Answer:
(a) the tangential speed of a point at the edge is 3.14 m/s
(b) At a point halfway to the center of the disc, tangential speed is 1.571 m/s
Explanation:
Given;
angular speed of the disc, ω = 500 rev/min
diameter of the disc, 120 mm
radius of the disc, r = 60 mm = 0.06 m
(a) the tangential speed of a point at the edge is calculated as follows;
[tex]\omega = 500 \ \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1 \min}{60 \ s} = 52.37 \ rad/s[/tex]
Tangential speed, v = ωr
v = 52.37 rad/s x 0.06 m
v = 3.14 m/s
(b) at the edge of the disc, the distance of the point = radius of the disc
at half-way to the center, the distance of the point = half the radius.
r₁ = ¹/₂r = 0.5 x 0.06 m = 0.03 m
The tangential velocity, v = ωr₁
v = 52.37 rad/s x 0.03 m
v = 1.571 m/s
Explain briefly how solar energy is used to generate electricity
This when the energy from the sun is trapped by either sun panel or others. It passes through conventions then we get the energy out as electricity.
Answer:
Solar radiation may be converted directly into electricity by solar cells (photovoltaic cells). In such cells, a small electric voltage is generated when light strikes the junction between a metal and a semiconductor (such as silicon) or the junction between two different semiconductors.
Explanation:
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Solar radiation may be converted directly into electricity by solar cells or photovoltaic cells. This is how solar energy is used to generate electricity.
What is solar energy?Solar energy is the radiant light and heat from the Sun that is captured and used in a variety of technologies, including solar power to generate electricity, solar thermal energy, and solar architecture.
When the sun shines on a solar panel, the photovoltaic cells in the panel absorb the energy from the sun. When light strikes the junction of a metal and a semiconductor (such as silicon) or the junction of two different semiconductors, a small electric voltage is generated. This energy generates electrical charges that move in response to an internal electrical field in the cell, resulting in the flow of electricity.
Therefore, solar energy is used to generate electricity.
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A motorcycle of mass 160 kg accelerates from rest to 53 m/s in 9 seconds. Ignore air resistance. Assuming there's no slipping between the wheels and the pavement of the road.
Required:
a. What is the average horizontal component of the force that the road exerts on the wheels (total force on all two wheels, not the force on one wheel)?
b. How far does the motorcycle travel in 9 seconds?
c. In the point-particle analysis of this situation, what is the work done by this force?
d. For the real system, how much work is done by the force of the road on the wheels?
Answer:
a) [tex]F=940.8N[/tex]
b) [tex]S=234.14m[/tex]
c) [tex]W=2.2*10^5J[/tex]
d) [tex]W=0[/tex]
Explanation:
Mass [tex]m=160kg[/tex]
Velocity [tex]v=53m/s[/tex]
Time [tex]t=9seconds[/tex]
a)
Generally the Newton's equation for motion is mathematically given by
[tex]a=\frac{v}{t}[/tex]
[tex]a=\frac{53}{9}[/tex]
[tex]a=5.9m/s^2[/tex]
Therefore
F=ma
[tex]F=160*5.88[/tex]
[tex]F=940.8N[/tex]
b)
Generally the Newton's equation for motion is mathematically given by
[tex]S=0.5at^2[/tex]
[tex]S=0.5*5.9*9^2[/tex]
[tex]S=234.14m[/tex]
c)
Generally the Newton's equation for work done is mathematically given by
[tex]W=Fd[/tex]
[tex]W=940.8*238.14[/tex]
[tex]W=2.2*10^5J[/tex]
d)
Generally the Newton's equation for work done by the force of the road on the wheels is mathematically given by
[tex]W=Fdcos\theta[/tex]
[tex]W=0[/tex]
A billiard ball moving at 5 m/s strikes a stationary ball of the same mass. After the collision, the original ball moves at a velocity of 4.35 m/s at an angle of 30 o below its original motion. Find the velocity and angle of the second ball after the collision.
A) 1.25 m/s at 31.2o
B) 1.44 m/s at 60.0o
C) 2.16 m/s at 30.0o
D) 2.47 m/s at 61.9o
90 degrees - 30 = 60 degrees
Velocity = (5m/s - 4.35m/s x cos(30)) / cos(60)
Velocity = 2.47 m/s
The answer is D) 2.47 m/s at 61.9 degrees
The velocity and angle of the second ball after the collision are (A.) 1.25 m/s at 31.2° below the horizontal. Option A
How to calculate velocityUse the conservation of momentum and conservation of kinetic energy to solve this problem.
Let's denote the velocity and angle of the second ball after the collision as v₂ and θ₂, respectively.
Thus:
Conservation of momentum: [tex]m_1v_1 = m_1v_1'cos(30^o) + m_2v_2cos(\theta_2)[/tex]
Conservation of kinetic energy: [tex](1/2)m_1v_1^2 = (1/2)m_1v_1'^2 + (1/2)m_2v_2^2[/tex]
where m₁ and m₂ are the masses of the first and second balls, respectively.
Since the masses and initial velocity are the same, we can simplify the equations to:
m₁v₁ = m₁v₁'cos(30°) + m₂v₂cos(θ₂)
[tex]v_1^2 = v_1'^2 + v_2^2[/tex]
Substitute in the given values
[tex](1)(5) = (1)(4.35)cos(30^o) + (1)(v_2)cos(\theta_2)\\5^2 = 4.35^2 + v2^2\\v_2 = 1.25 m/s\\\theta2 = 31.2^o[/tex]
Therefore, the velocity and angle of the second ball after the collision are approximately 1.25 m/s at an angle of 31.2° below the horizontal.
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the roque requried to turn the crank on an ice cream maker is 4.50 N.m how much work does it take to turn the crank through 300 full turns
Answer:
the work required to turn the crank at the given revolutions is 8,483.4 J
Explanation:
Given;
torque required to turn the crank, T = 4.50 N.m
number of revolutions, = 300 turns
The work required to turn the crank is given as;
W = 2πT
W = 2 x 3.142 x 4.5
W = 28.278 J
1 revolution = 28.278 J
300 revlotions = ?
= 300 x 28.278 J
= 8,483.4 J
Therefore, the work required to turn the crank at the given revolutions is 8,483.4 J
Which object has the most thermal energy?
A. 2 kg of liquid oxygen at -225°C
B. 5 kg of liquid oxygen at -220°C
C. 2 kg of oxygen gas at 20°C
D. 5 kg of oxygen gas at 30°C
5 kg of liquid oxygen at -220°C will have the most thermal energy due to its higher mass and temperature.
What is thermal energy?Thermal energy is the total heat required to raise the entire mass of a given substance.
Q = mcΔθ
where;
Q is thermal energym is mass of the substancec is specific heat capacityΔθ is change in temperatureThus, 5 kg of liquid oxygen at -220°C will have the most thermal energy due to its higher mass and temperature.
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What is the resistance of a copper wire 200 ft long and
0.01 in. in diameter (T 20°C)?
Answer:
R = 20.21 ohms
Explanation:
Given that,
The length of the wire, l = 200 ft
The diameter of the wire, d = 0.01 in
Radius, r = 0.005 in
200 ft = 60.96 m
0.005 in = 0.000127 m
The resistivity of copper is, [tex]\rho=1.68\times 10^{-8}\ \Omega-m[/tex]
So, the resistance of a wire is given by :
[tex]R=\rho \dfrac{l}{A}\\\\R=\rho \dfrac{l}{\pi r^2}\\\\R=1.68\times 10^{-8}\times \dfrac{60.96 }{\pi \times (0.000127 )^2}\\\\R=20.21\ \Omega[/tex]
So, the resistance of the copper wire is 20.21 ohms.
why the change of the pressure and temperature affect the velocity of the sound
Air pressure has no effect at all in an ideal gas approximation. This is because pressure and density both contribute to sound velocity equally, and in an ideal gas the two effects cancel out, leaving only the effect of temperature. Sound usually travels more slowly with greater altitude, due to reduced temperature.
State what is meant by a gravitational potential at point A is -1·70 × 109 J kg-1.
Answer:
The energy stored in a body due to either it's position or change in shape is called gravitational potential energy.
what is the critical angle of light traveling from vegetable oil into water
56.1∘
Question: A glass is half-full of water, with a layer of vegetable oil (n = 1.47) floating on top. A ray of light traveling downward through the oil is incident on the water at an angle of 56.1∘ .
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Some copper wire has a resistance of 200 ohms at 20 degrees C . A current is then passed through the same wire and the temperature rises to 90 degrees C. Determine the resistance of the wire at 90 degrees correct to the nearest ohm assuming the coefficient of resistance is 0.004/degree C at 0 degrees
Answer:
256 ohms
Explanation:
Applying,
R = R'[1+α(T-T')]............. Equation 1
Where R = Final resistance of the wire, R' = Initial resistance of the wire, T = Final temperature, T' = Initial temperature, α = Temperature coefficient of resistance
From the question,
Given: R' = 200 ohms, T = 90 degrees, T' = 20 degrees, α = 0.004/degree
Substitute these values into equation 1
R = 200[1+0.004(90-20)]
R = 200[1+0.28]
R = 200(1.28)
R = 256 ohms
The resistance of the wire at 90 °C correct to the nearest ohm assuming the coefficient of resistance is 0.004 °C¯¹ is 256 ohm
Data obtained from the question Original resistance (R₁) = 200 ohmOriginal temperature (T₁) = 20 °C Coefficient of resistivity (α) = 0.004 °C¯¹New temperature (T₂) = 90 °C New resistance (R₂) =? How to determine the new resistanceα = R₂ – R₁ / R₁(T₂ – T₁)
0.004 = R₂ – 200 / 200(90 – 20)
0.004 = R₂ – 200 / 200(70)
0.004 = R₂ – 200 / 14000
Cross multiply
R₂ – 200 = 0.004 × 14000
R₂ – 200 = 56
Collect like terms
R₂ = 56 + 200
R₂ = 256 ohm
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An capacitor consists of two large parallel plates of area A separated by a very small distance d. This capacitor is connected to a battery and charged until its plates carry charges Q and - Q, and then disconnected from the battery. If the separation between the plates is now doubled, the potential difference between the plates will
Answer:
Will be doubled.
Explanation:
For a capacitor of parallel plates of area A, separated by a distance d, such that the charges in the plates are Q and -Q, the capacitance is written as:
[tex]C = \frac{Q}{V} = e_0\frac{A}{d}[/tex]
where e₀ is a constant, the electric permittivity.
Now we can isolate V, the potential difference between the plates as:
[tex]V = \frac{Q}{e_0} *\frac{d}{A}[/tex]
Now, notice that the separation between the plates is in the numerator.
Thus, if we double the distance we will get a new potential difference V', such that:
[tex]V' = \frac{Q}{e_0} *\frac{2d}{A} = 2*( \frac{Q}{e_0} *\frac{d}{A}) = 2*V\\V' = 2*V[/tex]
So, if we double the distance between the plates, the potential difference will also be doubled.
when we jump on a concrete surface,the feet get injured.Why
Answer:
Explanation:
Bhjb
Explanation:
its because a concrete surface is a hard surface which doesn't absorb the energy of gravitation when we fall down so we get hurt more badly..
hope this helps
A 215 N sign is supported by two ropes. One rope pulls up and to the right 1=29.5∘ above the horizontal with a tension 1 , and the other rope pulls up and to the left 2=44.5∘ above the horizontal with a tension 2 , as shown in the figure. Find the tensions 1 and 2 .
The sign is held in equilibrium. Using Newton's second law, we set up the equations of the net forces acting on the sign in the horizontal and vertical directions:
∑ F (horizontal) = T₁ cos(29.5°) - T₂ cos(44.5°) = 0
(right is positive, left is negative)
∑ F (vertical) = T₁ sin(29.5°) + T₂ sin(44.5°) - 215 N = 0
(up is positive, down is negative)
Solve the system of equations. I use elimination here:
• Multiply the first equation by sin(29.5°) and the second by cos(29.5°):
sin(29.5°) (T₁ cos(29.5°) - T₂ cos(44.5°)) = 0
cos(29.5°) (T₁ sin(29.5°) + T₂ sin(44.5°) - 215 N) = 0
T₁ cos(29.5°) sin(29.5°) - T₂ cos(44.5°) sin(29.5°) = 0
T₁ cos(29.5°) sin(29.5°) + T₂ cos(29.5°) sin(44.5°) = (215 N) cos(29.5°)
• Subtract the first equation from the second to eliminate T₁ :
T₂ cos(29.5°) sin(44.5°) - (- T₂ cos(44.5°) sin(29.5°)) = (215 N) cos(29.5°)
• Solve for T₂ :
T₂ (cos(29.5°) sin(44.5°) + cos(44.5°) sin(29.5°)) = (215 N) cos(29.5°)
T₂ sin(74.0°) = (215 N) cos(29.5°)
… … … (using the fact that sin(x + y) = sin(x) cos(y) + cos(y) sin(x))
T₂ = (215 N) cos(29.5°) / sin(74.0°)
T₂ ≈ 195 N
• Solve for T₁ :
T₁ cos(29.5°) - T₂ cos(44.5°) = 0
T₁ cos(29.5°) = T₂ cos(44.5°)
T₁ = T₂ cos(44.5°) / cos(29.5°)
T₁ ≈ 160. N
? What is the difference between the Primitive cell and convectional cell
* 1a Average speed
Carl Lewis runs the 100 m sprint in about 10 s.
His average speed in units of m/s would be:
of
Answer:
Explanation:
[tex] \implies v_{av} = \dfrac{total \: displacement}{total \: time} [/tex]
[tex] \implies v_{av} = \dfrac{100}{10} [/tex]
[tex]\implies v_{av} =10 \: {ms}^{ - 1} [/tex]
The diagram below shows snapshots of an oscillator at different times. What is the amplitude of oscillation?
Answer:
the amplitude of the oscillation of the given mass is 0.1 m.
Explanation:
The amplitude of an oscillation is the maximum displacement of the object from the equilibrium position.
The equilibrium position of the given mass in question is at the zero (0) mark.
The maximum displacement of the object from the equilibrium position is 0.1 m upwards or 0.1 m downwards.
Therefore, the amplitude of the oscillation of the given mass is 0.1 m.
Projectile Problems – Type 2:
1. The photograph below shows a basic projectile at several locations on its trajectory.
a) List the location(s) where the vertical component of the velocity would be zero.
b) What is the vertical component of the acceleration at location #3?
c) What is the horizontal acceleration of the projectile?
d) Identify the location where the vertical displacement would be zero.
e) Identify the location with the maximum displacement.
f) Rank each location in terms of the projectile's speed (highest to lowest).
Answer:
Explanation:
( a ) At top position , vertical component of velocity will be zero .
So answer is position no (3)
b )
At position (3) which is the topmost position , acceleration is acting due to gravity , so it will be downwards.
c )
Horizontal component of acceleration at all points will be zero because gravity acts vertically downwards.
d )
Vertical displacement will be zero at position ( 1 ) and ( 5 )
e )
Displacement is maximum at extreme position , ie at position ( 5 )
f )
Speed is highest at position (1) and it is lowest at position ( 3 )
From highest to lowest
( 1 ) , ( 2 ) , ( 3 )
explain relative velocity briefly
Answer:
Explanation:
Relative velocity is defined as the velocity of an object B in the rest frame of another object A.
What is the kinetic energy of a 6.00 kg toy car that is going at 1.75 m/s across the floor?
Answer:
9.1875 Joules
Explanation:
The kinetic energy of an object can be given by the equation [tex]EK = \frac{1}{2} mv^{2}[/tex], in which m is the mass and v is the velocity. Plugging in the mass and velocity values into the equation, we get:
[tex]EK = \frac{1}{2} (6kg)(1.75m/s)^2\\EK = 9.1875 J[/tex]
Answer:
[tex]\boxed {\boxed {\sf 9.19 \ Joules}}[/tex]
Explanation:
Kinetic energy is the energy an object possesses due to motion. The following formula is used to calculated kinetic energy.
[tex]E_k= \frac{1}{2} mv^2[/tex]
In this formula, m is the mass and v is the velocity.
The car has a mass of 6.00 kilograms and a velocity of 1.75 meters per second.
m= 6.00 kgv=1.75 m/sSubstitute the values into the formula.
[tex]E_k= \frac{1}{2} (6.00 \ kg )(1.75 \ m/s)^2[/tex]
Solve the exponent.
(1.75 m/s)²= 1.75 m/s *1.75 m/s = 3.0625 m²/s²[tex]E_k= \frac{1}{2}(6.00 \ kg )(3.0625 \ m^2/s^2)[/tex]
Multiply the numbers together.
[tex]E_k=\frac {1}{2} (18.375 \ kg*m^2/s^2)[/tex]
[tex]E_k= 9.1875 \ kg*m^2/s^2[/tex]
The original measurements of mass and velocity both have 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredths place. The 7 in the thousandths place to the right (9.1875) tells us to round to 8 up to a 9.
[tex]E_k \approx 9.19 \ kg*m^2/s^2[/tex]
1 kilogram meter squared per second squared is equal to 1 Joule. Our answer of 9.19 kg*m²/² is equal to 9.19 Joules.
[tex]E_k \approx 9.19 \ J[/tex]
The kinetic energy of the toy care is approximately 9.19 Joules.
When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and arm–shoulder combination?
Complete question:
The recoil of a shotgun can be significant. Suppose a 3.6-kg shotgun is held tightly by an arm and shoulder with a combined mass of 15.0 kg. When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and arm–shoulder combination?
Answer:
The recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s
Explanation:
Given;
combined mass of the shotgun and arm–shoulder, m₁ = 15 kg
mass of the projectile, m₂ = 0.04 kg
speed of the projectile, u₂ = 380 m/s
let the recoil velocity of the shotgun and arm–shoulder combination = u₁
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = 0
m₁u₁ = - m₂u₂
[tex]u_1 = -\frac{m_2u_2}{m_1} \\\\u_1 = - \frac{0.04\times 380}{15} \\\\u_1 =-1.013 \ m/s\\\\u_1 = 1.013 \ m/s \ \ \ in \ opposite \ direction[/tex]
Therefore, the recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s
Expresar por notación científica:
846 000 000
Answer:
[tex]8.46*10^8[/tex]
A boy at a football field practice kicked a 0.500-kg ball with a force of 100N. How fast will the ball move after reaching a distance of 7m?
Answer:
v(7) = 52.915 m/s
Explanation:
First, find the value for acceleration.
F = ma
100 = .5 * a
a = 200 m/s²
Next find the velocity at x = 7 using kinematic equations.
v² = v₀² + 2a(Δx)
v² = (0)² + 2(200)(7)
v = [tex]\sqrt{2800}[/tex]
v = 52.915 m/s
In a race, Usain Bolt accelerates at
1.99 m/s2 for the first 60.0 m, then
decelerates at -0.266 m/s2 for the final
40.0 m. How much time did the race take?
(Unit = s)
Answer:
65.87 s
Explanation:
For the first time,
Applying
v² = u²+2as.............. Equation 1
Where v = final velocity, u = initial velocity, a = acceleration, s = distance
From the question,
Given: u = 0 m/s (from rest), a = 1.99 m/s², s = 60 m
Substitute these values into equation 1
v² = 0²+2(1.99)(60)
v² = 238.8
v = √238.8
v = 15.45 m/s
Therefore, time taken for the first 60 m is
t = (v-u)/a............ Equation 2
t = (15.45-0)/1.99
t = 7.77 s
For the final 40 meter,
t = (v-u)/a
Given: v = 0 m/s(decelerates), u = 15.45 m/s, a = -0.266 m/s²
Substitute into the equation above
t = (0-15.45)/-0.266
t = 58.1 seconds
Hence total time taken to cover the distance
T = 7.77+58.1
T = 65.87 s
Which is a product of nuclear fusion?
Answer:
Creation of energy by joining the nuclei of two hydrogen atoms to form helium
When a liquid is cooled,the kinetic energy of the particles______. The force of attraction between the particles______,the space between the particles_______
When a liquid is cooled,the kinetic energy of the particles decreases. The force of attraction between the particles increases, the space between the particles decreases.
How much work does a supermarket checkout attendant do on a can of soup he pushes 0.420 m horizontally with a force of 4.60 N? Express your answer in joules and kilocalories.
We know
[tex]\boxed{\sf Work\:Done=Force\times Displacement} [/tex]
[tex]\\ \rm\longmapsto Work\:done=0.420\times 4.60[/tex]
[tex]\\ \rm\longmapsto Work\:done=1.932J[/tex]
Estimate the average power of a water wave when it hits the chest of an adult standing in the water at the seashore. Assume that the amplitude of the wave is 0.56 m , the wavelength is 2.0 m , and the period is 3.4 s . Assume that the area of the chest is 0.14 m^2.
Answer:
[tex]P=45.2W[/tex]
Explanation:
From the question we are told that:
Amplitude [tex]A=0.56m[/tex]
Period [tex]T=3.4s[/tex]
Wavelength [tex]\lambda=2.0[/tex]
Area [tex]a=0.14m^2[/tex]
Generally the equation for Power is mathematically given by
[tex]Power = 2 \pi ^2 \rho a(\frac{\lambda}{T})(\frac{1}{T})*A[/tex]
[tex]P= 2 3.142^2 1025 0.14(\frac{2.0}{3.4})(\frac{1}{3.4})^2*0.56^2[/tex]
[tex]P=45.2W[/tex]
Thermodynamic Processes: An ideal gas is compressed isothermally to one-third of its initial volume. The resulting pressure will be
Answer:
The resulting pressure is 3 times the initial pressure.
Explanation:
The equation of state for ideal gases is described below:
[tex]P\cdot V = n \cdot R_{u}\cdot T[/tex] (1)
Where:
[tex]P[/tex] - Pressure.
[tex]V[/tex] - Volume.
[tex]n[/tex] - Molar quantity, in moles.
[tex]R_{u}[/tex] - Ideal gas constant.
[tex]T[/tex] - Temperature.
Given that ideal gas is compressed isothermally, this is, temperature remains constant, pressure is increased and volume is decreased, then we can simplify (1) into the following relationship:
[tex]P_{1}\cdot V_{1} = P_{2}\cdot V_{2}[/tex] (2)
If we know that [tex]\frac{V_{2}}{V_{1}} = \frac{1}{3}[/tex], then the resulting pressure of the system is:
[tex]P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)[/tex]
[tex]P_{2} = 3\cdot P_{1}[/tex]
The resulting pressure is 3 times the initial pressure.
While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 6.63 m/s. The stone subsequently falls to the ground, which is 14.5 m below the point where the stone leaves his hand.
At what speed does the stone impact the ground? Ignore air resistance and use =9.81 m/s2 for the acceleration due to gravity.
impact speed:
18.1151
m/s
How much time is the stone in the air?
elapsed time:
Answer:
Explanation:
You final velocity is correct but not to the correct number of significant digits. The actual answer should be 18.1 m/s. We will use that to find the total time the stone was in the air in the equation:
v = v₀ + at
18.1 = 6.63 + (-9.81)t and
11.5 = -9.81t so
t = 1.17 seconds.
We know this was how long the stone was in the air (as compared to the time that the stone took to reach its max height or some other height) because we used the velocity with which the stone hit the ground to find the total time the rock was in the air before it hit the ground going at that velocity.
What is your angular position 75 seconds after the wheel starts turning, measured counterclockwise from the top? Express your answer as an angle between 0∘ and 360∘. Express your answer in degrees.
Complete Question
A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the wheel starts turning, you are at the very top.
What is your angular position 75 seconds after the wheel starts turning, measured counterclockwise from the top? Express your answer as an angle between 0∘ and 360∘. Express your answer in degrees.
Answer:
[tex]\phi=123.75[/tex]
Explanation:
From the question we are told that:
Height [tex]h=27m[/tex]
Period [tex]T=32sec[/tex]
Time [tex]t=75sec[/tex]
Generally the equation for angular velocity is mathematically given by
[tex]\omega=\frac{2 \pi}{T}[/tex]
[tex]\omega=\frac{2 \pi}{32}[/tex]
[tex]\omega=0.196rad/s[/tex]
Therefore
[tex]\theta=\omega t[/tex]
[tex]\theta=0.196rad/s*75sec[/tex]
[tex]\theta=843.75 \textdegree[/tex]
Therefore
[tex]\phi=\theta-2(360)[/tex]
[tex]\phi=123.75[/tex]
