Given,
The radius of the loop of the roller coaster, r=22.8 m
The forces that are acting on the roller coaster when it is at the top of the loop are the centripetal force directed upwards and the weight of the roller coaster including the passengers directed downwards.
For the passengers to stay in the seat, the centripetal force must be, at the least, equal to the weight of the passengers and the rollercoaster.
That is,
[tex]\frac{Mv^2}{r}=Mg[/tex]Where M is the combined mass of the rollercoaster and the passengers, v is the minimum speed of the roller coaster when it is at the top of the loop, and g is the acceleration due to gravity.
On simplifying the above equation,
[tex]v=\sqrt[]{gr}[/tex]On substituting the known values,
[tex]\begin{gathered} v=\sqrt[]{9.8\times22.8} \\ =14.95\text{ m/s} \end{gathered}[/tex]Thus the minimum speed that the roller coaster must have when it is at the top of the loop so that the passengers stay in contact with the seats is 14.95 m/s.
A car is going at a speed of 25m/s when the driver puts her foot on the gas pedal. The carfeels a net force of 2000N for 50m. The car's mass is 1000kg.How much kinetic energy does the car have initially?
The initial kinetic energy of the car = 312.5 kJ
Explanation:The initial volume of the car, v = 25 m/s
The mass of the car, m = 1000 kg
The initial kinetic energy is given by the formula
[tex]\begin{gathered} KE=\frac{1}{2}mv^2 \\ \end{gathered}[/tex]Substitute m = 1000 kg, and v = 25 m/s into the formula
[tex]\begin{gathered} KE=\frac{1}{2}\times1000\times25^2 \\ KE=500\times625 \\ KE=312500J \\ KE=312.5kJ \end{gathered}[/tex]The initial kinetic energy of the car = 312.5 kJ
What laboratory equipment is used to carry or grip a test tube after it has been heated or cooled?
Test tube holder
Explanations:When a test tube is hot or need to be properly held to avoid the spillage of its contents, a test tube holder is the laboratory equipment that is used to grip or hold the test tube.
Therefore, the test tube holder is used for holding or gripping a test tube after it has been heated or cooled
According to Wikipedia, as of November 1, 2021, 4,864 extrasolar planets have been identified. One of the closest multiple-planet solar systems to our own is around the star Gliese 876, about 15 light-years away, and it contains four planets. One takes 63.8 Earth days to revolve, at a distance of 3.07 x 107 kilometers from Gliese 876. Another planet takes 130 Earth days to revolve. How far is this second planet from Gliese 876?
Using kepler's law:
[tex]\frac{T1^2}{r1^2}=\frac{T2^2}{r2^2}[/tex]Where:
T1 = Planet's period of the first planet
T2 = Planet's period of the second planet
r1 = Average distance to Gliese of the first planet
r2 = Average distance to Gliese of the second planet
First, we need to do a conversion:
[tex]\begin{gathered} T1=63.8days\times\frac{24h}{1day}\times\frac{60min}{1h}\times\frac{60s}{1min}=5512320s \\ T2=130days\frac{24h}{1day}\times\frac{60m\imaginaryI n}{1h}\times\frac{60s}{1m\imaginaryI n}=11232000s \end{gathered}[/tex]Now, solving for r2:
[tex]\begin{gathered} r2=\sqrt{\frac{r1^2\cdot T2^2}{T1^2}} \\ r2=\sqrt{\frac{(3.07\times10^7)^2(11232000)^2}{(5512320)^2}} \\ r2\approx62554858.93km \end{gathered}[/tex]Answer:
62554858.93 km
Draw In figure what the compass needles would show if that current is as shown in the figure. Make the arrow the “north” direction. (Please refer to picture)
The direction of the magnetic field around a current-carrying conductor is given by the right-hand thumb rule. In the figure, the direction of the flow of the current is into the plane of the paper.
Therefore the needles of the compass would align as,
In the above figure, the arrowheads indicate the north direction.
Ball A with diameter d and ball B with diameter 2d are dropped from the same height. When the two balls have the same speed, what is the ratio of the drag force on ball A to the drag force on ball B?
Ball A with diameter d and ball B with diameter 2d are dropped from the same height. When the two balls have the same speed, the ratio of the drag force on ball A to the drag force on ball B will be F1 : F2 = 1 : 4
When objects travel through fluids (a gas or a liquid), they will undoubtedly encounter resistive forces called drag forces.
The drag force always acts in the opposite direction to fluid flow. If the body’s motion exists in the fluid-like air, it is called aerodynamic drag.
formula to calculate drag force is = F(d) = 1/2 * C * rho*A * [tex]v^{2}[/tex]
C = drag coefficient
A = area of object
rho = density in which object is moving
v = velocity of object
A = area of the object
F1 ( drag force on ball A ) = 1/2 * C * rho * area of ball A * [tex]v^{2}[/tex]
F2 (drag force on ball A ) = 1/2 * C * rho * area of ball B * [tex]v^{2}[/tex]
since , both the balls have same speed and falling in same environment hence , density and speeds are the same , the only difference is in area of both the balls
F1/F2 = area of ball A / area of ball B = 4 * pi * [tex]r1^{2}[/tex] / 4 * pi * [tex]r2^{2}[/tex]
= [tex]r1^{2}[/tex] / [tex]r2^{2}[/tex]
= [tex](\frac{d}{2} )^{2}[/tex]/ [tex](\frac{2d}{2}) ^{2}[/tex]
= 1/4
F1 : F2 = 1 : 4
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wat is the mass of the car that has kinetic energy of 2400J and is moving with a speed of 20 m\s
Given,
The kinetic energy of the car, E=2400 J
The speed of the car, v=20 m/s
Kinetic energy is the energy that is possessed by an object due to its motion.
It is given by,
[tex]E=\frac{1}{2}mv^2[/tex]Where m is the mass of the car.
On substituting the known values in the above equation,
[tex]\begin{gathered} 2400=\frac{1}{2}\times m\times20^2 \\ m=\frac{2\times2400}{20^2} \\ =\frac{4800}{400} \\ =12\text{ kg} \end{gathered}[/tex]Thus the mass of the car is 12 kg
21. An object m is tied to one end of a string, moves in a circle with a constant speed v
on a horizontal frictionless table. The second end of the string is connected to a big
mass M and goes through a small hole in the table. What is the value of M if it stays
in equilibrium?
I
(B) v²/rmg
(A) mv²/rg
(C) rg/mv²
(D) mv²r/g
The value of M that goes through a small hole in the table if it stays in equilibrium is mv²/rg ( A )
The force acting on object m is the centripetal force.
[tex]F_{c}[/tex] = m v² / r
Mass = m
Velocity = v
Radius = r
The force acting on object M is the gravitational force,
[tex]F_{g}[/tex] = M g
g = Acceleration due to gravity
Since the system is at equilibrium,
[tex]F_{c}[/tex] - [tex]F_{g}[/tex] = 0
m v² / r = M g
M = m v² / r g
Therefore, the value of M if it stays in equilibrium is mv²/rg ( A )
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A locomotive enters a station with an initial velocity of 19 m/s and slows down at a rate of as it goes through. If the station is 175 m long, how fast is it going when the nose leaves the station?
ANSWER
[tex]9m\/s[/tex]EXPLANATION
Parameters given:
Initial velocity, u = 19 m/s
Acceleration, a = -0.8m/s² (the train is slowing down)
Distance traveled, s = 175 m
To find the final velocity of the train, we have to apply one of Newton's equations of motion:
[tex]v^2=u^2+2as[/tex]where v = final velocity
Substituting the given values into the equation, the final velocity of the train as its nose leaves the station is:
[tex]\begin{gathered} v^2=19^2+(2\cdot-0.8\cdot175) \\ v^2=361-280 \\ v^2=81 \\ \Rightarrow v=\sqrt[]{81} \\ v=9m\/s \end{gathered}[/tex]That is the answer.
The Student Council conducted a vote to determine whether the homcoing dance should have live muscis ora DJ. The number of students voting for live music was 215. The number of students voting of a DJ was 645.What percent of the students voted for the DJ?(Hint: Find the total first)
Divide the number of students who voted for the DJ over the total number of students to find the percent of the students that voted for the DJ.
Since 215 students voted for live music and 645 students voted for the DJ, the total number of students was:
[tex]215+645=860[/tex]Then, the percent of the students who voted for the DJ was:
[tex]\frac{645}{860}=0.75=75\text{ \%}[/tex]Therefore, the answer is: 75%
An Alaskan rescue plane traveling 37 m/s
drops a package of emergency rations from
a height of 153 m to a stranded party of
explorers.
The acceleration of gravity is 9.8 m/s².
Where does the package strike the ground
relative to the point directly below where it
was released?
Throught the given statement, the package strike the ground at the vilocity = -57.04 m/s
What is the meaning of velocity?The direction of the movement of the body or the object is defined by its velocity. Most of the time, speed is a scalar quantity. In its purest form, velocity is a vector quantity. It measures how quickly a distance changes. It is the rate at which displacement is changing.
we get,
y = 0m
y0 = 166m
v0y = 0 m/s
g = 9.8 m/s^2
t = ?
Now, solving for t,
y = y0 + v0y×t - (0.5)gt²
0 = 166 - (0.5)(9.8)t²
t = 5.82 s
Now that we have time, we can use the equation to solve for the range.
x = vx(t)
x = (40)(5.82)
x = 232.8 m
v = v0y - gt
v = 0 - (9.8)(5.82)
v = -57.04 m/s
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For a convex lens to form a virtual image the object must be located at some distance less than the focal length. Is this true or false?
For a convex lens to form a virtua
What is the efficiency of a block and tackle if you pull 20 m of rope with a force of 600 N to raise 200kg piano 5 m?
W out = mass * distance * gravity
E in = Force * distance
Efficiency = W out / E in = (200 kg * 5m*9.8N) / (600N * 20m) = 0.81
0.81 x 100 = 81 %
A 27.6 kg block is pulled along a rough level surface at constant velocity by the force . The figure shows the free-body diagram for the block. FN represents the normal force on the block; and f represents the force of kinetic friction. The coefficient of kinetic friction is 0.30, what is the strength of P?
The strength of P, the pulling force is 352 N.
What is friction?Friction is the force that acts to oppose the motion of an object moving over another object at their surfaces of contact.
The force of friction depends on the following:
mass/weight of the objectnature of the surface of contactConsidering the free-body diagram of the given object, the forces acting on the object are as follows:
FN = normal force acting in the opposite direction and equal to the weight of the object, mgmg = the weight of the object acting downwardsP = the pulling force in the forward directionf = frictional force acting in an opposite direction to PSolving for P:
P = f + mg
μ = f/N
f = μ * N
N = mg
f = μ * m * g
f = 0.3 * 27.6 * 9.81
f = 81.2 N
P = 81.2 N + 27.6 * 9.81
P = 352 N
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7. Which of the following measurement tools would you need to
determine the temperature of boiling water?
Answer:Laboratory thermometer
Explanation:
3 Fig. 2.1 is a head-on view of an airliner flying at constant speed in a circular horizontal path. The centre of the circle is to the left of the diagram. Fig. 2.1 II (a) On Fig. 2.1, draw the resultant force acting on the airliner. Explain your answer.
Answer:
Part (a)
The airliner is moving at a constant speed in a circular path, so it follows a circular motion where the net force is the centripetal force that points to the center of the circle. So, we can draw the resultant force as
Part (b)
The weight and aerodynamic lift force is represented in the following free body diagram
Then, we can calculate the net vertical force and the net horizontal force.
The net vertical force is
Fnety = (1.39 x 10⁶ N)cos(30) - (1.20 x 10⁶ N)
Fnety = 1.20 x 10⁶ N - 1.20 x 10⁶ N
Fnety = 0
The net horizontal force
Fnetx = (1.39 x 10⁶ N)sin(30)
Fnetx = 6.95 x 10⁵ N to the left.
Therefore, the net force has a component only in the horizontal direction which is equivalent to the direction shown in part (a)
Part (c)
By the second law of Newton, if there is a net force, there is acceleration. So, there is a change in the velocity. In this case, the speed is constant but the velocity is changing constantly because the airliner is changing its direction to follow the circular path.
A 1500 kg car traveling at 10 m/s suddenly runs out of gaswhile approaching the valley shown in FIGURE EX10.11. The alertdriver immediately puts the car in neutral so that it will roll.What will be the car’s speed as it coasts into the gas station onthe other side of the valley?
If the car does not brake, its mechanical energy will be conserved along all its trajectory.
Then, the sum of the kinetic energy (K) of the car and its gravitational potential energy (U) will be the same for any two points in the trajectory:
[tex]K_1+U_1=K_2+U_2[/tex]The kinetic energy of an object with mass m and speed v is:
[tex]K=\frac{1}{2}mv^2[/tex]The gravitational potential energy of an object with mass m at a height h above a reference level, is:
[tex]U=mgh[/tex]Where g is the acceleration of gravity.
In the beginning, the car has a height of 10m and a speed of 10m/s. In the end, the car reaches the gas station at a height 15m and a unknown speed v.
Then, the initial speed and height are known, as well as the final height. Use the equation for the conservation of mechanical energy to isolate v_2, the unknown final speed of the car:
[tex]\begin{gathered} K_2+U_2=K_1+U_1 \\ \Rightarrow\frac{1}{2}mv_2^2+mgh_2=\frac{1}{2}mv_1^2+mgh_1 \\ \Rightarrow\frac{1}{2}v_2^2+gh_2=\frac{1}{2}v_1^2+gh_1 \\ \Rightarrow\frac{1}{2}v_2^2=\frac{1}{2}v_1^2+gh_1-gh_2 \\ \Rightarrow\frac{1}{2}v_2^2=\frac{1}{2}v_1^2+g\mleft(h_1-h_2\mright) \\ \Rightarrow v_2^2=v_1^2+2g\mleft(h_1-h_2\mright) \\ \\ \therefore v_2=\sqrt{v_1^2+2g(h_1-h_2)} \end{gathered}[/tex]Replace v_1=10m/s, h_1=10m, h_2=15m and g=9.8m/s^2 to find the speed of the car as it reaches the gas station:
[tex]\begin{gathered} v_2=\sqrt{(10\frac{m}{s})^2+2(9.8\frac{m}{s^2})(10m-15m)} \\ =1.4142...\frac{m}{s} \\ \approx1.4\frac{m}{s} \end{gathered}[/tex]Therefore, the car's speed as it coasts into the gas station on the other side of the valley will be approximately 1.4 meters per second.
A baseball is rolling along a tabletop with avelocity of 3.9 m/s to the right. The tabletopis 1.1 m above the ground. The ball rolls offthe edge of the table and falls to theground.A.) What is the ball's final vertical Velocity?B.) How long does the ball take to fall?C.) how far from the table does the ball land?
To answer this question we need to notice that once the ball starts falling we have a projectile motion; which means that horizontally we have a rectilinear motion and vertically we have an uniformly accelerated motion.
Then we can use the following equations for each direction:
[tex]\begin{gathered} \text{ Horizontal motion:} \\ x=x_0+v_{0x}t \\ \text{ Vertical motion:} \\ a=\frac{v_f-v_0}{t} \\ y=y_0+v_0t+\frac{1}{2}at^2 \\ v_f^2-v_0^2=2a(y-y_0) \end{gathered}[/tex]Since the ball is moving down in the vertical direction we will think that down is the positive direction vertically.
a)
We know that the ball is rolling to the right when it rolls off the edge of the table, this means that vertically the initial velocity is zero; we also know that the ball will fall for 1.1 m and that the acceleration is the gravitational acceleration. Then we can use the third vertical motion equation to find the final velocity, plugging the values we know we have that:
[tex]\begin{gathered} v_f^2-0^2=2(9.8)(1.1) \\ v_f=\sqrt{2(9.8)(1.1)} \\ v_f=4.64 \end{gathered}[/tex]Therefore, the final vertical velocity is 4.64 m/s.
b)
To determine the time we can use the second vertical equation with the values we know:
[tex]\begin{gathered} 1.1=0+0t+\frac{1}{2}(9.8)t^2 \\ 4.9t^2=1.1 \\ t^2=\frac{1.1}{4.9} \\ t=\sqrt{\frac{1.1}{4.9}} \\ t=0.474 \end{gathered}[/tex]Therefore, it takes 0.474 s for the ball to fall.
c)
While the ball is falling it is also moving horizontally, in this direction we know the initial velocity is 3.9 m/s; using the horizontal equations we have:
[tex]\begin{gathered} x=0+(3.9)(0.474) \\ x=1.85 \end{gathered}[/tex]Therefore, the ball lads 1.85 m from the table.
A baseball of mass 1.23 kg is thrown at a speed of 65.8 mi/h. What is its kinetic energy?
Given:
The mass of the ball is
[tex]m=1.23\text{ kg}[/tex]The speed of the ball is
[tex]\begin{gathered} v=65.8\text{ mi/h} \\ \end{gathered}[/tex]Required: calculate the kinetic energy of the baseball
Explanation: to calculate the kinetic energy of a body we will use the formula as
[tex]K.E=\frac{1}{2}mv^2[/tex]first, we convert velocity from mi/h into m/s.
we know that
[tex]1\text{ mi=1609.34 m}[/tex]and
[tex]1\text{ h=3600 sec}[/tex]then the velocity is
[tex]\begin{gathered} v=\frac{65.8\times1602.34\text{ m}}{3600\text{ s}} \\ v=29.29\text{ m/s} \end{gathered}[/tex]now plugging all the values in the above formula, we get
[tex]\begin{gathered} K.E=\frac{1}{2}mv^2 \\ K.E=\frac{1}{2}\times1.23\text{ kg}\times(29.29\text{ m/s})^2 \\ K.E=527.61\text{ J} \end{gathered}[/tex]Thus, the kinetic energy of the baseball is
[tex]527.61\text{ J}[/tex]What is the kinetic energy of a 2.0 kg object moving at 5 m/s?
Let's put the given values into the formula below and get the result;
[tex]K.E=\frac{1}{2}mv^2[/tex]We know the numerical value of speed. We also know the mass. We can jump straight to the conclusion.
[tex]K.E=\frac{1}{2}(2kg)(5m/s)^2[/tex][tex]=\frac{1}{2}(2kg)(25m^2/s^2)[/tex][tex]=25m^2/s^2=25J[/tex]The Kinetic Energy is 25J.
If we want to accelerate an object, we must apply force on it, after applying Force some work has to be done in which energy should be transferred to the object. The energy is known as kinetic energy. The kinetic energy always depends on the mass and the velocity. It is denoted as K and the Si unit is Joules (J).
To calculate the Kinetic Energy,
K= 1/2 mv^2
m = mass
v = velocity
K = kinetic energy
In solving the above equation,
K = 1/2 x2 x 5^2
K = 5x5
K = 25
The Kinetic Energy is 25J.
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The chart shows data for four moving objects.
Object
W
X
Y
Z
Which object has the greatest acceleration?
W
Initial Velocity
(m/s)
Ο Ζ
11
10
12
20
Final Velocity
(m/s)
29
34
40
28
Change in
Time (s)
6
12
7
8
Object Y has the greatest acceleration = 4m/s2
What is an acceleration ?
acceleration: the rate at which the speed and direction of a moving object vary over time. A point or object going straight ahead is accelerated when it accelerates or decelerates. Even if the speed is constant, motion on a circle accelerates because the direction is always shifting. Both effects contribute to the acceleration for all other motions.
Acceleration = (final velocity - initial velocity ) /time
For object W
Acceleration = ( 29-11)/6 = 3m/s2
For object X
Acceleration = (34-10)/12 = 2m/s2
For object Y
Acceleration = ( 40-12)/7 = 4m/s2
For object Z
Acceleration = (28-20)/8 = 1m/s2
Object Y has the greatest acceleration = 4m/s2
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An AC generator produces an output voltage E = 200 sin96.084 t volts. What is the frequency of the voltage ?
We can write the output voltage E in the following form:
E = E0*sin(ωt), where ω is angular period.
ω = 96.084
We also know that ω = 2πf, where f is frequency.
96.084 = 2πf; f = 96.084/2π
f = 15.2922 Hz
The wave shown below is headed towards the end to the right. What will happen to the wave when it reaches the end of the string?-The wave will be absorbed by the support.-The wave will be reflected but inverted.-The wave will stop at the end of the string. -The wave will be reflected but not inverted.
Answer:
-The wave will be reflected but inverted
Explanation:
This is a transverse wave because the wave is moving to the right and the particles are moving up and down. When a transverse wave reaches the end, it is reflected and inverted so the crest becomes through and the through becomes valleys. So, the answer is
-The wave will be reflected but inverted.
Because when a wave finds a fixed end, the wave is reflected, which means that there will be a wave with the same speed and amplitude but in the opposite direction.
I need help with this question.The answer choices for each one is eitherAB C D E Any type of help will be appreciated even if it’s just a hint!
Conductor that carries electrons:
From the pic we can conclude that the conductor is the wire, in this case it would be A.
Load that transforms energy:
From the pic, it is the bulb which transform the electric energy into light and heat, so it would be B
Insulation that prevents electrons from flowing:
It is the cable sheath, from the pic it is C
Area of low potential energy for electrons:
It is the negative part of the battery, from the pic it is E
Area of high potential energy for electrons:
It is the positive part of the battery, it is D
Question 24 of 25What disadvantage of analog signals is overcome by sending digital signals?A. The waves used to transmit analog signals carry more energy.B. The waves used to transmit analog signals are more dangerous.dC. Noise decreases the quality of analog signals.O0D. Noise decreases the loudness of analog signals.SUBMIT
The correct answer is option C, "Noise decreases the quality of the analog signals."
The anlog signals q
20. A 15.9kg rock is dropped from a height of 113m. Calculate its potential energy. What is the rock'skinetic energy right before it hits the ground? What is the rock's velocity right before it hits the ground?
Kinetic energy = 1/2 * m * v^2
Potential energy = m*g*h
m = 15.9 kg
h= 113 m
Potential energy = 15.9 kg * 9.8 N * 113 m = 17,607.66 J
Kinetic energy right before it hits the ground
PE = KE
mgh = 1/2 m v^2
Masses cancel out
gh = 1/2v^2
9.8 N * 113m = 1/2 v^2
Solve for v
1,107.4 Nm = 1/2 v^2
1,107.4 Nm / (1/2) = v^2
2,214.8 Nm = v^2
√2,214 = v
47.06 m/s = V
Kinetic energy : 1/2 * m * v^2 = 1/2 * 15.9 * 47.06^2 = 17,606.41 J
Kathleen is rowing her canoe northward with an average force of 19.1 N. The wind is blowing directly south with an average force of 14.4 N. What is the net force on the canoe?
4.7 N north
33.5 N south
33.5 N north
4.7 N south
The net force on the canoe rowed northward with an average force of 19.1 N is 4.7 N North if the wind is blowing directly south with an average force of 14.4 N.
Net force = Force due to rowing - Force due to wind
Net force = 19.1 - 14.4
Net force = 4.7 N
The direction is towards North because the force due to rowing is greater than the force due to wind.
If two force act in opposite directions, the resultant force is the difference of these two forces and the resultant direction is the direction in which the greater force is applied.
Therefore, the net force on the canoe is 4.7 N North
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17. A material that makes energy transfer difficult is one that is a good...Select one:a. radiator.b. insulator.c. conductor.d. convector.
b. insulator.
The others describe a type of transfer of energy
Convert the following number from scientific notation to standard notation1.9300 x 10^-30.000 193 000.001 93000.001931930
The number converted to the standard notation as
[tex]1.93\times10^{-3}=0.00193[/tex]Hence, the correct answer is 0.00193
If the electric intensity between two parallel plates, placed 1cm apart is 104 NC-1and the direction of the field of this intensity is vertically upward. Find the force on an electron in this field and compare the force on the electron with the electron’s weight
The force of a charge in an electric field is:
[tex]\vec{F}_e=q\vec{E}[/tex]In this case we know the electric field is:
[tex]\vec{E}=104\hat{j}[/tex]and that the charge is that of the electron, then we have:
[tex]\begin{gathered} \vec{F}_e=-1.6\times10^{-19}(104\hat{j}) \\ \vec{F}_e=-1.664\times10^{-17}\text{ N} \end{gathered}[/tex]Therefore, the magnitude of the force is
[tex]1.664\times10^{-17}\text{ N}[/tex]and in points down.
The weight of the electron is:
[tex]\begin{gathered} W=1.67\times10^{-27}(9.98) \\ W=1.6366\times10^{-26} \end{gathered}[/tex]Making the quotient between the force we have:
[tex]\frac{1.664\times10^{-17}}{1.6366\times10^{-26}}=1.02\times10^9[/tex]Therefore, the electric force is approximately 1e9 times the weight.
Two equal charges q1=q2= -6uC are on the y-axis at y1=3cm and y2= -3cm. What is the magnitude and direction of the electric field on the x-axis at x=4cm. If a test charge q0=2uC is placed at x =4cm find the force the test charge experiences?
The electric field charges, q₁, and q₂, which are each -6×10⁻⁶ μC, gives:
First part:
The magnitude of the electric field at x = 4 cm is -3.456×10⁷ N/CThe direction of the electric field is towards the origin, along the x-axisSecond part:
The force experienced by the charge is 69.12 NWhat is an electric field?An electric field is the field around a particle that is electrically charged and which exerts a force on charged particles within the field.
The given information are:
The electric charges, q₁ = q₂ = -6 μC
The location of the charge q₁ = y₁ = 3 cm on the y-axis
Location of the charge q₂ = y₂ = -3 cm
First part:
The required location of the point where the electric field magnitude and direction is required is x = 4 cm
The electric field formula is: [tex]\displaystyle{E = \frac{k\cdot q}{r^2}[/tex]
Where:
k = The electrostatic constant ≈ 9 × 10⁹ N·m²/C²
The distances, r, of the charges from the required point are therefore obtained using Pythagorean theorem as follows:
r = √(3² + 4²) = 5
r = 5 cm = 0.05 m
Which gives;
[tex]\displaystyle{E = \frac{9 \times 10^9\times (-6) \times 10^{-6}}{(0.05)^2} = -2.16\times 10^{7}[/tex]
Given that the magnitude of the electric field along the y-axis cancel out, the magnitude of the electric field along the x-axis is found as follows:
[tex]E_x = 2 \times -2.16\times 10^{7}\times \dfrac{4}{5} = -3.456 \times 10^7[/tex]
The magnitude of the electric field at x = 4 is -3.456 × 10⁷ N/C
Second part: The magnitude of the test charge is q₀ = 2 × 10⁻⁶ μC
The force of an electric field, F = E × q
The force experienced by the test charge is therefore:
F = -3.456 × 10⁷ × 2 × 10⁻⁶ = -69.12
The force the test charge experiences is 69.12 N acting towards the origin from the point x = 4 cm.
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