"Longitudinal wave" is the wave where the difference between the coils increases as well as decreases.
Generating waves whenever the form of communication being displaced in a similar direction as well as in the reverse way of the wave's designated points, could be determined as Longitudinal waves.A wave running the length of something like a Slinky stuffed animal, which expands as well as reduces the spacing across spindles, produces a fine image or graphic.
Thus the above answer is correct.
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We will determine the amount of electric energy stored in a capacitor by discharging it through a light bulb. Light bulbs are rated according to their power output at a given voltage. Considering that power is the rate that energy is converted from one form to another (or, equivalently, work is done) per unit time, the energy stored in an initially-charged capacitor that is hooked up to the light bulb through which the capacitor discharges is approximately
A. the power rating of the light bulb divided by the time that the bulb remains lit.
B. simply the time that the light bulb remains lit.
C. the product of the power rating of the light bulb and the time that it remains lit.
D. the time that the light bulb remains lit divided by the power rating of the bulb.
Answer:
C. the product of the power rating of the light bulb and the time that it remains lit.
Explanation:
The power rating of the light is bulb is defined as the energy supplied to the light bulb divided by the time the bulb is lit up. Therefore,
[tex]P = \frac{E}{t}\\\\E = Pt[/tex]
where,
E = Energy Supplied to the bulb = Energy stored in capacitor = ?
P = Power rating of the bulb
t = time the bulb is lit up
Hence, the correct option is:
C. the product of the power rating of the light bulb and the time that it remains lit.
Army is standing still on the ground; Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward. How fast does Bill see Carlos moving and in what direction?
a. 10 mis eastward
b. 5 m/s eastward
c. 15 m/s westward
d. 20 m/s westward
e. 10 m/s westward
Explanation:
Given that,
Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward.
Taking eastward as positive direction, we have:
[tex]v_B=+5\ m/s[/tex]is the velocity of Bill with respect to Amy (which is stationary)
[tex]v_c=15\ m/s[/tex] is the velocity of Carlos with respect to Amy.
Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is
[tex]v_B=+5\ m/s[/tex]
Therefore, Carlos velocity in Bill's reference frame will be
[tex]v_c'=-15\ m/s-(+5\ m/s)\\\\=-20\ m/s[/tex]
So, the magnitude is 20 m/s and the direction is westward (negative sign).
Categorize each statement as true or false.
1. Electric field lines radiate away from positive charges and towards negative charges.
2. Electric field is always perpendicular to equipotential lines.
3. The electric field points in the direction of increasing electric potential.
4. The electric field inside a parallel plate capacitor decreases as it approaches the negative plate.
5. The units of electric field are either newtons per coulomb or volts per meter.
Answer:
1. True
2. True
3. False
4. True
5. True
1. A turtle and a rabbit are to have a race. The turtle’s average speed is 0.9 m/s. The rabbit’s average speed is 9 m/s. The distance from the starting line to the finish line is 1500 m. The rabbit decides to let the turtle run before he starts running to give the turtle a head start. If the rabbit started to run 30 minutes after the turtle started, can he win the race? Explain.
Answer:no
Explanation:because 0.9*(30*60)=0.9*1800=1620
The turtle has already won the race
Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours
What will be the speed of the rabbit and the turtle?It is given
[tex]V_{t} = 0.9 \frac{m}{s}[/tex]
[tex]V_{r} = 9 \frac{m}{s}[/tex]
[tex]D=1500 m[/tex]
Time taken by turtle
[tex]T= \dfrac{D}{V_{t} }=\dfrac{1500}{0.9_{} }[/tex]
[tex]T=1666 minutes= 27 hours[/tex]
Time taken by rabbit
[tex]T= \dfrac{D}{V_{r} }=\dfrac{1500}{9_{} }[/tex]
[tex]T=166 minutes[/tex]
since rabbit started 30 minutes after turtle then
[tex]T= 136+30=196 minutes[/tex]
[tex]T= 3.2 hours[/tex]
Hence Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours
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A series of pulses, each of amplitude 0.1 m, is sent down a string that is attached to a post at one end. The pulses are reflected at the post and travel back along the string without loss of amplitude. What is the net displacement at a point on the string where two pulses are crossing
Answer:
A_resulting = 0.2 m
Explanation:
Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.
With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is
A_res = 2A
A_resultant = 2 .01
A_resulting = 0.2 m
What would you expect to happen to the velocity of the bobber if the mass of the washers in the cylinder remained the same and the radius was doubled?
Answer:
The velocity becomes [tex]v\sqrt 2[/tex].
Explanation:
The force acting on the bobber is centripetal force.
The centripetal force is given by
[tex]F =\frac{mv^2}{r}[/tex]
when mass remains same, radius is doubled and the force is same, so the velocity is v'.
[tex]F =\frac{mv^2}{r}=\frac{mv'^2}{2r}\\\\v'=v\sqrt 2[/tex]
Principal axis is the:________
a. straight line drawn from the center of curvature to the mid point of the mirror.
b. straight line drawn from the center of curvature to a point on the outer edge of the mirror.
c. straight line drawn from the center of curvature to any point of the mirror.
d. straight line joining any two points on the mirror.
e. None of the other answers given is correct.
Answer:
d. straight line joining any two points on the mirror
Explanation:
Principal axis is the straight line that passes through the center of a mirror, which is also perpendicular to the surface of the mirror.
The principal axis connects the principal focus and the center of curvature of the mirror. In other words, it is a straight line or axis on which the center of curvature and principal focus can be found. The principal axis joins these two points; the center of curvature and principal focus.
Therefore, the correct option is "D"
d. straight line joining any two points on the mirror
The pan flute is a musical instrument consisting of a number of closed-end tubes of different lengths. When the musician blows over the open ends, each tube plays a different note. The longest pipe is 0.31 m long.
What is the frequency of the note it plays? Assume room temperature of 20∘C.
Answer:
f = 276.6 Hz
Explanation:
This musical instrument can be approximated to a tube system where each tube has one end open and the other closed.
In the closed part there is a node and in the open part a belly or antinode. Therefore the wavelength is
L = λ/ 4
speed is related to wavelength and frequency
v = λ f
λ = v / f
we substitute
L = v / 4f
f = v / 4L
the speed of sound at 20ºC is
v = 343 m / s
let's calculate
f = [tex]\frac{343 }{4 \ 0.31}[/tex]
f = 276.6 Hz
A 0.40-kg mass attached to the end of a string swings in a vertical circle having a radius of 1.8 m. At an instant when the string makes an angle of 40 degrees below the horizontal, the speed of the mass is 5.0 m/s. What is the magnitude of the tension in the string at this instant
Answer:
[tex]T=8.1N[/tex]
Explanation:
From the question we are told that:
Mass m=0.40
Radius r=1.8m
Angle Beneath the Horizontal \theta =40 \textdegree
Speed v=5.0m/s
The Tension Angle
[tex]\alpha=90-\theta\\\\\alpha=90-40[/tex]
[tex]\alpha=50 \textdegree[/tex]
Generally the equation for Tension is is mathematically given by
[tex]T=\frac{mv^2}{r}+mgcos \alpha[/tex]
[tex]T=\frac{0.40*5^2}{1.8}+0.40*5cos50[/tex]
[tex]T=8.1N[/tex]
Suppose that 2 J of work is needed to stretch a spring from its natural length of 34 cm to a length of 46 cm. (a) How much work is needed to stretch the spring from 36 cm to 41 cm
Answer:
0.83 J of work
Explanation:
2 J of work is required to stretch a spring from 34cm to 46cm
So that is 12cm stretched with 2 J of work
We can make that 6cm for 1 J of work
So, we need the find the work for stretching 36cm to 41cm
Which is 5cm
So, What is the work required to stretch 5cm?
1 J of work for 6cm
x work for 5cm
So, by proportion method
1 : 6 :: x : 5
6 * x = 1 * 5
6x = 5
x = 5/6
= 0.83
So to stretch 36cm to 41cm we need 0.83 J of work
Calculate the minimum area moment of inertia for a rectangular cross-section with side lengths 6 cm and 4 cm.
52 cm4
72 cm4
32 cm4
24 cm4
2 cm4
Answer:
Minimum Area of rectangle = 24 centimeter²
Explanation:
Given:
Length of rectangle = 6 centimeter
Width of rectangle = 4 centimeter
Find:
Minimum Area of rectangle
Computation:
Minimum Area of rectangle = Length of rectangle x Width of rectangle
Minimum Area of rectangle = 6 x 4
Minimum Area of rectangle = 24 centimeter²
find the exit angle relative to the horizontal in an isosceles triangle with 36 °
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Example 2.13 The acceleration a of a particle in a time t is given by the equation a = 2+ 5t^2. Find the instantaneous velocity after 3s. Solution
Answer:
the instantaneous velocity is 51 m/s
Explanation:
Given;
acceleration, a = 2 + 5t²
Acceleration is the change in velocity with time.
[tex]a = \frac{dv}{dt} \\\\a = 2 + 5t^2\\\\The \ acceleration \ (a) \ is \ given \ so \ we \ have \ to \ find \ the \ velocity \ (v)\\\\To \ find \ the \ velocity, \ integrate\ both \ sides \ of \ the \ equation\\\\2 + 5t^2 = \frac{dv}{dt} \\\\\int\limits^3_0 {(2 + 5t^2)} \, dt = dv\\\\v = [2t + \frac{5t^3}{3} ]^3_0\\\\v = 2(3) + \frac{5(3)^3}{3} \\\\v = 6 + 5(3)^2\\\\v = 6 + 45\\\\v = 51 \ m/s[/tex]
Therefore, the instantaneous velocity is 51 m/s
A 1030 kg car has four 12.0 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles
Answer:
The required fraction is 0.023.
Explanation:
Given that
Mass of a car, m = 1030 kg
Mass of 4 wheels = 12 kg
We need to find the fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles.
The rotational kinetic energy due to four wheel is
[tex]=4\times \dfrac{1}{2}I\omega^2\\\\=4\times \dfrac{1}{2}\times \dfrac{1}{2}mR^2(\dfrac{v}{R})^2\\\\=mv^2[/tex]
Linear kinetic Energy of the car is:
[tex]=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times Mv^2[/tex]
Fraction,
[tex]f=\dfrac{mv^2}{\dfrac{1}{2}Mv^2}\\\\f=\dfrac{m}{\dfrac{1}{2}M}\\\\f=\dfrac{12}{\dfrac{1}{2}\times 1030}\\\\=0.023[/tex]
So, the required fraction is 0.023.
Required information
You are designing a high-speed elevator for a new skyscraper. The elevator will have a mass limit of 2400 kg (including
passengers). For passenger comfort, you choose the maximum ascent speed to be 18.0 m/s, the maximum descent speed
to be 10.0 m/s, and the maximum acceleration magnitude to be 1.80 m/s2. Ignore friction.
What is the minimum upward force that the supporting cables exert on the elevator car?
KN
Answer:
19,224 N
Explanation:
The given parameters are;
The mass limit of the elevator = 2,400 kg
The maximum ascent speed = 18.0 m/s
The maximum descent speed = 10.0 m/s
The maximum acceleration = 1.80 m/s²
Given that the acceleration due to gravity, g ≈ 9.81 m/s²
The minimum upward force that the elevator cable exert on the elevator car, [tex]F_{min}[/tex] , is given in the downward motion as follows;
[tex]F_{min}[/tex] = m·g - m·a
∴ [tex]F_{min}[/tex] = 2,400 kg × 9.81 m/s² - 2,400 kg × 1.80 m/s² = 19,224 N
The minimum upward force that the elevator cable exert on the elevator car, [tex]F_{min}[/tex] = 19,224 N
Write a short note of the following
a) Reflection
b) Refraction
c) Diffraction
Answer:
a) Light that passes through the floor to reveal yourself (not shadow).
b) 2 rays of light that bounce between 2 transparent media.
c) I don't know what is Diffraction?
You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 690.0 kg and was traveling eastward. Car B weighs 520.0 kg and was traveling westward at 74.0 km/h. The cars locked bumpers and slid eastward with their wheels locked for 6.00 m before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.750. 1) How fast (in kilometer per hour) was car A traveling just before the collision
Answer:
The speed of car A before collision is 3.5 km/h.
Explanation:
Mass of car A = 690 kg eastwards
Mass of car B = 520 kg at 74 km/h west wards
Distance, s = 6 m
coefficient of friction = 0.75
Let the speed after collision is v.
Use third equation of motion
[tex]v^2 = u^2 + 2 as \\\\0 =v^2- 2 \times 0.75\times9.8\times 6\\\\v = 9.4 m/s = 33.84 km/h[/tex]
Let the initial speed of car A is v'.
Use conservation of momentum
690 x v' - 520 x 74 = (690 + 520) x 33.8
690 v' + 38480 = 40898
v' = 3.5 km/h
A mother is pulling a sled at constant velocity by means of a rope at 37°. The tension on the rope is 120 N. Mass of children plus sled is 55 kg. The mother has a mass of 61 kg. Find the static friction acting on the mother.
Answer:
f = 106.3 N
Explanation:
The force applied on the sled must be equal to the static frictional force to move the sled:
Tension Force Horizontal Component = Static Frictional Force
[tex]TCos\theta = \mu W\\TCos\theta = \mu mg[/tex]
where,
T = Tension = 120 N
θ = angle of rope = 37°
μ = coefficient of static friction = ?
m = mass of children plus sled = 55 kg
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex](120\ N)Cos\ 37^o = \mu (55\ kg)(9.81\ m/s^2)\\\\\mu = \frac{95.84\ N}{(55\ kg)(9.81\ m/s^2)}\\\\\mu = 0.18[/tex]
Now, the static friction acting on the mother will be:
[tex]f = \mu mg = (0.18)(61\ kg)(9.81\ m/s^2)\\[/tex]
f = 106.3 N
State how you agree or disagree with the following statement. A good circuit cannot have internal resistance.
Answer: I do
Explanation:
Resistance opposes current thereby reducing the amount of current that flows through a circuit. In other words, it leads to a loss of electrical energy.
Ideally speaking, a good circuit should have no internal resistance as this would lead to more energy having to be supplied to overcome that resistance. External resistance however, is not a bad thing. For instance, oxygen being removed from lightbulbs.
Two identical cars, each traveling at 16 m>s, slam into a concrete wall and come to rest. In car A the air bag does not deploy and the driver hits the steering wheel; in car B the driver contacts the deployed air bag. (a) Is the impulse delivered by the steering wheel to driver A greater than, less than, or equal to the impulse delivered by the air bag to driver B
Answer:
I = - m 16 the two impulses are the same,
Explanation:
The impulse is given by the relationship
I = Δp
I = p_f - p₀
in this case the final velocity is zero therefore p_f = 0
I = -p₀
For driver A the steering wheel impulse is
I = - m v₀
I = - m 16
For driver B, the airbag gives an impulse
I = - m 16
We can see that the two impulses are the same, the difference is that in the air bag more time is used to give this impulse therefore the force on the driver is less
What gauge pressure is required in the city water mains for a stream from a fire hose connected to the mains to reach a vertical height of 15.0 m
Answer:
The gauge pressure is equal to 147 kPa.
Explanation:
The pressure exerted by fluid is given by :
[tex]P=\rho gh[/tex]
Where
[tex]\rho[/tex] is density of water
h is height
So, put all the values,
[tex]P=1000\times 9.8\times 15\\\\P=147000\ Pa[/tex]
or
P = 147 kPa
So, the gauge pressure is equal to 147 kPa.
Answer:
The gauge pressure is 147000 Pa.
Explanation:
Height, h = 15 m
density of water, d= 1000 kg/m^3
gravity, g = 9.8 m/s^2
The gauge pressure is the pressure exerted by the fluid.
The pressure exerted by the fluid is given by
P = h d g
P = 15 x 1000 x 9.8 = 147000 Pa
A satellite measures a spectral radiance of 8 Watts/m2/um/ster at a wavelength of 10 microns. Assuming a surface emissivity of 0.90, what would be the estimated temperature
A stone dropped from the top of a 80m high building strikes the ground at 40 m/s after falling for 4 seconds. The stone's potential energy with respect to the ground is equal to its kinetic energy … (use g = 10 m/s 2)
A) at the moment of impact.
B) 2 seconds after the stone is released.
C) after the stone has fallen 40 m.
D) when the stone is moving at 20 m/s.
At the moment of impact both Kinetic Energy and Potential Energy should be 0, right? So it can't be A), right? Or is this wrong? Is it indeed A)? Please show work and explain it well.
Answer:
Explanation:
The answer is C because the building is 80 meters high. Before the stone is dropped, it has ONLY potential energy since kinetic energy involves velocity and a still stone has no velocity. At impact, there is no potential energy because potential energy involves the height of the stone relative to the ground and a stone ON the ground has no height; here there is ONLY kinetic.
From the First Law of Thermodynamics, we know that energy cannot be created or destroyed, it can only change form. Therefore, that means that at the halfway point of 40 meters, half of the stone's potential energy has been lost, and it has been lost to kinetic energy. Here, at 40 meters, there is an equality between PE and KE. It only last for however long the stone is AT 40 meters, which is probably a millisecond of time, but that's where they are equal.
At what angle torque is half of the max
Here we will use a simple example to demonstrate the difference between the rms speed and the average speed. Five ideal-gas molecules chosen at random are found to have speeds of 500, 600, 700, 800, and 900 m/s. Find the rms speed for this collection. Is it the same as the average speed of these molecules
Explanation:
Given that,
Five ideal-gas molecules chosen at random are found to have speeds of 500, 600, 700, 800, and 900 m/s.
The rms speed for this collection is as follows :
[tex]v_{rms}=\sqrt{\dfrac{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2}{5}} \\\\v_{rms}=\sqrt{\dfrac{500^2+600^2+700^2+800^2+900^2}{5}} \\\\v_{rms}=714.14[/tex]
The average speed of these molecules is :
[tex]v_{a}=\dfrac{v_1+v_2+v_3+v_4+v_5}{5}} \\\\v_{a}={\dfrac{500+600+700+800+900}{5}} \\\\v_{a}=700\ m/s[/tex]
So, the rms speed is 714.14 m/s abd the average speed is 700 m/s.
If you stand next to a wall on a frictionless skateboard and push the wall with a force of 44 N , how hard does the wall push on you
Answer:
44 N
Explanation:
Given that,
If you stand next to a wall on a frictionless skateboard and push the wall with a force of 44 N, then we need to find the force the wall push on you.
It is based on Newton's third law of motion which states that for an action there is an equal and opposite reaction. If the you push the wall with a force of 44 N, the wall push on you is 44 N also as it is based on Newton's third law of motion.
Need help! Need help! Need help! Need help! Need help! Need help!
Answer:
i can help you i know this answer
Answer: the side two are 50 then the other two are 140 i thank
Explanation:
If you double the current in a long straight wire, the magnetic field at a fixed point will... be cut in half. triple. double. quadruple.
Answer:
the magnetic field must double
Explanation:
For this exercise we use Ampere's law
∫ B . ds = μ₀ I
Where the bold indicate vectors
With this expression we can see that if we double the current, keeping the same trajectory, the magnetic field must double
Select the only true statement:
A)A beam in bending experiences tensile stresses on one side and compressive stresses on the other side.
B)A beam in bending experiences tensile stresses along the beam center and compressive stresses along the beam’s edges.
C)A beam in bending experiences only compressive stresses.
D)A beam in bending experiences only tensile stresses.
Answer:
Sorry I dont know this answer sorry
On the average, in a ferromagnetic domain, permanent atomic magnetic moments are aligned ____ to one another.a. antiparallel b. parallel c. perpendicular d. alternately parallel and antiparallel e. randomly relative
Answer:
b. parallel
Explanation:
Ferromagnetism is a magnetism that is associated with iron and cobalt and nickel. Ferromagnetisms material are magnetics easily and in strong magnetic fields are magnetized by a defined limit called a situation. The force keeps magnetic moments of many atoms parallel to each other.