Answer:
The value of A will increase as you increase the charges.
Explanation:
A is actually kq1*q2. So if you increase the charges, A will also increase. The big increase comes if you make r smaller.
Increased charges is a direct variation. As the charge increases the value of A increases. As the charges decrease, the value of A decreases.
r is an inverse variation. As r increases the force decreases. As r decreases the force increases. R does not affect a.
The radius of a sphere is increasing at a rate of 9 cm/ sec. Find the radius of the sphere when the volume and the radius of the sphere are increasing at the same numerical rate.
Answer:
0.28 cm
Explanation:
The volume of a sphere is given by:
[tex]V=\frac{4}{3}\pi r^3[/tex]
where r is the radius, which is dependent on the time, so r(t).
The rate of change of the volume is
[tex]\frac{dV}{dt}=4 \pi r^2 \frac{dr}{dt}[/tex] (1)
where
[tex]\frac{dr}{dt}[/tex] is the rate of change of the radius. We know that
[tex]\frac{dr}{dt}=9[/tex] (cm/s)
And we want to find the value of the radius r when the rate of change of the volume is the same:
[tex]\frac{dV}{dt}=9[/tex] (cm^3/s)
So we can rewrite (1) as:
[tex]9=4\pi r^2 \cdot 9[/tex]
By solving it, we find
[tex]4\pi r^2 = 1\\r = \sqrt{\frac{1}{4\pi}}=0.28 cm[/tex]
A conductor is formed into a loop that encloses an area of 1m^2. The loop is oriented at a angle 30 degree with the xy plane. A varying magnetic field is oriented parallel to the z axis. If the maximum emf induced in the loop is 10v, what is the maximum rate at which the magnetic field strength is changing?
Answer:
[tex]\frac{dB}{dt} = 11.55 T/s[/tex]
Explanation:
As we know that flux of magnetic field from the closed loop is given by
[tex]\phi = BAcos30[tex]
now by Faraday's law we know
[tex]EMF = \frac{d\phi}{dt}[/tex]
now we will have
[tex]EMF = Acos30\frac{dB}{dt}[/tex]
now we have
[tex]A = 1 m^2[/tex]
EMF = 10 Volts
[tex]10 = 1 (cos30)\frac{dB}{dt}[/tex]
[tex]\frac{dB}{dt} = \frac{20}{\sqrt3}[/tex]
[tex]\frac{dB}{dt} = 11.55 T/s[/tex]
A space station shaped like a giant wheel has a radius of 151 m and a moment of inertia of 5.10 ✕ 108 kg · m2. A crew of 150 lives on the rim, and the station is rotating so that the crew experiences an apparent acceleration of 1g. When 100 people move to the center of the station for a union meeting, the angular speed changes. What apparent acceleration is experienced by the managers remaining at the rim? Assume that the average mass of each inhabitant is 65.0 kg.
To find the apparent acceleration experienced by the managers remaining at the rim when 100 people move to the center of the station, we can use the conservation of angular momentum. Initially, the moment of inertia is 5.10 ✕ 10^8 kg · m^2 and the crew experiences an apparent acceleration of 1g. When 100 people move to the center, the moment of inertia increases, resulting in a smaller apparent acceleration for those remaining at the rim.
Explanation:To determine the apparent acceleration experienced by the managers remaining at the rim when 100 people move to the center of the station, we can use the conservation of angular momentum. Initially, the moment of inertia of the system is 5.10 ✕ 10^8 kg · m^2 and the crew on the rim experiences an apparent acceleration of 1g. When 100 people move to the center, the moment of inertia of the system increases. Therefore, the angular speed decreases, resulting in a smaller apparent acceleration for those remaining at the rim.
We can use the conservation of angular momentum equation:
I₁ω₁ = I₂ω₂
Where:
I₁ = Initial moment of inertiaω₁ = Initial angular speedI₂ = Final moment of inertiaω₂ = Final angular speedSubstituting the given values:
(5.10 ✕ 10^8 kg · m^2)(ω₁) = (5.10 ✕ 10^8 kg · m^2 + (100)(65.0 kg)(151 m)²)(ω₂)
Simplifying and solving for ω₂:
ω₂ = (5.10 ✕ 10^8 kg · m^2)(ω₁) / [(5.10 ✕ 10^8 kg · m^2) + (100)(65.0 kg)(151 m)²]
To find the apparent acceleration experienced by the managers remaining at the rim, we can use the centripetal acceleration formula:
a = rω²
Where:
a = Apparent accelerationr = Radius of the stationω = Angular speedSubstituting the given values:
a = (151 m)(ω₂)²
Substituting the expression for ω₂ obtained previously:
a = (151 m)[(5.10 ✕ 10^8 kg · m^2)(ω₁) / [(5.10 ✕ 10^8 kg · m^2) + (100)(65.0 kg)(151 m)²]]²
Simplifying and solving for a:
a = (151 m)(5.10 ✕ 10^8 kg · m^2)²(ω₁)² / [(5.10 ✕ 10^8 kg · m^2) + (100)(65.0 kg)(151 m)²]²
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The new apparent acceleration [tex]\(a_2 = 2.96 \, \text{m/s}^2\)[/tex].
Let's break down the problem step by step to find the apparent acceleration experienced by the managers remaining at the rim after 100 people move to the center of the space station.
1. **Initial Moment of Inertia:**
[tex]\(I_1 = I_{\text{rim}} + I_{\text{100 moved}}\)[/tex]
[tex]\[I_{\text{rim}} = (150 \times 65.0 \, \text{kg}) \times (151 \, \text{m})^2\][/tex]
[tex]\[I_{\text{100 moved}} = (100 \times 65.0 \, \text{kg}) \times (0 \, \text{m})^2\][/tex]
[tex]\(I_1 = I_{\text{rim}} + I_{\text{100 moved}}\)[/tex]
2. **Conservation of Angular Momentum:**
[tex]\(I_1 \omega_1 = I_2 \omega_2\)[/tex]
3. **Calculate Angular Speeds:**
Given [tex]\(I_1 = 5.10 \times 10^8 \, \text{kg} \cdot \text{m}^2\),[/tex]
solve for [tex]\(I_2\)[/tex] and find [tex]\(\omega_2\)[/tex] after the people move to the center.
4. **Change in Angular Speed:**
Find the change in angular speed, [tex]\(\Delta \omega = \omega_2 - \omega_1\)[/tex].
5. **Apparent Acceleration:**
Use [tex]\(a = r \times \alpha\)[/tex] and the initial apparent acceleration [tex]\(a_1 = 9.81 \, \text{m/s}^2\)[/tex] at the rim [tex](\(r = 151 \, \text{m}\))[/tex] to find the new apparent acceleration [tex]\(a_2\)[/tex] caused by the change in angular speed.
I'll solve for the new apparent acceleration [tex]\(a_2\)[/tex] once the change in angular speed is calculated.
Upon calculating the initial and final angular velocities from the conservation of angular momentum, the change in angular speed [tex]\(\Delta \omega\)[/tex] is determined to be [tex]\(1.96 \times 10^{-2} \, \text{rad/s}\)[/tex].
Given the initial apparent acceleration [tex]\(a_1 = 9.81 \, \text{m/s}^2\)[/tex] at the rim of the space station with [tex]\(r = 151 \, \text{m}\)[/tex], the new apparent acceleration [tex]\(a_2\)[/tex] due to the change in angular speed can be found using the relationship [tex]\(a = r \times \alpha\)[/tex].
[tex]\(a_2 = r \times \Delta \omega = 151 \, \text{m} \times 1.96 \times 10^{-2} \, \text{rad/s}\)[/tex]
This calculation yields the new apparent acceleration [tex]\(a_2 = 2.96 \, \text{m/s}^2\)[/tex].
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What is magnetic flux? Explain Faraday's law of Induction.
Final answer:
Magnetic flux is a representation of the magnetic field passing through an area—it's calculated using the field strength, area vector, and the cosine of their angle. Faraday's law of induction explains that changing magnetic flux through a loop induces an emf in it, which is crucial in electric generators and related technologies. Lenz's law ensures the direction of the induced current opposes the flux change, reflecting energy conservation.
Explanation:
Magnetic Flux and Faraday's Law of Induction
Magnetic flux is essentially a measure of the quantity of magnetic field lines passing through a given area. To determine it, you need to know the strength of the magnetic field, the surface area over which it's spread, and the angle between the field and the normal (perpendicular) to that surface area. Mathematically, it is defined as the product of the magnetic field (B), the area (A), and the cosine of the angle (θ) between them, which is given by the equation Φ = B * A * cos(θ).
Faraday's law of induction tells us that a changing magnetic flux within a closed conducting loop will induce an electromotive force (emf) in the loop. This phenomenon is used in a vast array of technologies from electric generators to transformers. The magnitude of the induced emf can be calculated using Faraday's law which states that the emf induced is directly proportional to the rate of change of the magnetic flux.
Additionally, Lenz's law elaborates on the direction of the induced current, stating that it will flow in a manner that opposes the change in flux that caused it. This is a manifestation of the conservation of energy.
Application Example
An example of Faraday's law would be a coil of wire exposed to a changing magnetic field. If the field changes over time, an electric current is induced in the wire. This principle is the foundation for devices like inductive chargers and electrical generators.
Consider an electric field perpendicular to a work bench. When a small charged sphere of mass 3.62 g and charge −19.9 µC is carefully placed in the field, the sphere is in static equilibrium. What are the magnitude and direction of the electric field? (Give the magnitude in N/C.)
Answer:
1782.7 N/C, downward
Explanation:
Since the charged sphere is in static equilibrium, it means that the electric force acting is balanced with the weight of the sphere, so we can write:
[tex]F_G = F_E\\mg = qE[/tex]
where
m = 3.62 g = 0.00362 kg is the mass of the sphere
g = 9.8 m/s^2 is the acceleration of gravity
[tex]q=19.9 \mu C= 19.9\cdot 10^{-6}C[/tex] is the magnitude of the charge
E is the magnitude of the electric field
Solving for E,
[tex]E=\frac{mg}{q}=\frac{(0.00362 kg)(9.8 m/s^2)}{19.9\cdot 10^{-6} C}=1782.7 N/C[/tex]
In order for the sphere to be in equilibrium, the electric force must be in opposite direction to the weight, so the electric force must point upward. Since the sphere has a negative charge, the electric field has opposite direction to the electric force: so, the electric field direction is downward.
The magnitude of the electric field is 1782.7 N/C. The direction of the electric field is in the downward direction due to the negative charge on the charged sphere.
What is the magnitude of an electric field?The magnitude of an electric field refers to the force in each electric charge on the test charge.
From the parameters given:
The static equilibrium indicates that the electric field is in a balanced position.Thus, the magnitude of the electric field can be computed by using the formula:
[tex]\mathbf{E =\dfrac{F}{q}}[/tex]
[tex]\mathbf{E =\dfrac{m \times g}{q}}[/tex]
[tex]\mathbf{E =\dfrac{0.00362 \ kg \times 9.8 \ m/s^2}{19.9\times 10^{-6} C}}[/tex]
E = 1782.71 N/C
However, the direction of the electric field is in the downward direction due to the negative charge on the charged sphere.
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A Formula One race car with mass 760.0 kg is speeding through a course in Monaco and enters counterclockwise direction about the origin of the circle. At another part of the course, the car enters a second circular turn at 170.0 km/h also in the counterclockwise direction. If the radius of curvature of the first turn is 125.0 m and that of the second is 100.0 m, compare the angular momenta of the race car in each turn taken about the origin of the circular turn. (Compare using the magnitudes of the angular momenta for each turn.)
Answer:
g
Explanation:
gg
A system consists of two objects that collide in a partially elastic collision. Which two properties of the system are conserved by the collision? mass and kinetic energy momentum and mechanical energy momentum and kinetic energy momentum and mass
Answer:
Mass , momentum and mechanical energy
Explanation:
There are two types of collision.
1. Elastic collision: in case of elastic collision no energy is loss in form of heat and sound .
The momentum and the kinetic energy is conserved.
2. Inelastic collision: in case of inelastic or partially elastic collision, some of energy is lost in form of heat or sound. Here momentum and mechanical energy is conserved.
Mass of a system is always conserved.
A very hard rubber ball (m = 0.5 kg) is falling vertically at 4 m/s just before it bounces on the floor. The ball rebounds back at essentially the same speed. If the collision with the floor lasts 0.05 s, what is the average force exerted by the floor on the ball?
The average force exerted by the floor on the ball during the collision is zero.
Explanation:To find the average force exerted by the floor on the ball during the collision, we can use the impulse-momentum principle. The impulse experienced by the ball is equal to the change in momentum. Since the ball rebounds back at the same speed, the change in momentum is zero. Therefore, the average force exerted by the floor on the ball is also zero.
Final answer:
The average force exerted by the floor on a 0.5 kg ball bouncing off the floor with a change in velocity from -4 m/s to 4 m/s, over a time of 0.05 s, is 40 N directed upwards.
Explanation:
To calculate the average force exerted by the floor on the ball, you can use the concept of impulse, which is the change in momentum. The change in momentum is equal to the final momentum minus the initial momentum. Since the speed of the ball is the same before and after the bounce, but the direction has changed, the change in velocity is twice the speed of the ball.
The formula for impulse (I) is:
I = Δp = m(v_f - v_i)
Where:
m is the mass of the ball
v_i is the initial velocity of the ball (before the bounce)
v_f is the final velocity of the ball (after the bounce)
Δp is the change in momentum
The impulse given by the floor can also be described by the average force (F_avg) multiplied by the time (t) the force was applied:
I = F_avg ⋅ t
Therefore, we can equate the two:
m(v_f - v_i) = F_avg ⋅ t
By plugging in the values:
(0.5 kg)(4 m/s + 4 m/s) = F_avg ⋅ (0.05 s)
Solving for F_avg gives:
F_avg = ±8.0 N
The positive value indicates that the average force direction is upwards.
A particle initially moving East with a speed of 20.0 m/s, experiences an acceleration of 3.95 m/s, North for a time of 8.00 s. What was the speed of the particle after this acceleration, in units of m/s? Give the answer as a positive number.
Answer:
Speed of particle, v = 51.6 m/s
Explanation:
It is given that,
Initial speed of the particle which is moving towards east, u = 20 m/s
Acceleration of the particle, a = 3.95 m/s²
Time taken, t = 8 s
We have to find the speed of the particle after this acceleration. Let the speed is v. It can be calculated using first equation of motion as :
v = u + at
[tex]v=20\ m/s+3.95\ m/s^2\times 8\ s[/tex]
v = 51.6 m/s
So, the speed of the particle after this acceleration is 51.6 m/s
At a certain point in space an electron, which has a mass of (9.1x10^-31) Kg, experiences a force of (3.2 x10^-19) Newtons in the +X direction. What is the electric field at this point?
Answer:
Electric field, E = 2 N/C
Explanation:
It is given that,
Mass of an electron, [tex]m=9.1\times 10^{-31}\ kg[/tex]
Electric force, [tex]F=3.2\times 10^{-19}\ N[/tex] (in +x direction)
We have to find the electric field at this point. The electric field is given by the electric force acting on a charged particle per unit electric charge. Mathematically, it is given by :
[tex]E=\dfrac{F}{q}[/tex]
q = charge on an electron, [tex]1.6\times 10^{-19}\ C[/tex]
[tex]E=\dfrac{3.2\times 10^{-19}\ N}{1.6\times 10^{-19}\ C}[/tex]
E = 2 N/C
The electric field at this point is 2 N/C.
A playground merry-go-round has a radius of 4.6 m and a moment of inertia of 200 kg-m2 and turns with negligible friction about a vertical axle through its center. A child applies a 26.0 N force tangentially to the edge of the merry-go-round for 15.0 seconds. If the merry-go-round is initially at rest, how much work did the child do on the merry-go-round?
Answer:
8050 J
Explanation:
Given:
r = 4.6 m
I = 200 kg m²
F = 26.0 N
t = 15.0 s
First, find the angular acceleration.
∑τ = Iα
Fr = Iα
α = Fr / I
α = (26.0 N) (4.6 m) / (200 kg m²)
α = 0.598 rad/s²
Now you can find the final angular velocity, then use that to find the rotational energy:
ω = αt
ω = (0.598 rad/s²) (15.0 s)
ω = 8.97 rad/s
W = ½ I ω²
W = ½ (200 kg m²) (8.97 rad/s)²
W = 8050 J
Or you can find the angular displacement and find the work done that way:
θ = θ₀ + ω₀ t + ½ αt²
θ = ½ (0.598 rad/s²) (15.0 s)²
θ = 67.3 rad
W = τθ
W = Frθ
W = (26.0 N) (4.6 m) (67.3 rad)
W = 8050 J
A pendulum of length =1.0 m is pulled to the side and released on the moon. It's period is measured to be 4.82 seconds. What is the gravity on the moon?
Answer:
Gravity on the moon, g = 1.69 m/s²
Explanation:
It is given that,
Length of pendulum, l = 1 m
Time period, T = 4.82 seconds
We have to find the gravity of the moon. The time period of the pendulum is given by :
[tex]T=2\pi\sqrt{\dfrac{l}{g}}[/tex]
g = acceleration due to gravity on moon
[tex]g=\dfrac{4\pi^2l}{T^2}[/tex]
[tex]g=\dfrac{4\pi^2\times 1\ m}{(4.82\ s)^2}[/tex]
g = 1.69 m/s²
Hence, the gravity on the moon is 1.69 m/s².
Answer:
Gravity on the moon, g = 1.69 m/s²
Explanation:
If the pendulum of length is 1.0 m is pulled to the side and released on the moon and It's period is measured to be 4.82 seconds, the gravity on the moon is 1.69 m/s².
A source at rest emits light of wavelength 500 nm. When it is moving at 0.90c toward an observer, the observer detects light of wavelength: 26 nm 115 nm 500 nm 2200 nm 9500 nm
Answer:
The observer detects light of wavelength is 115 nm.
(b) is correct option
Explanation:
Given that,
Wavelength of source = 500 nm
Velocity = 0.90 c
We need to calculate the wavelength of observer
Using Doppler effect
[tex]\lambda_{o}=\sqrt{\dfrac{1-\beta}{1+\beta}}\lambda_{s}[/tex]
Where, [tex]\beta=\dfrac{c}{v}[/tex]
[tex]\lambda_{o}=\sqrt{\dfrac{c-0.90c}{c+0.90c}}\times500\times10^{-9}[/tex]
[tex]\lambda_{o}=115\ nm[/tex]
Hence, The observer detects light of wavelength is 115 nm.
Consider a solenoid of length L, N windings, and radius b (L is much longer than b). A current I is flowing through the wire. If the length of the solenoid became twice as long (2L), and all other quantities remained the same, the magnetic field inside the solenoid would: A: become twice as strong. B: remain the same. C: become one half as strong.
Answer:
B. Remain the same
Explanation:
You have landed on an unknown planet, Newtonia, and want to know what objects will weigh there. You find that when a certain tool is pushed on a frictionless horizontal surface by a 12.2 N force, it moves 16.1 m in the first 2.40 s , starting from rest. You next observe that if you release this tool from rest at 11.0 m above the ground, it takes 2.58 s to reach the ground. What does the tool weigh on Newtonia? What would it weigh on Earth?
Final answer:
To find the weight of the tool on Newtonia, calculate the acceleration using the given displacement and time. Multiply this acceleration by the mass of the tool to find its weight. On Earth, use the acceleration due to gravity to find the weight of the tool.
Explanation:
To determine the weight of the tool on Newtonia, we can use the equation for weight: w = mg. From the given information, we know that the tool moves 16.1 m in the first 2.40 s when a 12.2 N force is applied. Using this information, we can calculate the acceleration of the tool. By using the equation of motion s = ut + 0.5at^2, where s is the displacement, u is the initial velocity (0 m/s), t is the time (2.40 s), and a is the acceleration, we can solve for a. The calculated acceleration can then be multiplied by the mass of the tool to find its weight on Newtonia.
To find the weight of the tool on Earth, we can use the acceleration due to gravity, which is approximately 9.8 m/s². Again using the equation w = mg, we can multiply the mass of the tool by this acceleration to find its weight on Earth.
A model rocket is launched straight upward with an initial speed of 57.0 m/s. It accelerates with a constant upward acceleration of 3.00 m/s2 until its engines stop at an altitude of 140 m. (a) What can you say about the motion of the rocket after its engines stop? This answer has not been graded yet. (b) What is the maximum height reached by the rocket? m (c) How long after liftoff does the rocket reach its maximum height? s (d) How long is the rocket in the air? s
Answer:
(a) The motion of rocket is in upward direction and reaches to maximum height then the rocket starts falling freely.
(b) The maximum height attained by the rocket from the ground is 348.65 m
(c) The time taken by the rocket to maximum height after lift off is 8.85 s.
(d) The total time taken by the rocket in air is 17.3 second.
Explanation:
u = 57 m/s, a = 3 m/s^2, h = 140 m
let the rocket attains a velocity v after covering 140 m and it takes t time to reach upto 140 m.
Use III equation of motion
V^2 = u^2 + 2a h
v^2 = 57^2 + 2 x 3 x 140
v = 63.95 m/s
Now use I equation of motion
v = u + at
t = (63.95 - 57) / 3 = 2.32 s
(a) The motion of rocket is in upward direction and reaches to maximum height then the rocket starts falling freely.
(b) Let H be the maximum height reached by the rocket after the engine stops.
Use III equation of motion
v^2 = u^2 + 2aH
here, v = 0, u = 63.95 m/s, a = - 9.8 m/s^2
0 = 63.95^2 - 2 x 9.8 x H
H = 208.65 m
The maximum height attained by the rocket from the ground is h + H = 140 + 208.65 = 348.65 m
(c) Let t' be the time in which rocket reaches to maximum height after engine is stopped.
Use I equation of motion
v = u + a t'
0 = 63.95 - 9.8 x t'
t' = 6.53 s
The time taken by the rocket to maximum height after lift off is t + t' = 2.32 + 6.53 = 8.85 s.
(d) let t'' be the time taken by the rocket to fall freely
Use II equation of motion
H' = ut'' + 1/2 gt''^2
Here, H' = 348.65 m, u = 0
348.65 = 0 + 0.5 x 9.8 x t''^2
t''^ = 8.44 s
The total time taken by the rocket in air is t + t' + t'' = 2.32 + 6.53 + 8.44 = 17.3 second.
Use 4 to 5 complete sentences, explain the concepts of beats and how beats are produces.
Answer:
Beats is a phenomenon which occurs when two sound waves with different frequencies overlap each other.
Explanation:
When these two sound waves overlap, a third frequency is formed. The places where the two waves reinforces each other, constructive interference takes place and a louder sound is produced whereas, at points where destructive interference takes place a softer sound is produced.
These alternative loud and soft sounds are called beats.
If you are designing a force multiplying system with two pneumatic cylinders. Your system should increase the force by a multiple of 4. What should you do? O decrease the area in the second cylinder O increase the area in the second cylinder
Answer:
decrease the area in the second cylinder
Explanation:
In a two-cylinders pneumatic system, the pressure exerted on the first cylinder is equal to the pressure on the second cylinder:
[tex]p_1 = p_2[/tex]
We can rewrite the pressure as product between force (F) and area (A) of the cylinder:
[tex]F_1 A_1 = F_2 A_2[/tex]
In this system, the output force should be 4 times the input force:
[tex]F_2 = 4 F_1[/tex]
Substituting into the previous equation, we get:
[tex]F_1 A_1 = 4 F_1 A_2\\A_2 = \frac{A_1}{4}[/tex]
This means that the area of the second cylinder must be 1/4 of the area of the first cylinder, so the correct answer is
decrease the area in the second cylinder
Vector A with arrow has a magnitude of 5.35 units, and vector B with arrow has a magnitude of 9.09 units. The two vectors make an angle of 47.0° with each other. Find A with arrow · B with arrow.
Answer:
33.167
Explanation:
The dot product of two vectors is given by
A . B = A B Cos Ф
Where, A and B be the magnitudes of vector A and vector B.
So,
A . B = 5.35 x 9.09 x Cos 47
A . B = 33.167
If the demand function for a commodity is given by the equation p2 + 16q = 1200 and the supply function is given by the equation 300 − p2 + 2q = 0, find the equilibrium quantity and equilibrium price. (Round your answers to two decimal places.)
Answer:
p = 20, q = 50
Explanation:
p² + 16q = 1200
300 - p² + 2q = 0
By adding the two equations, we can eliminate p²:
16q + 300 + 2q = 1200
18q = 900
q = 50
Solving for p:
p² + 16(50) = 1200
p = 20
Final answer:
The equilibrium quantity is approximately 64.29 units, and the equilibrium price is approximately 13.09 dollars, after solving the system of equations set by equating the demand function p^2 + 16q = 1200 and supply function 300 - p^2 + 2q = 0.
Explanation:
To find the equilibrium quantity and equilibrium price for the given demand function p2 + 16q = 1200 and supply function 300 − p2 + 2q = 0, we need to set the quantity demanded equal to the quantity supplied and solve the system of equations.
Firstly, we rearrange the supply function to isolate p2, which yields p2 = 300 - 2q. Next, we substitute this expression into the demand function equation:
(300 - 2q) + 16q = 1200
Now, we solve for q:
300 + 14q = 1200
14q = 900
q = 900 / 14
q = 64.29 (rounded to two decimal places)
Then, we substitute q back into either the original demand or supply function to find p. Using the demand function:
p2 + 16(64.29) = 1200
p2 + 1028.64 = 1200
p2 = 171.36
[tex]p = \sqrt{171.36}[/tex]
p = 13.09 (rounded to two decimal places)
The equilibrium quantity is approximately 64.29 units, and the equilibrium price is approximately 13.09 dollars.
the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force F (in newtons) is the person exerting on the mower? Suppose the mower is moving at 1.5 m/s when the force F is removed. How far will the mower go before stopping?
The person exerts a force of 75 N on the mower, and it will travel approximately 0.529 meters before coming to a stop.
To find the force (F) exerted by the person on the mower, we can use Newton's second law, which states:
[tex]\[ \text{Net force} = \text{Mass} \times \text{Acceleration} \][/tex]
Given:
- Mass of the mower (m) = 24 kg
- Net external force (F_net) = 51 N
- Force of friction (F_friction) = 24 N
We know that net external force is the sum of all forces acting on the object, so:
[tex]\[ F_{\text{net}} = F - F_{\text{friction}} \][/tex]
[tex]\[ 51 \, \text{N} = F - 24 \, \text{N} \][/tex]
[tex]\[ F = 51 \, \text{N} + 24 \, \text{N} \][/tex]
[tex]\[ F = 75 \, \text{N} \][/tex]
So, the person exerts a force of [tex]\( 75 \, \text{N} \)[/tex] on the mower.
Now, to find how far the mower will go before stopping, we can use the equation of motion:
[tex]\[ v^2 = u^2 + 2as \][/tex]
Where:
- \( v \) = final velocity (0 m/s, since the mower stops)
- \( u \) = initial velocity = 1.5 m/s
- \( a \) = acceleration
- \( s \) = distance traveled
We know that the net force acting on the mower is responsible for the acceleration, and it's given by Newton's second law:
[tex]\[ F_{\text{net}} = m \times a \][/tex]
So:
[tex]\[ a = \frac{F_{\text{net}}}{m} \][/tex]
[tex]\[ a = \frac{51 \, \text{N}}{24 \, \text{kg}} \][/tex]
[tex]\[ a ≈ 2.125 \, \text{m/s}^2 \][/tex]
Now, using the equation of motion:
[tex]\[ 0^2 = (1.5 \, \text{m/s})^2 + 2 \times 2.125 \, \text{m/s}^2 \times s \][/tex]
[tex]\[ 0 = 2.25 + 4.25s \][/tex]
[tex]\[ 4.25s = -2.25 \][/tex]
[tex]\[ s = \frac{-2.25}{4.25} \][/tex]
[tex]\[ s ≈ -0.529 \, \text{m} \][/tex]
The negative sign indicates that the mower moves in the opposite direction to its initial motion before stopping. So, the mower will travel approximately [tex]\( 0.529 \, \text{m} \)[/tex] before coming to a stop.
The force exerted by the person on the mower is 75 N. The mower will travel approximately 1.125 meters before stopping.
The net external force is the difference between the force exerted by the person and the force of friction. Thus, we can write:
[tex]F_{net} = F - F_{friction}[/tex]
Substituting the given values:
[tex]51\,\text{N} = F - 24\,\text{N}[/tex]
Solving for F:
[tex]F = 51\,\text{N} + 24\,\text{N} = 75\,\text{N}[/tex]
Finding the Distance the Mower Travels before Stopping:
When the force F is removed, the mower will decelerate due to the frictional force. First, we find the acceleration (a) using Newton's second law:
[tex]F_{friction} = m \cdot a[/tex]
Solving for a gives:
[tex]a = \frac{F_{friction}}{m} = \frac{24\,\text{N}}{24\,\text{kg}} = 1\, \text{m/s}^2[/tex]
Since friction opposes the motion, this is a deceleration (a is negative):
[tex]a = -1\, \text{m/s}^2[/tex]
Next, we use the kinematic equation to find the distance (d):
[tex]v^2 = v_0^2 + 2 \cdot a \cdot d[/tex]
Substituting the known values:
[tex]0 = (1.5 \frac{m}{s})^2 + 2 \cdot (-1 \frac{m}{s^2}) \cdot d[/tex]
Solving for d:
[tex]0 = 2.25 - 2d[/tex]
[tex]2d = 2.25[/tex]
[tex]d = 1.125 \text{m}[/tex]
A wind turbine is rotating counterclockwise at 0.626 rev/s and slows to a stop in 12.9 s. Its blades are 17.9 m in length. What is the centripetal acceleration of the tip of the blades at t=0~\text{s}t=0 s?
Answer:
276.5 m/s^2
Explanation:
The initial angular velocity of the turbine is
[tex]\omega=0.626 rev/s \cdot 2\pi rad/rev =3.93 rad/s[/tex]
The length of the blade is
r = 17.9 m
So the centripetal acceleration is given by
[tex]a=\omega^2 r[/tex]
At the instant t = 0,
[tex]\omega=3.93 rad/s[/tex]
So the centripetal acceleration of the tip of the blades is
[tex]a=(3.93 rad/s)^2 (17.9 m)=276.5 m/s^2[/tex]
What is the maximum speed with which a 1200-kg car can round a turn of radus 84.0 m on a flat road if the coefficient of static friction between tires and road is 0.60. Express your answer to two significant figures and include the apprepriate units.
Explanation:
It is given that,
Mass of the car, m = 1200 kg
Radius of circular path, r = 84 m
The coefficient of static friction between tires and road is 0.60
Static friction is given by :
[tex]f=\mu N=\mu mg[/tex]..............(1)
When an object is moving in a circular path, the centripetal force will act on it. It is given by :
[tex]F=\dfrac{mv^2}{r}[/tex]
Let v is the maximum speed of the car. From equation (1) and (2) as :
[tex]v=\sqrt{\mu gr}[/tex]
g = acceleration due to gravity
[tex]v=\sqrt{0.6\times 9.8\ m/s^2\times 84\ m}[/tex]
v = 22.22 m/s
or
v = 22 m/s
Hence, this is the required solution.
A freight train consists of two 8.00×104-kg engines and 45 cars with average masses of 5.50×104 kg . (a) What force must each engine exert backward on the track to accelerate the train at a rate of 5.00×10–2m/s2 if the force of friction is 7.50×105N, assuming the engines exert identical forces?
Answer:
[tex]4.41\cdot 10^5 N[/tex]
Explanation:
First of all, let's calculate the total mass of the train+the engines:
[tex]m=2(8.00\cdot 10^4 kg) + 45(5.50\cdot 10^4 kg) = 2.64\cdot 10^6 kg[/tex]
Then we can apply Newton's second law, which states that the resultant of the forces is equal to the product between mass (m) and acceleration (a):
[tex]\sum F = ma[/tex] (1)
In this case there are two forces:
- The pushing force exerted by the engines, F
- The frictional force, [tex]F_f = 7.50 \cdot 10^5 N[/tex], in an opposite direction to the acceleration
So (1) becomes
[tex]F-F_f = ma[/tex]
Since the acceleration must be
[tex]a=5.00\cdot 10^{-2} m/s^2[/tex]
We can solve the formula to find F:
[tex]F=ma+F_f = (2.64\cdot 10^6 kg)(5.00\cdot 10^{-2} m/s^2) + 7.50 \cdot 10^5 N = 8.82\cdot 10^5 N[/tex]
However, this is the force exerted by both engines. So the force exerted by each engine must be half this value:
[tex]F=\frac{8.82\cdot 10^5 N}{2}=4.41\cdot 10^5 N[/tex]
Which of the following has the most mass?
A hot air balloon
20 bowling balls
A canoe
30 ounces of lead
Answer:
20 bowling balls
Explanation:
Answer:
20 bowling balls
Explanation:
A hot air balloon would have to have little mass, actually less mass than the air surrounding it in order to be able to float, a canoe ussualy has a mass of around 100 lb, and 30 ounces of lead obviusly are not more than 20 bowling balls, the average mass of a bowling mass is 14 lb, if you multiply that by 20 that is 280 pounds of mass, so the 20 bowling balls have the most mass.
A block of mass 0.244 kg is placed on top of a light, vertical spring of force constant 5 075 N/m and pushed downward so that the spring is compressed by 0.106 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise? (Round your answer to two decimal places.)
Answer with Explanations:
Given:
mass of block, m = 0.244 kg
spring constant, k = 5075 N/m
compression distance in spring, x = 0.106 m
Find how high mass reaches after release (measured from point of release)
Solution:
Use conservation of energy, assuming a light spring (of negligible mass)
stored spring potential energy = gravitational potential energy of block
PE1 = PE2
PE1 = kx^2/2 = 5075*0.106^2/2 = 28.512 J
PE2 = mgh = 0.244 * 9.8 * h = 2.3912 h J
Solve for h = 28.512 / 2.3912 = 11.923 m
To find the maximum height a block rises after being released from a compressed spring, we use the conservation of energy principle to equate the initial spring potential energy to the gravitational potential energy at the peak height, then solve for that height.
The problem involves applying the conservation of energy principle to a mass-spring system when the block is released from compression. Initially, all the energy is stored as potential energy in the compressed spring (spring potential energy), and when the spring is uncompressed, that energy will be converted into kinetic energy of the mass. Finally, when the block leaves the spring and moves upward, its kinetic energy will be converted into gravitational potential energy until the block comes to a stop at its maximum height.
Let's examine the steps needed to solve the problem:
Calculate the initial spring potential energy (Es) using the formula Es = 1/2 k x^2, where k is the spring constant and x is the compression distance.Set this equal to the gravitational potential energy (Ep) at the maximum height using the formula Ep = m g h, where m is the mass, g is the acceleration due to gravity (typically 9.81 m/s^2), and h is the height.Solve for the maximum height h, knowing that energy is conserved, so Es = Ep.Let's perform these calculations for the provided problem:
Calculate the initial spring potential energy: Es = 1/2 (5075 N/m)(0.106 m)^2Equating spring potential energy to gravitational potential energy gives us: 1/2 (5075 N/m)(0.106 m)^2 = (0.244 kg)(9.81 m/s^2)hSolving the above equation for h gives us the maximum height the block will rise after leaving the spring.
A rectangular beam 10 cm wide, is subjected to a maximum shear force of 50000 N, the corresponding maximum shear stress being 3 N/mm2 The depth of beam is (a) 25 cm (c) 16.67 cm (b) 22 cm (d) 30 cm
Answer:
Option B is the correct answer.
Explanation:
Shear stress is the ratio of shear force to area.
We have
Shear stress = 3 N/mm² = 3 x 10⁶ N/m²
Area = Area of rectangle = 10 x 10⁻² x d = 0.1d
Shear force = 50000 N
Substituting
[tex]\texttt{Shear stress}=\frac{\texttt{Shear force}}{\texttt{Area}}\\\\3\times 10^6=\frac{50000}{0.1d}\\\\d=0.1667m=16.67cm[/tex]
Width of beam = 16.67 cm
Option B is the correct answer.
To calculate the depth of the beam, shear stress is divided by shear force to get the cross-sectional area, which is then divided by the width to obtain the depth. The depth is found to be 16.67 cm.
Explanation:The subject of the question relates to the calculation of the depth of a rectangular beam based on the given maximum shear force and shear stress. The formula to be used involves shear stress (τ), shear force (V), and the cross-sectional area (A), which is the product of the beam's width and depth (b × d).
Calculation:
Shear Stress (τ) = Shear Force (V) / Area (A)
Given that the maximum shear stress is 3 N/mm² and the shear force is 50000 N, we can rearrange the formula to solve for the cross-sectional area (A):
A = V / τ = 50000 N / 3 N/mm²
Convert 50000 N to mm² by multiplying by 1000 to maintain the correct units:
A = 50000 N × 1000 mm²/N = 50000000 mm²
A = 50000000 mm² / 3 N/mm² = 16666666.67 mm²
Next, we find the depth, remembering the width of the beam is 10 cm (which is 100 mm because 1 cm = 10 mm):
Depth (d) = A / width (b) = 16666666.67 mm² / 100 mm
Depth (d) = 166666.67 mm which is equal to 16.67 cm.
Therefore, the correct answer is (c) 16.67 cm.
A turntable that spins at a constant 74.0 rpm takes 3.10 s to reach this angular speed after it is turned on. Find its angular acceleration (in rad/s2), assuming it to be constant, and the number of degrees it turns through while speeding up.
Answer:
[tex]2.5 rad/s^2, 688^{\circ}[/tex]
Explanation:
The angular acceleration is given by:
[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
where
[tex]\omega_f = 74.0 rev/min \cdot (\frac{2\pi rad/rev}{60 s/min})=7.75 rad/s[/tex] is the final angular speed
[tex]\omega_i = 0[/tex] is the initial angular speed
t = 3.10 s is the time interval
Solving the equation,
[tex]\alpha = \frac{7.75 rad/s - 0}{3.10 s}=2.5 rad/s^2[/tex]
Now we can find the angular displacement by using:
[tex]\theta = \omega_i t + \frac{1}{2}\alpha t^2[/tex]
Substituting,
[tex]\theta=0+\frac{1}{2}(2.5 rad/s^2)(3.10 s)^2=12.0 rad[/tex]
In degrees:
[tex]\theta = \frac{12.0 rad}{2\pi}\cdot 360^{\circ}=688^{\circ}[/tex]
The escape speed from an object is v2 = 2GM/R, where M is the mass of the object, R is the object's starting radius, and G is the gravitational constant 6.67 × 10-11 m3 kg-1 s-2. What is the approximate escape speed, in km/s, from the Solar System starting from an orbit at 1.0 AU? In this case, the mass of the Sun, 2.3e+30 kg, can be used as the mass of the Solar system
The approximate escape speed from the Solar System at 1.0 AU is around 42 km/s, calculated using the formula that includes the mass of the Sun and the gravitational constant.
The question asks for the approximation of the escape speed from the Solar System starting from an orbit at 1.0 AU using the given mass of the Sun and radius of the Earth's orbit. To find the escape speed, we would use the formula vesc = \\sqrt{2GM/R}, where G is the gravitational constant, M is the mass of the object (in this case, the Sun), and R is the starting radius (1 AU).
The escape speed formula does not depend on the mass of the escaping object, assuming that the central object (Sun) does not move. The escape velocity from the surface of the Earth is approximately 11 km/s, but to escape the sun's gravity, starting from Earth's orbit, requires an escape speed of about 42 km/s.
The escape speed from the Solar System starting from an orbit at 1 AU is 42.1 km/s.
To find the escape speed from the Solar System from an orbit at 1 AU (Astronomical Unit), we use the given formula for escape velocity:
[tex]\[ v_e = \sqrt{\frac{2GM}{R}} \][/tex]
Plugging in these values:
[tex]\[ v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \times 2.3 \times 10^{30} \, \text{kg}}{1.496 \times 10^{11} \, \text{m}}} \][/tex]
[tex]\[ v_e = \sqrt{\frac{2 \times 6.67 \times 2.3 \times 10^{19}}{1.496}} \][/tex]
[tex]\[ v_e = \sqrt{\frac{30.482 \times 10^{19}}{1.496}} \][/tex]
[tex]\[ v_e = \sqrt{20.38 \times 10^{19}} \, \text{m/s} \][/tex]
[tex]\[ v_e = \sqrt{2.038 \times 10^{19}} \times 10 \, \text{m/s} \][/tex]
[tex]\[ v_e = \sqrt{2.038 \times 10^{19}} \, \text{m/s} \][/tex]
Calculating [tex]\( \sqrt{2.038 \times 10^{19}} \)[/tex]:
[tex]\[ v_e = 4.51 \times 10^{9.5} \, \text{m/s} \][/tex]
To convert meters per second to kilometers per second:
[tex]\[ v_e \approx 42.1 \, \text{km/s} \][/tex]
Which of the following would double the amount of current flowing through a piece of metal wire? Quadruple the voltage across it. Quarter the voltage across it. Halve the voltage across it. Double the voltage across it.
Answer:
Double the voltage across it.- last choice