Answer:
Let [tex]A[/tex], [tex]B[/tex], and [tex]k[/tex] denote three constants (with the requirement that [tex]k > 0[/tex].) The following assumes that the mass of this object is [tex]m[/tex]. Assume that [tex]x(t)[/tex] denotes the position of the object at time [tex]t[/tex].
(a) [tex]\displaystyle x^\prime(t) = -\sqrt{\frac{k}{m}}\, A\, \sin\left(t\, \sqrt{\frac{k}{m}}\right) + \sqrt{\frac{k}{m}}\, B\, \cos\left(t\, \sqrt{\frac{k}{m}}\right)[/tex].
(b) [tex]\displaystyle x^{\prime\prime}(t) = -\left(\frac{k}{m}\right)\, A\, \cos\left(t\, \sqrt{\frac{k}{m}}\right) - \left(\frac{k}{m}\right)^{2}\, B\, \sin\left(t\, \sqrt{\frac{k}{m}}\right)[/tex]
Explanation:
The differential equation for a simple harmonic motion might take the following form:
[tex]\displaystyle \frac{d^{2} x}{d t^{2}} = -\frac{k}{m}\, x[/tex].
The minus sign on the right-hand side highlights the fact that the displacement and acceleration of the object should be in opposite directions.
Notice how this equation is in the form of a homogeneous second-order ODE:
[tex]x^{\prime\prime}(t) + \underbrace{P(t)}_{0}\, x(t) + \underbrace{Q(x)}_{\sqrt{k / m}} = 0[/tex]
Let [tex]r[/tex] be a constant. One possible solution to this homogeneous second-order ODE would be in the form [tex]x(t) = e^{r\, t}[/tex], such that [tex]x^{\prime}(t) = r\, e^{r\, t}[/tex] whereas [tex]x^{\prime\prime}(t) = r^{2}\, e^{r\, t}[/tex].
Substitute into the original ODE to obtain:
[tex]\displaystyle \underbrace{r^{2}\, e^{r t}}_{x^{\prime\prime}(t)} + \frac{k}{m}\, \underbrace{e^{r t}}_{x(t)} = 0[/tex].
Rearrange the equation and solve for [tex]r[/tex].
[tex]\displaystyle e^{r t}\, \left(r^{2} + \frac{k}{m}\right) = 0[/tex]
Notice that [tex]e^{r\, t} > 0[/tex]. Hence, it must be true that [tex]\displaystyle r^{2} + \frac{k}{m} = 0[/tex]. Solve for [tex]r[/tex] given that [tex]k > 0[/tex]:
[tex]\displaystyle r_{1, 2} = \pm i\sqrt{\frac{k}{m}}[/tex], where [tex]i[/tex] is the imaginary unit.
The two particular solutions for the ODE would be:
[tex]x_1(t) = e^{\left(i\,\sqrt{k/m}\right)\, t}[/tex] and [tex]x_2(t) = e^{\left(-i\,\sqrt{k/m}\right)\, t}[/tex].
Apply Euler's Formula to rewrite both solutions in terms of trigonometric functions:
[tex]\displaystyle x_1(t) = e^{\left(i\,\sqrt{k/m}\right)\, t}= \sqrt{\frac{k}{m}} \left(\cos\left( \sqrt{\frac{k}{m}} t\right) + + i\, \sin\left( \sqrt{\frac{k}{m}} t\right)\right)[/tex].
[tex]\displaystyle x_2(t) = e^{\left(-i\,\sqrt{k/m}\right)\, t}= -\sqrt{\frac{k}{m}} \left(\cos\left( -\sqrt{\frac{k}{m}} t\right) + i\, \sin\left( -\sqrt{\frac{k}{m}} t\right)\right)[/tex].
The general solution would be in the form:
[tex]\displaystyle x(t) = C_1\, x_1(t) + C_2\, x_2(t) = A \cos\left(\sqrt{\frac{k}{m}} t\right) + B\, \left(i\, \sin\left(\sqrt{\frac{k}{m}} t\right)\right)\right)[/tex],
Where [tex]C_1[/tex] and [tex]C_2[/tex] are constants (not necessarily real numbers.)
Since position is supposed to assume a real value for any real [tex]t[/tex], set [tex]B[/tex] to a multiple of [tex]i[/tex] such that the general solution is real-valued:
[tex]\displaystyle x(t) = C_1\, x_1(t) + C_2\, x_2(t) = A \cos\left(\sqrt{\frac{k}{m}} t\right) + B\, \sin\left(\sqrt{\frac{k}{m}} t\right)\right)[/tex].
Differentiate to obtain general expressions for velocity (first derivative) and acceleration (second derivative.)
find the time taken by body whose rate and distance is 6 miles per hour and 2 miles respectively?
Answer:
12 miles per hour
Explanation:
time is equal to speed times distance
Saturn's volume is more than 10,000 times as large as Mercury's.
true
false?
Since Saturn's volume is equivalent to 763.59 earths and Mercury is only 0.055 of one earth, then Saturn's volume is over 10,000 times larger than Mercury's.
Answer: TRUE
Consider the table, graph, and equation below. Which of the three has the greater
rate of change, if any? Explain your reasoning
10 points
8
6
2
2
6
do
o
1
2
3
3
6
9
12
y=0.5x + 1
FASHIDYF2A ChamondBAN 40N6zdoldY2RENVKhulo YeOA/formResponse
Answer:
.
Explanation:
Changes in behavior of managers, coworkers, and subordinates can be a warning sign before a layoff. True or false
Answer:
The correct answer is - true.
Explanation:
Changes in the higher authorities or managers or coworkers can be a warning sign before a layoff, however, a company or organization needs to give an official notice before a particular time period before the layoff.
The layoff is a temporary suspension or permanent termination from the employment of an employee due to reasons related to business such as reducing manpower or staff and less work in the organization. These reasons can lead to a change in the behavior of the managers, coworkers, and subordinates.
The coefficients of friction between the load and the flatbed trailershown are μs = 0.40 and μk = 0.30. Knowing that the speed of the rigis 72 km/h, determine the shortest distance in which the rig can bebrought to a stop if the load is not to shift
Answer:
50.97 m
Explanation:
m = Mass of truck
[tex]\mu_s[/tex] = Coefficient of static friction = 0.4
v = Final velocity = 0
u = Initial velocity = 72 km/h = [tex]\dfrac{72}{3.6}=20\ \text{m/s}[/tex]
s = Displacement
Force applied
[tex]F=ma[/tex]
Frictional force
[tex]f=\mu_s mg[/tex]
Now these forces act opposite to each other so are equal. This is valid for the case when the load does not slide
[tex]ma=\mu_s mg\\\Rightarrow a=\mu_s g\\\Rightarrow a=0.4\times 9.81\\\Rightarrow a=3.924\ \text{m/s}^2[/tex]
Since the obect will be decelerating the acceleration will be [tex]-3.924\ \text{m/s}^2[/tex]
From the kinematic equations we have
[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-20^2}{2\times -3.924}\\\Rightarrow s=50.97\ \text{m}[/tex]
So, the minimum distance at which the car will stop without making the load shift is 50.97 m.
A 9800 N car traveling at 22 m/s strikes a concrete bridge support and comes to a complete halt in 0.5 sec. Calculate the impulse of the force.
Answer:
21977.56kg⋅m/s
Explanation:
Step one:
Given data
Weight of car= 9800N
The mass of the car is
m= w/g
m= 9800/9.81
m=998.98kg
Velocity of car= 22 m/s
time of impact= 0.5 seconds
Required
The impulse of the force F
Step two:
Ft= mv
Ft= 998.98*22
Ft=21977.56kg⋅m/s
The impulse is 21977.56kg⋅m/s
A ball dropped from a window strikes the ground 2.76 seconds later. How high is the window above the ground
Answer:
37.33m
Explanation:
Using the equation of motion
S = ut + 1/2gt^2
Time t = 2.76secs
g = 9.8m/s^2
S = 0 + 1/2(9.8)(2.76)^2
S = 4.9*7.6176
S = 37.33
Hence the window is 37.33m above the ground
A string, 0.28 m long and vibrating in its third harmonic, excites an open pipe that is 0.82 m long into its second overtone resonance. The speed of sound in air is 345 m/s. The speed of transverse waves on the string is closest to
Answer:
117.8 m/s
Explanation:
Given that:
String length, L = 0.28
pipe length, L' = 0.82
Speed of sound in air, v = 345 m/s
n = 3 (3rd harmonic)
Frequency, f of 3rd harmonic ;
f = (v*n) / 2L - - - - (1)
for the pipe: ; 3rd harmonic
f = (v*n) / 2L' - - - (2)
Equating (1) and (2)
(v*n) / 2L = (v*n) / 2L'
2L' * v * n = v* n * 2L
v = vL / L'
v = (345 * 0.28) / 0.82
v = 96.6 / 0.82
v = 117.80487
v = 117.8 m/s
In a chemical reaction atoms will do what?
Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respectively. Determine the electric field for (a) 0 < r < r1 , (b) r1 < r < r2 and (c) r > r2. (d) Under what conditions
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.
So,
a) 0 < r < r1 :
We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.
Hence, E = 0 for r < r1
b) r1 < r < r2:
Electric field =?
Let, us consider the Gaussian Surface,
E x 4 [tex]\pi[/tex] [tex]r^{2}[/tex] = [tex]\frac{Q1}{E_{0} }[/tex]
So,
Rearranging the above equation to get Electric field, we will get:
E = [tex]\frac{Q1}{E_{0} . 4 \pi. r^{2} }[/tex]
Multiply and divide by [tex]r1^{2}[/tex]
E = [tex]\frac{Q1}{E_{0} . 4 \pi. r^{2} }[/tex] x [tex]\frac{r1^{2} }{r1^{2} }[/tex]
Rearranging the above equation, we will get Electric Field for r1 < r < r2:
E= (σ1 x [tex]r1^{2}[/tex]) /([tex]E_{0}[/tex] x [tex]r^{2}[/tex])
c) r > r2 :
Electric Field = ?
E x 4 [tex]\pi[/tex] [tex]r^{2}[/tex] = [tex]\frac{Q1 + Q2}{E_{0} }[/tex]
Rearranging the above equation for E:
E = [tex]\frac{Q1+Q2}{E_{0} . 4 \pi. r^{2} }[/tex]
E = [tex]\frac{Q1}{E_{0} . 4 \pi. r^{2} }[/tex] + [tex]\frac{Q2}{E_{0} . 4 \pi. r^{2} }[/tex]
As we know from above, that:
[tex]\frac{Q1}{E_{0} . 4 \pi. r^{2} }[/tex] = (σ1 x [tex]r1^{2}[/tex]) /([tex]E_{0}[/tex] x [tex]r^{2}[/tex])
Then, Similarly,
[tex]\frac{Q2}{E_{0} . 4 \pi. r^{2} }[/tex] = (σ2 x [tex]r2^{2}[/tex]) /([tex]E_{0}[/tex] x [tex]r^{2}[/tex])
So,
E = [tex]\frac{Q1}{E_{0} . 4 \pi. r^{2} }[/tex] + [tex]\frac{Q2}{E_{0} . 4 \pi. r^{2} }[/tex]
Replacing the above equations to get E:
E = (σ1 x [tex]r1^{2}[/tex]) /([tex]E_{0}[/tex] x [tex]r^{2}[/tex]) + (σ2 x [tex]r2^{2}[/tex]) /([tex]E_{0}[/tex] x [tex]r^{2}[/tex])
Now, for
d) Under what conditions, E = 0, for r > r2?
For r > r2, E =0 if
σ1 x [tex]r1^{2}[/tex] = - σ2 x [tex]r2^{2}[/tex]
An ice-hockey player weighing 900 N is skating with a velocity of 8.0 m/s when they slam into a member of the opposing team who is standing still. They become tangled together and move down the ice with a velocity of 5.0 m/s. What is the mass of the other player
Answer:
Mass of player 2 is 55.044kg
Explanation:
Step one:
given data
Weight of player 1= 900 N
mass= weight/acceleration due to gravity
mass of player= 900/9.81
m1= 91.74kg
v1= 8m/s
mass of player 2
m2= ?
v2= 0m/s
Step two:
from the problem description, after impact, they both moved with a common velocity of V= 5m/s, hence the collision is an inelastic collision
the expression for inelastic collision is
m1v1+m2v2=(m1+m2)V
substitute
91.74*8+m2*0=(91.74+m2)5
733.92=458.7+5m2
275.22=5m2
divide both sides by 5
m2=275.22/5
m2=55.044kg
The weight of player two is
=55.044*9.81
=539.98N
Identify the part of the water cycle when ocean water may end up in the river.
Condensation.
Precipitation.
Collection.
Evaporation.
Answer:
Collection. or Precipitation.
Explanation:
b or d
The part of the water cycle when ocean water may end up in the river is to be considered as the condensation.
What is condensation?
It is the change with regard to the state of matter from the gas phase into the liquid phase. Also, it should be considered as the reverse of vaporization.
So based on this, we can say that The part of the water cycle when ocean water may end up in the river is to be considered as the condensation.
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A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 29.0 m/s. Then the truck travels for 25.0 s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.00 s.
(a) How long is the truck in motion?
(b) What is the average velocity of the truck for the motion described?
Answer:
44.5s ; 22.64 m/s
Explanation:
The motion of the truck could be separated into 3 different phases :
First :
Time of motion :
Initial Velocity, u = 0 ; final velocity, v = 29 m/s
Acceleration, a = 2 m/s²
Recall: acceleration = change in velocity / time
Time = change in velocity / acceleration
Time = (29 - 0) / 2 = 14.5 second
Distance traveled = ((29 + 0) /2) * 14.5 = 210.25 m
Second :
Time = 25 seconds at constant speed
29 m/s for 25 seconds
v*t = 29 * 25 = 725 m
Third:
5 seconds before coming to rest
((29 + 0) /2) * 5
14.5 * 5 = 72.5 m
A.)
Length of journey = (14.5+ 25 + 5) = 44.5 seconds
B.)
Average velocity = total distance / total time taken
Average velocity = (210.25 + 725 + 72.5) / 44.5
= 1007.75 / 44.5
= 22.646067
= 22.64 m/s
How much mass is required to exert a force of 25 Newtons, accelerating at 5 m/s2?
Answer:
5
Explanation:
5.
A constant force of 15 N in the negative y direction acts on a particle as it moves from the origin
to the point (3î +31 - 1k) m. How much work is done by the given force during this
displacement?
+45 J a.
-45 J b.
+30 J C.
-30 J d.
+75 J e.
The work done by the given force should be -45 joules
Calculation of the work done:
Since force acts in the negative y direction
So,
F = -15j
here the displacement vector is d = 3i+3j-1k
Now
work done is W = dot product of force and displacement
= [-15j ] . [ 3i+3j-1k]
=-45 joules
Therefore, The work done by the given force should be -45 joules
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How will you know which dye diffuses the most quickly?
Answer:
If there is a large difference between the concentration of one fluid and the concentration of the other, then the particles will diffuse faster. In the experiment, the dye particles in the food coloring are very concentrated.
Explanation:
ANSWER PLEASEEEEE!!!!!
Answer:
a ball sitting motionless on the ground
Explanation:
If a circler shape is balanced its going to have to be motionless
forces applied to the collision
Answer: I think this is the answer, In a collision, there is a force on both objects that causes an acceleration of both objects; the forces are equal in magnitude and opposite in direction. For collisions between equal-mass objects, each object experiences the same acceleration.
Explanation: I had a question similar to this, Hope this helps!
HELP URGENT WILL GIVE BRAINLIEST
Answer:
F • t= m•∆v
Explanation:
the impulse experienced by the object equals the change in momentum of the object. in equation form: F • t= m•∆v
Two charged particles, with charges q1=qq1=q and q2=4qq2=4q, are located on the x axis separated by a distance of 2.00cm2.00cm . A third charged particle, with charge q3=qq3=q, is placed on the x axis such that the magnitude of the force that charge 1 exerts on charge 3 is equal to the force that charge 2 exerts on charge 3.Find the position of charge 3 when qqq = 2.00 nCnC .
Answer: Charge 3 is located on the x-axis a distance of 0.67 cm from charge 1 and 1.33 cm from charge 2.
Explanation: Electrostatic Force is the force of repulsion or attraction between two charged particles. It's directly proportional to the charge of the particles and inversely proportional to the distance between them:
[tex]F=k\frac{|q||Q|}{r^{2}}[/tex]
k is an electrostatic constant
For the system of 3 particles, suppose distance from 1 to 3 is x meters, so, distance from 2 to 3 is (0.02-x) meters.
Force will be
[tex]F_{13}=F_{23}[/tex]
[tex]k\frac{q_{1}q_{3}}{r_{13}^{2}} =k\frac{q_{2}q_{3}}{r_{23}^{2}}[/tex]
[tex]\frac{q_{1}}{r_{13}^{2}} =\frac{q_{2}}{r^{2}_{{23}}}[/tex]
Substituting:
[tex]\frac{2.10^{-6}}{x^{2}} =\frac{8.10^{-6}}{(0.02-x)^{2}}[/tex]
[tex]8.10^{-6}x^{2}=2.10^{-6}(0.0004-0.04x+x^{2})[/tex]
[tex]4x^{2}=x^{2}-0.04x+0.0004[/tex]
[tex]3x^{2}+0.04x-0.0004=0[/tex]
Solving quadratic equation using Bhaskara:
[tex]x_{1}=\frac{-0.04+\sqrt{(0.04)^{2}+0.048} }{6}[/tex]
[tex]x_{2}=\frac{-0.04-\sqrt{(0.04)^{2}+0.048} }{6}[/tex]
x₂ will give a negative value and since distance can't be negative, use x₁
[tex]x_{1}=\frac{-0.04+\sqrt{0.0064} }{6}[/tex]
x₁ = 0.0067 m
The position of charge 3 is 0.67 cm from charge 1 and 1.33 cm from charge 2.
explain how are frequency, wavelength, energy, and penetrating ability of the waves change and differ from one another PLEASE HELP YOU WILL GET 56 POINTS IT HAS TO BE CORRECT PLEASE HELP YOU WILL GET 56 POINT IT HAS TO BE CORRECT
If we compare the energy of visible light to the energy of X-rays, we find that X-rays have a much higher frequency. Usually, electromagnetic radiation with higher frequency (energy) have a higher degree of penetration than those with low frequency.
Which is an example of convection currents?
marshmallows toasting over a campfire
a pot being heated by an electric burner
feet getting hot when stepping across sand
a radiator emitting warm air and drawing in cool air
Answer:
D
Explanation:
took the quiz!!! I picked B and got it wrong!
An example of convection currents is a radiator emitting warm air and drawing in cool air.
What is Convection?This is a method of heat transfer which is characterized by movement of a heated fluid such as air or water.
A radiator emitting warm air and drawing in cool air involves a fluid which is why option D was chosen as the most appropriate choice.
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A cube slides down the surface of a ramp at a constant velocity. What is the magnitude of the frictional force that acts on the cube due to the surface?
Answer:
[tex]f=mg*sin\theta[/tex]
The component of weight of the cube parallel to the plane
Explanation:
From the question we are told that
Cube slides at constant speed
Generally the equation for frictional force is given by
[tex]f=\mu.N[/tex]
where
[tex]f = frictional force\\\mu =coefficient of friction\\N=Normal force[/tex]
Generally the equation fo Normal force is mathematically given as
[tex]N=mg.cos\theta[/tex]
Therefore
[tex]f=\mu*mg*cos\theta[/tex]
Generally at constant velocity frictional force f will be
[tex]f=mg*sin\theta[/tex]
Therefore
The magnitude of the frictional force that acts on the cube due to the surface is the component of weight of the cube parallel to the plane
The magnitude of the frictional force that acts on the cube due to the surface is determined with [tex]\mu_ k \times mg \times cos(\theta)[/tex].
Let the mass of the cube = mLet the angle of inclination of the ramp = θThe normal force on the cube is calculated as;
[tex]F_n = W \times cos (\theta)\\\\F_n = mg \times cos (\theta)[/tex]
The frictional force acting on the cube due to the surface is calculated as follows;
[tex]F_f = \mu_k F_n\\\\F_f = \mu_ k \times mg \times cos(\theta)[/tex]
where;
[tex]\mu_k[/tex] is the coefficient of kinetic frictionThus, the magnitude of the frictional force that acts on the cube due to the surface is determined with [tex]\mu_ k \times mg \times cos(\theta)[/tex].
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If a planet has a non-circular orbit around a star, as the planet moves closer towards a star its orbital speed ….
A) remains the same.
B) lightyear.
C) decreases.
D) increases.
3. When a fo.0-g nugget of pure gold is heated from 49°C to 60.0°C, it absorbed 5339.01 of energy. What is
the specific heat of gold?
Complete Question:
When a 40.0-g nugget of pure gold is heated from 49°C to 60.0°C, it absorbed 5339.01 of energy. What is
the specific heat of gold?
Answer:
Specific heat capacity, c = 12.134J/KgC
Explanation:
Given the following data;
Mass = 40g
Initial temperature, T1 = 49°C
Final temperature, T2 = 60.0°C
Quantity of heat = 5339.01J
To find the specific heat capacity of gold;
Heat capacity is given by the formula;
[tex] Q = mcdt[/tex]
Where;
Q represents the heat capacity or quantity of heat. m represents the mass of an object. c represents the specific heat capacity of water. dt represents the change in temperature.dt = T2 - T1
dt = 60 - 49
dt = 11°C
Substituting the values into the equation, we have;
[tex] 5339.01 = 40*c*11 [/tex]
[tex] 5339.01 = 440c [/tex]
[tex] c = \frac {5339.01}{440} [/tex]
Specific heat capacity, c = 12.134J/KgC
A 1.25 m long string attached to a 60.0 Hz vibra
tor created a standing wave with 5 loops. What was the velocity of the waves in the string?
Unit= m/s
Answer:
The velocity of the waves in the string is 30 m/s.
Explanation:
Given;
frequency of the wave, F = 60.0 Hz
number of loops, n = 5
length of the string, L = 1.25 m
1 loop, L = λ/2
For 5 loops;
[tex]L = \frac{5\lambda}{2}[/tex]
where;
λ is the wavelength
[tex]L = \frac{5 \lambda}{2} \\\\5 \lambda = 2L\\\\\lambda = \frac{2L}{5} \\\\\lambda = \frac{2 \ \times \ 1.25}{5} \\\\\lambda = 0.5 \ m[/tex]
The velocity of the waves in the string is calculated as;
V = Fλ
V = (60)(0.5)
V = 30 m/s
Therefore, the velocity of the waves in the string is 30 m/s.
A 16.9 kg monkey is swinging on a 5.32 m long vine. It starts at rest, with the vine at a 43.0° angle. How fast is the monkey moving at the bottom of its swing? (Unit = m/s)
Answer:
v = 5.7554 m/s
Explanation:
First of all we need to know if the angle of the vine is measured in the horizontal or vertical.
To do this easier, let's assume the angle is measured with the horizontal. In this case, the innitial height of the monkey will be:
h₀ = h sinα
h₀ = 5.32 sin43° = 3.6282 m
As the monkey is dropping from the innitial point which is the suspension point, is also dropping from 5.32. Then the actual height of the monkey will be:
Δh = 5.32 - 3.63 = 1.69 m
In order to calculate the speed of the monkey we need to understand that the monkey has a potential energy. This energy, because of the gravity, is converted in kinetic energy, and the value will be the same. Therefore we can say that:
Ep = Ek
From here, we can calculate the speed of the monkey.
Ep = mgΔH
Ek = 1/2 mv²
The potential energy is:
Ep = 16.9 * 9.8 * 1.69 = 279.9
Now with the kinetic energy:
1/2 * (16.9) * v² = 279.9
v² = (279.9) * 2 / 16.9
v² = 33.12
v = √33.12
v = 5.7554 m/sHope this helps
A container with a mass of 5 kg is lifted to a height of 8m
Answer:
392 Nm or J
Explanation:
Work is equal to force times distance. The force required is the mass times acceleration - in this case F = 5kg x 9.81 ( gravity) = 49N. So work is 49N x 8m = 392 Nm or Joules.
Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 478 N/C. If the particles are free to move, what are their speeds (in m/s) after 52.4 ns
Answer:
the speed of electron is 4.42 x 10⁶ m/s
the speed of proton is 2406.7 m/s
Explanation:
Given;
electric field strength, E = 478 N/C
charge of the particles, Q = 1.6 x 10⁻¹⁹ C
mass of proton, Mp = 1.673 x 10⁻²⁷ kg
mass of electron Me = 9.11 x 10⁻³¹ kg
time of motion, t = 54.2 ns = 54.2 x 10⁻⁹ s
The magnitude of charge experienced by the particles is calculated as;
F = EQ
F = 478 x 1.6 x 10⁻¹⁹
F = 7.648 x 10⁻¹⁷ N
The speed of the particles is calculated as;
[tex]F = \frac{mv}{t} \\\\v = \frac{Ft}{m} \\\\v_e = \frac{(7.684 \times 10^{-17})(52.4\times 10^{-9})}{9.11\times 10^{-31}} \\\\v_e = 4.42 \ \times \ 10^6 \ m/s[/tex]
[tex]v_p = \frac{Ft}{m_p} \\\\v_p = \frac{(7.684 \times 10^{-17})(52.4\times 10^{-9})}{1.673\times 10^{-27}} \\\\v_p = 2406.7 \ m/s[/tex]
pls help
Tell me something you know about football. (i.e. how to play the game, a team, players)
some positions are quarterback, offence, defense, linebacker, and wider receiver hoped it helped :)
