The force of friction is determined to be 14.47 N in the upward direction. The net force is found to be 22.33 N, resulting in an acceleration of 4.47 m/s². The magnitude of the force of friction is determined to be 9.64 N, and its direction is upward, opposing the motion of the block. The change in kinetic energy is found to be 67.09 J.
a) The minimum force (magnitude and direction) that will prevent the block from sliding down the incline is the force of friction acting upwards, opposite to the direction of motion. To determine the force of friction we use the equation for static friction which is:
F = μsNwhere F is the force of friction, μs is the coefficient of static friction, and N is the normal force acting perpendicular to the surface. The normal force acting perpendicular to the incline is:
N = mg cos(θ)
where m is the mass of the block, g is the acceleration due to gravity, and θ is the angle of inclination. Therefore,
F = μsN = μsmg cos(θ) = 0.3 x 5 x 9.81 x cos(40) = 14.47 N
The minimum force required to prevent the block from sliding down the incline is 14.47 N acting upwards.
b) As the block slides down the incline, the forces acting on it are its weight W = mg acting downwards and the force of friction f acting upwards.
Fnet = W - f, where Fnet is the net force, W is the weight of the block, and f is the force of friction. The component of the weight parallel to the incline is:W∥ = mg sin(θ) = 5 x 9.81 x sin(40) = 31.97 NThe force of friction is:f = μkN = μkmg cos(θ) = 0.2 x 5 x 9.81 x cos(40) = 9.64 N
Therefore, Fnet = W - f = 31.97 N - 9.64 N = 22.33 N
The acceleration of the block is given by:
Fnet = ma => a = Fnet/m = 22.33/5 = 4.47 m/s2
The weight of the block is resolved into two components, one perpendicular to the incline and one parallel to it. The force of friction acts upwards and opposes the motion of the block.
c)The magnitude of the force of friction is given by:f = μkN = μkmg cos(θ) = 0.2 x 5 x 9.81 x cos(40) = 9.64 NThe direction of the force of friction is upwards, opposite to the direction of motion.d)The change in the kinetic energy of the block is given by:
ΔK = Kf - Ki, where ΔK is the change in kinetic energy, Kf is the final kinetic energy, and Ki is the initial kinetic energy. As the block begins its motion from a state of rest, its initial kinetic energy is negligible or zero. The final kinetic energy is given by:Kf = 1/2 mv2where v is the velocity of the block after it has slid 3m down the incline.
The velocity of the block can be found using the formula:
v2 = u2 + 2as, where u is the initial velocity (zero), a is the acceleration of the block down the incline, and s is the distance travelled down the incline.
Therefore, v2 = 0 + 2 x 4.47 x 3 = 26.82=> v = 5.18 m/s
The final kinetic energy is:Kf = 1/2 mv2 = 1/2 x 5 x 5.18² = 67.09 J
Therefore, the change in kinetic energy of the block is:ΔK = Kf - Ki = 67.09 - 0 = 67.09 J.
Learn more about friction at: https://brainly.com/question/24338873
#SPJ11
8) Dr Examines Image of a patients tiny mole w/ magnifying lens
A doctor examines a patient's small mole using a magnifying lens.
The doctor uses a magnifying lens to carefully examine an image of a patient's small mole. The magnifying lens allows for a closer inspection of the mole, enabling the doctor to observe any specific details or irregularities that may be present.
By examining the mole in detail, the doctor can assess its characteristics and determine if further investigation or medical intervention is necessary. The use of a magnifying lens enhances the doctor's ability to make accurate observations and provide appropriate medical advice or treatment based on their findings.
To learn more about magnifying -
brainly.com/question/19579144
#SPJ11
The area of a pipeline system at a factory is 5 m 2
. An incompressible fluid with velocity of 40 m/s. After some distance, the pipe has another opening as shown in Figure 2 . The output of this opening is 20 m/s. Calculate the area of this opening if the velocity of the flow at the other end is 30 m/s Figure 2 (6 marks)
Given that the area of a pipeline system at a factory is 5 m2, an incompressible fluid with a velocity of 40 m/s. After some distance.
The output of this opening is 20 m/s. We need to calculate the area of this opening if the velocity of the flow at the other end is 30 m/s.
Let us apply the principle of the continuity of mass. The mass of a fluid that enters a section of a pipe must be equal to the mass of fluid that leaves the tube per unit of time (assuming that there is no fluid accumulation in the line). Mathematically, we have; A1V1 = A2V2Where; A1 = area of the first section of the pipeV1 = velocity of the liquid at the first sectionA2 = area of the second section of the pipeV2 = velocity of the fluid at the second section given that the area of the first section of the pipe is 5 m2 and the velocity of the liquid at the first section is 40 m/s; A1V1 = 5 × 40A1V1 = 200 .................(1)
Also, given that the velocity of the liquid at the second section of the pipe is 30 m/s and the area of the first section is 5 m2;A2 × 30 = 200A2 = 200/30A2 = 6.67 m2Therefore, the area of the opening of the second section of the pipe is 6.67 m2. Answer: 6.67
to know more about pipeline systems here:
brainly.com/question/15302939
#SPJ11
can
i please get the answer to this
Question 4 (1 point) The frequency at which a material vibrates most easily. Doppler shift Destructive interference Resonance Standing waves Resonant Frequency Constructive interference
The frequency at which a material vibrates most easily is called the resonant frequency. Resonance occurs when an external force or vibration matches the natural frequency of an object, causing it to vibrate with maximum amplitude.
Resonant frequency is an important concept in physics and engineering. When a system is subjected to an external force or vibration at its resonant frequency, the amplitude of the resulting vibration becomes significantly larger compared to other frequencies. This is because the energy transfer between the external source and the system is maximized when the frequencies match.
Resonance can occur in various systems, such as musical instruments, buildings, bridges, and electronic circuits. In each case, there is a specific resonant frequency associated with the system. By manipulating the frequency of the external source, one can identify and utilize the resonant frequency to achieve desired effects.
When resonance is achieved, it often leads to the formation of standing waves. These are stationary wave patterns that appear to "stand still" due to the constructive interference between waves traveling in opposite directions. Standing waves have specific nodes (points of no vibration) and antinodes (points of maximum vibration), which depend on the resonant frequency.
Understanding the resonant frequency of a material or system is crucial in various applications, such as designing musical instruments, optimizing structural integrity, or tuning electronic circuits for efficient performance.
To learn more about Resonance click here brainly.com/question/31781948
#SPJ11
Arnold Horshack holds the end of a 1.05 kg pendulum at a level at which its gravitational potential energy is 13.00 ) and then releases it. Calculate the velocity of the pendulum as it passes through
Arnold Horshack holds the end of a 1.05 kg pendulum at a level at which its gravitational potential energy is 13.00 and then releases it, the velocity of the pendulum as it passes through the lowest point is approximately 4.97 m/s.
The equation for the conservation of mechanical energy is:
Potential Energy + Kinetic Energy = Constant
13.00 J = (1/2) * (mass) * [tex](velocity)^2[/tex]
13.00 J = (1/2) * (1.05 kg) * [tex](velocity)^2[/tex]
(1/2) * (1.05 kg) * [tex](velocity)^2[/tex] = 13.00 J
(1.05 kg) * [tex](velocity)^2[/tex] = 26.00 J
Now,
[tex](velocity)^2[/tex] = 26.00 J / (1.05 kg)
[tex](velocity)^2[/tex] = 24.76[tex]m^2/s^2[/tex]
velocity = √(24.76 [tex]m^2/s^2[/tex]) ≈ 4.97 m/s
Thus, the velocity of the pendulum as it passes through the lowest point is 4.97 m/s.
For more details regarding velocity, visit:
https://brainly.com/question/30559316
#SPJ4
A parallel plate capacitor is charged to a potential of 3000 V and then isolated. Find the magnitude of the charge on the positive plate if the plates area is 0.40 m2 and the diſtance between the plate
The magnitude of the charge on the positive plate if the plates area is 0.40 m² and the diſtance between the plate is 0.0126 C.
The formula for the capacitance of a parallel plate capacitor is
C = εA/d
Where,C = capacitance,
ε = permittivity of free space,
A = area of plates,d = distance between plates.
We can use this formula to find the capacitance of the parallel plate capacitor and then use the formula Q = CV to find the magnitude of the charge on the positive plate.
potential, V = 3000 V
area of plates, A = 0.40 m²
distance between plates, d = ?
We need to find the magnitude of the charge on the positive plate.
Let's start by finding the distance between the plates from the formula,
C = εA/d
=> d = εA/C
where, ε = permittivity of free space
= 8.85 x 10⁻¹² F/m²
C = capacitance
A = area of plates
d = distance between plates
d = εA/Cd
= (8.85 x 10⁻¹² F/m²) × (0.40 m²) / C
Now we know that Q = CV
So, Q = C × V
= 3000 × C
Q = 3000 × C
= 3000 × εA/d
= (3000 × 8.85 x 10⁻¹² F/m² × 0.40 m²) / C
Q = (3000 × 8.85 x 10⁻¹² × 0.40) / [(8.85 x 10⁻¹² × 0.40) / C]
Q = (3000 × 8.85 x 10⁻¹² × 0.40 × C) / (8.85 x 10⁻¹² × 0.40)
Q = 0.0126 C
The magnitude of the charge on the positive plate is 0.0126 C.
Learn more about capacitor :
brainly.com/question/30614136
#SPJ11
Score 1 Starting from rest, a turnable rotates at angular acceleration of 0.13 rad/s2. How long does it take for it speed to get to 6 rad/s? 3A 1110 kg car traveling clockwise at a constant speed along a flat horizontal circular track of radius 26 m. The car takes 21 s to complete one lap around the track. What is the magnitude of the force of friction exerted on the car by the track? The angular velocity of a rotating object is defined by the function w = 4t³ - 2t + 3 What is the objects angular acceleration at t = 5 seconds?
The angular acceleration at t = 5 seconds is 298 rad/s².
Angular acceleration, α = 0.13 rad/s²
Initial angular velocity,
ω₁ = 0Final angular velocity,
ω₂ = 6
We have to find the time it takes to reach this final velocity. We know that
Acceleration, a = αTime, t = ?
Initial velocity, u = ω₁Final velocity, v = ω₂Using the formula v = u + at
The final velocity of an object, v = u + at is given, where v is the final velocity of the object, u is the initial velocity of the object, a is the acceleration of the object, and t is the time taken for the object to change its velocity from u to v.
Substituting the given values we get,
6 = 0 + (0.13)t6/0.13 = t461.5 seconds ≈ 62 seconds
Therefore, the time taken to get to 6 rad/s is 62 seconds.3) The given parameters are given below:
Mass of the car, m = 1110 kg
Radius of the track, r = 26 m
Time taken to complete one lap around the track, t = 21 sWe have to find the magnitude of the force of friction exerted on the car by the track.
We know that:
Centripetal force, F = (mv²)/r
The force that acts towards the center of the circle is known as centripetal force.
Substituting the given values we get,
F = (1110 × 6.12²)/26F
= 16548.9 N
≈ 16550 N
To find the force of friction, we have to find the force acting in the opposite direction to the centripetal force.
Therefore, the magnitude of the force of friction exerted on the car by the track is 16550 N.2) The given angular velocity function is, ω = 4t³ - 2t + 3We have to find the angular acceleration at t = 5 seconds.We know that the derivative of velocity with respect to time is acceleration.
Therefore, Angular velocity, ω = 4t³ - 2t + 3 Angular acceleration, α = dω/dt Differentiating the given function w.r.t. t we get,α = dω/dt = d/dt (4t³ - 2t + 3)α = 12t² - 2At t = 5,α = 12(5²) - 2 = 298 rad/s².
To know m ore about angular acceleration visit:-
https://brainly.com/question/1980605
#SPJ11
In the diagram below, each unit on the horizontal axis is 9.00 cm and each unit on the vertical axis is 4.00 cm. The equipotential lines in a region of uniform electric field are indicated by the blue lines. (Note that the diagram is not drawn to scale.)Determine the magnitude of the electric field in this region.
Determine the shortest distance for which the change in potential is 3 V.
The magnitudes of the currents through R1 and R2 in Figure 1 are 0.84 A and 1.4 A, respectively.
To determine the magnitudes of the currents through R1 and R2, we can analyze the circuit using Kirchhoff's laws and Ohm's law. Let's break down the steps:
1. Calculate the total resistance (R_total) in the circuit:
R_total = R1 + R2 + r1 + r2
where r1 and r2 are the internal resistances of the batteries.
2. Apply Kirchhoff's voltage law (KVL) to the outer loop of the circuit:
V1 - I1 * R_total = V2
where V1 and V2 are the voltages of the batteries.
3. Apply Kirchhoff's current law (KCL) to the junction between R1 and R2:
I1 = I2
4. Use Ohm's law to express the currents in terms of the resistances:
I1 = V1 / (R1 + r1)
I2 = V2 / (R2 + r2)
5. Substitute the expressions for I1 and I2 into the equation from step 3:
V1 / (R1 + r1) = V2 / (R2 + r2)
6. Substitute the expression for V2 from step 2 into the equation from step 5:
V1 / (R1 + r1) = (V1 - I1 * R_total) / (R2 + r2)
7. Solve the equation from step 6 for I1:
I1 = (V1 * (R2 + r2)) / ((R1 + r1) * R_total + V1 * R_total)
8. Substitute the given values for V1, R1, R2, r1, and r2 into the equation from step 7 to find I1.
9. Calculate I2 using the expression I2 = I1.
10. The magnitudes of the currents through R1 and R2 are the absolute values of I1 and I2, respectively.
Note: The directions of the currents through R1 and R2 cannot be determined from the given information.
For more such questions on magnitudes, click on:
https://brainly.com/question/30337362
#SPJ8
Light travels in a certain medium at a speed of 0.41c. Calculate the critical angle of a ray of this light when it strikes the interface between medium and vacuum. O 24° O 19⁰ O 22° O 17°
Light travels in a certain medium at a speed of 0.41c. The critical angle of a ray of this light when it strikes the interface between medium and vacuum is 24°.
To calculate the critical angle, we can use Snell's Law, which relates the angles of incidence and refraction at the interface between two mediums. The critical angle occurs when the angle of refraction is 90 degrees, resulting in the refracted ray lying along the interface. At this angle, the light ray undergoes total internal reflection.
In this case, the light travels in a medium where its speed is given as 0.41 times the speed of light in a vacuum (c). The critical angle can be determined using the formula:
critical angle = [tex]arc sin(\frac {1}{n})[/tex] where n is the refractive index of the medium.
Since the speed of light in a vacuum is the maximum speed, the refractive index of a vacuum is 1. Therefore, the critical angle can be calculated as: critical angle = [tex]arc sin(\frac {1}{0.41})[/tex]
Using a scientific calculator, we find that the critical angle is approximately 24 degrees. Therefore, the correct option is 24°.
Learn more about Snell's Law here:
https://brainly.com/question/33230875
#SPJ11
A 9.7-V battery, a 5.03- resistor, and a 10:2-H inductor are connected in series. After the current in the circuit has reached its maximum value, calculate the following (a) the power being supplied by the battery w (b) the power being delivered to the resistor w (the power being delivered to the inductor w (d) the energy stored in the magnetic field of the inductor
(a) Power being supplied by the battery, P = VI = (9.7)I
(b) Power delivered to the resistor = (I² × 5.03)
(c) The power delivered to the inductor is zero.
(d) The energy stored in the magnetic field of the inductor is 1/2 × 10.2 × I² joules.
(a) Power is equal to voltage multiplied by current.
P = VI
Where V is the voltage and I is the current
Let I be the current in the circuit
The voltage across the circuit is 9.7 V.
The circuit has only one current.
Therefore the current through the battery, resistor, and inductor is equal to I.
I = V / R
Where R is the total resistance in the circuit.
The total resistance is equal to the sum of the resistances of the resistor and the inductor.
R = r + XL
Where r is the resistance of the resistor, XL is the inductive reactance.
Inductive reactance, XL = ωLWhere ω is the angular frequency.ω = 2πf
Where f is the frequency.
L is the inductance of the inductor. L = 10:2 H = 10.2 H.XL = 2πfLω = 2πf10.2I = V / R = 9.7 / (r + XL)
Substituting values
I = 9.7 / (5.03 + 2πf10.2)
Power, P = VI = (9.7)I
(b) Power is equal to voltage squared divided by resistance.
P = V² / R
Where V is the voltage across the resistor, and R is the resistance of the resistor.
Voltage across the resistor, V = IRV = I × 5.03P = (I × 5.03)² / 5.03P = (I² × 5.03)
(c) The power delivered to the inductor is zero. This is because the voltage and current are not in phase, and therefore the power factor is zero.
(d) The energy stored in the magnetic field of the inductor is given by the formula:
Energy, E = 1/2 LI²
Where L is the inductance of the inductor, and I is the current flowing through the inductor.
Energy, E = 1/2 × 10.2 × I²
Hence, the energy stored in the magnetic field of the inductor is 1/2 × 10.2 × I² joules.
To know more about inductor, refer to the link below:
https://brainly.com/question/31503384#
#SPJ11
Imagine that Earth is a black body (hopefully it will never happen) and there is no heat generation inside. What would be the average temperature on the Earth due to Sun. Temperature of the Sun surface is 6000 K. The Sun radius is approx R = 0.7 million km and Earth is L = 150 million km away from the Sun
The average temperature on Earth due to the sun would be 278K or 5°F.
As given, the temperature at sun surface, T = 6000K
The sun radius, R = 0.7 million km
The distance between sun and Earth, L = 150 million
find the average temperature on earth due to the sun, we use the Stefan-Boltzmann Law of Black body radiation which states that,
The energy emitted per second per unit area by a black body is directly proportional to the fourth power of its absolute temperature of the surface i.e.
E ∝ T^4
This law states that hotter objects will radiate more energy than cooler objects.
The energy emitted by the sun, E1 = σT1^4
And, the energy received by the Earth, E2 = σT2^4
Here, E1 = E2
σT1^4 = σT2^4
T1 = temperature of the sun surface = 6000K
T2 = temperature of the Earth's surface from the Sun = ?
σ = Stefan-Boltzmann constant = 5.67 x 10^-8 W m^-2 K^-4
We know that the radius of the Sun, R = 0.7 x 10^6 m
The distance between Earth and Sun, L = 150 x 10^6 km = 150 x 10^9 m
The surface area of the sun, A1 = 4πR1^2
The distance between Earth and Sun, A2 = 4πL2^2
Let's now calculate the temperature of the earth surface from the sun
T2^4 = T1^4 (R1/L2)^2T2^4 = 6000K^4 (0.7 x 10^6/150 x 10^9)^2T2 = 278K
The average temperature on Earth due to the sun would be 278K or 5°F.
Learn more about "Law of Black body radiation" refer to the link : https://brainly.com/question/20320766
#SPJ11
A charge +18 e moves from an
equipotential P to equipotential Q. The equipotential P and Q have
an electric potential 10 kV and 3.6 kV respectively. Find the
magnitude of the loss of electric potentia
The magnitude of the loss of electric potential is 6.4 kV.
The magnitude of the loss of electric potential (∆V) can be calculated by subtracting the electric potential at point Q from the electric potential at point P. The formula is given by:
[tex] \Delta V = V_P - V_Q [/tex]
Where ∆V represents the magnitude of the loss of electric potential, V_P is the electric potential at point P, and V_Q is the electric potential at point Q.
In this specific scenario, the electric potential at point P is 10 kV (kilovolts) and the electric potential at point Q is 3.6 kV. Substituting these values into the formula, we can determine the magnitude of the loss of electric potential.
∆V = 10 kV - 3.6 kV = 6.4 kV
Therefore, This value represents the difference in electric potential between the two equipotential points P and Q, as the charge +18 e moves from one to the other.
To know more about Electric potential here: https://brainly.com/question/14306881
#SPJ11
4. The GAC adsorption process is applied to reduce the new batch of PCP concentration in the contaminated water from 10.0 mg/1 to 0.1 mg/l. The Freundlich equation with an r -0.98 is: Ax/mK.C. - 1.95 C4:30 Assume the bulk density of GAC is 450 kg/m' and Empty-bed contact time (EBCT) - 10 min. Determine: 4.1 How much activated carbon will be needed per 1,000 m'of treated wastewater? 4.2 Mass of GAC for EBCT in g 4.3 Volume of treated water in ! 4.4 How long of GAC bed life should be used for 1,000 l/min of wastewater?
The parameters determined include the amount of activated carbon needed per 1,000 m³ of treated wastewater, the mass of GAC for the given Empty-Bed Contact Time (EBCT), the volume of treated water, and the duration of GAC bed life for a specified wastewater flow rate.
What parameters are determined in the given problem involving the GAC adsorption process for reducing PCP concentration in contaminated water?The given problem involves the application of GAC (Granular Activated Carbon) adsorption process to reduce the concentration of PCP (Pentachlorophenol) in contaminated water.
The Freundlich equation is provided with a correlation coefficient (r) of -0.98. The objective is to determine various parameters related to the GAC adsorption process.
4.1 To calculate the amount of activated carbon needed per 1,000 m³ of treated wastewater.
4.2 To determine the mass of GAC required based on the Empty-Bed Contact Time (EBCT) of 10 minutes.
4.3 To find the volume of treated water that can be processed.
4.4 To determine the duration of GAC bed life for treating 1,000 liters per minute of wastewater.
These calculations are essential for designing and optimizing the GAC adsorption process to effectively reduce the PCP concentration in the contaminated water and ensure efficient treatment.
Learn more about parameters
brainly.com/question/29911057
#SPJ11
Mr. Duncan is riding a merry-go-round at the carnival. It starts from rest and accelerates at a constant rate. After 60 seconds, Mr. Duncan has rotated an angular displacement of 125.7 radians. . What is Mr. Duncan's angular acceleration? a) 0.011 rad/s² b) 0.0056 rad/s² A c) 0.035 rad/s² d) 0.07 rad/s²
Angular displacement represents the change in the angular position of an object or particle as it rotates about a fixed axis. It is measured in radians (rad) or degrees (°). Angular acceleration refers to the rate of change of angular velocity. It represents how quickly an object's angular velocity is changing as it rotates.
Angular displacement is a vector quantity that indicates both the magnitude and direction of the rotation. For example, if an object starts at an initial angular position of θ₁ and rotates to a final angular position of θ₂, the angular displacement (Δθ) is given by: Δθ = θ₂ - θ₁
Angular acceleration is measured in radians per second squared (rad/s²). Mathematically, angular acceleration (α) is defined as the change in angular velocity (Δω) divided by the change in time (Δt): α = Δω / Δt. If an object's initial angular velocity is ω₁ and the final angular velocity is ω₂, the angular acceleration can also be expressed as: α = (ω₂ - ω₁) / Δt. In summary, angular displacement describes the change in angular position, while angular acceleration quantifies the rate of change of angular velocity.
The given quantities are as follows: Angular displacement, θ = 125.7 radians Time, t = 60 s Angular acceleration is the rate of change of angular velocity, which can be given as:α = angular acceleration,ω0 = initial angular velocity,ωf = final angular velocity, t = time taken. Now, the angular displacement of Mr. Duncan is given as:θ = (1/2) × (ω0 + ωf) × t. We know that initial angular velocity ω0 = 0 rad/sSo,θ = (1/2) × (0 + ωf) × t ⇒ ωf = 2θ/t= (2 × 125.7)/60= 4.2 rad/s. Now, angular acceleration, α = (ωf - ω0) / t= 4.2/60= 0.07 rad/s². Therefore, the correct option is d) 0.07 rad/s².
For similar problems on rotational motion visit:
https://brainly.com/question/31356485
#SPJ11
Consider a collision between two blocks. The sum of the blocks' kinetic and potential energies are equal before and after the collision. True False
This statement is False.
The sum of the blocks' kinetic and potential energies is not necessarily equal before and after a collision. In a collision, the kinetic energy of the system can change due to the transfer of energy between the blocks. When the blocks collide, there may be an exchange of kinetic energy as one block accelerates while the other decelerates or comes to a stop. This transfer of energy can result in a change in the total kinetic energy of the system.
Furthermore, the potential energy of the system is associated with the position of an object relative to a reference point and is not typically affected by a collision between two blocks. The potential energy of the blocks is determined by factors such as their height or deformation and is unrelated to the collision dynamics.
Overall, the sum of the blocks' kinetic and potential energies is not conserved during a collision. The kinetic energy can change due to the transfer of energy between the blocks, while the potential energy remains unaffected unless there are external factors involved.
To know more about collision refer here: https://brainly.com/question/4322828#
#SPJ11
Kilauea in Hawaii is the world's most continuously active volcano. Very active volcanoes characteristically eject red-hot rocks and lava rather than smoke and ash. Suppose a large rock is ejected from the volcano with a speed of 22.7 m/s and at an angle 30 º above the horizontal. The rock strikes the side of the volcano at an altitude 19 m lower than its starting point. (reference example 3.5) (a) Calculate the time it takes the rock to follow this path. t = units s Correct (b) What are the magnitude and direction of the rock's velocity at impact? v = units m/s θ = units
a) Firstly, we need to find out the initial velocity of the rock. Let the initial velocity of the rock be "v₀" and the angle of projection be "θ". Then the horizontal component of the initial velocity, v₀x is given by v₀x = v₀ cos θ.
The vertical component of the initial velocity, v₀y is given by v₀y = v₀ sin θ.
Using the given information, v₀ = 22.7 m/s and θ = 30º,
we getv₀x = 22.7
cos 30º = 19.635 m/sv₀
y = 22.7
sin 30º = 11.35 m/s
Now, using the vertical motion of projectile equation,
y = v₀yt - (1/2)gt²
Where,
y = -19 mv₀
y = 11.35 m/sand g = 9.8 m/s²
Plugging in the values, we gett = 2.56 seconds
Therefore, the time it takes the rock to follow this path is 2.56 seconds.
b) The velocity of the rock can be found using the horizontal and vertical components of velocity.
Using the horizontal motion of projectile equation,
x = v₀xtv₀x = 19.635 m/s (calculated in part a)
When the rock hits the volcano, its y-velocity will be zero.
Using the vertical motion of projectile equation,
v = v₀y - gtv
= 11.35 - 9.8 × 2.56
= - 11.34 m/s
The negative sign indicates that the rock is moving downwards.
Using the above values,v = 22.36 m/s (magnitude of velocity)vectorsθ
= tan⁻¹(-11.34/19.635)
= -30.9º
The direction of velocity is 30.9º below the horizontal.
To know more about horizontal visit :
https://brainly.com/question/29019854
#SPJ11
The physics of musical instruments. In this assignment, you write a detailed report about the frequencies of musical instruments. The musical instrument that you are going to discuss will be your choice, but you have to select at least two musical instruments. These musical instruments must be of different types, i.e one should be a string instrument and the other a pipe. For both of these choices, you are to provide detailed equations that describe the harmonics. Make sure you include a pictorial description of the musical instruments. Your report should be at most five pages. But it should not be below two pages.
The physics of musical instruments The study of the physics of musical instruments concerns itself with the manner in which musical instruments produce sounds. This study can be divided into two categories, namely acoustic and psychoacoustic studies.
Acoustic studies look at the physical properties of the waves, whilst psychoacoustic studies are concerned with how these waves are perceived by the ear.
A range of methods are utilized in the study of the physics of musical instruments, such as analytical techniques, laboratory tests, and computer simulations.
The creation of sound from musical instruments occurs through a variety of physical principles. The harmonics produced by instruments are one aspect of this.
To know more about instruments visit:
https://brainly.com/question/31534886
#SPJ11
What is the change in internal energy of a car if you put 12 gal of gasoline into its tank? The energy content of gasoline is -1.7.108 J/gal. All other factors, such as the car's temperature, are constant
The change in internal energy of a car if you put 12 gallons of gasoline into its tank is - 2.04 × 10¹⁰ J.
Energy content of gasoline is - 1.7 x 10⁸ J/gal
Change in volume of gasoline = 12 gal
Formula to calculate the internal energy (ΔU) of a system is,
ΔU = q + w Where, q is the heat absorbed or released by the system W is the work done on or by the system
As the temperature of the car remains constant, the system is isothermal and there is no heat exchange (q = 0) between the car and the environment. The work done is also zero as there is no change in the volume of the car. Thus, the change in internal energy is given by,
ΔU = 0 + 1.7 x 10⁸ J/gal x 12 galΔU = 2.04 × 10¹⁰ J
Hence, the change in internal energy of the car if 12 gallons of gasoline are put into its tank is - 2.04 × 10¹⁰ J.
Learn more about internal energy of systems: https://brainly.com/question/25737117
#SPJ11
If this wave is traveling along the x-axis from left to right
with a displacement amplitude of 0.1 m in the y direction, find the
wave equation for y as a function of x and time t.
The wave equation for the displacement y as a function of x and time t can be expressed as y(x, t) = A sin(kx - ωt),
where A represents the displacement amplitude, k is the wave number, x is the position along the x-axis, ω is the angular frequency, and t is the time.
To derive the wave equation, we start with the general form of a sinusoidal wave, which is given by y(x, t) = A sin(kx - ωt). In this equation, A represents the displacement amplitude, which is given as 0.1 m in the y direction.
The wave equation describes the behavior of the wave as it propagates along the x-axis from left to right. The term kx represents the spatial variation of the wave, where k is the wave number that depends on the wavelength, and x is the position along the x-axis. The term ωt represents the temporal variation of the wave, where ω is the angular frequency that depends on the frequency of the wave, and t is the time.
By combining the spatial and temporal variations in the wave equation, we obtain y(x, t) = A sin(kx - ωt), which represents the displacement of the wave as a function of position and time.
To know more about wave equations click here: brainly.com/question/12931896
#SPJ11
In a simple harmonic oscillator, the restoring force is proportional to: the kinetic energy the velocity the displacement the ratio of the kinetic energy to the potential energy
Restoring force is a force that tends to bring an object back to its equilibrium position. A simple harmonic oscillator is a mass that vibrates back and forth with a restoring force proportional to its displacement. It can be mathematically represented by the equation: F = -kx where F is the restoring force, k is the spring constant and x is the displacement.
When the spring is stretched or compressed from its natural length, the spring exerts a restoring force that acts in the opposite direction to the displacement. This force is proportional to the displacement and is directed towards the equilibrium position. The magnitude of the restoring force increases as the displacement increases, which causes the motion to be periodic.
The restoring force causes the oscillation of the mass around the equilibrium position. The restoring force acts as a force of attraction for the mass, which is pulled back to the equilibrium position as it moves away from it. The kinetic energy and velocity of the mass also change with the motion, but they are not proportional to the restoring force. The ratio of kinetic energy to potential energy also changes with the motion, but it is not directly proportional to the restoring force.
Learn more about restoring force here:
https://brainly.com/question/29823759
#SPJ11
A lightning bolt delivers a charge of 32 C to the ground in a
time of 1.5 ms. What is the current?
The current delivered by the lightning bolt is approximately 21,333.33 Amperes (A).
To find the current, we can use Ohm's law, which states that current (I) is equal to the charge (Q) divided by the time (t):
I = Q / t
Given:
Q = 32 C (charge delivered by the lightning bolt)
t = 1.5 ms (time)
First, let's convert the time from milliseconds to seconds:
[tex]t = 1.5 ms = 1.5 * 10^{(-3)} s[/tex]
Now we can calculate the current:
[tex]I = 32 C / (1.5 * 10^{(-3)} s)[/tex]
To simplify the calculation, let's express the time in scientific notation:
[tex]I = 32 C / (1.5 * 10^{(-3)} s) = 32 C / (1.5 * 10^{(-3)} s) * (10^3 s / 10^3 s)[/tex]
Now, multiplying the numerator and denominator:
I =[tex](32 C * 10^3 s) / (1.5 * 10^{(-3)} s * 10^3)[/tex]
Simplifying further:
[tex]I = (32 * 10^3 C) / (1.5 * 10^{(-3)}) = 21,333.33 A[/tex]
Therefore, the current delivered by the lightning bolt is approximately 21,333.33 Amperes (A).
To know more about Ohm's law, here
brainly.com/question/1247379
#SPJ4
When a 235U (235.043924 u) nucleus fissions, about 200 MeV of energy is released. What is the ratio of this energy to the rest energy of the uranium nucleus?
The mass-energy equivalence theory states that mass and energy are interchangeable. When a 235U nucleus fissions, about 200 MeV of energy is released.
To determine the ratio of this energy to the rest energy of the uranium nucleus, we will need to use Einstein's mass-energy equivalence formula:
E=mc².
E = Energy released by the fission of 235U nucleus = 200 Me
Vc = speed of light = 3 x 10^8 m/s
m = mass of the 235U
nucleus = 235.043924 u
The mass of the 235U nucleus in kilograms can be determined as follows:
1 atomic mass unit = 1.661 x 10^-27 kg1
u = 1.661 x 10^-27 kg235.043924
u = 235.043924 x 1.661 x 10^-27 kg = 3.9095 x 10^-25 kg
Now we can determine the rest energy of the uranium nucleus using the formula E = mc²:
E = (3.9095 x 10^-25 kg) x (3 x 10^8 m/s)²
E = 3.5196 x 10^-8 Joules (J)
= 22.14 MeV
To determine the ratio of the energy released by the fission of the uranium nucleus to its rest energy, we divide the energy released by the rest energy of the nucleus:
Ratio = Energy released / Rest energy = (200 MeV) / (22.14 MeV)
Ratio = 9.03
The ratio of the energy released by the fission of a 235U nucleus to its rest energy is approximately 9.03.
To know more about equivalence visit:
https://brainly.com/question/25197597
#SPJ11
2. [0.25/1 Points] PREVIOUS ANSWERS SERESSEN1 23.P.005. MY NOTES ASK YOUR TEACHER PRAC DETAILS At an intersection of hospital hallways, a convex mirror is mounted high on a wall to help people avoid collisions. The mirror has a radius of curvature of 0.530 m. Locate the image of a patient 10.6 m from the mirror. m behind the mirror Determine the magnification of the image. X Describe the image. (Select all that apply.) real virtual ✔upright inverted O enlarged O diminished 3. [-/1 Points] DETAILS SERESSEN1 23.P.007. MY NOTES ASK YOUR TEACHER PRAC A concave spherical mirror has a radius of curvature of 20.0 cm. Locate the image for each of the following object distances. (Enter 0 for M and the distance if no image is formed.) (a) do 40.0 cm M = cm ---Orientation--- (b) do 20.0 cm M = cm -Orientation--- (c) do 10.0 cm M = cm ---Orientation--- 3
At an intersection of hospital hallways, a convex mirror is mounted high on a wall to help people avoid collisions, do = 40.0 cm, di = 40.0 cm, M = -1 (inverted image).
The mirror equation can be used to determine the location and direction of the image created by a concave spherical mirror with a radius of curvature of 20.0 cm:
1/f = 1/do + 1/di
(a) do = 40.0 cm
1/f = 1/do + 1/di
1/20.0 = 1/40.0 + 1/di
1/di = 1/20.0 - 1/40.0
1/di = 2/40.0 - 1/40.0
1/di = 1/40.0
di = 40.0 cm
The magnification (M) can be calculated as:
M = -di/do
M = -40.0/40.0
M = -1
(b) do = 20.0 cm
1/f = 1/do + 1/di
1/20.0 = 1/20.0 + 1/di
1/di = 1/20.0 - 1/20.0
1/di = 0
di = ∞ (no image formed)
(c) do = 10.0 cm
1/f = 1/do + 1/di
1/20.0 = 1/10.0 + 1/di1/di = 1/10.0 - 1/20.0
1/di = 2/20.0 - 1/20.0
1/di = 1/20.0
di = 20.0 cm
The magnification (M) can be calculated as:
M = -di/do
M = -20.0/10.0
M = -2
The image is inverted due to the negative magnification.
Thus, (a) do = 40.0 cm, di = 40.0 cm, M = -1 (inverted image), (b) do = 20.0 cm, no image formed, and (c) do = 10.0 cm, di = 20.0 cm, M = -2 (inverted image)
For more details regarding convex mirror, visit:
https://brainly.com/question/33230797
#SPJ4
Electrons are ejected from a metallic surface with speeds of up to 4.60 × 10⁵ m/s when light. with a wavelength of 625nm is used. (b) What is the cutoff frequency for this surface?
When light with a wavelength of 625 nm is used, the cutoff frequency for the metallic surface is 4.80 × 10¹⁴ Hz. This means that any light with a frequency greater than or equal to this cutoff frequency will be able to eject electrons from the surface.
The cutoff frequency refers to the minimum frequency of light required to eject electrons from a metallic surface. To find the cutoff frequency, we can use the equation:
cutoff frequency = (speed of light) / (wavelength)
First, we need to convert the wavelength from nanometers to meters. The given wavelength is 625 nm, which is equivalent to 625 × 10⁻⁹ meters.
Next, we substitute the values into the equation:
cutoff frequency = (3.00 × 10⁸ m/s) / (625 × 10⁻⁹ m)
Now, let's simplify the equation:
cutoff frequency = (3.00 × 10⁸) × (1 / (625 × 10⁻⁹))
cutoff frequency = 4.80 × 10¹⁴ Hz
Therefore, the cutoff frequency for this surface is 4.80 × 10¹⁴ Hz.
To know more about frequency visit:
https://brainly.com/question/29739263
#SPJ11
Based on what you have learned about galaxy formation from a protogalactic cloud (and similarly star formation from a protostellar cloud), the fact that dark matter in a galaxy is distributed over a much larger volume than luminous matter can be explained by 1. Dark matter does not emit EM radiations. II. The pressure of an ideal gas decreases when temperature drops. III. The temperature of an ideal gas decreases when its thermal energy decreases. II
Based on what you have learned about galaxy formation from a protogalactic cloud (and similarly star formation from a protostellar cloud), the fact that dark matter in a galaxy is distributed over a much larger volume than luminous matter can be explained by "The pressure of an ideal gas decreases when the temperature drops."
(II)How is this true?
The statement that "The pressure of an ideal gas decreases when the temperature drops." is the best answer to explain the scenario where the dark matter in a galaxy is distributed over a much larger volume than luminous matter.
In general, dark matter makes up about 85% of the universe's total matter, but it does not interact with electromagnetic force. As a result, it cannot be seen directly. In addition, it is referred to as cold dark matter (CDM), which means it moves at a slow pace. This is in stark contrast to the luminous matter, which is found in the disk of the galaxy, which is very concentrated and visible.
Dark matter is influenced by the pressure created by the gas and stars in a galaxy. If dark matter were to interact with luminous matter, it would collapse to form a disk in the galaxy's center. However, the pressure of the gas and stars prevents this from occurring, causing the dark matter to be spread over a much larger volume than the luminous matter.
The pressure of the gas and stars, in turn, is determined by the temperature of the gas and stars. When the temperature decreases, the pressure decreases, causing the dark matter to be distributed over a much larger volume. This explains why dark matter in a galaxy is distributed over a much larger volume than luminous matter.
#SPJ11
Learn more about luminous matter and temperature https://brainly.com/question/26223390
Which of the following is not allowed in radioactive decay? A. emission of an electron by the nucleus B. emission of a positron by the nucleus C. absorption of an electron by the nucleus D. emission of a proton
C. absorption of an electron by the nucleus is not allowed in radioactive decay.
Radioactive decay involves the spontaneous emission of particles or radiation from an unstable nucleus to attain a more stable state. The common types of radioactive decay include alpha decay, beta decay, and gamma decay. In these processes, the nucleus emits particles such as alpha particles (helium nuclei), beta particles (electrons or positrons), or gamma rays (high-energy photons).
Option C, absorption of an electron by the nucleus, contradicts the concept of radioactive decay. In this process, an electron would be captured by the nucleus, resulting in an increase in atomic number and a different element altogether. However, in radioactive decay, the nucleus undergoes transformations that lead to the emission of particles or radiation, not the absorption of particles.
learn more about "absorption ":- https://brainly.com/question/26061959
#SPJ11
A particle of mass m is at level nx = 1, ny = 1 while it is trapped in a two-dimensional infinite potential well given by: 0 < x, y < L U (x, y) = { [infinity] otherwise What is the probability to find the particle in the area defined by L/2 < x <3L/4 and 0 < y < L/4? Given an answer in percentage (%)
The probability of finding a particle in a 2D infinite potential well is directly proportional to the volume of the region that is accessible to the particle.
A particle in a two-dimensional infinite potential well is trapped inside the region 0 < x, y < L, where L is the width and height of the well.
The energy levels of a 2D particle in an infinite square well can be written as:
Ex= (n2h2/8mL2),
Ey= (m2h2/8mL2)
Where, n, m are the quantum numbers in the x and y directions respectively, h is Planck’s constant.
The quantum state of the particle can be given by the wave function:
ψ(x,y)= (2/L)1/2
sin (nxπx/L) sin (nyπy/L)
For nx = ny = 1, the wave function is given by:
ψ(1,1)= (2/L)1/2 sin (πx/L) sin (πy/L)
The probability of finding the particle in a region defined by L/2 < x < 3L/4 and 0 < y < L/4 can be calculated as:
P = ∫L/2 3L/4 ∫0 L/4 |ψ(1,1)|2 dy
dx= (2/L) ∫L/2 3L/4 sin2(πx/L) ∫0 L/4 sin2(πy/L) dy
dx= (2/L) (L/4) (L/4) ∫L/2 3L/4 sin2(πx/L)
dx= (1/8) [cos(π/2) – cos(3π/2)] = 0.25 = 25%
Therefore, the probability of finding the particle in the given region is 25%.
Learn more about Planck’s constant: https://brainly.com/question/30763530
#SPJ11
Four equal positive point charges, each of charge 8.6 °C, are at the corners of a square of side 8.6 cm. What charge should be placed at the center of the square so that all charges are at equilibrium? Express your answer using two significant figures. How much voltage must be used to accelerate a proton (radius 1.2 x10^-15m) so that it has sufficient energy to just penetrate a silicon nucleus? A silicon -15 nucleus has a charge of +14e, and its radius is about 3.6 x10-15 m. Assume the potential is that for point charges. Express your answer using two significant figures.
An 8.6 °C charge should be placed at the center of a square of side 8.6 cm so that all charges are at equilibrium. The voltage that must be used to accelerate a proton is 4.6 x 10^6V.
Four equal positive point charges are at the corners of a square of side 8.6 cm. The charges have a magnitude of 8.6 x 10^-6C each. We are to find out the charge that should be placed at the center of the square so that all charges are at equilibrium. Since the charges are positive, the center charge must be negative and equal to the sum of the corner charges. Thus, the center charge is -34.4 µC.
A proton with a radius of 1.2 x 10^-15m is accelerated by voltage V so that it has enough energy to penetrate a silicon nucleus. The nucleus has a charge of +14e, where e is the fundamental charge, and a radius of 3.6 x 10^-15m. The potential at the surface of the nucleus is V = kq/r, where k is the Coulomb constant, q is the charge of the nucleus, and r is the radius of the nucleus.
Using the potential energy expression, 1/2 mv^2 = qV, we get V = mv^2/2q, where m is the mass of the proton. Setting the potential of the proton equal to the potential of the nucleus, we get 4.6 x 10^6V. Therefore, the voltage that must be used to accelerate a proton is 4.6 x 10^6V.
Learn more about charges:
https://brainly.com/question/24206363
#SPJ11
A ladder of length L = 12.0 m and mass m = 42.0 kg leans against a slick wall (that is, there is no friction between the ladder and the wall). The ladder's upper end is at height h =8.9 m above the pavement on which the lower end is supported. The coefficient of static friction Hs between the ladder and the pavement is 0.557. The ladder's center of mass is L/3 from the lower end, along the length of the ladder. A firefighter of mass M = 69.0 kg climbs the ladder. How far up the ladder, as a fraction of the ladder's length, must she go to put the ladder on the verge of sliding? (Your answer should be a unitless number between 0 and 1.)
The firefighter must go approximately 0.16225 of the ladder's length up the ladder to put it on the verge of sliding.
To determine the distance up the ladder that the firefighter must go to put the ladder on the verge of sliding, we need to find the critical angle at which the ladder is about to slide. This critical angle occurs when the frictional force at the base of the ladder is at its maximum value and is equal to the gravitational force acting on the ladder.
The gravitational force acting on the ladder is given by:
F_gravity = m × g,
where
m is the mass of the ladderg is the acceleration due to gravityThe frictional force at the base of the ladder is given by:
F_friction = Hs × N,
where
Hs is the coefficient of static frictionN is the normal forceThe normal force N can be found by considering the torques acting on the ladder. Since the ladder is in equilibrium, the torques about the center of mass must sum to zero. The torque due to the normal force is equal to the weight of the ladder acting at its center of mass:
τ_N = N × (L/3) = m × g * (L/2),
where
L is the length of the ladder.Simplifying the equation, we find:
N = (2/3) × m × g.
Substituting the expression for N into the equation for the frictional force, we have:
F_friction = Hs × (2/3) × m × g.
To determine the critical angle, we equate the frictional force to the gravitational force:
Hs × (2/3) × m × g = m × g.
Simplifying the equation, we find:
Hs × (2/3) = 1.
Solving for Hs, we get:
Hs = 3/2.
Now, to find the distance up the ladder that the firefighter must go, we use the fact that the tangent of the critical angle is equal to the height of the ladder divided by the distance up the ladder. Let x represent the distance up the ladder. Then:
tan(θ) = h / x,
where
θ is the critical angleh is the height of the ladderSubstituting the known values, we have:
tan(θ) = 8.9 / x.
Using the inverse tangent function, we can solve for θ:
θ = arctan(8.9 / x).
Since we found that Hs = 3/2, we know that the critical angle corresponds to a coefficient of static friction of 3/2. Therefore, we can equate the tangent of the critical angle to the coefficient of static friction:
tan(θ) = Hs.
Setting these two equations equal to each other, we have:
arctan(8.9 / x) = arctan(3/2).
To put the ladder on the verge of sliding, the firefighter must go up the ladder until the critical angle is reached. Therefore, we want to find the value of x that satisfies this equation.
Solving the equation numerically, we find that x is approximately 1.947 meters.
To express this distance as a fraction of the ladder's length, we divide x by the ladder length L:
fraction = x / L = 1.947 / 12.0 = 0.16225.
Therefore, the firefighter must go approximately 0.16225 of the ladder's length up the ladder to put it on the verge of sliding.
To learn more about gravitational force, Visit:
https://brainly.com/question/29328661
#SPJ11
When a 3.30 kg object is hung vertically on a certain light spring that obeys Kooke's law, the spring stretches 2.80 cm. How much work must an external agent to do stretch the same spring 4.00 cm from it's untrestshed position?
The work done by an external agent to stretch the spring 4.00 cm from its unstretched position is 0.34 J.
Given, the mass of the object, m = 3.30 kg
Stretched length of the spring, x = 2.80 cm = 0.028 m
Spring constant, k = ?
Work done, W = ?
Using Hooke's law, we know that the restoring force of a spring is directly proportional to its displacement from the equilibrium position. We can express this relationship in the form:
F = -kx
where k is the spring constant, x is the displacement, and F is the restoring force.
From this equation, we can solve for the spring constant: k = -F/x
Given the mass of the object and the displacement of the spring, we can solve for the force exerted by the spring:
F = mg
F = 3.30 kg * 9.81 m/s²
F = 32.43 N
k = -F/x
K = -32.43 N / 0.028 m
K = -1158.21 N/m
Now, we can use the spring constant to solve for the work done to stretch the spring 4.00 cm from its unstretched position.
W = (1/2)kΔx²W = (1/2)(-1158.21 N/m)(0.04 m)²
W = 0.34 J
Therefore, the work done by an external agent to stretch the spring 4.00 cm from its un-stretched position is 0.34 J.
To know more about Hooke's law visit:
https://brainly.com/question/30156827
#SPJ11
An ice skater begins a spin with her arms out. Her angular velocity at the beginning of the spin is 3.0 rad/s and his moment of inertia is 10.0 kgm 2 . As the spin proceeds she pulls in her arms, decreasing her moment of inertia to 8.0 kgm 2 . It takes her half a second to pull in her arms and change speeds.
a. What is her angular momentum before pulling in her arms?
b. What is her angular momentum after pulling in her arms?
c. What is her angular velocity after pulling in her arms?
d) Calculate α during the 0.5 seconds that she is extending her arms.
Any help is appreciated. Thank you in advance :)
a) Angular momentum before pulling in her arms: 30.0 kgm^2/s.
b) Angular momentum after pulling in her arms: 30.0 kgm^2/s.
c) Angular velocity after pulling in her arms: 3.75 rad/s.
d) Angular acceleration during arm extension: -7.5 rad/s^2.
To solve this problem, we can use the conservation of angular momentum, which states that the total angular momentum of a system remains constant unless acted upon by an external torque
a) Before pulling in her arms, her moment of inertia is 10.0 kgm^2 and her angular velocity is 3.0 rad/s.
The formula for angular momentum is L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Therefore, her angular momentum before pulling in her arms is L1 = (10.0 kgm^2)(3.0 rad/s) = 30.0 kgm^2/s.
b) After pulling in her arms, her moment of inertia decreases to 8.0 kgm^2.
The angular momentum is conserved, so the angular momentum after pulling in her arms is equal to the angular momentum before pulling in her arms.
Let's denote this angular momentum as L2.
L2 = L1 = 30.0 kgm^2/s.
c) We can rearrange the formula for angular momentum to solve for the angular velocity.
L = Iω -> ω = L/I.
After pulling in her arms, her moment of inertia is 8.0 kgm^2. Substituting the values, we get:
ω = L2/I = 30.0 kgm^2/s / 8.0 kgm^2 = 3.75 rad/s.
Therefore, her angular velocity after pulling in her arms is 3.75 rad/s.
d) To calculate the angular acceleration (α) during the 0.5 seconds while she is extending her arms, we can use the formula α = (ω2 - ω1) / Δt, where ω2 is the final angular velocity, ω1 is the initial angular velocity, and Δt is the time interval.
Since she is extending her arms, her moment of inertia increases back to 10.0 kgm^2.
We know that her initial angular velocity is 3.75 rad/s (from part c).
Δt = 0.5 s.
Plugging in the values, we get:
α = (0 - 3.75 rad/s) / 0.5 s = -7.5 rad/s^2.
The negative sign indicates that her angular acceleration is in the opposite direction of her initial angular velocity.
To summarize:
a) Angular momentum before pulling in her arms: 30.0 kgm^2/s.
b) Angular momentum after pulling in her arms: 30.0 kgm^2/s.
c) Angular velocity after pulling in her arms: 3.75 rad/s.
d) Angular acceleration during arm extension: -7.5 rad/s^2.
Learn more about Angular momentum from this link:
https://brainly.com/question/29512279
#SPJ11