Answer:
d. if any of the above occurs
Explanation:
That is The number of fringes per unit length on the screen can be doubled if
if the distance between the slits is doubled.
if the wavelength is changed to λ = λ/2.
And if the distance between the slits is quadruple the original distance and the wavelength is changed to λ = 2λ
Learning Goal:
To understand the use of Hooke's law for a spring.
Hooke's law states that the restoring force F on a spring when it has been stretched or compressed is proportional to the displacement x of the spring from its equilibrium position. The equilibrium position is the position at which the spring is neither stretched nor compressed.
Recall that F∝x means that F is equal to a constant times x . For a spring, the proportionality constant is called the spring constant and denoted by k. The spring constant is a property of the spring and must be measured experimentally. The larger the value of k, the stiffer the spring.
In equation form, Hooke's law can be written
F =−kx .
The minus sign indicates that the force is in the opposite direction to that of the spring's displacement from its equilibrium length and is "trying" to restore the spring to its equilibrium position. The magnitude of the force is given by F=kx, where x is the magnitude of the displacement.
In Haiti, public transportation is often by taptaps, small pickup trucks with seats along the sides of the pickup bed and railings to which passengers can hang on. Typically they carry two dozen or more passengers plus an assortment of chickens, goats, luggage, etc. Putting this much into the back of a pickup truck puts quite a large load on the truck springs.
Part A
A truck has springs for each wheel, but for simplicity assume that the individual springs can be treated as one spring with a spring constant that includes the effect of all the springs. Also for simplicity, assume that all four springs compress equally when weight is added to the truck and that the equilibrium length of the springs is the length they have when they support the load of an empty truck.
A 70 kg driver gets into an empty taptap to start the day's work. The springs compress 2.4×10−2 m . What is the effective spring constant of the spring system in the taptap?
2.9x10^4
Part B
After driving a portion of the route, the taptap is fully loaded with a total of 23 people including the driver, with an average mass of 70 kg per person. In addition, there are three 15-kg goats, five 3-kgchickens, and a total of 25 kg of bananas on their way to the market. Assume that the springs have somehow not yet compressed to their maximum amount. How much are the springs compressed?
Part C
Whenever you work a physics problem you should get into the habit of thinking about whether the answer is physically realistic. Think about how far off the ground a typical small truck is. Is the answer to Part B physically realistic?
Answer:
A) k = 2,858 10⁴ N / m , B) x_total = - 5,812 10⁻¹ m
C) it is very possible that the obtained value is not realistic for small vehicles
Explanation:
Part A
In this case we can use Hooke's law
F = - k x
force is the weight of the driver
F = W
mg = - k x
k = - mg / x
the springs are compressed x = - 2,4 10⁻² m
k = - 70 9.8 / (-2.4 10⁻²)
k = 2,858 10⁴ N / m
Part B
Since we have the spring constant we must find the complete weight, for this we look for the total masses
each mass is the number of element by the mass of an element
M = 23 70 + 3 15 + 5 3 + 1 25
M = 1695 kg
F = -k x
F = W = M g
Mg = - k x_total
x_total = -M g / k
x_total = -1695 9.8 / 2,858 10⁴
x_total = - 5,812 10⁻¹ m
The negative sign indicates that the springs are compressing
Part C
The truck has lowered 0.58 m = 58 cm
This drop is very large probably in a real vehicle with this drop it would be touching the ground or very close, therefore it is very possible that the obtained value is not realistic for small vehicles
a figures skater rotating at 5 rads with arms extended has a moment of inertia of 2.25 kg. if the arms are pulled in so the moment of inertia decrease to 1.8 what is the final angular speed
Answer:
The final angular speed is 6.25 rad/s
Explanation:
Given;
initial angular speed, ω₁ = 5 rad/s
initial moment of inertia, I₁ = 2.25 kg.m²
Final moment of inertia, I₂ = 1.8 kg.m²
final angular speed, ω₂ = ?
Based on conservation of angular momentum, we will have the following expression;
ω₁I₁ = ω₂I₂
ω₂ = (ω₁I₁ ) / I₂
ω₂ = (5 x 2.25) / 1.8
ω₂ = 6.25 rad/s
Therefore, the final angular speed is 6.25 rad/s
A segment of wire of total length 3.0 m carries a 15-A current and is formed into a semicircle. Determine the magnitude of the magnetic field at the center of the circle along which the wire is placed.
Answer:
4.9x10^-6T
Explanation:
See attached file
Jane is collecting data for a ball rolling down a hill. she measure out a set of different distances and then proceeds to use a stopwatch to find the time it takes the ball to roll each distance
Answer:
The Independent variable in this experiment is the time taken by the ball to roll down each distance.
The dependent variable is the distance through which the ball rolls
The control variables are: slope of hill, weight, of the ball, size of ball, wind speed, surface characteristics of the ball.
Explanation:
The complete question is
Jane is collecting data for a ball rolling down a hill. She measures out a set of different distances and then proceeds to use a stop watch to find the time it takes the ball to roll. What are the independent, dependent, and control variables in this experiment?
Independent variable have their values not dependent on any other variable in the scope of the experiment. The time for the ball to roll down the hill is not dependent on any other variable in the experiment. Naturally, some common independent variables are time, space, density, mass, fluid flow rate.
A dependent variable has its value dependent on the independent variable in the experiment. The value of the distance the ball rolls depends on the time it takes to roll down the hill.
The relationship between the dependent and independent variables in an experiment is given as
y = f(x)
where y is the output or the dependent variable,
and x is the independent variable.
Control variables are those variable that if not held constant could greatly affect the results of an experiment. For an experiment to be more accurate, control variables should be confined to a given set of value throughout the experiment.
The driver of a stationary car hears a siren of an approaching police car at a frequency of 280Hz. If the actual frequency of the siren is 240Hz, find the speed of the police car (speed of sound is 343m/s).
Answer:
The speed of the police car is 294 m/s
Explanation:
Given;
frequency of the siren in air, f = 280 Hz
speed of sound in air, v = 343 m/s
Determine the wavelength of the sound in air to the stationary car:
v = fλ
where;
λ is wavelength of the sound
λ = v/f
λ = 343 / 280
λ = 1.225 m
Now, determine the speed at which the police car is approaching the stationary car;
The actual frequency of the police car, F = 240 Hz
V = Fλ
Where;
V is speed of the police car
λ is the distance between the police car and the stationary car, (wavelength)
V = 240 x 1.225
V = 294 m/s
Therefore, the speed of the police car is 294 m/s
Based on the graph below, what prediction can we make about the acceleration when the force is 0 newtons? A. It will be 0 meters per second per second. B. It will be 5 meters per second per second. C. It will be 10 meters per second per second. D. It will be 15 meters per second per second.
PLZ HURRY WILL MARK BRAINLIEST IF CORRECT
Answer:
Option A
Explanation:
Acceleration will be obviously zero when Force = 0
That is how:
Force = Mass * Acceleration
So, If force = 0
0 = Mass * Acceleration.
Dividing both sides by Mass
Acceleration = 0/Mass
Acceleration = 0 m/s²
Answer:
[tex]\boxed{\mathrm{A. \: It \: will \: be \: 0 \: meters \: per \: second \: per \: second. }}[/tex]
Explanation:
[tex]\mathrm{force=mass \times acceleration}[/tex]
The force is given 0 newtons.
[tex]\mathrm{force=0 \: N}[/tex]
Plug force as 0.
[tex]\mathrm{0=mass \times acceleration}[/tex]
Divide both sides by mass.
[tex]\mathrm{\frac{0}{mass} =acceleration}[/tex]
[tex]\mathrm{0 =acceleration}[/tex]
[tex]\mathrm{acceleration= 0\: m/s/s}[/tex]
An electric field can be created by a single charge or a distribution of charges. The electric field a distance from a point charge has magnitude E = k|q'|/r^2.
The electric field points away from positive charges and toward negative charges. A distribution of charges creates an electric field that can be found by taking the vector sum of the fields created by individual point harges. Note that if a charge is placed in an electric field created by q', q will not significantly affect the electric field if it is small compared to q'. Imagine an isolated positive point charge with a charge Q (many times larger than the charge on a single electron).
1. There is a single electron at a distance from the point charge. On which of the following quantities does the force on the electron depend?
a. the distance between the positive charge and the electron
b. the charge on the electron
c. the mass of the electron
d. the charge of the positive charge
e. the mass of the positive charge
f. the radius of the positive charge
g. the radius of the electron
2. For the same situation as in Part A, on which of the following quantities does the electric field at the electron's position depend?
a. the distance between the positive charge and the electron
b. the charge on the electron
c. the mass of the electron
d. the charge of the positive charge
e. the mass of the positive charge
f. the radius of the positive charge
g. the radius of the electron
Answer:
a) true.
b) True
c) False. In the equation above the mass does not appear
d) True
e) False. Mass does not appear in the equation
f) False. The load even when distributed in the space can be considered concentrated in the center
Explanation:
1. The electric force is given by the relation
F = k Q e / r2
where k is the Coulomb constant, Q the charge used, e the charge of the electron and r the distance between the two.
The strength depends on:
a) true.
b) True
c) False. In the equation above the mass does not appear
d) True
e) False. Mass does not appear in the equation
f) False. The load even when distributed in the space can be considered concentrated in the center
two.
a) True
b) Treu
c) Fail
f) false
For a single electron located at a distance from a positive charge, we have:
1. The force on the electron depends on the distance between it and the positive charge (option a) and the charge of both particles (option b and d).
2. The electric field at the electron's position depends on the distance between the positive charge and it (option a) and the charge of the positive particle (option d).
Part 1
The force on a single electron at a distance from the point charge is given by Coulomb's law:
[tex] F = \frac{Kq_{1}q_{2}}{r^{2}} [/tex] (1)
Where:
K: is the Coulomb's constant q₁: is the charge of the positive chargeq₂: is the charge of the electrond: is the distance between the positive charge and the electronAs we can see in equation (1), the force on the electron by the positive charge depends on both charges q₁ and q₂, and the distance, so the correct options are:
a. The distance between the positive charge and the electron
b. The charge on the electron
d. The charge of the positive charge
The other options (c, e, f, and g) are incorrect because the electric force does not depend on the particles' masses or their radii.
Part 2The electric field (E) at a distance "r" from a point charge is given by:
[tex] E = \frac{Kq_{1}}{r^{2}} [/tex] (2)
From equation (2), we can see that the electric field is directly proportional to the charge and inversely proportional to the distance of interest (r).
The electric field at the electron's position is given by the one produced by the positive charge, so the correct options are:
a. The distance between the positive charge and the electron
d. The charge of the positive charge
The other options (b, c, e, f, and g) are incorrect because the electric field is independent of the mass of the charges involved and their radii.
Therefore, the correct options for part 1 are a, b, and d and for part 2 are a and d.
Learn more about the electric field here:
brainly.com/question/13308086
I hope it helps you!
4. A 120 V potential difference is applied to a space heater whose resistance is 14 Ω when hot. a) At what rate is electrical energy transferred to thermal energy? b) What is the cost for 5.0 h at $0.05/kWh?
Answer:
a) P = 1028.6 W = 1.03 KW
b) Cost = $0.25
Explanation:
a)
The rate of electrical energy transfer or power of the heater is given as:
P = VI
where,
P = Rate of Electrical Energy Transferred = ?
V = Potential Difference = 120 V
I = Current
but, from Ohm's Law:
V = IR
I = V/R
Therefore,
P = V²/R
where,
R = Resistance = 14 Ω
Therefore,
P = 120²/14
P = 1028.6 W = 1.03 KW
b)
First we find energy used:
Energy = E = Pt
where,
t = time = 5 h
Therefore,
E = (1.03 KW)(5 h)
E = 5.14 KWh
Now, the cost is given as:
Cost = (E)(Unit Price)
Cost = (5.14 KWh)($0.05/KWh)
Cost = $0.25
You throw a stone vertically upward with a speed of 26.0 m/s. (a) How fast is it moving when it reaches a height of 15.0 m? (b) How much time is required to reach this height when it's falling down? a. 19.5 m/s , b. 4.51 s a. 17.9 m/s , b. 0.620 s a. 19.5 m/s , b. 0.800 s a. 17.9 m/s , b. 4.28 s a. 380 m/s , b. 8 s
Answer:
ok well
Explanation:
teghe
Answer:
v = 19.5 m/s
t = 4.51 s
Explanation:
a)
given:
height is 15m from the ground
initial velocity Vi = 26 m/s
acceleration a or g = 9.81 m/s²
formula: Vf² = Vi² + 2aΔy
26² = Vi² + 2 (9.81) 15
Vi = 19.5 m/s
now you can calculate the time by using the equations below:
Δy = 1/2 (Vi + Vf) t
Vf = Vi + a t
Δy = Vi t + 1/2 a t
time must be 4.51 s
23.15. Can an object carry a charge of 2.0 10-19 C?
Answer:
Ok, the minimal quantity of charge that we can find is on the electron or in the proton (the magnitude is the same, but the sign is different)
Where the charge of a single proton is:
C = 1.6x10^-19 C
Now, you need to remember that when we are working with charges, we are working with discrete math:
What means that?
If the minimum positive is the charge of one proton, then the consecutive charge will be the charge of two protons (there is no somethin in between)
So the consecutive charge will be:
C = 2*1.6x10^-19 C = 3.2x10^-19 C.
So, because we are working in discrete math, we can not have any object that has charge between 1.6x10^-19 C and 3.2x10^-19 C.
Particularly, 2.0x10^-19 C is in that range, so we can conclude that:
No, an object can not carry a charge of 2.0x10^-19 C.
a large crane has a mass of 8500kg calculate the weight of the crane
Answer:
Weight is 83 385 N
Explanation:
Weight is calculated by multiplying the mass by the gravitational acceleration constant
Weight = mass* gravity
Assuming that the gravitational constant is 9.81 m/s^2
Weight = mass* gravity
Weight of crane = (8500 kg)*(9.81 m/s^2)
Weight = 83 385 kg*m/s^2 or 83 385 N
You shine unpolarized light with intensity 54.0 W/m^2 on an ideal polarizer, and then the light that emerges from this polarizer falls on a second ideal polarizer. The light that emerges from the second polarizer has intensity 19.0 W/m^2. Find the angle between the polarizing axes of the two polarizers.°
Answer:
The angle between the polarizing axes of the two polarizers is 54°
Explanation:
Given;
intensity of unpolarized light, I₀ = 54.0 W/m²
intensity of light that emerges from second ideal polarizer, I₁ = 19.0 W/m²
The angle between the polarizing axes of the two polarizers is dtermined by applying Malus' law for intensity of a linearly polarized light passing through a polarizer.
I₁ = I₀Cos²θ
Cos²θ = I₁ / I₀
Cos²θ = 19 / 54
Cos²θ =0.3519
Cos θ = √0.3519
Cos θ = 0.5932
θ = Cos⁻¹(0.5932)
θ = 53.6°
θ = 54°
Therefore, the angle between the polarizing axes of the two polarizers is 54°
An insulating hollow sphere has inner radius a and outer radius b. Within the insulating material the volume charge density is given by rho(r)=αr,where α is a positive constant.
A). What is the magnitude of the electric field at a distance r from the center of the shell, where a
Express your answer in terms of the variables α, a, r, and electric constant ϵ0.
B) .A point charge
q is placed at the center of the hollow space, at r=0. What value must q have (sign and magnitude) in order for the electric field to be constant in the region a
Express your answer in terms of the variables α, a, and appropriate constants.
C). What then is the value of the constant field in this region?
Express your answer in terms of the variable αand electric constant ϵ0.
Answer:
E = α/2∈₀ [ 1 - a²/r² ]
Ф = α/2∈₀
Explanation:
Using Gauss Law:
ρ(r) = a/r, dA
= 4 π r²d r
Ф = [tex]\int\limits^r_a[/tex] ρ(r')dA
Ф[tex]_{encl}[/tex] = [tex]\int\limits^r_a[/tex] ρ(r')dA
= 4πα [tex]\int\limits^r_a[/tex] r'dr'
Ф[tex]_{encl}[/tex] = 4 π α 1/2(r²-a²)
E(4πr²) = [tex]2\pi\alpha (r^{2}-a^{2} )/[/tex]∈₀
= [tex]2\pi\alpha (r^{2}-a^{2} )/[/tex]∈₀(4πr²)
= α (r² - a²) / 2 ∈₀ (r²)
= α/2∈₀ [ r²/r² - a²/r² ]
E = α/2∈₀ [ 1 - a²/r² ]
Electric field of the point charge:
E[tex]_{q}[/tex] = q / 4π∈₀r²
[tex]E_{total}[/tex] = α / 2 ∈₀ - (α / 2 ∈₀ )(a² / r²) + q / 4 π ∈₀ r²
For [tex]E_{total}[/tex] to be constant:
- (αa²/ 2 ∈₀ ) + q / 4 π ∈₀ = 0 and q = 2παa²
-> α / 2 ∈₀ - αa²/ 2 ∈₀ + 2παa² / 4 π ∈₀
= α - αa² + αa² / 2 ∈₀
= α /2 ∈₀
Hence:
Ф = α/2∈₀
When static equilibrium is established for a charged conductor, the electric field just inside the surface of the conductor is
Answer:
The electric field just inside the charged conductor is zero.
Explanation:
Electric field is defined as the region where electrical force is experienced by an electric charge usually as a result of the presence of another electric charge. A charged conductor is said to be in electrostatic equilibrium when it is in an electrostatically balanced state. This simply means a state in which the free electrical charges in the charged conductor have stopped moving.
For any charged conductor that has attained electrostatic equilibrium, the electric field at any point below the surface of the charged conductor falls to zero. Hence the electric field just inside the charged conductor is zero.
A converging lens has the focal length of 25 cm. A 10-cm object is placed at 30 cm in front of the lens. How far is the image from the lens? What is the size of the image?
Explanation:
Given that,
Focal length of a converging lens, f = +25 cm
Size of the object, h = 10 cm
Object distance, u = -30 cm
We need to find the image distance and the size of the image.
Using lens formula, [tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}[/tex], v is image distance
[tex]\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{25}+\dfrac{1}{(-30)}\\\\v=150\ cm[/tex]
Magnification,
[tex]m=\dfrac{v}{u}=\dfrac{h'}{h}[/tex]
h' is the size of image
[tex]h'=\dfrac{vh}{u}\\\\h'=\dfrac{150\times 10}{(-30)}\\\\h'=-50\ cm[/tex]
So, the image distance is 150 cm the size of image is 50 cm.
What is dark energy?
Explanation:
Dark Energy. Dark Energy is a hypothetical form of energy that exerts a negative, repulsive pressure, behaving like the opposite of gravity. It has been hypothesised to account for the observational properties of distant type Ia supernovae, which show the universe going through an accelerated period of expansion
15. The
of a sine wave is the time it takes to complete one cycle of the wave.
O A. maximum amplitude
O B. minimum amplitude
O C. average value
O D. wavelength
That TIME is called the "period" of the wave.
(It's not one of the choices.)
When an object has a net force of zero, then it is said to be in ____________.
Answer:
Equilibrium
Explanation:
Physics terminology I guess? Equilibrium means that an object isn't moving.
Answer:
Balanced Forces
Explanation:
When forces are in balance, acceleration is zero. Velocity is constant and there is no net or unbalanced force. A plane will fly at constant velocity if the acceleration is zero.
A long straight wire carries a conventional current of 0.7 A. What is the approximate magnitude of the magnetic field at a location a perpendicular distance of 0.053 m from the wire due to the current in the wire
Answer:
2.64 x 10⁻⁶T
Explanation:
The magnitude of the magnetic field produced by a long straight wire carrying current is given by Biot-Savart law as follows: "The magnetic field strength is directly proportional to the current on the wire and inversely proportional to the distance from the wire". This can be written mathematically as;
B = (μ₀ I) / (2π r) ----------------(i)
B is magnetic field
I is current through the wire
r is the distance from the wire
μ₀ is the magnetic constant = 4π x 10⁻⁷Hm⁻¹
From the question;
I = 0.7A
r = 0.053m
Substitute these values into equation (i) as follows;
B = (4π x 10⁻⁷ x 0.7) / (2π x 0.053)
B = 2.64 x 10⁻⁶T
Therefore the approximate magnitude of the magnetic field at that location is 2.64 x 10⁻⁶T
Two very long parallel wires are a distance d apart and carry equal currents in opposite directions. The locations where the net magnetic field due to these currents is equal to zero are
Answer:
Its not zero anywhere
Explanation:
The magnetic field B at a distance r due to a long conductor carrying current I is given as
B= μol/2pi r
so the net magnetic field due to the current is not zero anywhere
On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m 3 kg/m3 , length 81.2 cmcm and diameter 2.60 cmcm from a storage room to a machinist. Calculate the weight of the rod, www. Assume the free-fall acceleration is ggg = 9.80 m/s2m/s2 .
Answer:
The weight of the rod is 32.87 N
Explanation:
Density of the rod = 7800 kg/m
length of the rod = 81.2 cm = 0.812 m
diameter of rod = 2.60 cm = 0.026 m
acceleration due to gravity = 9.80 m/s^2
The rod can be assumed to be a cylinder.
The volume of the rod can be calculated as that of a cylinder, and can be gotten as
V = [tex]\frac{\pi d^{2} l}{4}[/tex]
where d is the diameter of the rod
l is the length of the rod
V = [tex]\frac{3.142* 0.026^{2}* 0.812}{4}[/tex] = 4.3 x 10^-4 m^3
We know that the mass of a substance is the density times the volume i.e
mass m = ρV
where ρ is the density of the rod
V is the volume of the rod
m = 4.3 x 10^-4 x 7800 = 3.354 kg
The weight of a substance is the mass times the acceleration due to gravity
W = mg
where g is the acceleration due to gravity g = 9.80 m/s^2
The weight of the rod W = 3.354 x 9.80 = 32.87 N
A piece of electronic equipment that is surrounded by packing material is dropped so that it hits the ground with a speed of 4 m/s. After impact, the equipment experiences an acceleration of a = 2kx, where k is a constant and x is the compression of the packing material. If the packing material experiences a maximum compression of 20 mm, determine the maximum acceleration of the equipment.
Answer:
Maximum acceleration is 800m/s^2
Explanation:
See attached file
A ballistic pendulum consists of a 4 kg wooden block originally at rest at θ = 0o . When a 2 g bullet strikes and becomes embedded it, it is observed that the block swings upward to a maximum angle of θ = 6o . Estimate the speed of the bullet just before the impact.
Answer:
733m/s
Explanation:
Assuming that Just after impact:
given that at lowest point Is
T2 + V2 = T3 + V3
1 /2(4 + 0.002) (vB)²2 + 0 = 0 + (4 + 0.002)(9.81)(1.25)(1 -cos 6°)
(vB)2 = 0.3665
m>s For the system of bullet and block: ( S+ ) Σmv1 = Σmv2 0.002(vB)1 = (4 + 0.002)(0.3665) (vB)1 = 733 m/s
Current folw in which dirction
A piece of thin uniform wire of mass m and length 3b is bent into an equilateral triangle so that each side has a length of b. Find the moment of inertia of the wire triangle about an axis perpendicular to the plane of the triangle and passing through one of its vertices.
Answer:
Mb²/2
Explanation:
Pls see attached file
In practice, a good insulator In practice, a good insulator A. slows heat flow. B. speeds negative heat flow. C. stops heat flow. D. all of the above
Answer:
The answer is A. slows heat flow.Explanation:
An insulator is a material that impedes the movement of heat or electric current from flowing.
Theoretically good heat insulators stops the movement of heat, while practically this insulation can only be slowed down.
Hence from the options listed the correct answer practically is
A. slows heat flow.Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously.
Now, there is some information missing to this problem, since generally you will be given a figure to analyze like the one on the attached picture. The whole problem should look something like this:
"Beam AB has a negligible mass and thickness, and supports the 200kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously."
Answer:
[tex]\mu_{sB}=0.126[/tex]
[tex]\mu_{sC}=0.168[/tex]
Explanation:
In order to solve this problem we will need to draw a free body diagram of each of the components of the system (see attached pictures) and analyze each of them. Let's take the free body diagram of the beam, so when analyzing it we get:
Sum of torques:
[tex]\sum \tau_{A}=0[/tex]
[tex]N(3m)-W(1.5m)=0[/tex]
When solving for N we get:
[tex]N=\frac{W(1.5m)}{3m}[/tex]
[tex]N=\frac{(1962N)(1.5m)}{3m}[/tex]
[tex]N=981N[/tex]
Now we can analyze the column. In this case we need to take into account that there will be no P-ycomponent affecting the beam since it's a slider and we'll assume there is no friction between the slider and the column. So when analyzing the column we get the following:
First, the forces in y.
[tex]\sum F_{y}=0[/tex]
[tex]-F_{By}+N_{c}=0[/tex]
[tex]F_{By}=N_{c}[/tex]
Next, the forces in x.
[tex]\sum F_{x}=0[/tex]
[tex]-f_{sB}-f_{sC}+P_{x}=0[/tex]
We can find the x-component of force P like this:
[tex]P_{x}=360N(\frac{4}{5})=288N[/tex]
and finally the torques about C.
[tex]\sum \tau_{C}=0[/tex]
[tex]f_{sB}(1.75m)-P_{x}(0.75m)=0[/tex]
[tex]f_{sB}=\frac{288N(0.75m)}{1.75m}[/tex]
[tex]f_{sB}=123.43N[/tex]
With the static friction force in point B we can find the coefficient of static friction in B:
[tex]\mu_{sB}=\frac{f_{sB}}{N}[/tex]
[tex]\mu_{sB}=\frac{123.43N}{981N}[/tex]
[tex]\mu_{sB}=0.126[/tex]
And now we can find the friction force in C.
[tex]f_{sC}=P_{x}-f_{xB}[/tex]
[tex]f_{sC}=288N-123.43N=164.57N[/tex]
[tex]f_{sC}=N_{c}\mu_{sC}[/tex]
and now we can use this to find static friction coefficient in point C.
[tex]\mu_{sC}=\frac{f_{sC}}{N}[/tex]
[tex]\mu_{sC}=\frac{164.57N}{981N}[/tex]
[tex]\mu_{sB}=0.168[/tex]
The greater the frequency of the waves, the ____________ the pitch.
Answer:
Higher.
Explanation:
The greater the frequency the bigger the amplitude gets and the greater pitch gets.
Think - more energy, bigger waves, more waves, and higher sound
The maximum velocity of a particle executing simple harmonic motion with amplitude 7.0 mm is 4.4 m/s. The period of oscillation is? A. 0.5 s B. 0.1 s C. 100 s D. 0.01 s E. 10 s Reset Selection
Answer:
D. 0.01 s
Explanation:
The maximum speed is the amplitude times the angular frequency.
v = Aω
4.4 m/s = (0.0070 m) ω
ω = 628.6 rad/s
The period is:
ω = 2π / T
T = 2π / ω
T = 2π / (628.6 rad/s)
T = 0.01 s
3. Identify the mathematical relationship that exists between pressure and volume, when temperature and quantity are held constant, as being directly proportional or inversely proportional. Explain your answer and write an equation that relates pressure and volume to a constant, using variables
Answer:
P = cte / V
therefore pressure and volume are inversely proportional
Explanation:
For this exercise we can join the ideal gases equation
PV = n R T
they indicate that the amount of matter and the temperature are constant, therefore
PV = cte
P = cte / V
therefore pressure and volume are inversely proportional