × t²
Divide both side by 4.9
t² = 186/4.9
Take the square root of both side
t = √(186/4.9)
t = 6.16 s
Thus, the time taken for the rock to get to the ground is 6.16 s
Finally, we shall determine the horizontal distance travelled by the rock as follow:
Initial velocity (u) = 12 m/s
Time (t) = 6.16 s
Horizontal distance (d) =?
Horizontal distance (d) = Initial velocity (u) × Time (t)
d = u × t
d = 12 × 6.16
d = 73.92 ≈ 74 m
Therefore, the horizontal distance travelled by the rock is 74 m
A tired worker pushes a heavy (100-kg) crate that is resting on a thick pile carpet. The coefficients of static and kinetic friction are 0.6 and 0.4, respectively. The worker pushes with a horizontal force of 490 N. The frictional force exerted by the surface is
Answer:
The friction force exerted by the surface is 490 newtons.
Explanation:
From Physics, we remember that static friction force ([tex]f_{s}[/tex]), measured in newtons, for a particle on a horizontal surface is represented by the following inequation:
[tex]f_{s} \leq \mu_{s}\cdot m \cdot g[/tex] (Eq. 1)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.
[tex]m[/tex] - Mass of the crate, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
If [tex]f_{s} = \mu_{s}\cdot m \cdot g[/tex], then crate will experiment an imminent motion. The maximum static friction force is: ([tex]\mu_{s} = 0.6[/tex], [tex]m = 100\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex])
[tex]f_{s} = (0.6)\cdot (100\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]f_{s} = 588.42\,N[/tex]
From Newton's Laws we get that current force of friction as reaction to the pulling force done by the worker on the crate is:
[tex]\Sigma F = F-f = 0[/tex]
[tex]f = F[/tex] (Eq. 2)
Where:
[tex]F[/tex] - Horizontal force done by the worker, measured in newtons.
[tex]f[/tex] - Static friction force, measured in newtons.
If [tex]F = 490\,N[/tex], then the static friction force exerted by the surface is:
[tex]f = 490\,N[/tex]
Given that [tex]f < f_{s}[/tex], the crate does not change its state of motion. The friction force exerted by the surface is 490 newtons.
a 214kg boat is sinking in the ocean.its acceleration towards the sea floor is 6.12 m/s^2. the force of gravity that draws the boat down is partially offset by the buoyant force of the water, so what is the force pulling on the boat as it sinks?
How much is the spring stretched, in meters, by an object with a mass of 0.49 kg that is hanging from the spring at rest
Answer:
1.6m
Explanation:
Using the spring equation, which is as follows:
F = Kx
Where; F = force applied on the spring (N)
K = spring constant (3kg/s²)
x = extension of spring (m)
However, Force applied by object = mass × acceleration (9.8m/s²)
F = 0.49kg × 9.8m/s²
F = 4.802N
Since, F = 4.802N
F = Kx
4.802 = 3 × x
4.802 = 3x
x = 4.802/3
x = 1.6006
x = 1.6m
The spring is stretched i.e. the extension of the spring, by 1.6m
will give brainliest pls help
Answer:
a and c
Explanation:
1. a. Describe the type of energy that is observed when a musician's finger plucks a guitar string. does it give your ear kenetic or potential energy .
Answer:
kinetic
Explanation:
Answer:
kinetic energy.
Explanation:
A dentist’s drill starts from rest. After 7.28 s of constant angular acceleration it turns at a rate of 16740 rev/min. Find the drill’s angular acceleration. Answer in units of rad/s 2 .
Answer:
[tex]\alpha =240.79\ rad/s^2[/tex]
Explanation:
Given that,
Initial angular velocity, [tex]\omega_i=0[/tex]
Final angular velocity, [tex]\omega_f=16740\ rev/min = 1753\ rad/s[/tex]
We need to find the angular acceleration of the drill. It is given by the formula as follows :
[tex]\alpha =\dfrac{\omega_f-\omega_i}{t}\\\\\alpha =\dfrac{1753-0}{7.28}\\\\\alpha =240.79\ rad/s^2[/tex]
So, the angular acceleration of the drill is [tex]240.79\ rad/s^2[/tex].
What height does a frictionless playground slide need so that a 35 kg child reaches the bottom at a speed of 6.5 m/s?
Answer:
h = 2.15 m
Explanation:
Given that,
Mass of the child, m = 35 kg
The child reaches the bottom at a speed of 6.5 m/s
We need to find the height slide by a frictionless playground slide. Let it is equal to h. Using the conservation of energy to find it as follows :
[tex]\dfrac{1}{2}mv^2=mgh\\\\h=\dfrac{v^2}{2g}\\\\h=\dfrac{(6.5)^2}{2\times 9.8}\\\\h=2.15\ m[/tex]
So, it will slide to a height of 2.15 m
The height slid by the frictionless playground can be calculated by equating the kinetic and potential energy equations which is 2.16 meters
Mass of child = 35 kg Velocity, v = 6.5 m/s g = 9.8 m/s²K.E = P.E
K. E = 0.5mv² ; P.E = mgh
0.5mv² = mgh
gh = 0.5v²
h = [(0.5v²) ÷ g]
h = [(0.5 × 6.5²) ÷ 9.8]
h = (21.125 ÷ 9.8)
h = 2.155 m
Therefore, the height slid by a frictionless playground is 2.16 meters
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A 80-kg person stands at rest on a scale while pulling vertically downwards on a rope that is above them. Use g = 9.80 m/s2.
With what magnitude force must the tension in the rope be pulling on the person so that the scale reads 25% of the persons weight?
Answer:
The tension in the rope is 588 N
Explanation:
I have attached a free body diagram of the problem.
First we calculate the force applied downwards by the person's weight:
[tex]F_{weight} =m*a = 80 kg * 9.8 \frac{m}{s^{2}} = 784 N[/tex]
If the scale reads a 25% of the person' weight:
[tex]F_{Scale} = F_{weight}*0.25=784N*0.25=196N[/tex]
which is the same value (but opposite) to the effective weight of the persons ([tex]F_{Person}=-F_{Scale}[/tex])
The other 75% of the total weight is the force of the rope pulling the person upwards.
[tex]F_{Pull}= F_{weight}*0.75=784N*0.75=588N[/tex]
To verify, we can use 1st Newton's law
∑F = 0
[tex]F_{Rope}-F_{Pull}+F_{Person}-F_{Scale}=0 \\588N-588N+196N-196N=0[/tex]
A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 35 m in 3.5 min, starting and ending at rest. The elevator's counterweight has a mass of only 940 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable
Answer:
425.1 W
Explanation:
We are given;
Counter mass of elevator; m_c = 940 kg
Cab mass of elevator; m_d = 1200 kg
Distance from rest upwards; d = 35 m
Time to cover distance; t = 3.5 min
Now, this elevator will have 3 forces acting on it namely;
Force due to the counter weight of the elevator; F_c
Force due to the cab weight on the elevator; F_d
Force exerted by the motor; F_m
Now, from Newton's 2nd law of motion,
The force exerted by the motor on the elevator can be given by the relationship;
F_m = F_d - F_c
Now,
F_d = m_d × g
F_d = 1200 × 9.81
F_d = 11772 N
F_c = m_c × g
F_c = 940 × 9.81
F_c = 9221.4 N
Thus;
F_m = 11772 - 9221.4
F_m = 2550.6 N
Now, the average power required of the force the motor exerts on the cab via the cable is given by;
P_m = F_m × v
Where v is the velocity of the elevator.
The velocity is calculated from;
v = distance/time
v = 35/3.5
v = 10 m/min
Converting to m/s gives;
v = 10/60 m/s = 1/6 m/s
Thus;
P_m = 2550.6 × 1/6
P_m = 425.1 W
How much force (in N) is exerted on one side of an 35.3 cm by 50.0 cm sheet of paper by the atmosphere
Answer:
A force of 1.788 newtons is exerted on one side of the sheet of paper by the atmosphere.
Explanation:
From definition of pressure, we get that force exerted by the atmosphere ([tex]F[/tex]), measured in newtons, on one side of the sheet of paper is given by the expression:
[tex]F = P_{atm}\cdot w\cdot l[/tex] (Eq. 1)
Where:
[tex]P_{atm}[/tex] - Atmospheric pressure, measured in pascals.
[tex]w[/tex] - Width of the sheet of paper, measured in meters.
[tex]l[/tex] - Length of the sheet of paper, measured in meters.
If we know that [tex]P_{atm} = 101325\,Pa[/tex], [tex]w = 0.353\,m[/tex] and [tex]l = 0.05\,m[/tex], then the force exerted on one side of the sheet of paper is:
[tex]F = (101325\,Pa)\cdot (0.353\,m)\cdot (0.05\,m)[/tex]
[tex]F = 1.788\,N[/tex]
A force of 1.788 newtons is exerted on one side of the sheet of paper by the atmosphere.
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is rotating at 4.40 rev/s; 60.0 revolutions later, its angular speed is 15.0 rev/s. Calculate (a) the angular acceleration (rev/s2), (b) the time required to complete the 60.0 revolutions, (c) the time required to reach the 4.40 rev/s angular speed, and (d) the number of revolutions from rest until the time the disk reaches the 4.40 rev/s angular speed.
Answer:
The answer is below
Explanation:
a) Using the formula:
[tex]\omega^2=\omega_o^2+2\alpha \theta\\\\\omega=final\ velocity=15\ rev/s,w_o=initial\ velocity=4.4\ rev/s, \\\theta=distance=60\ rev\\\\Substituting:\\\\15^2=4.4^2+2(60)\alpha\\\\2(60)\alpha=15^2-4.4^2\\\\2(60)\alpha=205.64\\\\\alpha=1.71\ rev/s^2[/tex]
b) The disk is initially at rest. Using the formula:
[tex]\theta=\omega_ot+\frac{1}{2}\alpha t^2 \\\\but\ \omega_o=0(rest), \theta=60\ rev,\alpha=1.71\ rev/s^2\\\\Subsituting:\\\\60=0+\frac{1}{2}(1.71) t^2\\\\t=8.4\ s[/tex]
c)
[tex]\omega=\omega_o+\alpha t \\\\but\ \omega_o=0(rest), \omega=4.4 \ rev/s\ rev,\alpha=1.71\ rev/s^2\\\\Subsituting:\\\\4.4=1.71t\\\\t=2.6\ s[/tex]
d)
[tex]\omega^2=\omega_o^2+2\alpha \theta\\\\\omega=final\ velocity=4.4\ rev/s,w_o=initial\ velocity=0\ rev/s, \\\alpha=1.71\ rev/s^2\\\\Substituting:\\\\4.4^2=0+2(1.71)\theta\\\\19.36=3.42\theta\\\\\theta=5.66\ rev[/tex]