Answer:
Approximately [tex]0.97\; \rm N[/tex]. This force would point away from the center of the square (to the left at [tex]45^\circ[/tex] above the horizontal direction.)
Explanation:
Coulomb's constant: [tex]k \approx 8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2}[/tex].
By Coulomb's Law, the electrostatic force between two point charges [tex]q_1[/tex] and [tex]q_2[/tex] that are separated by [tex]r[/tex] (vacuum) would be:
[tex]\displaystyle F = \frac{k \cdot q_1 \cdot q_2}{r^2}[/tex].
Consider the charge on the top-left corner of this square.
Apply Coulomb's Law to find the electrostatic force between this charge and the charge on the lower-left corner.
Convert quantities to standard units:
[tex]q_1 = q_2 = 3 \times 10^{-6}\; \rm C[/tex].
[tex]r = 0.40\; \rm m[/tex].
[tex]\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^2} \\ &\approx \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times (3 \times 10^{-6}\; \rm C)^{2}}{(0.40\; \rm m)^{2}} \\ &\approx 0.506\; \rm N\end{aligned}[/tex].
As the two charges are of the same sign, the electrostatic force on each charge would point away from the other charge. Hence, for the charge on the top-left corner of the square, the electrostatic force from the charge below it would point upwards.
Similarly, the charge to the right of this charge would exert an electrostatic force with the same magnitude (approximately [tex]0.506\; \rm N[/tex]) that points leftwards.
For the charge to the lower-right of the top-left charge, [tex]r = \sqrt{2} \times 0.40\; \rm m[/tex]. Therefore:
[tex]\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^2} \\ &\approx \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times (3 \times 10^{-6}\; \rm C)^{2}}{(\sqrt{2} \times 0.40\; \rm m)^{2}} \\ &\approx 0.253 \; \rm N\end{aligned}[/tex].
This force would point to the top-left of the top-left charge, which is [tex]45^\circ[/tex] above the horizontal direction. Decompose this force into two components that are normal to one another:
Horizontal component: approximately [tex]\sin(45^\circ) \times 0.253\; \rm N \approx 0.179\; \rm N[/tex].Vertical component: approximately [tex]\cos(45^\circ) \times 0.253\; \rm N \approx 0.179\; \rm N[/tex]Consider the net force on the top-left charge in two components:
Horizontal component: approximately [tex]0.506\; \rm N[/tex] from the charge on the top-right corner and approximately [tex]0.179\; \rm N[/tex] from the charge on the lower-right corner. Both components point to the left-hand side. [tex]F_x \approx 0.506\; \rm N + 0.179\; \rm N = 0.685\;\rm N[/tex] (to the left).Vertical component: approximately [tex]0.506\; \rm N[/tex] from the charge on the lower-left corner and approximately [tex]0.179\; \rm N[/tex] from the charge on the lower-right corner. Both components point upwards. [tex]F_y \approx 0.506\; \rm N + 0.179\; \rm N = 0.685\;\rm N[/tex] (upward).Combine these two components to find the magnitude of the net force on this charge:
[tex]\begin{aligned}F &= \sqrt{{F_x}^{2} + {F_y}^{2}} \\ &\approx \sqrt{0.685^2 + 0.685^2 }\; \rm N \\ &\approx 0.97\; \rm N\end{aligned}[/tex].
This force would point to the top-left of this charge (also at [tex]45^\circ[/tex] above the horizontal direction, away from the center of the square) because its horizontal and vertical components have the same magnitude.
What is the total number of moles of products involved in the following reaction?
CaCO3 (s) + 2HCl (aq) - CaCl2 (aq) + CO2 (g) + H20 (g)
O 6
2.
3
5
Answer:
3
Explanation:
You must first make sure the equation is balanced. This one is. Then, you simply add up the coefficients of each compound on the products side of the equation. When the coefficient is not specified, you can assume it is 1 mole. So, in this equation, there is 1 mole of CaCl₂, 1 mole of CO₂, and 1 mole of H₂O = 3 moles.
The reactant side of the equation also has three moles:
1 mole of CaCO₃ and 2 moles of HCl.
A proton traveling due west in a region that contains only a magnetic field experiences a vertically upward force (away from the surface of the earth). What is the direction of the magnetic field?
South
Explanation:
The magnetic force F on a point charge moving with a velocity v in the presence of a magnetic field B is given by
[tex]\vec{\textbf{F}} = q\vec{\textbf{v}}\textbf{×}\vec{\textbf{B}}[/tex]
and according to the right-hand rule, an upward magnetic force on a proton moving westward is only possible if the magnetic field is directed southward.
The gravitational field strength due to its planet is 5N/kg What does it mean?
Answer:
The weight of an object is the force on it caused by the gravity due to the planet. The weight of an object and the gravitational field strength are directly proportional. For a given mass, the greater the gravitational field strength of the planet, the greater its weight.
Weight can be calculated using the equation:
weight = mass × gravitational field strength
This is when:
weight (W) is measured in newtons (N)
mass (m) is measured in kilograms (kg)
gravitational field strength (g) is measured in newtons per kilogram (N/kg)
Name the electrolyte in the chemical method of generating electricity
g A mass of 2.0 kg traveling at 3.0 m/s along a smooth, horizontal plane hits a relaxed spring. The mass is slowed to zero velocity when the spring has been compressed by 0.15 m. What is the spring constant of the spring
By the work-energy theorem, the total work done on the mass by the spring is equal to the change in the mass's kinetic energy:
W = ∆K
and the work done by a spring with constant k as it gets compressed a distance x is -1/2 kx ²; the work it does is negative because the restoring force of the spring points opposite the direction in which it's getting compressed.
So we have
-1/2 k (0.15 m)² = 0 - 1/2 (2.0 kg) (3.0 m/s)²
Solve for k to get k = 800 N/m.
A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?
Answer:
The correct answer is "[tex]4.49\times 10^{10} \ joules[/tex]".
Explanation:
According to the question,
The work will be:
⇒ [tex]Work=-\frac{kQq}{R}[/tex]
[tex]=-\frac{1}{4 \pi \varepsilon \times (18-30)\times 3\times 0.2}[/tex]
[tex]=-\frac{1}{4 \pi \varepsilon \times (-12)\times 3\times 0.2}[/tex]
[tex]=\frac{0.3978}{\varepsilon }[/tex]
[tex]=4.49\times 10^{10} \ joules[/tex]
Thus the above is the correct answer.
We have that the workdone is mathematically given as
W=4.49*10e10 J
From the question we are told
A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?WorkdoneGenerally the equation for the workdone is mathematically given as
W=-kQq/R
Therefore
0.3978/ε0 =-1/(4πε0*(18-30)*3*0.2
Hence
W=4.49*10e10 JFor more information on Charge visit
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If 5.4 J of work is needed to stretch a spring from 15 cm to 21 cm and another 9 J is needed to stretch it from 21 cm to 27 cm, what is the natural length (in cm) of the spring
Answer:
the natural length of the spring is 9 cm
Explanation:
let the natural length of the spring = L
For each of the work done, we set up an integral equation;
[tex]5.4 = \int\limits^{21-l}_{15-l} {kx} \, dx \\\\5.4 = [\frac{1}{2}kx^2 ]^{21-l}_{15-l}\\\\5.4 = \frac{k}{2} [(21-l)^2 - (15-l)^2]\\\\k = \frac{2(5.4)}{(21-l)^2 - (15-l)^2} \ \ \ -----(1)[/tex]
The second equation of work done is set up as follows;
[tex]9 = \int\limits^{27-l}_{21-l} {kx} \, dx \\\\9 = [\frac{1}{2}kx^2 ]^{27-l}_{21-l}\\\\9 = \frac{k}{2} [(27-l)^2 - (21-l)^2] \\\\k = \frac{2(9)}{(27-l)^2 - (21-l)^2} \ \ \ -----(2)[/tex]
solve equation (1) and equation (2) together;
[tex]\frac{2(9)}{(27-l)^2 - (21-l)^2} = \frac{2(5.4)}{(21-l)^2 - (15-l)^2}\\\\\frac{2(9)}{2(5.4)} = \frac{(27-l)^2 - (21-l)^2}{(21-l)^2 - (15-l)^2}\\\\\frac{9}{5.4} = \frac{(729 - 54l+ l^2) - (441-42l+ l^2)}{(441-42l+ l^2) - (225 -30l+ l^2)} \\\\\frac{9}{5.4 } = \frac{288-12l}{216-12l} \\\\\frac{9}{5.4 } =\frac{12}{12} (\frac{24-l}{18 -l})\\\\\frac{9}{5.4 } = \frac{24-l}{18 -l}\\\\9(18-l) = 5.4(24-l)\\\\162-9l = 129.6-5.4l\\\\162-129.6 = 9l - 5.4 l\\\\32.4 = 3.6 l\\\\l = \frac{32.4}{3.6} \\\\[/tex]
[tex]l = 9 \ cm[/tex]
Therefore, the natural length of the spring is 9 cm
An electric lamp consumes 60W at 220 volts. How many dry cells of 1.5 V and internal resistance 1 Ohm are required to glow the lamp?
Answer:
1. Number of dry cells of 1.5 V required is 40.
2. Number of internal resistance of 1 ohm required is 807
Explanation:
We'll begin by calculating the resistance. This can be obtained as follow:
Power (P) = 60 W
Voltage (V) = 220 V
Resistance (R) =?
P = V²/R
60 = 220² / R
Cross multiply
60 × R = 220²
60 × R = 48400
Divide both side by 60
R = 48400 / 60
R ≈ 807 Ohm
1. Determination of the number of dry cells of 1.5 V required.
Voltage (V) = 220
Dry Cells = 1.5 V
Number of dry cells (n) =?
n = Voltage / Dry cells
n = 60 / 1.5
n = 40
2. Determination of the number of internal resistance of 1 ohm required.
Resistance (R) = 807 Ohm
Internal resistance (r) = 1 ohm
Number of internal resistance (n) =?
n = R/r
n = 807 / 1
n = 807
SUMMARY:
1. Number of dry cells of 1.5 V required is 40.
2. Number of internal resistance of 1 ohm required is 807
Explore the Prisms screen to see how your understanding of refraction applies when light travels through a medium like glass. Give specific examples and images from the simulation to explain how your understanding applies
Explanation:
https://tse2.mm.bing.net/th?id=OGC.b52c959ac810db1177599a161631c917&pid=Api&rurl=https%3a%2f%2fupload.wikimedia.org%2fwikipedia%2fcommons%2fthumb%2ff%2ff5%2fLight_dispersion_conceptual_waves.gif%2f266px-Light_dispersion_conceptual_waves.gif&ehk=TdcWPzr5xGP8xUOSOqZXauGOS1jHDMu7WnxPzkl7esw%3d
a vessel with mass 10kg intially moving withthe velocicity 12m s along the x axis explodes into three exactly identical pieces Just after the explosion one piece moves with speed 10 m s along the x axis and asecond piece moves with speed 10 m s along the y axis What iis the magnitude of the component of velocity of the third piece along the y axiss
Answer:
Explanation:
Apply law of conservation of momentum along y-axis.
Initially there was no momentum along y-axis. So there will be nil momentum along y-axis again finally.
Let the mass of each piece after breaking be m .
Momentum of piece moving along positive y-axis
= m x 10 = 10m .
Let the component of velocity of third piece along y-axis be v .
Its momentum along the same direction = m v .
Total momentum along y -axis = 10 m + m v
According to law of conservation of momentum
10 m + mv = 0
v = - 10 m/s .
Component of velocity of the third piece along y-axis will be - 10 m/s .
In other words it will be along negative y-axis with speed of 10 m/s.
Can you think of reasons why the charge on each ball decreases over time and where the charges might go
Answer:
By the principle of corona discharge.
Explanation:
The charge on each ball will decreases over time due to the electrical discharge in air.
According to the principle of corona discharge, when the curvature is small, the discharge of the charge takes placed form the pointed ends.
A wave moves in a rope with a certain wavelength. A second wave is made to move in the same rope with twice the wavelength of the first wave. The frequency of the second wave is _______________ the frequency of the first wave.
Answer:
The frequency of the second wave is half of the frequency of first one.
Explanation:
The wavelength of the second wave is double is the first wave.
As we know that the frequency is inversely proportional to the wavelength of the velocity is same.
velocity = frequency x wavelength
So, the ratio of frequency of second wave to the first wave is
[tex]\frac{f_2}{f_1} =\frac{\lambda _1}{\lambda _2}\\\\\frac{f_2}{f_1} =\frac{\lambda _1}{2\lambda _1}\\\\\frac{f_2}{f_1} =\frac{1}{2}\\\\[/tex]
The frequency of the second wave is half of the frequency of first one.
1.a machine gun fires a ball with an initial velocity of 600m/s with an elevation of 30° with respect to the ground neglecting air resistance calculate:
a.the maximum height that can be reached?
b.the time of flight of the bullet?
c.the maximum horizontal displacement of the ired bullet?
Answer:
See explanation
Explanation:
a) maximum height of a projectile = u sin^2θ/2g
H= 600 × (sin 30)^2/2 × 10
H= 7.5 m
b) Time of flight
t= 2u sinθ/g
t= 2 × 600 sin 30/10
t= 60 seconds
Range
R= u^2sin2θ/g
R= (600)^2 × sin2(30)/10
R= 31.2 m
An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m. (a) What is his deceleration
Answer:
a= -80.357 m/s
Explanation:
use the formula
vf^2=vi^2+2a(xf-xi)
Plug in givens
0=(7.50)^2+2a(0.350m)
solve for acceleration
a= -80.357 m/s
A point charge is positioned in the center of a hollow metallic shell of radius R. During four experiments the value of the point charge and the total charge on shell were, respectively:
Complete question is;
A point charge is positioned in the center of a hollow metallic shell of radius R. During four experiments the value of point charge and charge of the shell were respectively:
+5q; 0
-6q; +2q
+2q; -3q
-4q; +12q
Rank the results of experiments according to the charge on the inner surface of the shell, most positive first:
a. 2, 3, 1, 4
b. 1, 2, 3, 4
c. 2, 4, 3, 1
d. 1, 3, 4, 2
Answer:
c. 2, 4, 3, 1
Explanation:
In this question, we can say that;
q_in = q_b
Where;
q_in is the charge on the inner surface of the shell
q_b is the point charge on the shell.
Thus q_in = -q_b was written because, as the shell is conducting, it means that the electric field would have a value of zero and thus the radius inside will be zero.
Thus;
- For +5q; 0:
q_in = -(+5q)
q_in = -5q
- For -6q; +2q :
q_in = - (-6q)
q_in = +6q
- For +2q; -3q :
q_in = -(+2q)
q_in = -2q
- For -4q; +12q:
q_in = -(-4q)
q_in = +4q
Ranking the most positive to the least positive ones, we have;
+6q, +4q, -2q, -5q
This corresponds to options;
2, 4, 3, 1
Which simple machine is shown in the diagram?
a wedge
a screw
an inclined plane
a wheel and axle
Answer:
Wheel and axle
Explanation:
Which simple machine is shown in the diagram?
a wheel and axle
From the given diagram, the machine shown is actually a wheel and axle
Description of wheel and axle
The wheel and axle is a machine consisting of a wheel attached to a smaller axle so that these two parts rotate together in which a force is transferred from one to the other.
Answer:
Wheel and axle
Explanation:
A body initially at rest travels a distance 100 m in 5 s with a constant acceleration. calculate
(i) Acceleration
(ii) Final velocity at the end of 5 s.
Answer:
(i)8m/s²(ii)40m/s
Explanation:
according to the formula
½at²=s.
then substituting the data
½a•5²=100
a=8m/s²
v=at=8•5=40m/s
Answer:
(I)
[tex]{ \bf{s = ut + \frac{1}{2} a {t}^{2} }} \\ 100 = (0 \times 5) + \frac{1}{2} \times a \times {5}^{2} \\ 200 = 25a \\ { \tt{acceleration = 8 \: m {s}^{ -2} }}[/tex]
(ii)
[tex]{ \bf{v = u + at}} \\ v = 0 + (8 \times 5) \\ { \tt{final \: velocity = 40 \: m {s}^{ - 1} }}[/tex]
Differences between angle of twist and angle of shear
Answer:
idek
Explanation:
why is unit of power is called derived unit?
Distance travelled by a body in unit time is called speed. it is a scalar quantity because it can be specified only by magnitude.
A hot air balloon is a sphere of volume 2210 m3. The density of the hot air inside is 1.13 kg/m3, while the air outside has a density of 1.29 kg/m3. The balloon itself has a mass of 240 kg. What is the TOTAL NET force acting on the balloon?
[?]N
The total net force acting on the balloon will be 24498 Newtons
Given that
Volume of the balloon = 2210 cubic meter
Density of the air inside the balloon = 1.13 kg/m3
What will be the net force exerted on the balloon ?Here force on the balloon will be equal to the weight of the air displaced by balloon
[tex]F= mass of air displaced\times gravity[/tex]
[tex]F= Density \times volume \times gravity[/tex]
[tex]F=1.13 \times 2210 \times 9.81[/tex]
[tex]F=24498 N[/tex]
The total net force acting on the balloon will be 24498 Newtons
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A 200-lb man carries a 10-lb can of paint up a helical staircase that encircles a silo with radius 30 ft. If the silo is 60 ft high and the man makes exactly two complete revolutions, how much work is done by the man against gravity in climbing to the top
Answer:
17.07 kJ
Explanation:
The work done against gravity by the man W equals the potential energy change of the man and can of paint, ΔU
W = ΔU = mgΔy where m = mass of man and can of paint = 200 lb + 10 lb = 210 lb = 210 × 1 kg/2.205 lb, g = acceleration due to gravity = 9.8 m/s² and Δy = height of silo = 60 ft = 60 × 1m/3.28 ft
Since W = mgΔy, we substitute the values of the variables into the equation.
So,
W = mgΔy
W = 210 lb × 1 kg/2.205 lb × 9.8 m/s² × 60 ft × 1m/3.28 ft
W = 123480/7.2324 J
W = 17073.2 J
W = 17.0732 kJ
W ≅ 17.07 kJ
Physical quantities expresed only by their magnitude is
Answer:
Scalar quantity is the Physical quantity expresed only by their magnitude.
A gymnast falls from a height onto a trampoline. For a moment, both the gymnast’s kinetic energy and gravitational potential energy are zero. How is the gymnast’s mechanical energy stored for that moment? Question 12 options: rest energy chemical energy elastic energy thermal energy
Answer:
elastic energy
Explanation:
When a gymnast falls on a trampoline from a height, after coming in contact with the trampoline, both the gymnast and the trampoline start to move down due to the elastic property of the trampoline.
During this stretching of the trampoline there comes a maximum point up to which the trampoline is stretched. At this point, both the kinetic energy and the gravitational potential energy of the gymnast are zero due to zero speed and zero height, respectively.
The only energy stored in the gymnast's body at this point is the elastic potential energy due to stretching of the trampoline. Hence,the correct option is:
elastic energy
Which level of government relies the most on income tax?
OA.
federal
state
OC.
local
Answer:
Its the Federal government
A 1,760 W toaster, a 1,420 W electric frying pan, and an 85 W lamp are plugged into the same outlet in a 15 A, 120 V circuit. (The three devices are in parallel when plugged into the same socket.) (a) What current (in A) is drawn by each device
Answer:
Toaster = I = 14.67 A
Frying Pan = 11.83 A
Lamp = 0.71 A
Explanation:
The electric power is given as:
[tex]P = VI\\\\I = \frac{P}{V}[/tex]
where,
I = current
P = Power
V = Voltage = 120 V
FOR TOASTER:
P = 1760 W
Therefore,
[tex]I = \frac{1760\ W}{120\ V}[/tex]
I = 14.67 A
FOR FRYING PAN:
P = 1420 W
Therefore,
[tex]I = \frac{1420\ W}{120\ V}[/tex]
I = 11.83 A
FOR LAMP:
P = 85 W
Therefore,
[tex]I = \frac{85\ W}{120\ V}[/tex]
I = 0.71 A
When a rigid body rotates about a fixed axis, all the points in the body have the same Group of answer choices linear displacement. angular acceleration. centripetal acceleration. tangential speed. tangential acceleration.
Answer:
angular acceleration.
Explanation:
Newton's law of universal gravitation states that the force of attraction (gravity) acting between the Earth and all physical objects is directly proportional to the Earth's mass, directly proportional to the physical object's mass and inversely proportional to the square of the distance separating the Earth's center and that physical object.
Generally, when a rigid body is made to rotate about a fixed axis, all the points in the body would typically have the same angular acceleration, angular displacement, and angular speed.
A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140 N. Fp is parallel to the displacement of the block. The final speed of the block is 2.35 m/s.
a) How much work was converted to thermal energy? What work did friction do on the box?
b) What is the coefficient of friction?
Answer:
The answer is "151.25 J and -547.64 J".
Explanation:
[tex]u = 0\\\\v = 2.35\ \frac{m}{sec}\\\\d = 5.0 \ m\\\\[/tex]
Using formula:
[tex]v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\[/tex]
[tex]= 0.55 \ \frac{m}{sec^2}\\\\[/tex]
[tex]F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\[/tex]
Calculating the Work by net force
[tex]W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\[/tex]
The above work is converted into thermal energy.
Now,
[tex]F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\ by\ friction = -547.64 \ J[/tex]
Write one advantage of MKS system over CGS system.
g Three masses are located in the x- y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m, 0 m), and a mass of 2 kg is located at (0 m, 3 m). Where is the center of mass of the system
Answer:
Xcm = (6 * 0 + 4 & 3 + 2 * 0) / 12 = 1
Ycm = (6 * 0 + 4 * 0 + 2 * 3) / 12 = 1/2
(Xcm , Ycm) = (1 , 1/2)
Using definition of center of mass
An automobile engine has an efficiency of 22.0% and produces 2510 J of work. How much heat is rejected by the engine
Answer:
If efficiency is .22 then W = .22 * Q where Q is the heat input
Heat Input Q = 2510 / .22 = 11,400 J
Heat rejected = 11.400 - 2510 = 8900 J of heat wasted
Also, 8900 J / (4.19 J / cal) = 2120 cal
An efficiency is the measure of productivity of an engine. The heat rejected by the engine is 8900 Joules.
What is efficiency?An efficiency of a heat engine is the ratio of the work done and heat supplied.
Given is the automobile engine has the efficiency 22% and Work done is 2510 Joules.
The efficiency is written as,
η= W / Qs.
The work done is W= Qs - Qr, where Qr is the rejected heat.
The heat rejected can be represented as
Qr = W ( 1/η -1)
Substituting the value into the equation, we get the rejected heat.
Qr = 2510 (1/0.22 -1)
Qr = 8900 Joules.
Thus, the heat rejected by the engine is 8900 Joules.
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