Answer:
yes
Explanation:
gl m8
how do you find the weight of an object on an incline in physics?
Refers to the attachment for the answer.
Let us assume that the object of mass m is kept on the Inclined plane.
Now, there will act one force called as Component of the Weight along the Incline which is given by the Relation,
mgsinθ,
where θ is the angle which the incline makes with the surface or Angle of the Incline.
Now, If there will be no friction and the object is moving along the incline
Force = mgsinθ
⇒ ma = mgsinθ
∴ a = gsinθ
This case is valid when the angle of the Incline is greater than the angle of repose, which means the object is moving with no cause or acing of the force.
But sometimes when the object does not move without the action of force, I mean that the angle of repose is greater than the angle of the incline, then we need to apply the force so that the object can move then,
Force applied = mgsinθ
∴ a = gsinθ
It will change the cases when friction is involved.
Now, For velocity, It can be found by using the equation of Motions. Time, Distance or initial velocity, etc must be given if the question will be asked related to the velocity. So by using them, you can find that.
A 10 kg box initially at rest is pulled with a 50 N horizontal force for 4 m across a level surface. The force of friction acting on the box is a constant 20 N. How much work is done by the normal force pushing up on the box from the ground?
A. 10 J
B. 0 J
C. 100 J
D. 50 J
Answer: 0 j
Explanation:
It is given that,
Mass of the box, m = 10 kg
Force with which the box is pulled, F = 50 N
It is moved a distance of 4 m
Force of friction acting on the box, f = 20 N
We need to find the initial kinetic energy the box have. It is clear that the box is at rest initially. As there is no motion in the box at that time. The formula for the kinetic energy of the box is given by :
As v = 0
So, the initial kinetic energy of the box is 0. Hence, this is the required solution.
The modern model of the atom describes electrons in a little less specific detail than earlier models did. Why is it that being less sure about the placement of electrons in an atom is actually an improvement over earlier models?
The plum pudding model of the atom states that
Answer:
It is because one cannot know exactly the position of the electron within the atom.
One formulation of Heisenberg's Uncertainty Principle tells us that one cannot know simultaneously the position and momentum of the electron, so one cannot specify exactly either coordinate because the other would be infinite.
Bohr specified the most probable position of the electron at its lowest energy level in hydrogen and the product of the two would be about the Heisenberg value.
can someone help me with science:
The cells within a multicellular organism are of a uniform shape.
True
or False
Answer:
trur
Explanation:
gvidttcyhjddgdhjfjdjsjs
Select the correct answer.
Which theory was first proposed by Albert Einstein?
OA. the theory of special relativity
O B. the theory of universal gravitation
c. the theory of inertia
OD. the theory of radioactivity
Answer:
A. the theory of special relativity.
Explanation:
The theory of special relativity was first proposed theory by Albert Einstein.
Hope it will help :)❤
The plates of a vacuum-gap parallel plate capacitor have a 100.0 mm2 area, a vacuum gap of 5.00 mm and are connected to a 1.5-volt battery. After the capacitor is charged, the battery is disconnected from the capacitor. After the battery is disconnected, the plates are pulled apart until the vacuum gap is 7.50 mm. a. What are the initial and final energies stored in the capacitor
Answer:
E₀ = 2.0*10⁻¹¹ J = 0.2 pJ
Ef = 3.0*10⁻¹¹ J = 0.3 pJ
Explanation:
The energy stored between the plates of a parallel plate capacitor can be expressed in terms of the capacitance C and the potential difference between plates V as follows:[tex]E = \frac{1}{2} * C * V^{2} (1)[/tex]
When the capacitor is fully charged, the potential difference between plates must be equal to the voltage of the battery, 1.5 V.In a parallel plate capacitor, the value of the capacitance is independent of the applied voltage, and depends only on geometric constants and the dielectric constant of the medium between plates, as follows:[tex]C = \frac{\epsilon_{o}*A}{d} (2)[/tex]
We can find the initial value of C replacing in (2) by the givens below:A = 100.0 mm2d= 5.00 mmε₀ = 8.85*10⁻¹² F/m[tex]C_{o} = \frac{\epsilon_{o}*A}{d} = \frac{(8.85*(10)^{-12} F/m)*(10^{-4} m2)}{5.0*(10)^{-3}m} = 1.77*10^{-13} F (3)[/tex]With this value of C₀, and the value of the initial potential difference between plates (1.5 V), we can find the initial charge on the capacitor, starting from the definition of capacitance:[tex]C =\frac{Q}{V} (4)[/tex]Solving for Q in (4):[tex]Q = C_{o}* V = 1.77*10^{-13} F * 1.5 V = 2.65*10^{-13} C (5)[/tex]Finally, we can find the initial energy stored in the capacitor, replacing (3) and V in (1):[tex]E_{o} = \frac{1}{2} * C_{o} * V_{o} ^{2} = \frac{1}{2} * 1.77*10^{-13}F*(1.5V)^{2} = 0.2 pJ (6)[/tex]
If we pull apart the plates until the vacuum gap is 7.50 mm, we will change the expression of C in (2), decreasing its value due to the expanded gap.Replacing in (2) the new value of the gap (7.50 mm), we can find the new value of C, as follows:[tex]C = \frac{\epsilon_{o}*A}{d} = \frac{(8.85*10^{-12}F/m)*10^{-4} m2}{7.5*10^{-3}m} = 1.18*10^{-13} F (7)[/tex] In order to find the final energy stored in the capacitor, we need also the value of the final potential difference between plates.Once disconnected from the battery, the charge on any of the plates must remain the same, due to the principle of conservation of the charge.So, since we have the value of Q from (5) and the new value of C from (7), we can find the new potential difference between plates as follows:[tex]V_{f} = \frac{Q}{C_{f}} = \frac{2.7*10^{-13}C}{1.18*10^{-13}F} = 2.25 V (8)[/tex]With the values of Vf and Cf, we can find the value of the final energy stored in the capacitor, replacing these values in (1):[tex]E_{f} = \frac{1}{2} * C_{f} * V_{f} ^{2} = \frac{1}{2} * 1.18*10^{-13}F*(2.25V)^{2} = 0.3 pJ (9)[/tex]
An ant crawls in a straight line at a constant speed of 0.24 m/s for a distance of 3.0 m, beginning in the corner of a square classroom. It then turns exactly 90 degrees to the right, and proceeds an additional 4.0 m, reaching the far corner of the same wall from which it began. If the second leg of the journey was crawled in half the amount of time as the first, what was the ant's average speed for the whole trip?
Answer:
vavg = 0.37 m/s
Explanation:
The average speed is just the relationship between the total distance traveled, and the total time required for that travel , as follows:[tex]v_{avg} = \frac{\Delta x}{\Delta t} (1)[/tex]
We know that for the first leg of the journey, the ant crawls at a constant speed of 0.24 m/s, moving 3.0 m.We can find the time required for this part, just applying the definition of average velocity, and solving for the time t (which we will call t₁), as follows:[tex]t_{1} =\frac{x_{1}}{v_{1} } = \frac{3.0m}{0.24m/s} = 12.5 s (2)[/tex]
From the givens, we know that the time for the second part is exactly the half of the value found in (2), so we can write the total time Δt as follows:[tex]\Delta t = t_{1} + \frac{t_{1} }{2} = 12.5 s + 6.25 s = 18.75 s (3)[/tex]
We also know that in the second leg of the journey, the ant traveled 4.0 m, which adds to the 3.0 m of the first part, making a total distance of 7.0 m.Per definition of average speed, we can write the following expression as in (1) replacing Δx and Δt by their values, as follows:[tex]v_{avg} = \frac{\Delta x}{\Delta t} = \frac{7.0m}{18.75m} = 0.37 m/s (4)[/tex]
The body mass of Asaiah is 70 Kg.
(a) What is his weight on Earth?
(b) If he goes to the Moon,
(i) What is his mass?
(ii) What is his weight?
Answer:
A I hope its not wrong I hope u do good
A submarine is 309 m horizontally from the shore of a freshwater lake and 105 m beneath the surface of the water. A laser beam is sent from the submarine so that the beam strikes the surface of the water 227 m from the shore. A building stands on the shore, and the laser beam hits a target at the top of the building. The goal is to find the height of the target above sea level.
Required:
a. Draw a diagram of the situation, Identifying the two triangles that are important in finding the solution.
b. Find the angle of incidence of the beam striking the water-air interface.
c. Find the angle of refraction.
d. What angle does the refracted beam make with the horizontal?
e. Find the height of the target above sea level.
Answer:
Explanation:
a) See the attached file for diagram . The two triangles are ABC and DEC .
b )
Angle of incidence = i
Tan i = ( 309 - 227 ) / 105
= 82 / 105 = .78
i = 38°
c )
Sin i / sinr = 1 / μ where μ is refractive index of water .
Sin i / sinr = 1 / μ = 1 / 1.33
sin 38 / sinr = .7518
.6156 / sinr = .7518
sinr = .6156 / .7518
.819
r = 55°
d )
angle the refracted beam makes with the horizontal = 90 - 55 = 35°
e )
h / 227 = Tan 35
h = 227 x Tan 35
= 159 m
Height required = 159 m .
You measure the pressure of the four tires of your car each to be 35.0 pounds per square inch (psi). You then roll your car forward so that each tire is upon a sheet of paper. You outline the surface area of contact between each tire and the paper, which you later measure to be 32.0 square inches. What is the weight of your car
Answer:
Weight of car = 36034.88 lb.ft/s²
Explanation:
We are told the pressure of the four tires of your car is; P = 35.0 psi
Also, the surface area of contact is; A = 32 in²
Thus;
Weight of tires = Pressure × Area
W_tires = 35 × 32
Weight_tires = 1120lb.
To get the weight of the car, we will multiply the tire weight by acceleration due to gravity.
It's value in ft/s² is g = 32.174 ft/s²
Thus;
Weight of car = 1120 × 32.174
Weight of car = 36034.88 lb.ft/s²
How can parents help children to gain friends?
Answer:
You could try finding a familiar peer to join the activity with your child. Or ask your child who their friends are at school, or what they look for in a friend at school.
Answer:
Let the parents their Children to play outside
Explanation:
I HOPE I HELP YOU
What kind of weather would MOST LIKELY lead to hurricane formation?
Answer:
The recipe for a hurricane is a combination of warm, humid wind over tropical waters. The temperature of tropical waters must be at least 80 degrees F for up to 165 feet below the ocean’s surface. As this warm water meets the wind that blows west from Africa across the ocean, it causes the water to vaporize. The water vapor then rises into the atmosphere, where it cools and liquefies.
Explanation:
I hope this helps
A 4.0 kg circular disk slides in the x- direction on a frictionless horizontal surface with a speed of 5.0 m/s as shown in the adjacent Figure. It collides with an identical disk that is at rest before the collision. The collision is elastic. Disk 1 goes off at an 60 5.0 m/s angle of 60 with respect to the x-direction. Disk 2 g 30 goes off at an angle of 30 with respect to the x-direction. What best describes the speeds of the disks after the collision?
Solution :
Let [tex]$m_1=m_2=4$[/tex] kg
[tex]$u_1 = 5$[/tex] m/s
Let [tex]$v_1$[/tex] and [tex]$v_2$[/tex] are the speeds of the disk [tex]$m_1$[/tex] and [tex]$m_2$[/tex] after the collision.
So applying conservation of momentum in the y-direction,
[tex]$0=m_1 .v_1_y -m_2 .v_2_y $[/tex]
[tex]$v_1_y = v_2_y$[/tex]
[tex]$v_1 . \sin 60=v_2. \sin 30$[/tex]
[tex]$v_2 = v_1 \times \frac{\sin 60}{\sin 30}$[/tex]
[tex]$v_2=1.732 \times v_1$[/tex]
Therefore, the disk 2 have greater velocity and hence more kinetic energy after the collision.
Now applying conservation of momentum in the x-direction,
[tex]$m_1.u_1=m_1.v_1_x+m_2.v_2_x$[/tex]
[tex]$u_1=v_1_x+v_2_x$[/tex]
[tex]$5=v_1. \cos 60 + v_2 . \cos 30$[/tex]
[tex]$5=v_1. \cos 60 + 1.732 \times v_1 \cos 30$[/tex]
[tex]$v_1 = 2.50$[/tex] m/s
So, [tex]$v_2 = 1.732 \times 2.5$[/tex]
= 4.33 m/s
Therefore, speed of the disk 2 after collision is 4.33 m/s
Equation for Specific latent heat
How much resistance is required to limit the current from a 12 V battery to 3.6 mA?
Answer:
3333.3 ohm
Explanation:
V=IR so R =V/I (in Ampere) so convert 3.6 mA to 3.6*10^-3 A
so R =12/3.6*10^-3 =3333.3
Which of the following statements is NOT correct about sea breezes?
A) In a sea breeze, the sand is much warmer than the water, therefore creating a low pressure above the sand.
B) The breeze/wind moves from the sea to the land.
C) The breeze/wind moves from the land to the sea.
D) There is high pressure above the sea, and low pressure above the sand.
The breeze/wind moves from the land to the sea. This statement is NOT correct about sea breezes.
What is sea breeze?Local wind patterns known as sea breezes flow from the sea to land during the day. When there is no strong large-scale wind system and it is very hot or very cold during the day or at night, sea breezes and land breezes alternate along the coasts of large lakes or oceans.
Because the sea breeze's surface flow ends over land, an area of low-level air convergence is created. Locally, this convergence frequently causes air to rise, which promotes the formation of clouds. Showers over land in the afternoon may result from such clouds.
Hence, the sea breeze/wind moves from the land to the sea. This statement is NOT correct about sea breezes.
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In Bohr's model, the position of the electron (with the monochromatic light on)... Group of answer choices is restricted to orbits or certain radii, but the electron may be observed between these orbits when it moves. is restricted to orbits of certain radii around the nucleus. The position may change, but it must be in one of these orbits. is not constrained. The electron may be found anywhere outside of the nucleus.
Answer:
True. it is restricted to the orbits of certain radii around the nucleus.
The position can change, but it must be in one of these orbits
Explanation:
In this exercise some affirmations are given and you must select which ones are correct, for this we review the Bohr atomic model that has the following postulates:
* the orbits are circular
* Only certain orbits are stable, stationary state
* the radiation emitted is the difference in energy between two stable orbits
* the size of the orbit is given by the quantization of the angular momentum
L = n [tex]\hbar[/tex]
When reviewing the different statements, the correct one is:
* it is restricted to the orbits of certain radii around the nucleus.
The position can change, but it must be in one of these orbits
An ice cube in a glass of water is pushed to the bottom of the glass and held there with a straw. Consequently, the buoyant force on the ice cube is now a. the same as when the cube was floating at the top. b. exactly balanced by the weight of the ice cube. c. exactly balanced by the force exerted by the straw. d. greater than when the cube was floating at the top.
Answer:d
Explanation:
Given
Initially, the ice cube is floating over the water
When the cube is pressed to touch the bottom, it is submerged fully
Therefore more buoyant force is acting on it
At first, a part of the volume is submerged in the water, so the buoyant force is less, but as the entire cube is immersed in the water, the buoyant force increases.
A car's bumper is designed to withstand a 7.20 km/h (2.0 m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force (in N) on a bumper that collapses 0.195 m while bringing a 830 kg car to rest from an initial speed of 2.0 m/s.
Answer:
8512 N
Explanation:
From the work energy theorem we know that: The net work done on a particle equals the change in the particles kinetic energy:
[tex]W=\Delta K=K_{f}-K_{i} \\ \\qquad \begin{array}{r} W=F \cdot d, \Delta K=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2} \\ F \cdot d=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}[/tex]
Where:
-W is the work done by the force.
- F is the force actin on the.
- d is the distance travelled.
- m is the mass of the car.
- [tex]v_{f}, v_{i}[/tex] are the final and the initial velocity of the car
[tex]K_{f}, K_{i}[/tex] are the final and the kinetic energy of the car.
Givens: [tex]m=830 \mathrm{~kg}, v_{i}=2 \mathrm{~m} / \mathrm{s}, v_{f}=0 \mathrm{~km} / \mathrm{h}, d=0.195 \mathrm{~m}[/tex]
Plugging known information to get:
[tex]F \cdot d &=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2} \\ F &=\frac{\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}}{d} \\ &=\frac{0-\frac{1}{2} \times 830 \times 2^{2}}{0.195} \\ &=8.512 \times 10^{3} \\ F &=8.512 \times 10^{3} \mathrm{~N}[/tex]
Why is the answer (A)?
In details please
Explanation:
resistance of wire can be determined by this equation
R = Rho . L / A
L = the length of wire
A = the Area, A = πr²
Rho = density of resictance
we can do some rational
R1/R2 = ( Rho L1/A1) / (Rho L2/A2)
L2 is the condition when the length is doubled and
L2 = 2 x L
A2 is the condition when the diameter is doubled
A2 = 4A1
R/R2 = (L1/A1)/(L2/A2)
R/R2 = (L/A) / (2L/4A)
R/R2 = 1 / ½
R2 = ½ R.
then we use formula of Voltage
V = I R
1st condition
V = 0.3 R
2nd condition ( V is constant)
V = I2 R2
0.3 R = I2 x ½R
i2 = 0.3/ ½
i2 = 0.6 Ampere
30 POINTS PLS HELP NO LINKS PLS :/
In a wave, each particle of the medium vibrates, or oscillates, around a _____________.
A. Corner
B. Noise
C. Variable position
D. Fixed position
Answer:
C
Explanation:
trust me
D. Fixed position ,mean position , Equilibrium position.
What is wave motion?Wave motion is the transfer of energy and momentum from one point of the medium to another point of the medium without actual transport of matter between two points.
The particles of the medium vibrate about their equilibrium position in a direction parallel to the direction of propagation of the wave is called a longitudinal waves.
Characteristics of Wave Motion
In wave motion, the disturbance travels through the medium due to repeated periodic oscillations of the particles of the medium about their mean position (or) Equilibrium position.
Energy and momentum are transferred from one point to another without any actual transfer of the particles of the medium.
There is a regular phase difference between the particles of the medium because each particle receives disturbance little later than its preceding particle.
Therefore,
In a wave, each particle of the medium vibrates, or oscillates, around a fixed position.
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A 3.0-kg and a 1.0-kg box rest side-by-side on a smooth, level floor. A horizontal force of 32 N is applied to the 1.0-kg box pushing it against the 3.0-kg box, and, as a result, both boxes slide along the floor. How hard do the two boxes push against each other
Considering both boxes as one body, it would have a total mass of 4.0 kg. By Newton's second law, the 32 N force applies an acceleration a such that
∑ F = 32 N = (4.0 kg) a → a = 8.0 m/s²
and both boxes share this acceleration. (There is no friction, so the given force is the only one involved in the direction of the boxes' motion.)
Now consider just the smaller box. It is feeling the effect of the 32 N push in one direction and, as it comes into contact with the larger box, a normal force that points in the opposite direction. Let n be the magnitude of this normal force; this is what you want to find. By Newton's second law,
∑ F = 32 N - n = (1.0 kg) (8.0 m/s²)
n = 32 N - 8.0 N
n = 24 N
Just to make sure that this is consistent: by Newton's third law, the larger box feels the same force but pointing in the opposite direction. On the smaller box, n opposes the pushing force, so points backward. So from the larger box's perspective, n acts on it in the forward direction. This is the only force acting on the larger box, so Newton's second law gives
∑ F = 24 N = (3.0 kg) (8.0 m/s²)
Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. Suppose that the temperature of a cup of coffee obeys Newton's law of cooling. If the coffee has a temperature of 185 degrees Fahrenheit when freshly poured, and 3 minutes later has cooled to 172 degrees in a room at 78 degrees, determine when the coffee reaches a temperature of 147 degrees.
Answer:
6.77 minutes
Explanation:
172 degree - 78 degree = (185 degree - 78 degree)e−2 k
=> 94 = 107
e−2 k => 94 ÷ 107
k => ln (94÷107) / 2
147 - 78 = (185 - 78)e ^[ln (94÷107) / 2]
=> 69 = 107 e^ [ln (94÷107) / 2]
e^[ln (94÷107) / 2] =69 ÷ 107
=> t = [ln (69 ÷ 107)] ÷ [ln (94÷107) / 2]
t=> -0.4387 ÷ -0.0648
t => 6.77 minutes.
Therefore, the final answer to the question is 6.77 minutes.
Why is the city of Hoboken, NJ (20 minutes from Newark) and other coastal cities in the United suing ExxonMobile?
Answer:
is that youuuuuuu? fine.
A pendulum has a length of 2 m and a 30 kg mass hanging on the end. What is the period of the
pendulum?
Answer:
T = 2.83701481512 seconds
Explanation:
Hi!
The formula that you will want to use to solve this question is:
[tex]T = 2\pi *\sqrt{\frac{L}{g} }[/tex]
T--> period
L --> length of the pendulum
g --> acceleration due to gravity (9.8m/s^2)
since we know that the mass of the bob at the end of the pendulum does not affect the period of the pendulum, we can go ahead and ignore that bit of information (unless, of course, the weight causes the pendulum to stretch)
so now we can plug in our given info into the formula above and solve!
T = 2*pi * sqrt(2/9.8)
T = 2.83701481512 seconds
*Note*
- I used 3.14 to pi, if you need to use a different value for pi (a longer version, etc) your answer will be slightly different
I hope this helped!
A net force of –8750N is used to stop of 1250.kg car travelling 25m/s. What braking distance is needed to bring the car to a halt?
Answer:
d = 44.64 m
Explanation:
Given that,
Net force acting on the car, F = -8750 N
The mass of the car, m = 1250 kg
Initial speed of the car, u = 25 m/s
Final speed, v = 0 (it stops)
The formula for the net force is :
F = ma
a is acceleration of the car
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{-8750}{1250}\\\\a=-7\ m/s^2[/tex]
Let d be the breaking distance. It can be calculated using third equation of motion as :
[tex]v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-7)}\\\\d=44.64\ m[/tex]
So, the required distance covered by the car is 44.64 m.
Background information:
We know that power is the rate that work is done; what that
really means is that power tells up how much work is done every
second. To calculate power we just divide the amount of work that
was done by the time it took to do it (Power=work/time). Work is
measured in Joules and time is measured in seconds, we end up with Joules/sec when we calculate power. A Joule/sec is a Watt; we’ll use Watts to measure and compare power.
We also know that the work done on an object is the amount of energy it has gained. For this activity, you’ll be raising your body up as you do pushups which means you’ll be giving your body gravitational potential energy (GPE). To calculate the amount of GPE we multiply mass time gravity times the height raised (GPE=mgh).
What to do (record everything in the table):
1. You need to have a decent estimate of your mass in kilograms: On earth, every
kilogram weighs 2.2 pounds. Either measure your weight on a scale or just
estimate it (in pounds) then divide by 2.2 to get your mass in kilograms.
2. Measure or estimate the length of your upper arm from your elbow to your
shoulder in centimeters...this is how high you raise yourself for every pushup.
3. Decide if you’ll do regular pushups (on your toes) or simpler pushups (knees). If
you are doing regular pushups, multiply your mass by 0.68 because you’ll only be lifting about 68% of your mass each time; if you’re doing simpler pushups, multiply your mass by 0.52 because you’ll only be lifting about 52% of your mass each time.
4. Measure the time it takes for you to do 10 pushups.
Answer:
Power is the rate at which work is done. It is the work/time ratio. Mathematically, it is computed using the following equation. The standard metric unit of power is the Watt.
Explanation:
Brainliest!!! Write: Forces are all around us. Imagine that your teacher has asked you to teach a lesson to your peers about forces. Explain, in detail, how you experience forces in your everyday life. Give 3 examples
If you drag a bag across floor, you are experiencing a friction force
if you throw a paper or feather up, it floats side ways slowly. It is called air Resistance.
if you push or pull a Door, it is Normal force.
I hope this helped!
have an amazing Day!!
identify the types of motion in each activity.1.walking a long a hallway. 2.motion of the blades of the fan. 3.earths rotation 4.ball moving on the ground. 5.soldiers marching.
Answer:
1) Linear motion
2) Rotational motion
3) Rotational Motion
4) Random Motion ( The ball can be rolling in any direction)
5) Linear motion
The types of motion in each activity include the following:
Walking a long a hallway- Linear motion Motion of the blades of the fan- Rotational motionEarths rotation- Rotational motion Ball moving on the ground- Random motion Soldiers marching- Linear motion.What is Motion?
This involves an object or a body changing position over time. There are
different types of motion with different examples as can be seen above in
this scenario.
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An electron is accelerated through a potential difference of 3.5 kV and directed into a region between two parallel plates separated by 29 mm with a potential difference of 100 V between them. The electron is moving perpendicular to the electric field when it enters the region between the plates. What magnetic field is necessary perpendicular to both the electron path and the electric field so that the electron travels in a straight line
Answer: [tex]0.985\times 10^{-4}\ T[/tex]
Explanation:
Given
Electron is accelerated 3.5 kV potential difference
Distance between plates d=29 mm
The potential difference between plates is V=100 V
here, the kinetic energy of an electron is acquired through a potential difference of 3.5 kV
[tex]\Rightarrow \dfrac{1}{2}m_ev^2=e\times 3.5\times 10^3\\\\\Rightarrow v=\sqrt{\dfrac{2\times 3.5\times 10^3e}{m_e}}=\sqrt{\dfrac{7\times 10^3\times 1.6\times 10^{-19}}{9.1\times 10^{-31}}}\\\\\Rightarrow v=\sqrt{1.23\times 10^{15}}=3.5\times 10^7\ m/s[/tex]
To move in a straight line Force due to magnetic field must be balanced by force due to charge
[tex]\Rightarrow F_B=F_q\\\\\Rightarrow evB=eE\\\\\Rightarrow B=\dfrac{E}{v}\\\\\Rightarrow B=\dfrac{\frac{V}{d}}{3.5\times 10^7}=\dfrac{\frac{100}{0.029}}{3.5\times 10^7}\\\\\Rightarrow B=\dfrac{3.448\times 10^3}{3.5\times 10^7}=0.985\times 10^{-4}\ T[/tex]