Answer:
The speed of the light signal as viewed from the observer is c.
Explanation:
Recall the basic postulate of the theory of relativity that the speed of light is the same in ALL inertial frames. Based on this, the speed of light is independent of the motion of the observer.
Three cars (car 1, car 2, and car 3) are moving with the same velocity when the driver suddenly slams on the brakes, locking the wheels. The most massive car is car 1, the least massive is car 3, and all three cars have identical tires. For which car does friction do the largest amount of work in stopping the car
Answer:
Car 3
Explanation:
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
what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed
Answer:
Final energy = Uf = initial energy × d₂/d₁
Explanation:
Energy is the ability to do work.
capacitor is an electronic device that store charges
where
V is the potential difference
d is the distance of seperation between the two plates
ε₀ is the dielectric constant of the material used in seperating the two plates, e.g., paper, mica, glass etc.
A = cross sectional area
U =¹/₂CV²
C =ε₀A/d
C × d=ε₀A=constant
C₂d₂=C₁d₁
C₂=C₁d₁/d₂
charge will 'q' remains same in the capacitor, if the capacitor was disconnected from the electric potential source (v) before the separation of the plates was replaced
Energy=U =(1/2)q²/C
U₂C₂ = U₁C₁
U₂ =U₁C₁ /C₂
U₂ =U₁d₂/d₁
Final energy = Uf = initial energy × d₂/d₁
•• A metal sphere carrying an evenly distributed charge will have spherical equipotential surfaces surrounding it. Suppose the sphere’s radius is 50.0 cm and it carries a total charge of (a) Calculate the potential of the sphere’s surface. (b)You want to draw equipotential surfaces at intervals of 500 V outside the sphere’s surface. Calculate the distance between the first and the second equipotential surfaces, and between the 20th and 21st equipotential surfaces. (c) What does the changing spacing of the surfaces tell you about the electric field?
Answer:
Explanation:
For this exercise we will use that the potential is created by the charge inside the equinoctial surface and just like in Gauss's law we can consider all the charge concentrated in the center.
Therefore the potential on the ferric surface is
V = k Q / r
where k is the Coulomb constant, Q the charge of the sphere and r the distance from the center to the point of interest
a) On the surface the potential
V = 9 10⁹ Q / 0.5
V = 18 10⁹ Q
Unfortunately you did not write the value of the load, suppose a value to complete the calculations Q = 1 10⁻⁷ C, with this value the potential on the surfaces V = 1800 V
b) The equipotential surfaces are concentric spheres, let's look for the radii for some potentials
for V = 1300V let's find the radius
r = k Q / V
r = 9 109 1 10-7 / 1300
r = 0.69 m
other values are shown in the following table
V (V) r (m)
1800 0.5
1300 0.69
800 1,125
300 3.0
In other words, we draw concentric spheres with these radii and each one has a potential difference of 500V
C) The spacing of the spheres corresponds to lines of radii of the electric field that have the shape
E = k Q / r²
An electromagnetic ware has a maximum magnetic field strength of 10^-8 T at a specific place in vacuum. What is the intensity of the light at that place. μ0=4πx10^-7 WbA/m g
Answer:
[tex]I=1.19\times 10^{-2}\ W/m^2[/tex]
Explanation:
It is given that,
Maximum value of magnetic field strength, [tex]B=10^{-8}\ T[/tex]
We need to find the intensity of the light at that place.
The formula of the intensity of magnetic field is given by :
[tex]I=\dfrac{c}{2\mu _o}B^2[/tex]
c is speed of light
So,
[tex]I=\dfrac{3\times 10^8}{2\times 4\pi \times 10^{-7}}\times (10^{-8})^2\\\\I=1.19\times 10^{-2}\ W/m^2[/tex]
So, the intensity of the light is [tex]1.19\times 10^{-2}\ W/m^2[/tex].
(a) Find the magnitude of an earthquake that has an intensity that is 37.25 (that is, the amplitude of the seismograph reading is 37.25 cm). (Round your answer to one decimal place.)
Answer:
The magnitude of an earthquake is 5.6.
Explanation:
The magnitude of an earthquake can be found as follows:
[tex] M = log(\frac{I}{S}) [/tex]
Where:
I: is the intensity of the earthquake = 37.25 cm
S: is the intensity of a standard earthquake = 10⁻⁴ cm
Hence, the magnitude is:
[tex]M = log(\frac{I}{S}) = log(\frac{37.25}{10^{-4}}) = 5.6[/tex]
Therefore, the magnitude of an earthquake is 5.6.
I hope it helps you!
Three resistors, 6.0-W, 9.0-W, 15-W, are connected in parallel in a circuit. What is the equivalent resistance of this combination of resistors?
Answer:
2.9Ω
Explanation:
Resistors are said to be in parallel when they are arranged side by side such that their corresponding ends are joined together at two common junctions. The combined resistance in such arrangement of resistors is given by;
1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn
Where;
Req refers to the equivalent resistance and R1, R2, R3 .......Rn refers to resistance of individual resistors connected in parallel.
Note that;
R1= 6.0Ω
R2 = 9.0Ω
R3= 15.0 Ω
Therefore;
1/Req = 1/6 + 1/9 + 1/15
1/Req= 0.167 + 0.11 + 0.067
1/Req= 0.344
Req= (0.344)^-1
Req= 2.9Ω
The equivalent resistance of this combination of resistors is 2.9Ω.
Calculation of the equivalent resistance:The combined resistance in such arrangement of resistors is provided by;
1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn
here.
Req means the equivalent resistance and R1, R2, R3
.Rn means the resistance of individual resistors interlinked in parallel.
Also,
R1= 6.0Ω
R2 = 9.0Ω
R3= 15.0 Ω
So,
1/Req = 1/6 + 1/9 + 1/15
1/Req= 0.167 + 0.11 + 0.067
1/Req= 0.344
Req= (0.344)^-1
Req= 2.9Ω
learn more about resistance here: https://brainly.com/question/15047345
A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at t = 2.15 s.
Answer:
206.67NExplanation:
The sum of force along both components x and y is expressed as;
[tex]\sum Fx = ma_x \ and \ \sum Fy = ma_y[/tex]
The magnitude of the net force which is also known as the resultant will be expressed as [tex]R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}[/tex]
To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.
Given the position of the object along the x-component to be x = 6t² − 4;
[tex]a_x = \frac{d^2 x }{dt^2}[/tex]
[tex]a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}[/tex]
Similarly,
[tex]a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2[/tex]
[tex]\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N[/tex]
[tex]R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N[/tex]
Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N
Estimate the distance (in cm) between the central bright region and the third dark fringe on a screen 6.3 m from two double slits 0.49 mm apart illuminated by 739-nm light. (give answer in millimeters)
Answer:
Explanation:
distance of third dark fringe
= 2.5 x λ D / d
where λ is wavelength of light , D is screen distance and d is slit separation
putting the given values
required distance = 2.5 x 739 x 10⁻⁹ x 6.3 / .49 x 10⁻³
= 23753.57 x 10⁻⁶
= 23.754 x 10⁻³ m
= 23.754 mm .
Consider the uniform electric field E = (8.0ĵ + 2.0 ) ✕ 103 N/C. What is its electric flux (in N · m2/C) through a circular area of radius 9.0 m that lies in the xy-plane? (Enter the magnitude.)
Answer:
5.09 x 10⁵ Nm²/C
Explanation:
The electric flux φ through a planar area is defined as the electric field Ε times the component of the area Α perpendicular to the field. i.e
φ = E A
From the question;
E = (8.0j + 2.0k) ✕ 10³ N/C
r = radius of the circular area = 9.0m
A = area of a circle = π r² [Take π = 3.142]
A = 3.142 x 9² = 254.502m²
Now, since the area lies in the x-y plane, only the z-component of the electric field is responsible for the electric flux through the circular area.
Therefore;
φ = (2.0) x 10³ x 254.502
φ = 5.09 x 10⁵ Nm²/C
The electric flux is 5.09 x 10⁵ Nm²/C
Current folw in which dirction
Which scientist proved experimentally that a shadow of the circular object illuminated 18. with coherent light would have a central bright spot?
A. Young
B. Fresnel
C. Poisson
D. Arago
Answer:
Your answer is( D) - Arago
A square coil of wire with side 8.0 cm and 50 turns sits in a uniform magnetic field that is perpendicular to the plane of the coil. The coil is pulled quickly out of the magnetic field in 0.2 s. If the resistance of the coil is 15 ohm and a current of 12 mA is induced in the coil, calculate the value of the magnetic field.
Answer:
Explanation:
area of the coil A = .08 x .08 = 64 x 10⁻⁴ m ²
flux through the coil Φ = area of coil x no of turns x magnetic field
= 64 x 10⁻⁴ x 50 x B where B is magnetic field
emf induced = dΦ / dt = ( 64 x 10⁻⁴ x 50 x B - 0 ) / .2
= 1.6 B
current induced = emf induced / resistance
12 x 10⁻³ = 1.6 B / 15
B = 112.5 x 10⁻³ T .
Two sound waves W1 and W2, of the same wavelength interfere destructively at point P. The waves originate from two in phase speakers. W1 travels 36m and W2 travels 24m before reaching point P. Which of the following values could be the wave length of the sound waves?
a. 24m
b. 12m
c. 6m
d. 4m
Answer:
a. 24 m
Explanation:
Destructive interference occurs when two waves arrive at a point, out of phase. In a completely destructive interference, the two waves cancel out, but in a partially destructive interference, they produce a wave with a time varying amplitude, but maintain a wavelength the wavelength of one of the original waves. Since the two waves does not undergo complete destructive interference, then the possible value of the new wave formed can only be 24 m, from the options given.
Suppose a space vehicle with a rest mass of 150 000 kg travels past the International Space Station at a constant speed of 2.6 x 108 m/s with respect to the I.S.S. When an observer on the I.S.S. measures the moving vehicle, her measurement of the space vehicle length is 25.0 m. Determine the relativistic mass of the space vehicle. Determine the length of the space vehicle as measured by an astronaut on the space vehicle.
Answer:
m = 300668.9 kg
L₀ = 12.47 m
Explanation:
The relativistic mass of the space vehicle is given by the following formula:
[tex]m = \frac{m_{0}}{\sqrt{1-\frac{v^{2} }{c^{2}} } }[/tex]
where,
m = relativistic mass = ?
m₀ = rest mass = 150000 kg
v = relative speed = 2.6 x 10⁸ m/s
c = speed of light = 3 x 10⁸ m/s
Therefore
[tex]m = \frac{150000kg}{\sqrt{1-\frac{(2.6 x 10^{8}m/s)^{2} }{(3 x 10^{8}m/s)^{2}} } }[/tex]
m = 300668.9 kg
Now, for rest length of vehicle:
L = L₀√(1 - v²/c²)
where,
L = Relative Length of Vehicle = 25 m
L₀ = Rest Length of Vehicle = ?
Therefore,
25 m = L₀√[1 - (2.6 x 10⁸ m/s)²/(3 x 10⁸ m/s)²]
L₀ = (25 m)(0.499)
L₀ = 12.47 m
5.
Find the equation of the circle tangential to the line 3x-4y+1=0 and with
centre at (4,7).
20
Answer: (x - 4)² + (y - 7)² = 9
Explanation:
The equation of a circle is: (x - h)² + (y - k)² = r² where
(h, k) is the centerr is the radiusGiven: (h, k) = (4, 7)
Find the intersection of the given equation and the perpendicular passing through (4, 7).
3x - 4y = -1
-4y = -3x - 1
[tex]y=\dfrac{3}{4}x-1[/tex]
[tex]m=\dfrac{3}{4}[/tex] --> [tex]m_{\perp}=-\dfrac{4}{3}[/tex]
[tex]y-y_1=m_{\perp}(x-x_1)\\\\y-7=-\dfrac{4}{3}(x-4)\\\\\\y=-\dfrac{4}{3}x+\dfrac{16}{3}+7\\\\\\y=-\dfrac{4}{3}x+\dfrac{37}{3}[/tex]
Use substitution to find the point of intersection:
[tex]x=\dfrac{29}{5}=5.8,\qquad y=\dfrac{23}{5}=4.6[/tex]
Use the distance formula to find the distance from (4, 7) to (5.8, 4.6) = radius
[tex]r=\sqrt{(5.8-4)^2+(4.6-7)^2}\\\\r=\sqrt{3.24+5.76}\\\\r=\sqrt9\\\\r=3[/tex]
Input h = 4, k = 7, and r = 3 into the circle equation:
(x - 4)² + (y - 7)² = 3²
(x - 4)² + (y - 7)² = 9
Use I=∫r2 dm to calculate I of a slender uniform rod of length L, about an axis at one end perpendicular to the rod. note: a "slender rod" often refers to a rod of neglible cross sectional area, so that the volume is the Length, and the mass density X Length.
Answer:
The moment of inertia of a slender uniform rod of length L about an axis at one end perpendicular to the rod is [tex]I = \frac{1}{3}\cdot m \cdot L^{2}[/tex].
Explanation:
Let be an uniform rod of length L whose origin is located at one one end and axis is perpendicular to the rod, such that:
[tex]\lambda = \frac{dm}{dr}[/tex]
Where:
[tex]\lambda[/tex] - Linear density, measured in kilograms per meter.
[tex]m[/tex] - Mass of the rod, measured in kilograms.
[tex]r[/tex] - Distance of a point of the rod with respect to origin.
Mass differential can translated as:
[tex]dm = \lambda \cdot dr[/tex]
The equation of the moment of inertia is represented by the integral below:
[tex]I = \int\limits^{L}_{0} {r^{2}} \, dm[/tex]
[tex]I = \lambda \int\limits^{L}_{0} {r^{2}} \, dr[/tex]
[tex]I = \lambda \cdot \left(\frac{1}{3}\cdot L^{3} \right)[/tex]
[tex]I = \frac{1}{3}\cdot m \cdot L^{2}[/tex] (as [tex]m = \lambda \cdot L[/tex])
The moment of inertia of a slender uniform rod of length L about an axis at one end perpendicular to the rod is [tex]I = \frac{1}{3}\cdot m \cdot L^{2}[/tex].
A proton that is initially at rest is accelerated through an electric potential difference of magnitude 500 V. What speed does the proton gain? (e = 1.60 × 10-19 C , mproton = 1.67 × 10-27 kg)
Answer:
[tex]3.1\times 10^{5}m/s[/tex]
Explanation:
The computation of the speed does the proton gain is shown below:
The potential difference is the difference that reflects the work done as per the unit charged
So, the work done should be
= Potential difference × Charge
Given that
Charge on a proton is
= 1.6 × 10^-19 C
Potential difference = 500 V
[tex]v= \sqrt{\frac{2.q.\Delta V}{m_{p}}} \\\\\\= \sqrt{\frac{2\times 1.6\times 10^{-19}\times 5\times 10^{2}}{1.67\times 10^{-27}}}[/tex]
[tex]v= \sqrt{9.58\times 10^{10}}m/s \\\\= 3.095\times 10^{5}m/s\\\\\approx 3.1\times 10^{5}m/s[/tex]
Simply we applied the above formulas
Ohm’s Law
pls answer this photos
Answer:
Trial 1: 2 Volts, 0 %
Trial 2: 2.8 Volts, 0%
Trial 3: 4 Volts, 0 %
Explanation:
Th experimental values are given in the table, while the theoretical value can be found by using Ohm/s Law:
V = IR
TRIAL 1:
V = IR
V = (0.1 A)(20 Ω)
V = 2 volts
% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%
% Difference = |(2 - 2)/2| x 100%
% Difference = 0 %
TRIAL 2:
V = IR
V = (0.14 A)(20 Ω)
V = 2.8 volts
% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%
% Difference = |(2.8 - 2.8)/2.8| x 100%
% Difference = 0 %
TRIAL 3:
V = IR
V = (0.2 A)(20 Ω)
V = 4 volts
% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%
% Difference = |(4 - 4)/4| x 100%
% Difference = 0 %
In a shipping yard, a crane operator attaches a cable to a 1,390 kg shipping container and then uses the crane to lift the container vertically at a constant velocity for a distance of 33 m. Determine the amount of work done (in J) by each of the following.
a) the tension in the cable.
b) the force of gravity.
Answer:
a) A = 449526 J, b) 449526 J
Explanation:
In this exercise they do not ask for the work of different elements.
Note that as the box rises at constant speed, the sum of forces is chorus, therefore
T-W = 0
T = W
T = m g
T = 1,390 9.8
T = 13622 N
Now that we have the strength we can use the definition of work
W = F .d
W = f d cos tea
a) In this case the tension is vertical and the movement is vertical, so the tension and displacement are parallel
A = A x
A = 13622 33
A = 449526 J
b) The work of the force of gravity, as the force acts in the opposite direction, the angle tea = 180
W = T x cos 180
W = - 13622 33
W = - 449526 J
From mechanics, you may recall that when the acceleration of an object is proportional to its coordinate, d2xdt2=−kmx=−ω2x , such motion is called simple harmonic motion, and the coordinate depends on time as x(t)=Acos(ωt+ϕ), where ϕ, the argument of the harmonic function at t=0, is called the phase constant. Find a similar expression for the charge q(t) on the capacitor in this circuit. Do not forget to determine the correct value of ϕ based on the initial conditions described in the problem. Express your answer in terms of q0 , L, and C. Use the cosine function in your answer.
Answer:
q = q₀ sin (wt)
Explanation:
In your statement it is not clear the type of circuit you are referring to, there are two possibilities.
1) The circuit of this problem is a system formed by an Ac voltage source and a capacitor, in this case all the voltage of the source is equal to the voltage at the terminals of the capacitor
ΔV = Δ[tex]V_{C}[/tex]
we assume that the source has a voltage of the form
ΔV = ΔV₀o sin wt
The capacitance of a capacitor is
C = q / ΔV
q = C ΔV sin wt
the current in the circuit is
i = dq / dt
i = c ΔV₀ w cos wt
if we use
cos wt = sin (wt + π / 2)
we make this change by being a resonant oscillation
we substitute
i = w C ΔV₀ sin (wt + π/2)
With this answer we see that the current in capacitor has a phase factor of π/2 with respect to the current
2) Another possible circuit is an LC circuit.
In this case the voltage alternates between the inductor and the capacitor
V_{L} + V_{C} = 0
L di / dt + q / C = 0
the current is
i = dq / dt
they ask us for a solution so that
L d²q / dt² + 1 / C q = 0
d²q / dt² + 1 / LC q = 0
this is a quadratic differential equation with solution of the form
q = A sin (wt + Ф)
to find the constant we derive the proposed solution and enter it into the equation
di / dt = Aw cos (wt + Ф)
d²i / dt²= - A w² sin (wt + Ф)
- A w² + 1 /LC A = 0
w = √ (1 / LC)
To find the phase factor, for this we use the initial conditions for t = 0
in the case of condensate for t = or the charge is zero
0 = A sin Ф
Ф = 0
q = q₀ sin (wt)
Suppose Young's double-slit experiment is performed in air using red light and then the apparatus is immersed in water. What happens to the interference pattern on the screen?
Answer:
The bright fringes will appear much closer together
Explanation:
Because λn = λ/n ,
And the wavelength of light in water is smaller than the wavelength of light in air. Given that the distance between bright fringes is proportional to the wavelength
A 5.00-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of 1.60 s. Find the force constant of the spring.
Answer:A7.50kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of2.30s. Find the force constant of the spring.
N/m
Explanation:
Find the absolute value of the change of the gravitational potential energy (GPE) of the puck-Earth system from the moment the puck begins to move to the moment it hits the spring. Use 0.253 m for the displacement of the puck along the ramp and 9.80 m/s2 for the acceleration due to gravity. Assume that the mass of the puck is 0.180 kg. Express your answer using SI units to three significant figures.
Answer:
0.16joules
Explanation:
Using the relation for The gravitational potential energy
E= Mgh
Where,
E= Potential energy
h = Vertical Height
M = mass
g = Gravitational Field Strength
To find the vertical component of angle of launch Where the angle is 22°
h= sin theta
So E = mghsintheta
= 0.18 x 0.98 x 0.253 sin22
=0.16joules
Explanation:
If the speed of a "cheetah" is 150 m / s. How long does it take to cover 800 m?
Answer:
5.33333... seconds
Explanation:
800 divided by 150 is equal to 5.33333... because it is per second that the cheetah moves at 150miles, the answer is 5.3333.....
A tiger leaps horizontally out of a tree that is 3.30 m high. He lands 5.30 m from the base of the tree. (Neglect any effects due to air resistance.)
Calculate the initial speed. (Express your answer to three significant figures.)
m/s Submit
Answer:
The initial velocity is [tex]v_h = 8.66 \ m/s[/tex]
Explanation:
From the question we are told that
The height of the tree is [tex]h = 3.30\ m[/tex]
The distance of the position of landing from base is [tex]d = 5.30 \ m[/tex]
According to the second equation of motion
[tex]h = u_o * t + \frac{1}{2} at^2[/tex]
[tex]Where\ u_o[/tex] is the initial velocity in the vertical axis
a is equivalent to acceleration due to gravity which is positive because the tiger is downward
So
[tex]3 = 0 + 0.5 * 9.8 *t^2[/tex]
=> [tex]t = \frac{3 }{9.8 * 0.5}[/tex]
[tex]t = 0.6122\ s[/tex]
Now the initial velocity in the horizontal direction is mathematically evaluated as
[tex]v_h = \frac{5.30}{0.6122}[/tex]
[tex]v_h = 8.66 \ m/s[/tex]
Charge of uniform density (40 pC/m2) is distributed on a spherical surface (radius = 1.0 cm), and a second concentric spherical surface (radius = 3.0 cm) carries a uniform charge density of 60 pC/m2. What is the magnitude of the electric field at a point 4.0 cm from the center of the two surfaces?
Answer:
4.1 N/C
Explanation:
First of all, we know from maths that the surface area of a sphere = 4πr²
Charge on inner sphere ..
Q(i) = 40.0*10^-12C/m² x 4π(0.01m)²
Q(i) = 5.03*10^-14 C
Charge on outer sphere
Q(o) = 60*10^-12 x 4π(0.03m)²
Q(o) = 6.79*10^-13 C
Inner sphere has a - 5.03*10^-14C charge (-Qi) on inside of the outer shell. As a result, there is a net zero charge within the outer shell.
For the outer shell to show a NET charge +6.79*10^-13C, it's must have a +ve charge
= +6.79*10^-13C + (+5.03*10^-14C)
= +7.29*10^-13 C
Now again, we have
E = kQ /r²
E = (9.0*10^9)(+7.29*10^-13 C) / (0.04)²
E = 6.561*10^-3 / 1.6*10^-3
E = 4.10 N/C
Thus, the magnitude of the electric field is 4.1 N/C
A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-axis is the electric potential zero?
Answer:
y = 10.2 m
Explanation:
It is given that,
Charge, [tex]q_1=-3\ nC[/tex]
It is placed at a distance of 9 cm at x axis
Charge, [tex]q_2=+4\ nC[/tex]
It is placed at a distance of 16 cm at x axis
We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,
[tex]\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0[/tex]
Here,
[tex]r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}[/tex]
So,
[tex]\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}[/tex]
Squaring both sides,
[tex]3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m[/tex]
So, at a distance of 10.2 m on the y axis the electric potential equals 0.
According to the question,
Charge,
[tex]q_1 = -3 \ nC[/tex] (9 cm at x-axis)[tex]q_2 = +4 \ nC[/tex] (16 cm at x-axis)Now,
→ [tex]\frac{kq_1}{r_1} +\frac{kq_2}{r_2} =0[/tex]
or,
→ [tex]\frac{kq_1}{r_1} =-\frac{kq_2}{r_2}[/tex]
→ [tex]\frac{q_1}{r_1} = \frac{q_2}{r_2}[/tex]
here,
[tex]r_1 = \sqrt{y^2+81}[/tex]
[tex]r^2 = \sqrt{y^2+225}[/tex]
By substituting the values,
→ [tex]\frac{-3 }{\sqrt{y^2+225} } = -\frac{4}{\sqrt{y^2+225} }[/tex]
By applying cross-multiplication,
[tex]3\times \sqrt{y^2+225} = 4\times \sqrt{y^2+81}[/tex]
By squaring both sides, we get
→ [tex]9(y^2+225) = 16(y^2+81)[/tex]
[tex]9y^2+2025 = 16 y^2+1296[/tex]
[tex]2025-1296=7y^2[/tex]
[tex]7y^2=729[/tex]
[tex]y = 10.2 \ m[/tex]
Thus the solution above is correct.
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3) If a ball launched at an angle of 10.0 degrees above horizontal from an initial height of 1.50 meters has a final horizontal displacement of 3.00 meters, what is its launch velocity
Answer:
35.6 m
Explanation:
The given ball possesses a projectile motion from its initial height. So, the required launch velocity of the ball is 6.55 m/s.
What is launch velocity?The horizontal component of velocity during the projection of an object is known as launch velocity. It is obtained when the horizontal range is known.
Given data -
The angle of projection is, [tex]\theta = 10.0 {^\circ}[/tex].
The initial height of the projection is, h = 1.50 m.
The horizontal displacement is, R = 3.00 m.
The mathematical expression for the horizontal displacement (Range) of the projectile is given as,
[tex]R = \dfrac{u^{2} \times sin2 \theta}{g}[/tex]
here,
u is the launch velocity.
g is the gravitational acceleration.
Solving as,
[tex]u =\sqrt{\dfrac{R \times g}{sin2 \theta}}\\\\\\u =\sqrt{\dfrac{1.50 \times 9.8}{sin(2 \times 10)}}\\\\\\u =\sqrt{\dfrac{1.50 \times 9.8}{sin20^\circ}}\\\\\\u=6.55 \;\rm m/s[/tex]
Thus, we can conclude that the required launch velocity of the ball is 6.55 m/s.
Learn more about the projectile motion here:
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A circular loop of wire of radius 10 cm carries a current of 6.0 A. What is the magnitude of the magnetic field at the center of the loop
Answer:
3.77x10^-5T
Explanation:
Magnetic field at center of the loop is given as
B=uo*I/2r =(4pi*10-7)*6/2*0.1
B=3.77*10-5Tor 37.7 uTi