Answer:
The answer is option A.
NuclearHope this helps you
Answer:
Answer is A Nuclear
Explanation:
Just answered this question on my test
What force is required so that a particle of mass m has the position function r(t) = t3 i + 7t2 j + t3 k?
Answer:
[tex]F(t)=m\,\,a(t)=6\,m\,t\,\hat i+14\,m\,\hat j+6\,m\,t\,\hat k\\F(t)=\,(6\,m\,t,14\,m,6\,m\,t)[/tex]
Explanation:
Recall that force is defined as mass times acceleration, and acceleration is the second derivative with respect to time of the position. Since the position comes in terms of time, and with separate functions for each component in the three dimensional space, we first calculate the velocity (with the first derivative, and then the acceleration as the second derivative:
[tex]r(t)=t^3\,\hat i+7\,t^2\,\hat j+t^3\,\hat k\\v(t)=3\,t^2\,\hat i+14\,t\,\hat j+3\,t^2\,\hat k\\a(t)=6\,t\,\hat i+14\,\hat j+6\,t\,\hat k[/tex]
Therefore, the force will be given by the product of this acceleration times the mass "m":
[tex]F(t)=m\,\,a(t)=6\,m\,t\,\hat i+14\,m\,\hat j+6\,m\,t\,\hat k[/tex]
Alternating Current In Europe, the voltage of the alternating current coming through an electrical outlet can be modeled by the function V 230 sin (100t), where tis measured in seconds and Vin volts.What is the frequency of the voltage
Answer:
[tex]\frac{50}{\pi }[/tex]Hz
Explanation:
In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;
V(t) = V sin (ωt + Ф) -----------------(i)
Where;
V = amplitude value of the voltage
ω = angular frequency = 2 π f [f = cyclic frequency or simply, frequency]
Ф = phase difference between voltage and current.
Now,
From the question,
V(t) = 230 sin (100t) ---------------(ii)
By comparing equations (i) and (ii) the following holds;
V = 230
ω = 100
Ф = 0
But;
ω = 2 π f = 100
2 π f = 100 [divide both sides by 2]
π f = 50
f = [tex]\frac{50}{\pi }[/tex]Hz
Therefore, the frequency of the voltage is [tex]\frac{50}{\pi }[/tex]Hz
Suppose the current in a conductor decreases exponentially with time according to the equation I(t) = I0e-t/τ, where I0 is the initial current (at t = 0), and τ is a constant having dimensions of time. Consider a fixed observation point within the conductor.
Required:
a. How much charge passes this point between t = 0 and t = τ?
b. How much charge passes this point between t = 0 and t = 10 τ?
c. What If ? How much charge passes this point between t = 0 and t = [infinity]?
Answer:
Pls see attached file
Explanation:
When the pivot point of a balance is not at the center of mass of the balance, how is the net torque on the balance calculated
Answer:
It is calculated as Force × perpendicular distance.
Explanation:
Torque is a rotational force and twisting force that can cause an object to rotate in it's axis. This cause angular rotation.
The torque due to gravity on a body about its centre of mass is zero because the centre of mass is the that point of the body at which the force acts by the gravity that is mg.
But if the pivot point of a balance is not at the centre of mass of the balance, it will be FORCE × PERPENDICULAR DISTANCE because at that point, there is no centre in which the force act on a body by gravity. The distance and force will be use to calculate.
What is the change in internal energy of an engine if you put 15 gallon of gasoline into its tank? The energy content of gasoline is 1.5 x 106 J/gallon. All other factors, such as the engine’s temperature, are constant. How many hours the engine can work if the power of the engine’s motor is 600 W? (8 marks)
Answer:
ΔU = 2.25 x 10⁸ J
t = 104.17 s
Explanation:
The change in internal energy of the engine can be given by the following formula:
ΔU = (Mass of Gasoline)(Energy Content of Gasoline)
ΔU = (1.5 x 10⁶ J/gallon)(15 gallon)
ΔU = 2.25 x 10⁸ J
Now, for the time of operation, we use the following formula of power.
P = W/t = ΔU/t
t = ΔU/P
where,
t = time of operation = ?
ΔU = Change in internal energy = 2.25 x 10⁸ J
P = Power of motor = 600 W
Therefore,
t = (2.25 x 10⁸ J)/(600 W)
t = (375000 s)(1 h/3600 s)
t = 104.17 s
Consider a conducting rod of length 31 cm moving along a pair of rails, and a magnetic field pointing perpendicular to the plane of the rails. At what speed (in m /s) must the sliding rod move to produce an emf of 0.75 V in a 1.75 T field?
Answer:
The speed of the rod is 1.383 m/s
Explanation:
Given;
length of the conducting rod, L = 31 cm = 0.31 m
induced emf on the rod, emf = 0.75V
magnetic field around the rod, B = 1.75 T
Apply the following Faraday's equation for electromagnetic induction in a moving rod to determine the speed of the rod.
emef = BLv
where;
B is the magnetic field
L is length of the rod
v is the speed of the rod
v = emf / BL
v = (0.75) / (1.75 x 0.31)
v = 1.383 m/s
Therefore, the speed of the rod is 1.383 m/s
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.75 A out of the junction. How many electrons per second move past a point in wire 3?
Answer:
number of electrons = 2.18*10^18 e
Explanation:
In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.
Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:
[tex]i_1+i_3=i_2[/tex] (1)
i1 = 0.40 A
i2 = 0.75 A
you solve the equation i3 from the equation (1):
[tex]i_3=i_2-i_1=0.75A-0.40A=0.35A[/tex]
Next, you take into account that 1A = 1C/s = 6.24*10^18
Then, you have:
[tex]0.35A=0.35\frac{C}{s}=0.35*\frac{6.24*10^{18}e}{s}=2.18*10^{18}\frac{e}{s}[/tex]
The number of electrons that trough the wire 3 is 2.18*10^18 e/s
6. Two forces of 50 N and 30 N, respectively, are acting on an object. Find the net force (in
N) on the object if
the forces are acting in the same direction
b. the forces are acting in opposite directions.
Answer:
same direction = 80 (n)
opposite direction = 20 (n) going one direction
Explanation:
same direction means they are added to each other
and opposite means acting on eachother
a heat engine with an efficiency of 30.0% performs 2500 j of work. how much heat is discharged to the lower temperature reservoir
Answer:
Q₂ = 5833.33 J
Explanation:
First we need to find the energy supplied to the heat engine. The formula for the efficiency of the heat engine is given as:
η = W/Q₁
where,
η = efficiency of engine = 30% = 0.3
W = Work done by engine = 2500 J
Q₁ = Heat supplied to the engine = ?
Therefore,
0.3 = 2500 J/Q₁
Q₁ = 2500 J/0.3
Q₁ = 8333.33 J
Now, we find the heat discharged to lower temperature reservoir by using the formula of work:
W = Q₁ - Q₂
Q₂ = Q₁ - W
where,
Q₂ = Heat discharged to the lower temperature reservoir = ?
Therefore,
Q₂ = 8333.33 J - 2500 J
Q₂ = 5833.33 J
The electric field at the surface of a charged, solid, copper sphere with radius 0.220 mm is 4200 N/CN/C, directed toward the center of the sphere. What is the potential at the center of the sphere, if we take the potential to be zero infinitely far from the sphere?
Answer:
The potential at the center of the sphere is -924 V
Explanation:
Given;
radius of the sphere, R = 0.22 m
electric field at the surface of the sphere, E = 4200 N/C
Since the electric field is directed towards the center of the sphere, the charge is negative.
The Potential is the same at every point in the sphere, and it is given as;
[tex]V = \frac{1}{4 \pi \epsilon_o} \frac{q}{R}[/tex] -------equation (1)
The electric field on the sphere is also given as;
[tex]E = \frac{1}{4 \pi \epsilon _o} \frac{|q|}{R^2}[/tex]
[tex]|q |= 4 \pi \epsilon _o} R^2E[/tex]
Substitute in the value of q in equation (1)
[tex]V = \frac{1}{4 \pi \epsilon_o} \frac{-(4 \pi \epsilon _o R^2E)}{R} \ \ \ \ q \ is \ negative\ because \ E \ is\ directed \ toward \ the \ center\\\\V = -RE\\\\V = -(0.22* 4200)\\\\V = -924 \ V[/tex]
Therefore, the potential at the center of the sphere is -924 V
An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal.
a) What is the maximum height reached by the object?
b) What is the total flight time (between launch and touching the ground) of the object?
c) What is the horizontal range (maximum x above ground) of the object?
d) What is the magnitude of the velocity of the object just before it hits the ground?
Answer:
(a) max. height = 3.641 m
(b) flight time = 1.723 s
(c) horizontal range = 31.235 m
(d) impact velocity = 20 m/s
Above values have been given to third decimal. Adjust significant figures to suit accuracy required.
Explanation:
This problem requires the use of kinematics equations
v1^2-v0^2=2aS .............(1)
v1.t + at^2/2 = S ............(2)
where
v0=initial velocity
v1=final velocity
a=acceleration
S=distance travelled
SI units and degrees will be used throughout
Let
theta = angle of elevation = 25 degrees above horizontal
v=initial velocity at 25 degrees elevation in m/s
a = g = -9.81 = acceleration due to gravity (downwards)
(a) Maximum height
Consider vertical direction,
v0 = v sin(theta) = 8.452 m/s
To find maximum height, we find the distance travelled when vertical velocity = 0, i.e. v1=0,
solve for S in equation (1)
v1^2 - v0^2 = 2aS
S = (v1^2-v0^2)/2g = (0-8.452^2)/(2*(-9.81)) = 3.641 m/s
(b) total flight time
We solve for the time t when the vertical height of the object is AGAIN = 0.
Using equation (2) for vertical direction,
v0*t + at^2/2 = S substitute values
8.452*t + (-9.81)t^2 = 3.641
Solve for t in the above quadratic equation to get t=0, or t=1.723 s.
So time for the flight = 1.723 s
(c) Horiontal range
We know the horizontal velocity is constant (neglect air resistance) at
vh = v*cos(theta) = 25*cos(25) = 18.126 m/s
Time of flight = 1.723 s
Horizontal range = 18.126 m/s * 1.723 s = 31.235 m
(d) Magnitude of object on hitting ground, Vfinal
By symmetry of the trajectory, Vfinal = v = 20, or
Vfinal = sqrt(v0^2+vh^2) = sqrt(8.452^2+18.126^2) = 20 m/s
Two children are balanced on a seesaw, but one child weighs twice as much as the other child. The heavier child is sitting half as far from the pivot as is the lighter child. Since the seesaw is balanced, the heavier child is exerting on the seesaw:_______.
a. a force that is less than the force the lighter child is exerting.
b. a force that is equal in amount but oppositely directed to the force the lighter child is exerting.
Answer:
B. A force that is equal in amount but oppositely directed to the force the lighter child is exerting.
Explanation:
If they are sitting at the same distance away from the pivot yet the seesaw is balanced, the only conclusion is the heavier child is exerting a lower force. This causes the pivot exertion and balances to be equal. The equilibrium of the pivot-seesaw is not affected by the weight because of force exertion.
Water flowing through a garden hose of diameter 2.76 cm fills a 20.0-L bucket in 1.45 min. (a) What is the speed of the water leaving the end of the hose
Answer:
v = 31.84 cm/s or 0.318 m/s
the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s
Explanation:
Given;
Diameter of hose d = 2.76 cm
Volume filled V = 20.0 L = 20,000 cm^3
Time t = 1.45 min = 105 seconds
The volumetric flow rate of water is;
F = V/t = 20,000cm^3 ÷ 105 seconds
F = 190.48 cm^3/s
The volumetric flow rate is equal the cross sectional area of pipe multiply by the speed of flow.
F = Av
v = F/A
Area A = πd^2/4
Speed v = F/(πd^2/4)
v = 4F/πd^2 ......1
Substituting the given values;
v = (4×190.48)/(π×2.76^2)
v = 31.83767439628 cm/s
v = 31.84 cm/s or 0.318 m/s
the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s
The compressor of an air conditioner draws an electric current of 16.2 A when it starts up. If the start-up time is 1.45 s long, then how much electric charge passes through the circuit during this period
Answer:
Q = 23.49 C
Explanation:
We have,
Electric current drawn by the air conditioner is 16.2 A
Time, t = 1.45 s
It is required to find the electric charge passes through the circuit during this period. We know that electric current is defined as the electric charge flowing per unit time. So,
[tex]I=\dfrac{q}{t}\\\\q=It\\\\q=16.2\times 1.45\\\\q=23.49\ C[/tex]
So, the charge of 23.49 C is passing through the circuit during this period.
find the value of k for which the given pair of vectors are not equal
2ki +3j and 8i + 4kj
Answer:
5
Explanation:
Particle A has charge qA and particle B has charge qB. When they are separated by a distance ri, they experience an attractive force Fi. The particles are moved without altering their charges. Now they experience an attractive force with a magnitude of 36Fi. Find an expression for their new separation.
Answer:
[tex]r_f=\frac{1}{6}r_i[/tex]
Explanation:
To find the new separation of the charges, you first take into account the formula for the electric force, when the force are separated a distance of ri.
You use the following expression:
[tex]F_i=k\frac{q_Aq_B}{r_i^2}[/tex] (1)
k: Coulomb's constant
qA: charge of A particle
qB: charge of B particle
When the charges are separated to a new distance rf, the new force is 36Fi, if the charges have not changed, you have:
[tex]F_f=36F_i=k\frac{q_Aq_B}{r_f^2}[/tex] (2)
To find the new separation you replace the expression for Fi of the equation (1) into the equation (2) and solve for rf in terms of ri:
[tex]36F_i=36k\frac{q_Aq_B}{r_i^2}=k\frac{q_Aq_B}{r_f^2}\\\\\frac{36}{r_i^2}=\frac{1}{r_f^2}\\\\r_f=\frac{1}{6}r_i[/tex]
The new separation of the charges is 1/6 times of the initial separation
Consider a skateboarder who starts from rest at the top of ramp that is inclined at an angle of 18.0 ∘ to the horizontal.
Assuming that the skateboarder's acceleration is gsin 18.0 ∘, find his speed when he reaches the bottom of the ramp in 3.50 s .
Answer:
Explanation:
v= u + at
v is final velocity , u is initial velocity . a is acceleration and t is time
Initial velocity u = 0 . Putting the given values in the equation
v = 0 + g sin 18 x 3.5
= 10.6 m /s
For a skateboarder who starts from the rest, the speed when he reaches the bottom of the ramp will be 10.6 m/s.
What are Velocity and Acceleration?The term "velocity" refers to a vector measurement of the rate and direction of motion. Velocity is the rate of movement in a single direction, to put it simply. Velocity can be used to determine how fast a rocket is heading into space and how fast a car is moving north on a congested motorway.
There are several types of velocity :
Instantaneous velocityAverage VelocityUniform VelocityNon-Uniform VelocityThe pace at which a person's velocity changes is known as acceleration. This implies that an object is accelerating if its velocity is rising or falling. An object that is accelerating won't have a steady change in location every second like an item moving at a constant speed does.
According to the question, the given values are :
Time, t = 3.50 sec
Initial Velocity, u = 0 m/s
Use equation of motion :
v = u+at
v = 0+ g sin 18 × 3.5
v = 10.6 m/s.
So, the final velocity will be 10.6 m/s.
To get more information about Velocity and Acceleration :
https://brainly.com/question/14683118
#SPJ2
The speed of a sound wave in air is 343m/s. If the density of the air is 1.2kg/m3, find the bulk modulus.
Answer:
141178.8
Explanation:
use : density x velocity²
1.2 x 343² = 141178.8 pa
what is the difference between a good conductor and a good insulator?
Answer:
Explanation:
In a conductor, electric current can flow freely, in an insulator it cannot.
Metals such as copper typify conductors, while most non-metallic solids are said to be good insulators, having extremely high resistance to the flow of charge through them.
Most atoms hold on to their electrons tightly and are insulators.
An automobile accelerates from zero to 30 m/s in 6 s. The wheels have a diameter of 0.4 m. What is the angular acceleration of each wheel
Answer:
12.5 rad/s²
Explanation:
Angular Acceleration: This can be defined as the ratio of linear acceleration and radius. The S.I unit is rad/s²
From the question,
a = αr................... Equation 1
Where a = linear acceleration, α = angular acceleration, r = radius.
But,
a = (v-u)/t.............. Equation 2
Where v = final velocity, u = initial velocity, t = time.
Substitute equation 2 into equation 1
(v-u)/t = αr
make α the subject of the equation
α = (v-u)/tr................. Equation 3
Given: v = 30 m/s, u = 0 m/s, t = 6 s, r = 0.4 m
Substitute into equation 3
α = (30-0)/(0.4×6)
α = 30/2.4
α = 12.5 rad/s²
A bicycle tire pump has a piston with area 0.43 In2. If a person exerts a force of 16 lb on the piston while Inflating a tire, what pressure does this produce on the air in the pump?
Answer:
The pressure produced on the air in the pump is 37.209 pounds per square inch.
Explanation:
By definition, the pressure is the force exerted on the piston divided by its area. Given that, force is distributed uniformly on the piston area, the formula to determine the pressure is:
[tex]p = \frac{F}{A}[/tex]
Where:
[tex]p[/tex] - Pressure, measured in pounds per square inch.
[tex]F[/tex] - Force exerted on the piston, measured in pounds.
[tex]A[/tex] - Piston area, measured in square inches.
If [tex]F = 16\,lb[/tex] and [tex]A = 0.43\,in^{2}[/tex], the pressure produced on the air in the pump is:
[tex]p = \frac{16\,lb}{0.43\,in^{2}}[/tex]
[tex]p = 37.209\,psi[/tex]
The pressure produced on the air in the pump is 37.209 pounds per square inch.
A particle is released as part of an experiment. Its speed t seconds after release is given by v (t )equalsnegative 0.4 t squared plus 2 t, where v (t )is in meters per second. a) How far does the particle travel during the first 2 sec? b) How far does it travel during the second 2 sec?
Answer:
a) 2.933 m
b) 4.534 m
Explanation:
We're given the equation
v(t) = -0.4t² + 2t
If we're to find the distance, then we'd have to integrate the velocity, since integration of velocity gives distance, just as differentiation of distance gives velocity.
See attachment for the calculations
The conclusion of the attachment will be
7.467 - 2.933 and that is 4.534 m
Thus, The distance it travels in the second 2 sec is 4.534 m
The density of a sample of metal was measured to be 8.91 g/cm3. An X-ray diffraction experiment measures the edge of a face-centered cubic cell as 352.4 pm. Part APart complete What is the atomic weight of the metal
Answer:
The atomic weight of the metal is 58.7 g/mol
Explanation:
Given;
density of the metal sample, ρ = 8.91 g/cm³
edge length of the face centered cubic cell, α = 352.4 pm = 352.4 x 10⁻¹⁰ cm
Volume of the unit cell of the metal;
V = α³
V = (352.4 x 10⁻¹⁰ cm)³
V = 4.376 x 10⁻²³ cm³
Mass of the metal in unit cell
mass = density x volume
mass = 8.91 g/cm³ x 4.376 x 10⁻²³ cm³
mass = 3.899 x 10⁻²² g
Atomic weight, based on 4 atoms per unit cell;
4 atoms = 3.899 x 10⁻²² g
6.022 x 10²³ atoms = ?
= (6.022 x 10²³atoms x 3.899 x 10⁻²² g) / (4 atoms)
= 58.699 g/mol
= 58.7 g/mol (this metal is Nickel)
Theerefore, the atomic weight of the metal is 58.7 g/mol
please help In a video game, a ball moving at 0.6 meter/second collides with a wall. After the collision, the velocity of the ball changed to -0.4 meter/second. The collision takes 0.2 seconds to occur. What’s the acceleration of the ball during the collision? Use . a= v-u/t
Answer:
the acceleration during the collision is: - 5 [tex]\frac{m}{s^2}[/tex]
Explanation:
Using the formula:
[tex]a=\frac{\Delta\,v}{\Delta\,t}[/tex]
we get:
[tex]a=\frac{-0.4-0.6}{0.2} \,\frac{m}{s^2} =\frac{-1}{0.2} \,\frac{m}{s^2} =-5\,\,\frac{m}{s^2}[/tex]
Briefly describe the relationship between an equipotential surface and an electric field, and use this to explain why we will plot equipotential lines.
Answer:
E = - dV/dx
Explanation:
Las superficies equipòtenciales son superficie donde el potencial eléctrico es constante por lo cual nos podemos desplazaren ella sin realizar nigun trabajo.
El campo electrico es el campo que existen algún punto en el espacio creado por alguna ddistribucion de carga.
De los antes expuesto las dos magnitudes están relacionadas
E = - dV/dx
por lo cual el potenical es el gradiente del potencial eléctrico.
Como el campo eléctrico sobre un superficie equipotenciales constante, podemos colocar una punta de prueba con un potencial dado y seguir la linea que de una diferencia de potencial constar, lo cual permite visualizar las forma de cada linea equipotencial
Calculate the ideal banking angle in degrees for a gentle turn of 1.88 km radius on a highway with a 136.3 km/hr speed limit, assuming everyone travels at the speed limit.
Answer:
Ф = 4.4°Explanation:
given:
radius (r) = 1.88 km
velocity (v) = 136.3 km/hr
required:
banking angle ∡ ?
first:
convert 1.88 km to m = 1.88km * 1000m / 1km
r = 1880 m
convert velocity v = 136.3 km/hr to m/s = 136.3 km/hr * (1000 m/ 3600s)
v = 37.86 m/s
now.. calculate the angle
Ф = inv tan (v² / r * g) we know that gravity = 9.8 m/s²
Ф = inv tan (37.86² / (1880 * 9.8))
Ф = 4.4°
According to Newton, when the distance between two interacting objects doubles, the gravitational force is
Answer:
1/4 of its original value
Explanation:
Newton's law of universal gravitation states that when two bodies of masses M₁ and M₂ interact, the force of attraction (F) between these bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance (r) between these bodies. i.e
F ∝ [tex]\frac{M_1 M_2}{r^2}[/tex] ------------(i)
From the equation above, it can be deduced that;
F ∝ [tex]\frac{1}{r^2}[/tex]
=> F = G [tex]\frac{1}{r^2}[/tex] -----------(ii)
Where;
G = constant of proportionality called the gravitational constant
Equation (ii) can be re-written as
Fr² = G
=> F₁r₁² = F₂r₂² -----------(iii)
Where;
F₁ and r₁ are the initial values of the force and distance respectively
F₂ and r₂ are the final values of the force and distance respectively
From the question, if the distance doubles i.e;
r₂ = 2r₁,
Then the final value of the gravitational force F₂ is calculated as follows;
Substitute the value of r₂ = 2r₁ into equation (iii) as follows;
F₁r₁² = F₂(2r₁)²
F₁r₁² = 4F₂r₁² [Divide through by r₁²]
F₁ = 4F₂ [Make F₂ subject of the formula]
F₂ = F₁ / 4 [Re-write this]
F₂ = [tex]\frac{1}{4} F_1[/tex]
Therefore the gravitational force will be 1/4 of its original value when the distance between the bodies doubles.
An elastic band is hung on a hook and a mass is hung on the lower end of the band. When the mass is pulled downward and then released, it vibrates vertically. The equation of motion is s = 9 cos(t) + 9 sin(t), t ≥ 0, where s is measured in centimeters and t in seconds. (Take the positive direction to be downward.) (a) Find the velocity and acceleration at time t.
Answer:
v(t) = s′(t) = −9sin(t)+9cos(t)
a(t) = v′(t) = −9cos(t) −9sin(t)
Explanation:
Given that
s = 9 cos(t) + 9 sin(t), t ≥ 0
Then acceleration and velocity is
v(t) = s′(t) = −9sin(t)+9cos(t)
a(t) = v′(t) = −9cos(t) −9sin(t)
Consider a sound wave modeled with the equation s(x, t) = 3.00 nm cos(3.50 m−1x − 1,800 s−1t). What is the maximum displacement (in nm), the wavelength (in m), the frequency (in Hz), and the speed (in m/s) of the sound wave?
Answer:
- maximum displacement = 3.00nm
- λ = 1.79m
- f = 286.47 s^-1
Explanation:
You have the following equation for a sound wave:
[tex]s(x,t)=3.00nm\ cos(3.50m^{-1}x- 1,800s^{-1} t)[/tex] (1)
The general form of the equation of a sound wave can be expressed as the following formula:
[tex]s(x,t)=Acos(kx-\omega t)[/tex] (2)
A: amplitude of the wave = 3.00nm
k: wave number = 3.50m^-1
w: angular frequency = 1,800s^-1
- The maximum displacement of the wave is given by the amplitude of the wave, then you have:
maximum displacement = A = 3.00nm
- The wavelength is given by :
[tex]\lambda=\frac{2\pi}{k}=\frac{2\pi}{3.50m^{-1}}=1.79m[/tex]
The values for the wavelength is 1.79m
- The frequency is:
[tex]f=\frac{\omega}{2\pi}=\frac{1,800s^{-1}}{2\pi}=286.47s^{-1}[/tex]
The frequency is 286.47s-1
a 5.0 charge is placed at the 0 cm mark of a meterstick and a -4.0 charge is placed at the 50 cm mark. what is the electric field at the 30 cm mark
Answer:
-1748*10^N/C
Explanation:
See attached file