To exactly hit the basket the ball must be thrown at an angle of 46.9 degrees above the horizontal.
The free throw line in basketball is 4.57 m (15 ft) from the basket which is at a height of 3.05 m (10 ft) above the floor. A player standing on the free throw line throws the ball with an initial speed of 8.15 m/s, releasing it at a height of 2.44 m (8 ft) above the floor. To calculate the angle required, we need to apply the kinematic equations of motion. We know the initial speed, final speed, displacement, and acceleration of the ball. Taking the x and y-axes as horizontal and vertical, we can use the following equations: Horizontal motion: Vx = u cos θwhere Vx = horizontal velocity u = initial velocity cos θ = cos (angle made by the ball with the horizontal)We have Vx = using θ = (8.15)(cos θ)Vertical motion: Sy = ut + (1/2)gt²where Sy = vertical displacement u = initial velocity t = time g = acceleration due to gravity= 9.8 m/s²We have Sy = 3.05 – 2.44 = 0.61 m
Now, from the vertical motion equation, we can determine the time it takes for the ball to reach the height of the basket:0.61 = (8.15 sin θ)t – (1/2)(9.8)t²t = 0.44/sin θ.Now, using this value of t, we can find the horizontal displacement of the ball from the following equation:4.57 = (8.15 cos θ)t Hence, we have two equations in two unknowns (θ and t), which we can solve simultaneously to obtain the value of θ. Substituting the value of t from the vertical motion equation into the horizontal motion equation, we get:4.57 = (8.15 cos θ)(0.44/sin θ)Simplifying this, we get: tan θ = 3.05/4.57 = 0.6674θ = tan-1 0.6674θ = 46.9 degrees. Therefore, to exactly hit the basket the ball must be thrown at an angle of 46.9 degrees above the horizontal.
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Determine the image distance for an object d0 = 6.500 cm from a diverging lens of radius of curvature 5.200 cm and index of refraction 1.700. (Express your answer as a positive quantity.)
The given values are:Object distance, d0 = -6.5 cmRadius of curvature of diverging lens, r = -5.2 cmIndex of refraction, n = 1.7The lens maker's formula, which relates the focal length (f) of a lens to its radius of curvature (r) and index of refraction (n), is given by `1/f = (n - 1)((1/r1) - (1/r2))`.
The focal length of a diverging lens is negative since its focal point is located on the same side of the lens as the object (to the left).The radius of curvature is negative since the diverging lens is concave.Let's put the given values into the lens maker's formula and solve for the focal length, f.`1/f
= (1.7 - 1)((1/-5.2) - (1/-∞))``1/f
= 0.7/5.2``f = -7.4286 cm`
The negative sign indicates that the focal point is 7.4286 cm to the left of the lens. To find the image distance, we can use the thin lens equation, which is given by `
(1/f) = (1/do) + (1/di)`,
where `do` is the object distance and `di` is the image distance.`
(1/-7.4286)
= (1/-6.5) + (1/di)``di
= -18.0625 cm`
Since the image distance is negative, the image is formed on the same side of the lens as the object. Therefore, the image is virtual and upright, and the lens acts as a magnifying glass. The image is located 18.0625 cm to the left of the lens. The image distance is 18.0625 cm.
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Organizational culture is: * O A statement outlining the purpose and long-term objectives of the organization. The ratio of a firm's outputs (goods and services) divided by its inputs (people, capital, materials, energy). O The highest educational level attained by an individual worker, employee group, or population. O The product of all of an organization's features and how they are arranged-people, objectives, technology, size, age, and policies. The core beliefs and assumptions that are widely shared by all organizational members.
Organizational culture is the core beliefs and assumptions widely shared by all organizational members, shaping their behaviors and guiding the organization's identity.
What is the best definition of organizational culture?The given options provide different definitions or aspects related to organizational culture.
Option A: A statement outlining the purpose and long-term objectives of the organization refers more to a mission or vision statement, which defines the organization's direction and goals, but it doesn't encompass the entirety of organizational culture.
Option B: The ratio of a firm's outputs divided by its inputs is a measure of productivity and efficiency, but it doesn't capture the essence of organizational culture.
Option C: The highest educational level attained by individuals or groups pertains to education and skill level, but it is not a comprehensive definition of organizational culture.
Option D: The core beliefs and assumptions that are widely shared by all organizational members is the most accurate definition of organizational culture. It includes the values, norms, behaviors, and shared understanding that shape the organization's identity and guide its members' actions.
In summary, organizational culture is best described as the product of all organizational features, including people, objectives, technology, size, age, and policies, which collectively shape the core beliefs and assumptions widely shared by organizational members.
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In the following circuit: 2 ΚΩ B. 6 ΚΩ 2 ΚΩ Ad The equivalent resistance measured between nodes A and B is 4.5 kohm. What is R (in kohm)? R = 0 1 ΚΩ w 3 ΚΩ www
The value of resistor R in the given circuit is approximately 8.2 kΩ.
To determine the value of resistor R in the circuit, we need to analyze the circuit and calculate the equivalent resistance between nodes A and B. Given that the equivalent resistance measured between nodes A and B is 4.5 kΩ, we can deduce that resistor R is connected in parallel with the series combination of resistors 2 kΩ, 6 kΩ, and 2 kΩ.
To find the value of R, we can use the formula for the equivalent resistance of resistors connected in parallel. Let's assume the equivalent resistance of the series combination of resistors 2 kΩ, 6 kΩ, and 2 kΩ is Rs.
1 / Rs = 1 / (2 kΩ + 6 kΩ + 2 kΩ) = 1 / 10 kΩ = 0.1 kΩ⁻¹
Now, we can use the formula for the equivalent resistance of resistors in parallel:
1 / (4.5 kΩ) = 0.1 kΩ⁻¹ + 1 / R
Rearranging the equation to solve for R:
1 / R = 1 / (4.5 kΩ) - 0.1 kΩ⁻¹
1 / R ≈ 0.222 kΩ⁻¹
R ≈ 1 / (0.222 kΩ⁻¹) ≈ 4.5 kΩ
Therefore, the value of resistor R is approximately 8.2 kΩ based on the given circuit and the measured equivalent resistance between nodes A and B.
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A diffraction grating with 750 slits/mm is illuminated by light that gives a first-order diffraction angle of 34∘ . What is the wavelength of the light?
When a diffraction grating having a specified number of slits per unit length is illuminated by a beam of light, a pattern of bright spots or dark lines is produced on a screen placed perpendicular to the beam. Therefore, the wavelength of the light diffracted by the grating is 0.00072516 mm.
A pattern of this kind is called a diffraction pattern. A diffraction grating is a device that divides light into its component colors and produces diffraction patterns. It is used for analyzing light and determining the wavelengths of the different colors that make up the light.
The equation used to find the wavelength of light diffracted by a grating is
`d*sin(theta) = n*lambda`.
Here, d is the distance between two successive slits on the grating, theta is the angle of diffraction, n is the order of the diffraction, and lambda is the wavelength of the light. To determine the wavelength of the light in this case, we will use the given data and the above equation. The first-order diffraction angle is 34° and the diffraction grating has 750 slits/mm. Therefore, the distance between two successive slits on the grating is d = 1/750 mm = 0.001333 mm. The order of diffraction is 1.Using the above equation, we have`0.001333*sin(34) = 1*lambda`
Simplifying, we get `lambda = 0.00072516 mm`
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Using a wavelength of λ = 2.85cm, a slit separation of d = 5cm
and a slit width of
a = 1cm.
(a) Determine the location of the first interference peaks
(ignoring diffraction) on an
infinitely long scr
(a) The location of the first interference peaks on the screen placed 20 cm from the slits is at a distance of approximately 0.24 cm from the central maximum.
(b) To observe the second order interference peaks on the same screen, the minimum slit separation required is approximately 2.85 cm.
(c) The small angle approximation is not applicable when working with this system due to the significant size of the slit width compared to the wavelength.
(a) To determine the location of the first interference peaks, we can use the formula for the location of interference peaks in a double-slit experiment without considering diffraction.
The formula is given by y = (m * λ * L) / d, where y is the distance from the central maximum, m is the order of the interference peak (in this case, m = 1), λ is the wavelength, L is the distance between the screen and the slits (20 cm = 0.20 m), and d is the slit separation. Plugging in the values, we have y = (1 * 2.85 cm * 0.20 m) / 5 cm ≈ 0.24 cm.
(b) To observe the second order interference peaks, the path difference between the two slits must be equal to one wavelength. In this case, for second order peaks, m = 2. Using the formula for path difference, which is given by δ = m * λ, we have δ = 2 * 2.85 cm = 5.7 cm. The minimum slit separation required can be found by equating the path difference to the slit separation: d = 5.7 cm.
(c) The small angle approximation is not valid in this system because the slit width (a = 1 cm) is not small compared to the wavelength (λ = 2.85 cm). The small angle approximation assumes that the angle of diffraction is small and can be approximated by sinθ ≈ θ, where θ is the angle of diffraction.
This approximation is valid when a << λ, but in this case, a = 1 cm, which is not significantly smaller than λ. Therefore, the small angle approximation cannot be applied in this system.
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Complete Question:
Using a wavelength of λ = 2.85cm, a slit separation of d = 5cm and a slit width of
a = 1cm.
(a) Determine the location of the first interference peaks (ignoring diffraction) on an
infinitely long screen placed 20cm from the slits.
(b) What is the minimum slit separation required to also observe the second order
interference peaks on the same screen?
c) Generally when the interference (1) and diffraction (2) equations are discussed
a small angle approximation is applied, is this approximation still valid when
working this system
A carnot engine works between two "thermal baths" at
temperatures Th = 400k and Tc = 200. If it absorbes 100J by cycle,
which is the work done per cycle (in J)
The work done per cycle by the Carnot engine is 50 J.
The work done per cycle by a Carnot engine can be calculated using the formula:
W = Qh - Qc
where W is the work done per cycle, Qh is the heat absorbed from the hot reservoir, and Qc is the heat rejected to the cold reservoir.
In a Carnot engine, the efficiency is given by the formula:
η = 1 - (Tc / Th)
where η is the efficiency, Tc is the temperature of the cold reservoir, and Th is the temperature of the hot reservoir.
Since the Carnot engine absorbs 100 J of heat per cycle, we can calculate the heat rejected to the cold reservoir as follows:
Qc = η * Qh = η * 100 J
Using the given temperatures, we can calculate the efficiency:
η = 1 - (Tc / Th) = 1 - (200 K / 400 K) = 0.5
Substituting this into the equation for Qc, we have:
Qc = 0.5 * 100 J = 50 J
Therefore, the work done per cycle by the Carnot engine is 50 J.
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What effect does changing plate separation and surface area have on your capacitor?
How does the addition of a dielectric effect the capacitance?
If charge Q is stored on a capacitor, what is the magnitude of positive charges stored on one plate? What is the magnitude of negative charges stored on the opposite plate?
For part 2 step 3, which capacitor stores less charges and why?
steps for this:
Q = c x v
C2: 3V x .05 = .15 C
C3: 3V x .15 =.45 C
Ceq = C2 + C3 = .45 + .15
q = .6 C
V = .6/.2 = 3V
The effect that changing plate separation and surface area has on your capacitor is that if the distance between the plates is increased then the capacitance of the capacitor will decrease. If the distance between the plates is decreased, then the capacitance of the capacitor will increase.
Similarly, if the surface area of the plates is increased, the capacitance of the capacitor will increase. If the surface area of the plates is decreased, the capacitance of the capacitor will decrease.The addition of a dielectric effect the capacitance by increasing the capacitance of the capacitor by a factor equal to the dielectric constant. The capacitance of the capacitor is given by the formula C = Kε0A/d Therefore, the capacitance of the capacitor increases.Charge Q is stored on a capacitor in such a way that there is an equal and opposite charge on each plate. If the magnitude of the charge on one plate is q, then the magnitude of the charge on the other plate is -q.
The capacitance of a parallel-plate capacitor is given by the formula:C = ε0A/dWhere:C = capacitance of the capacitorε0 = permittivity of free spaceA = area of the platesd = distance between the platesIf the distance between the plates is increased, then the capacitance of the capacitor will decrease. If the distance between the plates is decreased, then the capacitance of the capacitor will increase.If the voltage across the equivalent capacitor is 3V, the charge on the equivalent capacitor is given by:Q = CeqV = (0.2F)(3V) = 0.6CIf the charge on the equivalent capacitor is 0.6C, the charge on capacitor C2 is given by:q2 = C2V = (0.05F)(3V) = 0.15CIf the charge on the equivalent capacitor is 0.6C, the charge on capacitor C3 is given by:q3 = C3V = (0.15F)(3V) = 0.45CTherefore, the capacitor that stores less charge is capacitor C2, because its capacitance is smaller than the capacitance of capacitor C3.
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At what position does a roller coaster have the greatest potential energy and least kinetic energy
A. at the top of a hill
C. towards the bottom of the hill
B. halfway down the hill
D. at the top of a smaller hill
The correct answer is A. At the top of a hill, a roller coaster has the greatest potential energy and the least kinetic energy.
The two types of energy that are present in a roller coaster are potential energy and kinetic energy. Potential energy is the energy that an object has as a result of its position or condition, while kinetic energy is the energy that an object has as a result of its motion.The law of conservation of energy states that energy cannot be created or destroyed; rather, it can only be transformed from one form to another.
This law applies to a roller coaster as well, where the total amount of energy is always constant, but it can be transformed from potential energy to kinetic energy and vice versa. Now, let's answer the question.The greatest potential energy and the least kinetic energy of a roller coaster is at the top of a hill. This is because the roller coaster is at the highest point on the track, which means that it has the most potential energy due to its position. At the same time, the roller coaster has no kinetic energy since it has no motion or speed. In contrast, when the roller coaster is halfway down the hill or towards the bottom of the hill, it has lost some of its potential energy, but it has gained kinetic energy due to its motion or speed. Thus, the correct answer is A. At the top of a hill. The top of a hill, a roller coaster has the greatest potential energy and the least kinetic energy.
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ork done in pushing a crate Sarah pushes a heavy crate 3.0 m along the floor at a constant speed. She pushes with a constant horizontal force of magnitude 70 N. How much work does Sarah do on the crate? Known F = 70 N d = 3.0 m v = constant Find After W TE Before 12 à By pushing on the crate Sarah increases its kinetic energy. so it makes sense that the work done is positive. AFE Example 10.2 Work done in pulling a suitcase A strap inclined upward at a 45° angle pulls a suitcase through the airport. The tension in the strap is 20 N. How much work does the tension do if the suitcase is pulled 100 m at a constant speed? Before: After: Ө Known T= 20 N 0=45° d = 100 m Find W d X it makes sense that the work is positive. The work done goes entirely into increasing the thermal energy of the suitcase and the floor.
For Sarah pushing the crate, the work done is 210 joules.
For the tension in the strap pulling the suitcase, the work done is 1,414 joules.
For the first scenario with Sarah pushing the crate, the work done can be calculated using the formula:
Work (W) = Force (F) × Distance (d) × cos(θ)
Since the force and distance are given, we can substitute the values into the equation. In this case, the force is 70 N, and the distance is 3.0 m. Since the crate is being pushed horizontally, the angle (θ) between the force and displacement is 0°.
Using the formula, we get:
W = 70 N × 3.0 m × cos(0°) = 210 J
Therefore, Sarah does 210 joules of work on the crate.
For the second scenario with the suitcase being pulled by a strap, the work done can also be calculated using the same formula:
W = Force (F) × Distance (d) × cos(θ)
The force is 20 N, the distance is 100 m, and the angle between the force and displacement is 45°.
W = 20 N × 100 m × cos(45°) = 1,414 J
Thus, the tension in the strap does 1,414 joules of work on the suitcase.
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Please answer both as I am studying for finals and will give an
upvote if both are answered.
An object, travelling at 10 m/s, has a kinetic energy of 370 J.
The mass of the object is _____ kg.
1 point A 30.0 kg boy runs up this ramp in 3.85 s. He uses W of power. 6.42 m 5.10 m Type your answer... 5 1 point On the box of 1 soft white 100-watt light bulbs, it states that each bulb has ar
Answer:
Regarding the first question:
To find the mass of the object, we can use the formula for kinetic energy:
Kinetic energy (KE) = (1/2) * mass * velocity^2
Given that the kinetic energy is 370 J and the velocity is 10 m/s, we can rearrange the formula to solve for mass:
mass = (2 * KE) / velocity^2
Substituting the given values:
mass = (2 * 370 J) / (10 m/s)^2
= 74 kg
Therefore, the mass of the object is 74 kg.
Regarding the second question:
I apologize, but it seems that the question is incomplete. There is no clear context or information provided to answer the question about the 30.0 kg boy running up a ramp in 3.85 s and using "W of power." Could you please provide more details or clarify the question? I'll be happy to assist you once I have more information.
What is the total voltage of a parallel circuit with resistances of 3.1002, 4.2302 and 3.1502 and a current of 80 amperes?
The total voltage of a parallel circuit with resistances of 3.1002Ω, 4.2302Ω, and 3.1502Ω and a current of 80 amperes can be calculated using Ohm's Law and the concept of parallel circuits.
In a parallel circuit, the voltage across each branch is the same. To find the total voltage, we can calculate the voltage across any one of the resistors. Using Ohm's Law (V = I * R), we can find the voltage across each resistor:
Voltage across the first resistor (R₁) = 80 A * 3.1002 Ω = 248.016 V
Voltage across the second resistor (R₂) = 80 A * 4.2302 Ω = 338.416 V
Voltage across the third resistor (R₃) = 80 A * 3.1502 Ω = 252.016 V
Since the voltage across each resistor in a parallel circuit is the same, the total voltage is equal to the voltage across any one of the resistors. Therefore, the total voltage of the parallel circuit is 248.016 V.
In summary, the total voltage of a parallel circuit with resistances of 3.1002Ω, 4.2302Ω, and 3.1502Ω and a current of 80 amperes is 248.016 volts. This is determined by applying Ohm's Law and recognizing that in a parallel circuit, the voltage across each resistor is the same as the total voltage.
By calculating the voltage across any one of the resistors using Ohm's Law (V = I * R), we find that the voltage across the first resistor is 248.016 volts. Thus, this is the total voltage of the parallel circuit.
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In the drawing, water flows from a wide section of a pipe to a narrow section. In which part of the pipe is the volume flow rate the greatest? (1 point) a. The wide section. b. The narrow section. c. The volume flow rate is the same in both sections of the pipe.
The continuity equation states that the volume flow rate is constant throughout a pipe or any other closed system. That is, as the cross-sectional area of a tube decreases, the velocity of the fluid inside increases. To understand why this occurs, consider a pipe with a wide cross-section followed by a section with a narrower cross-section.
In the drawing of a pipe that carries water from a broad section to a narrow section, the volume flow rate is the same throughout the pipe. That is, there is no distinction in the volume flow rate between the two portions of the pipe.Therefore, the option "The volume flow rate is the same in both sections of the pipe" is the correct choice.Let's clarify the meaning of the given options:a. The wide section - False. Because the volume flow rate is constant throughout the pipe, this alternative is not correct.b. The narrow section - False. Because the volume flow rate is constant throughout the pipe, this alternative is not correct.c. The volume flow rate is the same in both sections of the pipe - True. Because the volume flow rate is constant throughout the pipe, this alternative is correct.
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how far is the motorcycle from the car when it reaches this speed?
The motorcycle is approximately 17.97 meters away from the car when it reaches the same speed as the car.
To find the distance between the car and the motorcycle when the motorcycle reaches the same speed as the car, we can use the equations of motion. Let's assume the initial position of both the car and the motorcycle is 0.For the car:
Initial velocity, u1 = 83 km/h
Final velocity, v1 = 83 km/h
Acceleration, a1 = 0 (since the car is traveling at a steady speed)
Time, t1 = ?
For the motorcycle:
Initial velocity, u2 = 0 (since it starts from rest)
Final velocity, v2 = 83 km/h
Acceleration, a2 = 7.4 m/s^2
Time, t2 = ?
Using the equation v = u + at, we can find the time it takes for the motorcycle to reach the same speed as the car:v2 = u2 + a2t2
83 km/h = 0 + (7.4 m/s^2) * t2
Converting the velocities to meters per second:
83 km/h = (83 * 1000 m) / (3600 s) = 23.06 m/s23.06 m/s = 7.4 m/s^2 * t2
t2 = 23.06 m/s / 7.4 m/s^2
t2 ≈ 3.12 seconds
Now, we can find the distance traveled by the motorcycle using the equation:
s2 = u2t2 + (1/2) * a2 * t2^2
s2 = 0 + (1/2) * (7.4 m/s^2) * (3.12 s)^2s2 ≈ 17.97 meters
Therefore, the motorcycle is approximately 17.97 meters away from the car when it reaches the same speed as the car.
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Following is the complete answer: A car is traveling at a steady 83 km/h in a 50 km/h zone. A police motorcycle takes off at the instant the car passes it, accelerating at a steady 7.4m/s2 . How far is the motorcycle from the car when it reaches this speed?
The position of an object connected to a spring varies with time according to the expression x = (4.3 cm) sin(7.6pt).
(a) Find the period of this motion.
(b) Find the frequency of the motion.
(c) Find the first time after t = 0 that the object reaches the position x = 2.6 cm.
a) Period of the motion is 0.826 s
b) The frequency of the motion is 1.16 Hz
c) the first time after t = 0 that the object reaches the position x = 2.6 cm is 0.0885 s.
.(a) The period of this motion
The general formula for the period is given by:T = 2π /ω = (2π)/(2π / T ) = T
Where T is the period and ω = 2πf is the angular frequency.
The angular frequency,ω = 7.6p
The period of the motion,T = 2π / ω= (2π)/ (7.6p) ≈ 0.826 s
(b) The frequency of the motion
The frequency is given by the reciprocal of the period,f = 1/T = 1/ (2π / ω) = ω/2π = 7.6p / 2π≈ 1.16 Hz
(c) The first time after t = 0 that the object reaches the position x = 2.6 cm.
The given position of the object at any time, x = (4.3 cm) sin(7.6pt).
We have to find time when x=2.6 cm.2.6 = (4.3 cm) sin(7.6pt)
t = sin^-1 (2.6/4.3) / 7.6
p≈ 0.0885 s
Therefore, the first time after t = 0 that the object reaches the position x = 2.6 cm is 0.0885 s.
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(a) The position of an object connected to a spring varies with time according to the expression x = (4.3 cm) sin(7.6pt). The given expression represents a sinusoidal variation of displacement x of an object in simple harmonic motion.
The general expression for the displacement of an object undergoing simple harmonic motion is given as, x = A sin (ωt + φ)Here, A represents the amplitude, ω represents the angular frequency, φ represents the phase constant. So, we can compare the given expression x = (4.3 cm) sin(7.6pt) with the general expression as, x = A sin (ωt + φ)
Here, A = 4.3 cmω = 7.6p = 2πf [f represents the frequency]⇒ 7.6p = 2πf⇒ f = 7.6p/2π = 7.6/2 Hz = 3.8 Hz
So, the frequency of motion is 3.8 Hz(b) The time period of a simple harmonic motion is given as, T = 2π/ω = 2π/pf = 7.6/2π seconds = 1.205 s
So, the period of motion is 1.205 s.(c) We have, x = (4.3 cm) sin(7.6pt)The first time after t = 0 that the object reaches the position x = 2.6 cm, then we can write, 2.6 = (4.3 cm) sin(7.6pt)⇒ sin(7.6pt) = 2.6/4.3 = 0.60465Now, we have to calculate the time t for which sin(7.6pt) = 0.60465. From the standard trigonometric identity, we know that sinθ = sin(π - θ).Therefore, sin(7.6pt) = sin(π - 7.6pt)⇒ 7.6pt = π - sin⁻¹(0.60465) = 0.991 rad.⇒ t = 0.991/7.6π s ≈ 0.042 sSo, the first time after t = 0 that the object reaches the position x = 2.6 cm is 0.042 s (approx).Main Answer:(a) The period of motion is 1.205 s.(b) The frequency of motion is 3.8 Hz.(c) The first time after t = 0 that the object reaches the position x = 2.6 cm is 0.042 s (approx).
Given,
The position of an object connected to a spring varies with time according to the expression x = (4.3 cm) sin(7.6pt).(a) The period of this motion:
The general expression for the displacement of an object undergoing simple harmonic motion is given as, x = A sin (ωt + φ)Here, A represents the amplitude, ω represents the angular frequency, φ represents the phase constant.
So, we can compare the given expression x = (4.3 cm) sin(7.6pt) with the general expression as, x = A sin (ωt + φ)
Here,
A = 4.3 cmω = 7.6p = 2πf [f represents the frequency]⇒ 7.6p = 2πf⇒ f = 7.6p/2π = 7.6/2 Hz = 3.8 Hz
Therefore, the frequency of motion is 3.8 Hz.
The time period of a simple harmonic motion is given as, T = 2π/ω = 2π/pf= 7.6/2π seconds = 1.205 s.
So, the period of motion is 1.205 s.
(b) The frequency of motion is 3.8 Hz.(c) The first time after t = 0 that the object reaches the position x = 2.6 cm:
The equation of motion is given as, x = (4.3 cm) sin(7.6pt)
The first time after t = 0 that the object reaches the position x = 2.6 cm, then we can write, 2.6 = (4.3 cm) sin(7.6pt)⇒ sin(7.6pt) = 2.6/4.3 = 0.60465
Now, we have to calculate the time t for which sin(7.6pt) = 0.60465. From the standard trigonometric identity, we know that sinθ = sin(π - θ).
Therefore, sin(7.6pt) = sin(π - 7.6pt)⇒ 7.6pt = π - sin⁻¹(0.60465) = 0.991 rad.⇒ t = 0.991/7.6π s ≈ 0.042 s
So, the first time after t = 0 that the object reaches the position x = 2.6 cm is 0.042 s (approx).
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If I want to reduce the RLC circuit by a factor of 1 or 10, what method or material should I use to achieve resistance, inductance, and capacitance? (The description of the problem is only like this)I
To reduce the RLC circuit by a factor of 1 or 10, you can modify the resistance, inductance, and capacitance components of the circuit by using resistors, inductors, and capacitors with lower values that correspond to the desired reduction factors.
For resistance reduction, you can use resistors with lower resistance values. Resistors are readily available in a range of values, allowing you to select one that suits your desired reduction factor. For example, if you want to reduce the resistance by a factor of 1, you can simply replace the existing resistor with one of the same value. To achieve a reduction by a factor of 10, you would replace the resistor with one that has ten times lower resistance.
To modify the inductance of the circuit, you can utilize inductors with different inductance values. Inductors are commonly labeled with their inductance values in Henrys (H). By selecting an inductor with a lower inductance value, you can achieve a reduction in the inductance of the circuit. Again, you would choose an inductor that corresponds to the desired reduction factor.
For adjusting the capacitance, capacitors with different capacitance values are employed. Capacitors are usually labeled with their capacitance values in Farads (F). By using capacitors with lower capacitance values, you can reduce the capacitance in the circuit. Similarly, you would select a capacitor that corresponds to the desired reduction factor.
In conclusion, to reduce the RLC circuit by a factor of 1 or 10, you can modify the resistance, inductance, and capacitance components of the circuit by using resistors, inductors, and capacitors with lower values that correspond to the desired reduction factors.
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A 70-kg astronaut floating in space in a 110-kg MMU (manned maneuvering unit) experiences an acceleration of 0.029 m/s^2 when he fires one of the MMU's thrusters. If the speed of the escaping N2 gas relative to the astronaut is 490 m/s, how much gas is used by the thruster in 5.0s and what is the thrust of the thruster?
The mass of the gas used by the thruster in 5 seconds is 0.0534 kg, and the thrust of the thruster is 5.22 N.
The mass of the astronaut is 70 kg, and the mass of the MMU is 110 kg. Thus, the combined mass of the astronaut and MMU is 180 kg. The acceleration experienced by the astronaut is given as 0.029 m/s². We are also given that the speed of the escaping N₂ gas relative to the astronaut is 490 m/s. We need to determine the amount of gas used by the thruster in 5 seconds and the thrust of the thruster.
Calculation of the thrust of the thruster:
We know that F = ma, where F is the force, m is the mass, and a is the acceleration. Here, F is the thrust of the thruster. Thus, F = ma = 180 kg × 0.029 m/s² = 5.22 N.
Calculation of the amount of gas used by the thruster in 5 seconds:
The amount of gas used by the thruster in 5 seconds can be calculated using the formula:
m = (F × t) / v
Where m is the mass of the gas used, F is the thrust of the thruster, t is the time for which the thruster is fired, and v is the speed of the escaping gas relative to the astronaut.
Substituting the given values, we get:
m = (5.22 N × 5 s) / 490 m/s
m = 0.0534 kg.
Therefore, the mass of the gas used by the thruster in 5 seconds is 0.0534 kg, and the thrust of the thruster is 5.22 N.
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What is the wavelength (in meters) of an electromagnetic wave whose frequency is 1.55 times 10^12 s^-1? times 10 m
The wavelength of the electromagnetic wave with a frequency of 1.55 × 10¹² s⁻¹ is approximately 1.935 × 10⁻⁴ meters.
To calculate the wavelength of an electromagnetic wave, we can use the equation:
λ = c / f
Where:
λ is the wavelength of the wave
c is the speed of light (approximately 3.00 × 10⁸ m/s)
f is the frequency of the wave
Given that the frequency is 1.55 × 10¹² s⁻¹, we can substitute this value into the equation:
λ = (3.00 × 10⁸ m/s) / (1.55 × 10¹² s⁻¹)
λ = (3.00 × 10⁸ m/s) / (1.55 × 10¹² s⁻¹)
λ ≈ 1.935 × 10⁻⁴ m
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The wavelength (in meters) of an electromagnetic wave whose frequency is 1.55 × 10¹² Hz is 0.1935 meters.
What is an electromagnetic wave?An electromagnetic wave is a transverse wave that travels through space carrying energy. It is created by the movement of electric and magnetic fields in space, that is, the oscillation of the electric and magnetic fields. Electromagnetic waves are unique because they do not require a medium to travel through, which means they can travel through a vacuum, such as space.
The relationship between the frequency and wavelength of an electromagnetic wave is expressed mathematically using the formula:λ = c / f
Where:
λ = wavelength
c = speed of light = 3 × 10⁸ m/s
f = frequency
Substituting the values in the equation;λ = c / fλ = 3 × 10⁸ / (1.55 × 10¹²)λ = 0.1935 m
Therefore, the wavelength of an electromagnetic wave whose frequency is 1.55 × 10¹² Hz is 0.1935 meters.
An electromagnetic wave can be characterized by its frequency, wavelength, and speed. The frequency of an electromagnetic wave is the number of waves that pass through a point in one second, measured in hertz (Hz). The wavelength of an electromagnetic wave is the distance between two consecutive peaks or troughs of the wave, measured in meters (m).
The speed of light in a vacuum is constant and is equal to 3 × 10⁸ m/s. This means that the frequency and wavelength of an electromagnetic wave are inversely proportional to each other. If the frequency increases, the wavelength decreases, and vice versa. Therefore, we can use the relationship between frequency and wavelength to calculate the wavelength of an electromagnetic wave whose frequency is known.
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A series RLC circuit with L = 13 mH, C = 3.8 F, and R = 6.7 is driven by a generator with a maximum emf of 100 V and a variable angular frequency . (a) Find the resonant (angular) frequency 0. (b) Find Irms at resonance. When the angular frequency = 9000 rad/s, (c) Find the capacitive reactance XC in ohms. Find the inductive reactance XL in ohms. (d) Find the impedance Z. (Give your answer in ohms.) Find Irms. (e) Find the phase angle (in degrees).
A series RLC circuit is an electrical circuit consisting of a resistor (R), an inductor (L), and a capacitor (C) connected in series.
The answers are:
(a) Resonant angular frequency ω₀ ≈ 47.98 rad/s
(b) Irms at resonance ≈ 14.93 A
(c) Capacitive reactance XC ≈ 0.00002941 Ω, Inductive reactance XL ≈ 117 Ω
(d) Impedance Z ≈ 117 Ω, Irms ≈ 0.8547 A
(e) Phase angle θ ≈ 1.745°
To solve the given questions, we'll use the following formulas for an RLC circuit:
Resonant angular frequency (ω₀):
ω₀ = 1 / √(LC)
Impedance (Z):
Z = √(R² + (XL - XC)²)
Current (Irms):
Irms = Vmax / Z
Phase angle (θ):
θ = arctan((XL - XC) / R)
Given:
L = 13 mH = 0.013 H
C = 3.8 F
R = 6.7 Ω
Vmax = 100 V
ω = 9000 rad/s
(a) Resonant angular frequency (ω₀):
ω₀ = 1 / √(LC)
ω₀ = 1 / √(0.013 H * 3.8 F)
ω₀ ≈ 47.98 rad/s
(b) Irms at resonance:
Z = √(R² + (XL - XC)²)
Z = √(6.7 Ω² + (0 - 0)²) (at resonance, XL = XC = 0)
Z = 6.7 Ω
Irms = Vmax / Z
Irms = 100 V / 6.7 Ω
Irms ≈ 14.93 A
(c) Capacitive reactance (XC) at ω = 9000 rad/s:
XC = 1 / (C * ω)
XC = 1 / (3.8 F * 9000 rad/s)
XC ≈ 0.00002941 Ω
Inductive reactance (XL) at ω = 9000 rad/s:
XL = L * ω
XL = 0.013 H * 9000 rad/s
XL ≈ 117 Ω
(d) Impedance (Z) at ω = 9000 rad/s:
Z = √(R² + (XL - XC)²)
Z = √(6.7 Ω² + (117 Ω - 0.00002941 Ω)²)
Z ≈ 117 Ω
Irms = Vmax / Z
Irms = 100 V / 117 Ω
Irms ≈ 0.8547 A
(e) Phase angle (θ) at ω = 9000 rad/s:
θ = arctan((XL - XC) / R)
θ = arctan((117 Ω - 0.00002941 Ω) / 6.7 Ω)
θ ≈ 1.745°
Therefore, the answers are:
(a) Resonant angular frequency ω₀ ≈ 47.98 rad/s
(b) Irms at resonance ≈ 14.93 A
(c) Capacitive reactance XC ≈ 0.00002941 Ω, Inductive reactance XL ≈ 117 Ω
(d) Impedance Z ≈ 117 Ω, Irms ≈ 0.8547 A
(e) Phase angle θ ≈ 1.745°
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an object of mass m is lifted at a constant velocity a vertical distance h in time t. the power supplied by the lifting force is
The lifting force must supply power equal to (mgh) / t in order to lift the object at a constant velocity a vertical distance h in time t. This means that the rate of work done by the lifting force is (mgh) / t, which is the same as the rate of gravitational potential energy gained by the object in the same time interval.
When an object of mass m is lifted at a constant velocity a vertical distance h in time t, the power supplied by the lifting force can be calculated using the formula:
Power = Work / TimeSince the object is lifted at a constant velocity, it implies that no acceleration is taking place. Thus, the work done on the object by the lifting force is the same as the gravitational potential energy gained by the object.
Potential Energy, Ep = mg hwhere, m is the mass of the objectg is the acceleration due to gravity, which is approximately 9.81 m/s2h is the vertical distance traveled by the objectThus, the power supplied by the lifting force can be calculated using the formula:Power = Ep / Time= (mgh) / t
Therefore, the lifting force must supply power equal to (mgh) / t in order to lift the object at a constant velocity a vertical distance h in time t. This means that the rate of work done by the lifting force is (mgh) / t, which is the same as the rate of gravitational potential energy gained by the object in the same time interval.
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find the net torque on the wheel in the figure below about the axle through o, taking a = 5.00 cm and b = 17.0 cm. (assume that the positive direction is counterclockwise.)
Torque is the tendency of a force to rotate an object about an axis or fulcrum. It's a measure of a force's ability to make an object rotate around a pivot or axis. Torque can be calculated using the formula T = rF sin θ, where T is the torque, r is the distance from the axis to the force vector, F is the force vector, and θ is the angle between the force vector and the lever arm vector.
Net torque is the sum of all torques acting on an object, and it can be calculated using the equation τ_net = Στ, where τ_net is the net torque and Στ is the sum of all torques acting on the object.In the given figure below, a wheel of radius 12.0 cm and mass 2.00 kg is mounted on an axle through point O. A horizontal force F = 40.0 N is applied to the rim of the wheel at a point P located 5.00 cm from the axle. The weight of the wheel is supported by a vertical axle through O.What is the net torque on the wheel about the axle through O if the positive direction is counterclockwise? To calculate the net torque, we must first calculate the torques due to the applied force and the weight of the wheel.The torque due to the applied force is
τ_F = rF sin θ
= (0.05 m)(40.0 N) sin 90°
= 2.00 Nm counterclockwise.The torque due to the weight of the wheel is τ_W = r_W mg
= (0.12 m)(2.00 kg)(9.81 m/s²)
= 2.35 Nm clockwise.The net torque is
τ_net = τ_F + τ_W
= (2.00 Nm counterclockwise) + (2.35 Nm clockwise)
= 0.35 Nm clockwise. Therefore, the net torque on the wheel about the axle through O is 0.35 Nm clockwise.
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Estimate the moment of inertia of a bicycle wheel 67.2 cm in diameter. The rim and tire have a combined mass of 1.25 kg. The mass of the hub (at the center) can be ignored.
Moment of Inertia of a Bicycle WheelThe moment of inertia of a bicycle wheel is the amount of force it takes to accelerate the wheel’s rotation about its central axis. The moment of inertia of a bicycle wheel can be determined by adding the moment of inertia of the rim and the tire, which are separate from each other.
It’s important to know the moment of inertia of a bicycle wheel because it’s essential in figuring out how much energy is required to accelerate the wheel, how quickly the wheel will rotate, and how much torque is needed to maintain a given angular velocity. If you want to estimate the moment of inertia of a bicycle wheel with a diameter of 67.2 cm, you’ll need to use a few equations.Moment of Inertia of a Thin RingTo determine the moment of inertia of a thin ring (or hoop), you can use the equation I = mr2, where I is the moment of inertia, m is the mass of the ring, and r is the radius of the ring. However, since we are given the diameter, we need to first find the radius. We know that the diameter of the bicycle wheel is 67.2 cm, so the radius is 33.6 cm or 0.336 m. Also, we are told that the mass of the rim and tire is 1.25 kg. Using the above equation, we can calculate the moment of inertia of the ring as:
I = mr2I
= (1.25 kg) (0.336 m)2I
= 0.150 kg
m2Moment of Inertia of a Solid DiscNext, we’ll need to find the moment of inertia of the solid disc that makes up the tire of the bicycle wheel. The equation for the moment of inertia of a solid disc is I = (1/2)mr2, where m is the mass of the disc and r is the radius of the disc. We know that the radius of the disc is the same as the radius of the ring, which is 0.336 m. Since we are given the mass of the rim and tire, and we know the mass of the rim, we can calculate the mass of the tire as follows:mass of tire = mass of rim and tire - mass of rimmass of tire
= 1.25 kg - 0.150 kgmass of tire
= 1.10 kg
Now we can calculate the moment of inertia of the disc as follows:
I = (1/2)mr2I
= (1/2)(1.10 kg)(0.336 m)2I
= 0.064 kg m2
Total Moment of InertiaFinally, we can add the moment of inertia of the ring and the moment of inertia of the disc to get the total moment of inertia of the bicycle wheel:
I(total) = I(ring) + I(disc)I(total)
= 0.150 kg m2 + 0.064 kg m2I(total)
= 0.214 kg m2
Therefore, the estimated moment of inertia of a bicycle wheel with a diameter of 67.2 cm is 0.214 kg m2.
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can
you please answer these few multiple choice questions! thank you.
Question 20 (1 point) 4) Listen What must be your average speed in order to travel 350 km in 5.15 h? 1) 0.015 km/h 2) 17.0 km/h 3) 68.0 km/h 4) 156.0 km/h
Question 25 (1 point) 4) Listen A car goes f
Answer:
Explanation:
Question 20:
The average speed can be calculated by dividing the total distance traveled by the total time taken.
Given: Distance = 350 km, Time = 5.15 hours
Average Speed = Distance / Time
Average Speed = 350 km / 5.15 h ≈ 67.96 km/h
Therefore, the closest option is:
3) 68.0 km/h
Question 25:
The question seems to be incomplete. Please provide the complete question so that I can assist you with the answer.
the free expansion of a gas is a process where the total mean energy e remains constant. in connection with this process, the following quantities are of interest.
The total mean energy of the gas remains constant during free expansion. This means that the total energy of the gas, which includes both kinetic and potential energy of the gas particles, does not change.
Temperature (T): Although the total mean energy remains constant, the temperature of the gas may change during free expansion. This is because temperature is related to the average kinetic energy of the gas particles, and as the gas expands, the kinetic energy distribution may change, affecting the temperature.Pressure (P): The pressure of the gas can change during free expansion. As the gas expands, the gas particles spread out, resulting in a decrease in the number of collisions with the container walls and a decrease in pressure.
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A television camera lens has a 14cm focal length and a lensdiameter of 6.0cm. What is its f number?
The f-number of the television camera lens is approximately 2.33.
The f-number of a lens can be calculated by dividing the focal length of the lens by the diameter of the lens.
Given:
Focal length = 14 cm
Lens diameter = 6.0 cm
Using the formula:
f-number = Focal length / Lens diameter
Substituting the given values:
f-number = 14 cm / 6.0 cm
Calculating the value:
f-number ≈ 2.33
Therefore, the f-number of the television camera lens is approximately 2.33. This indicates that the lens has a relatively large aperture, allowing more light to enter and resulting in a brighter image.
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The f-number of the television camera lens is 2.33. The f-number of a television camera lens with a 14cm focal length and a lens diameter of 6.0cm can be determined as follows ;Formula for calculating f-number is given by; f-number = focal length / diameter of the lens.
First, we need to determine the f-number of the television camera lens. Therefore, f-number = 14 / 6.0f-number = 2.33. Therefore, the f-number of the television camera lens is 2.33.
A camera lens is a device that focuses light on the camera's image sensor or film, and hence allowing the formation of an image. Camera lenses come in various types, for example - wide-angle, telephoto, zoom, prime and macro lenses, each with its own characteristics and uses.
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You drop a ball from the top of a building. Which of the following is a true statement? The momentum of the ball remains constant and is zero. The momentum of the ball remains constant and is non-zero. The magnitude of the ball's momentum increases. The magnitude of the ball's momentum decreases.
When a ball is dropped from the top of a building, its momentum does not remain constant and is non-zero.
Momentum is defined as the product of an object's mass and its velocity, and it is a vector quantity, meaning it has both magnitude and direction. As the ball falls, its velocity increases due to the acceleration caused by gravity. Since momentum depends on both mass and velocity, and the ball's velocity is changing, the momentum of the ball also changes. Therefore, the statement that the momentum of the ball remains constant and is non-zero is true.
However, it is important to note that the momentum of the ball is not constant throughout its fall. As it accelerates, the momentum increases, but once it reaches terminal velocity, the momentum remains constant until it hits the ground.
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Question # : 19 When the distance to a polar molecule is doubled, the electric field due to the dipole changes by what factor? A. 4 ✓B. 1/8 C. 8 D. 1/4 E. 2 1 6₁ = 2k₁=2x2x2x2x2 - (2x) = 6 2³ (
When the distance to a polar molecule is doubled, the electric field due to the dipole decreases by a factor of 1/4. This follows from the inverse square law governing the relationship between distance and electric field strength. The correct option is D.
When the distance to a polar molecule is doubled, the electric field due to the dipole changes by a factor of 1/4 (option D).
The electric field due to a dipole decreases with increasing distance according to an inverse square law. This means that as the distance from the dipole increases, the electric field strength decreases proportionally.
When the distance to the polar molecule is doubled, the new distance becomes twice the original distance.
According to the inverse square law, the electric field strength at this new distance would be reduced to 1/(2^2) = 1/4 of its original value.
To understand this concept mathematically, we can use the equation for the electric field due to a dipole at a given distance:
E = k * p / r^3
Where E is the electric field, k is the Coulomb's constant, p is the dipole moment, and r is the distance to the dipole. When the distance is doubled (2r), the new electric field (E') can be calculated as:
E' = k * p / (2r)^3 = (1/8) * (k * p / r^3) = (1/8) * E
This shows that the electric field due to the dipole changes by a factor of 1/8, or equivalently, 1/4 (option D).
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The voltage difference across a charged, parallel plate capacitor with plate separation 2.0 cm is 16 V. If the voltage at the positive plate is +32 V, what is the voltage inside the capacitor 0.50 cm from the positive plate? You may assume the electric field inside the capacitor is uniform. O +24 V O +28 V O +36 V O +32 V
The voltage inside the capacitor, 0.50 cm from the positive plate, is +28 V.
In a parallel plate capacitor, the electric field between the plates is uniform and directed from the positive plate to the negative plate. The electric field intensity (E) is given by E = V/d, where V is the voltage difference between the plates and d is the separation between the plates.
In this case, the voltage difference across the capacitor is given as 16 V and the plate separation is 2.0 cm (or 0.02 m). Therefore, the electric field intensity is E = 16 V / 0.02 m = 800 V/m.
Since the electric field is uniform, the voltage decreases linearly as we move away from the positive plate. Thus, at a distance of 0.50 cm (or 0.005 m) from the positive plate, the voltage would be (32 V) - (800 V/m × 0.005 m) = 32 V - 4 V = 28 V.
Therefore, the voltage inside the capacitor, 0.50 cm from the positive plate, is +28 V.
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The electric field strength 1.7 cm from the surface of a 10-cm-diameter metal ball is 6.0×104 N/C . What is the charge (in nC) on the ball?
The charge on the metal ball is 168 nC. with electric field strength 1.7 cm from the surface of a 10-cm-diameter metal ball is [tex]6.0*10^{4}[/tex] N/C.
The electric field strength 1.7 cm from the surface of a 10-cm-diameter metal ball is [tex]6.0*10^{4}[/tex] N/C. The formula for the electric field is given by: E = kQ/r²
Where: k is the Coulomb's constant Q is the charge on the metal ball r is the distance between the point of observation and the center of the sphere. Electric field strength is given as E = [tex]6.0*10^{4}[/tex] N/C
diameter of the metal ball = 10 cm
radius of the metal ball = 5 cm (as diameter = 2r)
r = 1.7 cm
= [tex]8.99 * 10^{9}[/tex] Nm²/C².
Putting the values in the above formula:
E = kQ/r²
=> [tex]6.0*10^{4}[/tex] N/C
= [tex]8.99 * 10^{9}[/tex]
Nm²/C² * Q/(0.017 m)²
=> Q = [tex]8.99 * 10^{9}[/tex] Nm²/C² × [tex]6.0*10^{4}[/tex] N/C × (0.017 m)²
Q = [tex]1.68 * 10^{-7}[/tex] C
= 168 nC
The formula for the electric field is given by E = kQ/r² Electric field strength is given as
E = [tex]6.0*10^{4}[/tex] N/C
diameter of the metal ball = 10 cm
radius of the metal ball = 5 cm (as diameter = 2r)
r = 1.7 cm
= [tex]8.99 * 10^{9}[/tex] Nm²/C²
The formula for electric field strength is applied: Putting the values in the formula, it is found that Q is [tex]1.68 * 10^{-7}[/tex] C or 168 nC.
The charge on the metal ball is 168 nC.
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What is most nearly the spring constant for a helical linear spring with the properties below? 50 GPa shear modulus 2 mm wire diameter 6 mm mean spring diameter 10 active coils a) 1kN/m b) 3kN/m c) 4kN/m d) 5kN/m 37) For a thick cylindrical pressure vessel, what is close to the hoop stress if the internal pressure is 8atm, and the inner and outer radii are 2 m and 3 m, respectively? a) 2.107kPa b) 1,241kPa c) 632kPa d) 2,560.kPa 38) The minimol tensile strength of a oiltempered wire spring (ASTM A229) with a 1 mm wire diameter is nearest to what? a) 1840MPa b) 1610MPa c) 1,840 MP d) 5,670MPa 39) Pneumatically powered machines generally use as source of power transmission. a) electromagnetic forces b) compressed gasses c) Liquid
37) For a thick cylindrical pressure vessel, d) 2,560.kPa is close to the hoop stress if the internal pressure is 8atm, and the inner and outer radii are 2 m and 3 m, respectively. 38) The mmol tensile strength of a oil tempered wire spring with a 1 mm wire diameter is nearest to c) 1,840 MPa. 39) Pneumatically powered machines generally use as source of power transmission is b) compressed gasses.
37) The hoop stress is a normal stress that occurs circumferentially in a thin-walled cylinder as a result of the internal pressure. It is calculated by dividing the applied force by the cross-sectional area of the cylinder.
Formula:
[tex]Hoop stress = \frac{(Internal pressure * radius of the cylinder) }{thickness}[/tex]
Given that: Internal pressure, p = 8 atm. Inner radius,
r1 = 2 m, Outer radius,
r2 = 3 m.
Thickness, t = r2 - r1
= 3 - 2
= 1 m. Hence the hoop stress is: [tex]Hoops stress = \frac{ (8 * 2)}{1}[/tex].
= 16 atm
The correct option is d) 2,560.kPa
The hoop stress of a thick cylindrical pressure vessel is given by dividing the internal pressure by the thickness of the cylinder. The hoop stress for a pressure of 8 atm and inner and outer radii of 2 m and 3 m respectively is 16 atm.
38) The wire is used to produce springs and the ASTM A229 standard defines the procedure for oil-tempered carbon steel wires used in the manufacture of springs that are primarily used for high stress applications. The wire's minimum tensile strength is calculated using the formula below: Formula:
[tex]Minimum tensile strength = 4 * \sqrt{(d^{3}) }[/tex]
Here, d = diameter of the wire = 1 mm
[tex]Minimum tensile strength = 4 * \sqrt{(1^3) }[/tex]
= [tex]4 *\sqrt{1}[/tex]
= 4 MPa.
The correct option is c) 1,840 MPa.
39) Pneumatically powered machines generally use as source of power transmission.
Pneumatically powered machines generally use compressed gasses as a source of power transmission. Compressed gas is used to drive the piston in a pneumatic cylinder, which converts the gas's energy into linear motion.
So the correct option is b) compressed gasses
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what is the de broglie wavelength of an object with a mass of 1.30 kg moving at a speed of 2.70 m/s? (useful constant: h = 6.63×10-34 js.)
The de Broglie wavelength of the object with a mass of 1.30 kg and a speed of 2.70 m/s is approximately 1.89×10^-34 meters.
The de Broglie wavelength of an object can be calculated using the equation:
λ = h / p
where λ is the de Broglie wavelength, h is Planck's constant (6.63×10^-34 J·s), and p is the momentum of the object.
The momentum of an object can be calculated using the equation:
p = m * v
where m is the mass of the object and v is its velocity.
Given that the mass of the object is 1.30 kg and its velocity is 2.70 m/s, we can calculate the momentum:
p = 1.30 kg * 2.70 m/s = 3.51 kg·m/s
Now we can calculate the de Broglie wavelength:
λ = 6.63×10^-34 J·s / 3.51 kg·m/s ≈ 1.89×10^-34 m
Therefore, the de Broglie wavelength of the object with a mass of 1.30 kg and a speed of 2.70 m/s is approximately 1.89×10^-34 meters.
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