Answer:
E = 12640.78 N/C
Explanation:
In order to calculate the electric field you can use the Gaussian theorem.
Thus, you have:
[tex]\Phi_E=\frac{Q}{\epsilon_o}[/tex]
ФE: electric flux trough the Gaussian surface
Q: net charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2
If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:
[tex]\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}[/tex]
r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m
Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:
[tex]Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C[/tex]
Finally, you obtain for E:
[tex]E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}[/tex]
hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C
Hellppppppppppp please thanks
Point C would the greatest
What is the impulse on a car (750 kg) that accelerates from rest to 5.0 m/s in 10 seconds
Explanation:
impulse J = m × (v2-v1) =750 × ( 5 - 0 ) =3750( N×s)
Answer:
3750Ns
Explanation:
Impulse is defined as Force × time
Force = mass × acceleration,
Hence impulse is;
mass × acceleration × time.
From Newton's second law
Force × time = mass × ∆velocity
750× 5 = 3750Ns
∆velocity = Vfinal-Vinitial ; the initial velocity is zero since the body starts from rest.
A cycler leaves home and rides due south for 65 km. She returns home on the same road. What is the cycler's
displacement?
Y
It doesn't matter what direction she travels, or how far, or whether she uses the same road in both directions.
If she ends up at the same place she started from, then her displacement for the ride is zero.
From the edge of a cliff, a 0.41 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum upward displacement from the launch point is 150 m. What are the (a) horizontal and (b) vertical components of its launch velocity
Answer:
v₀ₓ = 63.5 m/s
v₀y = 54.2 m/s
Explanation:
First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:
K.E = (0.5)(mv₀²)
where,
K.E = initial kinetic energy of projectile = 1430 J
m = mass of projectile = 0.41 kg
v₀ = launch velocity of projectile = ?
Therefore,
1430 J = (0.5)(0.41)v₀²
v₀ = √(6975.6 m²/s²)
v₀ = 83.5 m/s
Now, we find the launching angle, by using formula for maximum height of projectile:
h = v₀² Sin²θ/2g
where,
h = height of projectile = 150 m
g = 9.8 m/s²
θ = launch angle
Therefore,
150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)
Sin θ = √(0.4216)
θ = Sin⁻¹ (0.6493)
θ = 40.5°
Now, we find the components of launch velocity:
x- component = v₀ₓ = v₀Cosθ = (83.5 m/s) Cos(40.5°)
v₀ₓ = 63.5 m/s
y- component = v₀y = v₀Sinθ = (83.5 m/s) Sin(40.5°)
v₀y = 54.2 m/s
Bats use sound to sense objects by sending out short ultrasound pulses and listening for the echo off the object. A. Sketch the ultrasound pulse leaving the bat, reflecting off the object and returning to the bat. B. If a stationary bat is 43 m from an object, how much time elapses between when the bat emits the pulse and it hears the echo
Answer:
B. t = 0.250s
Explanation:
A. An image with the sketch of the bat emitting a sound, which reflects on a surface and return to the bat is attached below.
B. In order to calculate the time that the pulse emitted by the bat, return to the bat, you first calculate the time that pulse takes to arrive to the object.
You use the following formula:
[tex]x=vt[/tex] (1)
x: distance to the object = 43m
t: time = ?
v: speed of sound beat = 343 m/s
You solve the equation (1) for t:
[tex]t=\frac{x}{v}=\frac{43m}{343m/s}=0.125s[/tex]
The time on which the bat hears the echo is twice the value of t, that is:
[tex]t'=2(0.125s)=0.250s[/tex]
The time on which bat heart the echo of its sound, from the moment on which bat emitted it, is 0.250s
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.
Complete Question
Part of the question is shown on the first uploaded image
The rest of the question
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.
Answer:
The net force exerted on the third charge is [tex]F_{net}= 3.22*10^{-5} \ J[/tex]
Explanation:
From the question we are told that
The third charge is [tex]q_3 = 55 nC = 55 *10^{-9} C[/tex]
The position of the third charge is [tex]x = -1.220 \ m[/tex]
The first charge is [tex]q_1 = -16 nC = -16 *10^{-9} \ C[/tex]
The position of the first charge is [tex]x_1 = -1.650m[/tex]
The second charge is [tex]q_2 = 32 nC = 32 *10^{-9} C[/tex]
The position of the second charge is [tex]x_2 = 0 \ m[/tex]
The distance between the first and the third charge is
[tex]d_{1-3} = -1.650 -(-1.220)[/tex]
[tex]d_{1-3} = -0.43 \ m[/tex]
The force exerted on the third charge by the first is
[tex]F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}[/tex]
Where k is the coulomb's constant with a value [tex]9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.[/tex]
substituting values
[tex]F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}[/tex]
[tex]F_{1-3} = 4.28 *10^{-5} \ N[/tex]
The distance between the second and the third charge is
[tex]d_{2-3} = 0- (-1.22)[/tex]
[tex]d_{2-3} =1.220 \ m[/tex]
The force exerted on the third charge by the first is mathematically evaluated as
[tex]F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}[/tex]
substituting values
[tex]F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}[/tex]
[tex]F_{2-3} = 1.06*10^{-5} N[/tex]
The net force is
[tex]F_{net} = F_{1-3} -F_{2-3}[/tex]
substituting values
[tex]F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}[/tex]
[tex]F_{net}= 3.22*10^{-5} \ J[/tex]
The net force on the third charge is -3.216 x 10⁻⁵ N.
The given parameters:
Third charge, q3 = 55 nCFirst charge, q1 = -16 nCSecond charge q2 = 32 nCPosition of third charge, x3 = -1.22 mPosition of first charge, x1 = -1.65 mPosition of the second charge, x2 = 0 mThe force on the third charge due to the first charge is calculated as follows;
r = x1 - x3
r = -1.65 - (-1.22)
r = -0.43 m
[tex]F_{13} = \frac{9\times 10^9 \times (-16 \times 10^{-9}) \times (55 \times 10^{-9})}{(0.43)^2} \\\\F_{13} = -4.28 \times 10^{-5} \ N[/tex]
The force on the third charge due to the second charge is calculated as follows;
[tex]F_{23} = \frac{kq_2 q_3}{r^2} \\\\F_{23} = \frac{9\times 10^9 \times (32 \times 10^{-9}) \times (55 \times 10^{-9})}{(1.22)^2} \\\\F_{23} = 1.064 \times 10^{-5} \ N[/tex]
The net force on the third charge is calculated as follows;
[tex]F_{net} = F_{13} + F_{23}\\\\F_{net} = -4.28 \times 10^{-5} \ N \ + \ 1.064 \times 10^{-5} \ N\\\\F_{net} = -3.216 \times 10^{-5} \ N[/tex]
Learn more about net force here: https://brainly.com/question/14361879
Two cannonballs are dropped from a second-floor physics lab at height h above the ground. Ball B has four times the mass of ball A. When the balls pass the bottom of a first-floor window at height above the ground, the relation between their kinetic energies, KA and KB, is:_______.A) KA- 4KB B) KA 2KB C) KA KB D) KB 4KA.
Formula of kinetic energy
[tex]e = \frac{1}{2} m {v}^{2} [/tex]
Therefore Kinetic energy of ball B is 4 times more than ball A.
ans is KB=4KA
A frog hops at 2.45 m/s a distance of 2.11 m. How long does it take?
Answer:
Explanation:I don't say u must have to mark my ans as brainliest but if it has really helped u plz don't forget to thnk me...
Answer:
0.86s
Explanation:
How long it takes is the time required.
Time = distance /speed
Time =2.11/2.45=0.86s
HELP ASAP!
There is a lever with 5 m long. The fulcrum is 2 m from the right end. Each end hangs a box. The whole system is in balance. If the box hung to the right end is 12 kg, then what is the mass of the box hung to the left end?
Answer:
8kg
Explanation:
For the box to be in equilibrium. The clockwise moment ensued by the box on the right should be same as that ensued by the one on the right. Hence :
M ×3 = 12 ×2
M = 24/3 = 8kg
Note mass is used because trying to compute the weight by multiplying by the acceleration of free fall due to gravity on both sides will cancel out.
А
mass exerts force of
5.6 X 10^-10N on
another mass
when
seperated 93cm apart. If
one mass is the square root of
the other
Find
value of
the two masses.
Answer:
i hope it will be useful for you
Explanation:
F=5.6×10^-10N
R=93cm=0.93m
let take m1 and m2 =m²
according to newton's law of universal gravitation
F=m1m2/r²
F=m²/r²
now we have to find masses
F×r²=m²
5.6×10^10N×0.93m=m²
5.208×10^-9=m²
taking square root on b.s
√5.208×10^-9=√m²
so the two masses are m1=7.2×10^-5
and m2=7.2×10^-5
You have two balloons, one that has a spherical core of radius 4 cm and the other that is tubular with a radius of 0.5 cm and a length of 8 cm. Knowing that the force that you can initially apply to trying to expand each balloon is directly proportional to the volume of the balloon, show that the higher initial stress is achievable with the spherical balloon.
Answer:
Explanation:
Given That:
radius of spherical core r₁ = 4cm
radius of tubular r₂ = 0.5cm
length of tubular l = 8cm
Volume of spherical V₁
[tex]=\frac{4}{3} \pi r_1^3[/tex]
[tex]=\frac{4}{3} \pi(4)^3\\\\=\frac{4}{3} \pi 64\\\\=268.1cm^3[/tex]
Volume of tabular V₂
[tex]=\pi r ^2_2h[/tex]
[tex]=\pi(0.5)^2\times 8\\\\ =\pi 90.250\times8\\\\ =\pi 2\\\\=6.283cm^3[/tex]
F ∝ V
[tex]F_1 \propto V_1[/tex] and [tex]F_2 \propto V_2[/tex]
As V₁ is greater than V₂
⇒ F₁ is greater than F₂
F is force
V is volume
This is the required answer
Which of the following is NOT a type of electromagnetic wave?
Seismic waves
Visible light
Radio waves
Microwave
Answer:
Seismic waves
Explanation:
seismic waves are not represented by electromagnetic graphs, nor can they be reflected on an electromagnetic spectrum. Visible light, radio waves, and microwaves are all electromagnetic waves, which are represented by graphs and electromagnetic spectrums.
Answer:
Siesmic waves
Explanation:
A spaceship travels toward the Earth at a speed of 0.97c. The occupants of the ship are standing with their torsos parallel to the direction of travel. According to Earth observers, they are about 0.50 m tall and 0.50 m wide. Calculate what the occupants’ height and width according to the others on the spaceship?
Answer:
Explanation:
We shall apply length contraction einstein's relativistic formula to calculate the length observed by observer on the earth . For the observer , increased length will be observed for an observer on the earth
[tex]L=\frac{.5}{\sqrt{1-(\frac{.97c}{c})^2 } }[/tex]
[tex]L=\frac{.5}{.24}[/tex]
L= 2.05
The length will appear to be 2.05 m . and width will appear to be .5 m to the observer on the spaceship. . It is so because it is length which is moving parallel to the direction of travel. Width will remain unchanged.
Motion maps for two objects, Y and Z, are shown.
A motion map. The position line is a long black arrow pointing right with x as the reference point at left. Above the line are three dots, each with a vector pointed away from x back to back in a line labeled B. Above B, there are four dots, each with a shorter vector pointing away from x in a line labeled A starting closer to x .
Object Z passes object Y after how many seconds?
2
3
4
5
Answer: it takes 3 seconds (b)
Explanation:
Answer: B. 3
Explanation:
Each black point on the map represents one second. There are three black points with vectors representing Z's movement before Y begins to move.
A turtle and a rabbit are in a 150 meter race. The rabbit decides to give the turtle a 1 minute head start. The turtle moves at a constant speed of 0.500 m/s through the whole race (in fact the turtle even starts at a velocity of 0.500 m/s as while he was still approaching the starting line he was allowed to continue to keep going without stopping). The rabbit starts the race from rest and accelerates at a rate or 1.50 m/s2 until she reaches her top speed of 10 m/s. She then finishes the race running at a constant speed of 10 m/s. a) What is the turtle’s position when the rabbit starts to run (1 minute into the race)? b) How long does it take the turtle to finish the race? c) How long does it take the rabbit to reach max speed? d) What is the rabbit’s position when she reaches max speed? e) How long does it take the rabbit to finish the race? f) Who won?
Answer:
a) [tex]s_{T} = 30\,m[/tex], b) [tex]t = 5\,min[/tex], c) [tex]\Delta t = 6.667\,s[/tex], d) [tex]\Delta s_{R} = 33.333\,m[/tex], e) [tex]t' = 11.667\,s[/tex], f) The rabbit won the race.
Explanation:
a) As turtle moves at constant speed, its position is determined by the following formula:
[tex]s_{T} = v_{T}\cdot t[/tex]
Where:
[tex]t[/tex] - Time, measured in seconds.
[tex]v_{T}[/tex] - Velocity of the turtle, measured in meters per second.
[tex]s_{T}[/tex] - Position of the turtle, measured in meters.
Then, the position of the turtle when the rabbit starts to run is:
[tex]s_{T} = \left(0.5\,\frac{m}{s} \right)\cdot (60\,s)[/tex]
[tex]s_{T} = 30\,m[/tex]
The position of the turtle when the rabbit starts to run is 30 meters.
b) The time needed for the turtle to finish the race is:
[tex]t = \frac{s_{T}}{v_{T}}[/tex]
[tex]t = \frac{150\,m}{0.5\,\frac{m}{s} }[/tex]
[tex]t = 300\,s[/tex]
[tex]t = 5\,min[/tex]
The time needed for the turtle to finish the race is 5 minutes.
c) As rabbit experiments a constant acceleration until maximum velocity is reached and moves at constant speed afterwards, the time required to reach such speed is:
[tex]v_{R} = v_{o,R} + a_{R}\cdot \Delta t[/tex]
Where:
[tex]v_{R}[/tex] - Final velocity of the rabbit, measured in meters per second.
[tex]v_{o,R}[/tex] - Initial velocity of the rabbit, measured in meters per second.
[tex]a_{R}[/tex] - Acceleration of the rabbit, measured in [tex]\frac{m}{s^{2}}[/tex].
[tex]\Delta t[/tex] - Running time, measured in second.
[tex]\Delta t = \frac{v_{R}-v_{o,R}}{a_{R}}[/tex]
[tex]\Delta t = \frac{10\,\frac{m}{s}-0\,\frac{m}{s}}{1.50\,\frac{m}{s^{2}} }[/tex]
[tex]\Delta t = 6.667\,s[/tex]
The time taken by the rabbit to reach maximum speed is 6.667 s.
d) On the other hand, the position reached by the rabbit when maximum speed is reached is determined by the following equation of motion:
[tex]v_{R}^{2} = v_{o,R}^{2} + 2\cdot a_{R}\cdot \Delta s_{R}[/tex]
[tex]\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}[/tex]
[tex]\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}[/tex]
Where [tex]\Delta s_{R}[/tex] is the travelled distance of the rabbit from rest to maximum speed.
[tex]\Delta s_{R} = \frac{\left(10\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(1.50\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\Delta s_{R} = 33.333\,m[/tex]
The distance travelled by the rabbit from rest to maximum speed is 33.333 meters.
e) The time required for the rabbit to finish the race can be determined by the following expression:
[tex]t' = \frac{\Delta s_{R}}{v_{R}}[/tex]
[tex]t' = \frac{150\,m-33.333\,m}{10\,\frac{m}{s} }[/tex]
[tex]t' = 11.667\,s[/tex]
The time required for the rabbit from rest to maximum speed is 11.667 seconds.
f) The animal with the lowest time wins the race. Now, each running time is determined:
Turtle:
[tex]t_{T} = 300\,s[/tex]
Rabbit:
[tex]t_{R} = 60\,s + 6.667\,s + 11.667\,s[/tex]
[tex]t_{R} = 78.334\,s[/tex]
The rabbit won the race as [tex]t_{R} < t_{T}[/tex].
100picometer is equal to
Answer:
100picometer is equal to:
0.1 NanometreExplanation:
Hope this works !!When light with a wavelength of 225 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 2.98 × 10-19 J. Determine the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.
Answer:
The new wavelength is 112.5 nm.
Explanation:
It is given that,
When light with a wavelength of 225 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 2.98 × 10⁻¹⁹ J. We need to find the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.
The energy of incident electron is given by :
[tex]E=\dfrac{hc}{\lambda}[/tex]
New energy is 2 E and new wavelength is [tex]\lambda'[/tex]. So,
[tex]\dfrac{E}{2E}=\dfrac{\lambda'}{\lambda}\\\\\dfrac{1}{2}=\dfrac{\lambda'}{\lambda}\\\\\lambda'=\dfrac{\lambda}{2}\\\\\lambda'=\dfrac{225}{2}\\\\\lambda=112.5\ nm[/tex]
So, the new wavelength is 112.5 nm.
. An object whose mass is 375 lb falls freely under the influence of gravity from an initial elevation of 253 ft above the surface of Earth. The initial velocity is downward with a magnitude of 10 ft/s. The effect of air resistance is negligible. (a) Determine the speed, in ft/s, of the object just before it strikes the surface of Earth. (b) Calculate the kinetic energy of the object, in Btu, before it strikes the surface of Earth. Assume that the gravitational acceleration is g
Answer:
(a) Vf = 128 ft/s
(b) K.E = 122.8 Btu
Explanation:
(a)
In order to find the velocity of the object just before striking the surface of earth or the final velocity, we use 3rd equation of motion:
2gh = Vf² - Vi²
where,
g = 32.2 ft/s²
h = height = 253 ft
Vf = Final Velocity = ?
Vi = Initial Velocity = 10 ft/s
Therefore,
(2)(32.2 ft/s²)(253 ft) = Vf² - (10 ft/s)²
16293.2 ft²/s² + 100 ft²/s² = Vf²
Vf = √(16393.2 ft²/s²)
Vf = 128 ft/s
(b)
The kinetic energy of the object before it hits the surface of earth is given by:
K.E = (0.5)(m)(Vf)²
where,
m = mass of object = 375 lb
K.E = Kinetic energy of object before it strikes the surface of earth = ?
Therefore,
K.E = (0.5)(375 lb)(128 ft/s)²
K.E = 3073725 lb.ft²/s²
Now, converting this to Btu:
K.E = (3073725 lb.ft²/s²)(1 Btu/25037 lb.ft²/s²)
K.E = 122.8 Btu
Which statement best explains why the overall charge on an atom is zero?
ОООО
The positive charge of the neutrons in the nucleus equals the negative charge in the electron cloud.
The positive charge of the protons in the nucleus equals the negative charge in the electron cloud.
The negative charge of the neutrons in the nucleus equals the positive charge in the electron cloud.
The negative charge of the protons in the nucleus equals the positive charge in the electron cloud.
Answer:
B) The positive charge of the protons in the nucleus equals the negative charge in the electron cloud.
Explanation:
For every negative charge of an electron, there is an equal positively charged proton in the nucleus of the atom. This is why the overall charge on an atom is zero.
Answer:
B
Explanation:
This is described by Gauss a scientist.
The positive charge is found in the proton in the nucleus.
The neutron has no charge.
The positive charge radiates in all directions and a counter negative charge ensues.
In Parts A, B, C consider the following situation. In a baseball game the batter swings and gets a good solid hit. His swing applies a force of 12,000 N to the ball for a time of 0.70×10−3s. Part A Assuming that this force is constant, what is the magnitude J of the impulse on the ball?
Answer:
J = 8.4 kg*m/s
Explanation:
The magnitude of the impulse, J, on the ball can be calculated using the following equation:
[tex] J = F*t [/tex] (1)
Where:
F: is the force = 12000 N
t: is the time = 0.70x10⁻³ s
So, by entering the values above into equation (1) we can find the impulse on the ball:
[tex] J = F*t = 12000 N*0.70 \cdot 10^{-3} s = 8.4 kg*m/s [/tex]
Therefore, the impulse on the ball is 8.4 kg*m/s.
I hope it helps you!
The magnitude J of the impulse on the ball is 8.4 kg m/s.
Calculation of the impulse:Since it applies a force of 12,000 N to the ball for a time of 0.70×10−3s.
So,
The magnitude is
[tex]= Force \times time\\\\= 12,000 \times 0.70\times 10^{-3}[/tex]
= 8.4 kg m/s
Here we basically multiplied the force with the time to determine the magnitude.
Learn more about impulse here: https://brainly.com/question/16324614
Mention 4
applications of thermal expansion
thermak expansion is for cocluding the elctric value of solar panels
Answer:
thermometerscombustion enginesfitting or loosening of metallic partsproviding lift for hot-air balloonsExplanation:
The thermal expansion properties of liquids and metals are used in thermometers of many types. The liquid in a bulb thermometer expands to provide indication on a calibrated scale. Thermal expansion of a metal coil is used in dial thermometers and in thermostats of many kinds.
The thermal expansion of hot gas drives the cylinders or turbines in combustion engines, jet engines, and thermal cycle motors.
Thermal expansion of metal relative to glass can help remove a stuck jar lid. Similarly, machine parts can be expanded by heating to facilitate assembly or disassembly.
The expansion of warmer air in the atmosphere gives rise to updrafts and thermals that can be used by birds and bugs and people for gaining altitude. The expansion of air in the envelope of a hot-air balloon drives its lift as well.
A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.60 m/s. The car is a distance d away. The bear is 39.3 m behind the tourist and running at 5.00 m/s. The tourist reaches the car safely. What is the maximum possible value for d
Answer:
101 meters
Explanation:
Distance traveled by the tourist:
d = 3.60 m/s × t
Distance traveled by the bear:
d + 39.3 m = 5.00 m/s × t
Substitute:
3.6 t + 39.3 = 5 t
39.3 = 1.4 t
t = 28.1
d = 101
Two identical point charges q=71.0 pCq=71.0 pC are separated in vacuum by a distance of 2d=29.0 cm.2d=29.0 cm. Calculate the total electric flux ΦΦ through the infinite surface placed at a distance dd from each charge, perpendicular to the line on which the point charges are located.
Answer:
The electric flux at the infinite surface is ZERO
Explanation:
From the question we are told that
The point charge are identical and the value is [tex]q = 71.0 pC = 71 * 10^{-12} \ C[/tex]
The distance of separation is [tex]D = 29.0 \ cm = 0.29 \ m[/tex]
The distance of both from the infinite surface is d
Generally the electric force exerted by each of the charge on the infinite surface is
[tex]\phi = \frac{q}{\epsilon_o}[/tex]
Now given from the question that they are identical, it then means that the electric flux of the first charge on the infinite surface will be nullified by the electric flux of the second charge hence the electric flux at that infinite surface due to this two identical charges is ZERO
Consider the previous situation. Under what condition would the acceleration of the center of mass be zero? Keep in mind that F1x and F2x represent the components, of the corresponding forces. Consider the previous situation. Under what condition would the acceleration of the center of mass be zero? Keep in mind that and represent the components, of the corresponding forces. F1x=−F2x F1x=F2x m1=m2 m1≪m2
Answer:
a) m₁ = m₂ F₁ₓ = F₂ₓ
b) m₁ << m₂ F₂ₓ =0
Explanation:
This interesting exercise is unclear your statement, so that in a center of mass system has an acceleration of zero it is necessary that the sum of the forces on each axis is zero, to see this we write Newton's second law
∑ F = m a
for acceleration to be zero implies that the net force is zero.
we must write the expression for the center of mass
[tex]x_{cm}[/tex] = 1 / M (m₁ x₁ + m₂ x₂)
now let's use the derivatives
[tex]a_{cm}[/tex] = d² x_{cm}/dt² = 1 / M (m₁ a₁ + m₂a₂)
where M is the total mass M = m₁ + m₂
so that the acceleration of the center of mass is zero
0 = 1 / M (m₁ a₁ + m₂a₂)
m₁ a₁ = - m₂ a₂
In the case that we have components on the x axis, the modulus of the two forces are equal and their direction is opposite, therefore
F₁ₓ = -F₂ₓ
b)r when the two masses are equal , in the case of a mass greater than the other m₁ << m₂
acm = d2 xcm / dt2 = 1 / M (m1 a1 + m2a2)
so that the acceleration of the center of mass is zero
0 = 1 / M (m1 a1 + m2a2)
m1 a1 = - m 2 a2
with the initial condition, we can despise m₁, therefore
0 = m₂a₂
if we use Newton's second law
F₂ = 0
I tell you that in this case with a very high mass difference the force on the largest mass must be almost zero
Consider the binding energy of two stable nuclei, one with 60 nucleons and one with 200 nucleons. a. Is the total binding energy of the nucleus with 200 nucleons more than, less than, or equal to the total binding energy of the nucleus with 60 nucleons. Justify your reasoning.
Answer:
The total binding energy of the nucleus with 200 nucleons more than the total binding energy of the nucleus with 60 nucleons
Explanation:
Binding energy can be given by the formula:
Binding energy = Binding energy per nucleon * Number of nucleons
This means that if the binding energy per nucleon = x MeV
Where x is a positive real number
The nucleus with 60 nucleons will have Binding energy = 60x MeV
The nucleus with 200 nucleons will have binding energy = 200x MeV
For a +ve x, 200x > 60x
Binding energy is proportional (directly) to nucleon volume. A further explanation is provided below.
Binding energy
Binding energy involving 200 nucleons would've been greater than 60 nucleons because so many more nucleons result in what seems like a stronger reaction.
It makes absolutely no difference out whether nucleon seems to be a proton as well as a neutron because they just have a similar strong coupling relatively steady.
Thus the response above is correct.
Find out more information about binding energy here:
https://brainly.com/question/23942204
11. A vector M is 15.0 cm long and makes an angle of 20° CCW from x axis and another vector N is 8.0 cm long and makes an angle of 40° clockwise from the x axis. Find out resultant vector with its magnitude and direction using components method.
Answer:
The magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°
Explanation:
To find the resultant vector, you first calculate x and y components of the two vectors M and N. The components of the vectors are calculated by using cos and sin function.
For M vector you obtain:
[tex]M=M_x\hat{i}+M_y\hat{j}\\\\M=15.0cm\ cos(20\°)\hat{i}+15.0cm\ sin(20\°)\hat{j}\\\\M=14.09cm\ \hat{i}+5.13\ \hat{j}[/tex]
For N vector:
[tex]N=N_x\hat{i}+N_y\hat{j}\\\\N=8.0cm\ cos(40\°)\hat{i}+8.0cm\ sin(40\°)\hat{j}\\\\N=6.12cm\ \hat{i}+5.142\ \hat{j}[/tex]
The resultant vector is the sum of the components of M and N:
[tex]F=(M_x+N_x)\hat{i}+(M_y+N_y)\hat{j}\\\\F=(14.09+6.12)cm\ \hat{i}+(5.13+5.142)cm\ \hat{j}\\\\F=20.21cm\ \hat{i}+10.27cm\ \hat{j}[/tex]
The magnitude of the resultant vector is:
[tex]|F|=\sqrt{(20.21)^2+(10.27)^2}cm=22.66cm[/tex]
And the direction of the vector is:
[tex]\theta=tan^{-1}(\frac{10.27}{20.21})=29.93\°[/tex]
hence, the magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°
A very long, straight solenoid with a cross-sectional area of 2.22 cm2 is wound with 85.6 turns of wire per centimeter. Starting at t = 0, the current in the solenoid is increasing according to i(t)=(0.177A/s2)t2. A secondary winding of 5.0 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2A?
Answer:
The induce emf is [tex]\epsilon = 1.7966*10^{-5} V[/tex]
Explanation:
From the question we are told that
The cross-sectional area is [tex]A = 2,22 cm^3 = \frac{2.22}{10000} = 2.22*10^{-4} \ m^2[/tex]
The number of turn is [tex]N = 85.6 \ turns/cm = 85.6 \ \frac{turns }{\frac{1}{100} } = 8560 \ turns / m[/tex]
The starting time is [tex]t_o[/tex] = 0 s
The current increase is [tex]I(t) = (0.177A/s^2) t^2[/tex]
The number of turn of secondary winding is [tex]N_s = 5 \ turn s[/tex]
The current at the solenoid is [tex]I_(t) = 3.2 \ A[/tex]
at [tex]I_(t) = 3.2 \ A[/tex]
[tex]3.2 = 0.177* t^2[/tex]
=> [tex]t = \sqrt{ \frac{3.2}{0.177} }[/tex]
[tex]t = 4.25 s[/tex]
Generally Faraday's law of induction is mathematically represented as
[tex]\epsilon = A\mu_o N_s N * \frac{di}{dt}[/tex]
[tex]\epsilon = A\mu_o N_s N * \frac{d (0.177 t^2)}{dt}[/tex]
[tex]\epsilon = A\mu_o N_s N * (0.177)(2t)[/tex]
substituting values
[tex]\epsilon = (2.22*10^{-4}) * ( 4\pi * 10^{-7}) * 5 * [8560]* 0.177 * 2 * 4.25[/tex]
[tex]\epsilon = 1.7966*10^{-5} V[/tex]
An aluminium bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 30mm in diameter and is 100mm long. If the modulus of elasticity for the aluminium is 85GN/m2, calculate the total contraction on the bar due to a compressive load of 180kN.
Answer:
[tex]\delta = 0.385\,m[/tex] (Compression)
Explanation:
The aluminium bar is experimenting a compression due to an axial force, that is, a force exerted on the bar in its axial direction. (See attachment for further details) Under the assumption of small strain, the deformation experimented by the bar is equal to:
[tex]\delta = \frac{P\cdot L}{A \cdot E}[/tex]
Where:
[tex]P[/tex] - Load experimented by the bar, measured in newtons.
[tex]L[/tex] - Length of the bar, measured in meters.
[tex]A[/tex] - Cross section area of the bar, measured in square meters.
[tex]E[/tex] - Elasticity module, also known as Young's Module, measured in pascals, that is, newtons per square meter.
The cross section area of the bar is now computed: ([tex]D_{o} = 0.04\,m[/tex], [tex]D_{i} = 0.03\,m[/tex])
[tex]A = \frac{\pi}{4}\cdot (D_{o}^{2}-D_{i}^{2})[/tex]
Where:
[tex]D_{o}[/tex] - Outer diameter, measured in meters.
[tex]D_{i}[/tex] - Inner diameter, measured in meters.
[tex]A = \frac{\pi}{4}\cdot [(0.04\,m)^{2}-(0.03\,m)^{2}][/tex]
[tex]A = 5.498 \times 10^{-4}\,m^{2}[/tex]
The total contraction of the bar due to compresive load is: ([tex]P = -180\times 10^{3}\,N[/tex], [tex]L = 0.1\,m[/tex], [tex]E = 85\times 10^{9}\,Pa[/tex], [tex]A = 5.498 \times 10^{-4}\,m^{2}[/tex]) (Note: The negative sign in the load input means the existence of compressive load)
[tex]\delta = \frac{(-180\times 10^{3}\,N)\cdot (0.1\,m)}{(5.498\times 10^{-4}\,m^{2})\cdot (85\times 10^{9}\,Pa)}[/tex]
[tex]\delta = -3.852\times 10^{-4}\,m[/tex]
[tex]\delta = -0.385\,mm[/tex]
[tex]\delta = 0.385\,m[/tex] (Compression)
What is a major criticism of Maslow's hierarchy of needs? Select one: a. It is subjective. b. It does not take gender differences into account. c. It is humanistic. d. It only accounts for the objective world.
The correct answer is A. It is subjective
Explanation:
In 1943, the recognized psychologist Abraham Maslow proposed a theory to understand and classify human needs. The work of Maslow included five different categories to classify all basic needs, psychological needs, and self-esteem needs; additionally, in this, Maslow proposed individuals need to satisfy the needs of previous levels to satisfy more complex needs. For example, the first level includes physiological needs such as hunger and these are necessary to get to more complex needs such as the need for safety or self-satisfaction.
This hierarchy is still used all around the world to understand human needs; however, it was been widely criticized because the classification itself is related to Maslow's perspective as this was mainly based on Maslow's ideas about needs, which makes the hierarchy subjective. Also, due to its subjectivity, the hierarchy may apply only in some individuals or societies.
How many times can a three-dimensional object that has a radius of 1,000 units fit something with a radius of 10 units inside of it? How many times can something with a radius of 2,000 units fit something with a radius of 1 unit?
Answer:
# _units = 1000
Explanation:
This exercise we can use a direct proportion rule.
If a volume of radius r = 1 is one unit, how many units can fit in a volume of radius 10?
# _units = V₁₀ / V₁
The volume of a body of radius 1 is
V₁ = 4/3 π r₁³
V₁ = 4/3π
the volume of a body of radius r = 10
V₁₀ = 4/3 π r₂³
V10 = 4/3 π 10³
the number of times this content is
#_units = 4/3 π 1000 / (4/3 π 1)
# _units = 1000