Answer:
[tex]$\frac{d}{\lambda} = 1.54$[/tex]
Explanation:
Given :
The first dark fringe is for m = 0
[tex]$\theta_1 = \pm 19^\circ$[/tex]
Now we know for a double slit experiments , the position of the dark fringes is give by :
[tex]$d \sin \theta=\left(m+\frac{1}{2}\right) \lambda$[/tex]
The ratio of distance between the two slits, d to the light's wavelength that illuminates the slits, λ :
[tex]$d \sin \theta=\left(\frac{1}{2}\right) \lambda$[/tex] (since, m = 0)
[tex]$d \sin \theta=\frac{\lambda}{2}$[/tex]
[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin \theta}$[/tex]
[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin 19^\circ}$[/tex]
[tex]$\frac{d}{\lambda} = 1.54$[/tex]
Therefore, the ratio is [tex]$\frac{1}{1.54}$[/tex] or 1 : 1.54
A car hurtles off a cliff and crashes on the canyon floor below. Identify the system in which the net momentum is zero during the crash.
Solution :
It is given that a car ran off from a cliff and it crashes on canyon floor. Now the system of a car as well as the earth together have a [tex]\text{ net momentum of zero}[/tex] when the car crashes on the canyon floor, thus reducing the momentum of the car to zero. The earth also stops its upward motion and it also reduces the momentum to zero.
Gsjskebjwkksmndkkwksjdkdkskkskskkehdhjdj
Answer:
I DON'T UNDERSTAND
Explanation:
GUESS A MISUNDERSTANDING PLZ PUT A UNDERSTANDABLE QUESTION.
Which phase of matter makes up stars?
O liquid
O gas
O plasma
Answer:
The answer to this question is plasma
Answer:
Plasma
Explanation:
How do the magnitude and direction of the electric field on the left side of the dipole compare to the right side for the same distance
Answer:
The magnitude of the electric field is same while the direction at the left and at the right is opposite to each other.
Explanation:
The direction of the electric field due to the dipole on the axial line is same as the direction of dipole moment.
The magnitude of the electric field due to an electric dipole on its axial line is
[tex]E=\frac{2kp}{r^3}[/tex]
where, k is the constant, p is the electric dipole moment and r is the distance from the center of dipole.
The magnitude of the electric field is same while the direction at the left and at the right is opposite to each other.
An electrostatic paint sprayer has a 0.100 m diameter metal sphere at a potential of 30.0 kV that repels paint droplets onto a grounded object. (a) What charge (in C) is on the sphere?(b) What charge must a 0.100-mg drop of paint have to arrive at the object with a speed of 10.0 m/s?
Answer:
A) q = 1.67 × 10^(-7) C
B) q = 1.67 × 10^(-10) C
Explanation:
We are given;
Potential; V = 30 KV = 30000 V
Radius of sphere; r = diameter/2 = 0.1/2 = 0.05 m
A) To find the charge of the sphere, we will use the formula;
V = kq/r
Where;
q is the charge
k is electric force constant = 9 × 10^(9) N.m²/C²
Thus;
q = Vr/k
q = (30000 × 0.05)/(9 × 10^(9))
q = 1.67 × 10^(-7) C
B) Now, potential energy here is a formula; U = qV
However, for the drop of paint to move, the potential energy will be equal to the kinetic energy. Thus;
qV = ½mv²
q = mv²/2V
Where;
v is speed = 10 m/s
V = 30000 V
m = mass = 0.100 mg = 0.1 × 10^(-6) Thus;
q = (0.1 × 10^(-6) × 10²)/(2 × 30000)
q = 1.67 × 10^(-10) C
A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M and the table is 0.2, and also, M = 20 kg, m = 10 kg. How far will block m drop in the first seconds after the system is released?
How long will block M move during above time?
At the time, calculate the velocity of block M
Find out the deceleration of the block M, if the connected string is
removal by cutting after the first second. Then, calculate the time
taken to contact block M and pulley.
Answer:
a) y = 0.98 t², t=1s y= 0.98 m,
b) he two blocks must move the same distance
c) v = 1.96 m / s, d) a = -1.96 m / s², e) x = 0.98 m
Explanation:
For this exercise we can use Newton's second law
Big Block
Y axis
N-W = 0
N = M g
X axis
T- fr = Ma
the friction force has the expression
fr = μ N
fr = μ Mg
small block
w- T = m a
we write the system of equations
T - fr = M a
mg - T = m a
we add and resolved
mg- μ Mg = (M + m) a
a = [tex]g \ \frac{m - \mu M}{m+M}[/tex]
a = [tex]9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}[/tex]
a = 9.8 (6/30)
a = 1.96 m / s²
a) now we can use the kinematic relations
y = v₀ t + ½ a t²
the blocks come out of rest so their initial velocity is zero
y = ½ a t²
y = ½ 1.96 t²
y = 0.98 t²
for t = 1s y = 0.98 m
t = 2s y = 1.96 m
b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.
As the curda is in tension the two blocks must move the same distance
c) the velocity of the block M
v = vo + a t
v = 0 + 1.96 t
for t = 1 s v = 1.96 m / s
t = 2 s v = 3.92 m / s
d) the deceleration if the chain is cut
when removing the chain the tension becomes zero
-fr = M a
- μ M g = M a
a = - μ g
a = - 0.2 9.8
a = -1.96 m / s²
e) the distance to stop the block is
v² = vo² - 2 a x
0 = vo² - 2a x
x = vo² / 2a
x = 1.96² / 2 1.96
x = 0.98 m
the time to travel this distance is
v = vo - a t
t = vo / a
t = 1.96 /1.96
t = 1 s
To calculate the gravitational potential energy of a statue on a 10-meters-tall platform, you would have to know the statue's ______________
Answer:
mass
Explanation:
A car is moving at a speed of 60 mi/hr (88 ft/sec) on a straight road when the driver steps on the brake pedal and begins decelerating at a constant rate of 10ft/s2 for 3 seconds. How far did the car go during this 3 second interval?
Answer:
219 ft
Explanation:
Here we can define the value t = 0s as the moment when the car starts decelerating.
At this point, the acceleration of the car is given by the equation:
A(t) = -10 ft/s^2
Where the negative sign is because the car is decelerating.
To get the velocity equation of the car, we integrate over time, to get:
V(t) = (-10 ft/s^2)*t + V0
Where V0 is the initial velocity of the car, we know that this is 88 ft/s
Then the velocity equation is:
V(t) = (-10 ft/s^2)*t + 88ft/s
To get the position equation we need to integrate again, this time we get:
P(t) = (1/2)*(-10 ft/s^2)*t^2 + (88ft/s)*t + P0
Where P0 is the initial position of the car, we do not know this, but it does not matter for now.
We want to find the total distance that the car traveled in a 3 seconds interval.
This will be equal to the difference in the position at t = 3s and the position at t = 0s
distance = P(3s) - P(0s)
= ( (1/2)*(-10 ft/s^2)*(3s)^2+ (88ft/s)*3s + P0) - ( (1/2)*(-10 ft/s^2)*(0s)^2 + (88ft/s)*0s + P0)
= ( (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s + P0) - ( P0)
= (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s = 219ft
The car advanced a distance of 219 ft in the 3 seconds interval.
I need help with this physics question.
The acceleration will increase by 61.3%.
Explanation:
The centripetal acceleration [tex]a_c[/tex] is given by
[tex]a_c = \dfrac{v^2}{r}[/tex]
If the velocity of the object increases by 27.0%, then its new velocity v' becomes
[tex]v' = 1.270v[/tex]
The new centripetal acceleration [tex]a'_c[/tex] becomes
[tex]a'_c = \dfrac{(1.270v)^2}{r} = 1.613 \left(\dfrac{v^2}{r} \right)[/tex]
[tex]\:\:\:\:\:\:\:\:\:= 1.613a_c[/tex]
Based on the information in the table, what
is the acceleration of this object?
t(s) v(m/s)
0.0
9.0
1.0
4.0
2.0
-1.0
3.0
-6.0
A. -5.0 m/s2
B. -2.0 m/s2
C. 4.0 m/s2
D. 0.0 m/s2
Answer:
Option A. –5 m/s²
Explanation:
From the question given above, the following data were obtained:
Initial velocity (v₁) = 9 m/s
Initial time (t₁) = 0 s
Final velocity (v₂) = –6 m/s
Final time (t₂) = 3 s
Acceleration (a) =?
Next, we shall determine the change in the velocity and time. This can be obtained as follow:
For velocity:
Initial velocity (v₁) = 9 m/s
Final velocity (v₂) = –6 m/s
Change in velocity (Δv) =?
ΔV = v₂ – v₁
ΔV = –6 – 9
ΔV = –15 m/s
For time:
Initial time (t₁) = 0 s
Final time (t₂) = 3 s
Change in time (Δt) =?
Δt = t₂ – t₁
Δt = 3 – 0
Δt = 3 s
Finally, we shall determine the acceleration of the object. This can be obtained as follow:
Change in velocity (Δv) = –15 m/s
Change in time (Δt) = 3 s
Acceleration (a) =?
a = Δv / Δt
a = –15 / 3
a = –5 m/s²
Thus, the acceleration of the object is
–5 m/s².
2.
Select the correct answer.
Erica is working in the lab. She wants to remove the fine dust particles suspended in a sample of oil. Which method is she most likely to use?
Answer:
Reverse Osmosis
Explanation:
Reverse osmosis is a type of filtration that involves passing a solvent through a semipermeable membrane in the opposite direction that natural osmosis does. Separation is always enforced through the use of pressure in this process. Ions, fine dust particles, molecules, and larger particles are typically removed from solvents using this method. The technique is particularly popular in the treatment and purification of water.
Answer:
filtration is used to separate
What is the biggest planet in the solar system
Answer:
Jupiter
Explanation:
Answer:
The answer is Jupiter.
Explanation:
Jupiter is an orange/yellow colored planet.
Calculate the change in length of a 90.5 mm aluminum bar that has increased in temperature by from -14.4 oC to 154.6 oC
Take the coefficient of expansion to be 25 x 10-6 (oC)-1 . Write the answer in meters with three significant figures
Answer:
ΔL = 3.82 10⁻⁴ m
Explanation:
This is a thermal expansion exercise
ΔL = α L₀ ΔT
ΔT = T_f - T₀
where ΔL is the change in length and ΔT is the change in temperature
Let's reduce the length to SI units
L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m
let's calculate
ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))
ΔL = 3.8236 10⁻⁴ m
using the criterion of three significant figures
ΔL = 3.82 10⁻⁴ m
what is Friction
short note on friction
Answer:
Explanation:
Friction can be defined as a force that resists the relative motion of two objects when there surface comes in contact. Thus, it prevents two surface from easily sliding over or slipping across one another. Also, friction usually reduces the efficiency and mechanical advantage of machines but can be reduced through lubrication.
Generally, there are four (4) main types of friction and these includes;
I. Static friction.
II. Rolling friction.
III. Sliding friction.
IV. Fluid friction.
A 1.64 kg mass on a spring oscillates horizontal frictionless surface. The motion of the mass is described by the equation: X = 0.33cos(3.17t). In the equation, x is measured in meters and t in seconds. What is the maximum energy stored in the spring during an oscillation?
Answer:
[tex]K.E_{max}=0.8973J[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=1.64kg[/tex]
Equation of Mass
[tex]X=0.33cos(3.17t)[/tex]...1
Generally equation for distance X is
[tex]X=Acos(\omega t)[/tex]...2
Therefore comparing equation
Angular Velocity [tex]\omega=3.17rad/s[/tex]
Amplitude A=0.33
Generally the equation for Max speed is mathematically given by
[tex]V_{max}=A\omega[/tex]
[tex]V_{max}=0.33*3.17[/tex]
[tex]V_{max}=1.0461m/s[/tex]
Therefore
[tex]K.E_{max}=0.5mv^2[/tex]
[tex]K.E_{max}=0.5*1.64*(1.0461)^2[/tex]
[tex]K.E_{max}=0.8973J[/tex]
Give an example of a substance with an amorphous structure.
Answer:
Tempered glass
Explanation:
When warmed, an amorphous substance has a non-crystalline architecture that differentiates from its isochemical liquid, but this does not go through structural breakdown or the glass transition.
vector A has a magnitude of 8 unit make an angle of 45° with posetive x axis vector B also has the same magnitude of 8 unit along negative x axis find the magnitude of A+B?
Answer:
45 × 8 units = A + B as formular
If the potential (relative to infinity) due to a point charge is V at a distance R from this charge, the distance at which the potential (relative to infinity) is 2V is
A. 4R
B. 2R
C. R/2.
D. R/4
Answer:
R/2
Explanation:
The potential at a distance r is given by :
[tex]V=\dfrac{kq}{r}[/tex]
Where
k is electrostatic constant
q is the charge
The potential (relative to infinity) due to a point charge is V at a distance R from this charge. So,
[tex]\dfrac{V_1}{V_2}=\dfrac{r_2}{r_1}[/tex]
Put all the values,
[tex]\dfrac{V}{2V}=\dfrac{r_2}{R}\\\\\dfrac{1}{2}=\dfrac{r_2}{R}\\\\r_2=\dfrac{R}{2}[/tex]
So, the distance at which the potential (relative to infinity) is 2V is R/2.PLEASE HELP ME WITH THIS ONE QUESTION
What is the rest energy of a proton? (c = 2.9979 x 10^9 m/s, mp = 1.6726 x 10^-27)
Answer:
multiply mp and c^2
Explanation:
e=mc^2
Help me with my physics, please
A ball drops from a height, bounces three times, and then rolls to a stop when it reaches the ground the fourth time.
At what point is its potential energy greatest?
At what points does it have zero kinetic energy?
At what point did it have maximum kinetic energy?
Answer:
Greatest potential: moment before being dropped
Zero Kinetic: when it comes to rest
Greatest Kinetic: moment before first bounce
Explanation:
What would the separation between two identical objects, one carrying 4 C of positive charge and the other 4 C of negative charge, have to be if the electrical force on each was precisely 8 N
Answer:
7.46×10⁻⁶ m
Explanation:
Applying,
F = kqq'/r²............ Equation 1
make r the subject of the equation
r = √(F/kqq').......... Equation 2
From the question,
Given: F = 8 N, q' = q= 4 C
Constant: k = 8.98×10⁹ Nm²/C²
Substitute these values into equation 2
r = √[8/(4×4×8.98×10⁹)]
r = √(55.7×10⁻¹²)
r = 7.46×10⁻⁶ m
A solenoid 10.0 cm in diameter and 85.1 cm long is made from copper wire of diameter 0.100 cm, with very thin insulation. The wire is wound onto a cardboard tube in a single layer, with adjacent turns touching each other. What power must be delivered to the solenoid if it is to produce a field of 8.90 mT at its center
Answer:
P = 29.3 W
Explanation:
The magnetic field in a solenoid is
B = μ₀ n i
i = B /μ₀ n
where n is the density of turns
We can use a direct rule of proportions or rule of three to find the number of turns, 1 a turn has a diameter of 0.100 cm = 10⁻³ m, in the length of
L= 85.1 cm = 0.851 m how many turns there are
#_threads = 0.851 / 10⁻³
#_threads = 8.50 10³ turns
the density of turns is
n = # _threads / L
n = 8.51 103 / 0.851
n = 104 turn / m
the current that must pass through the solenoid is
i = 8.90 10-3 / 4pi 10-7 104
i = 0.70823 A
now let's find the resistance of the copper wire
R = ρ L / A
the resistivity of copper is ρ = 1.72 10⁻⁸ Ω m
wire area
A = π r²
A = π (5 10⁻⁴)
A = 7,854 10⁻⁷ m²
let's find the length of wire to build the coil, the length of a turn is
Lo = 2π r = ππ d
Lo = π 0.100
Lo = 0.314159 m / turn
With a direct proportion rule we find the length of the wire to construct the 8.5 103 turns
L = Lo #_threads
L = 0.314159 8.50 10³
L = 2.67 10³ m
resistance is
R = 1.72 10⁻⁸ 2.67 10₃ / 7.854 10⁻⁷
R = 5,847 10¹
R = 58.47 ohm
The power to be supplied to the coil is
P = VI = R i²
P = 58.47 0.70823²
P = 29.3 W
Two balls of known masses hang from the ceiling on massless strings of equal length. They barely touch when both hang at rest. One ball is pulled back until its string is at 45 ∘, then released. It swings down, collides with the second ball, and they stick together.The problem can be divided into three parts: (1) from when the first ball is released and to just before it hits the stationary ball, (2) the two balls collide, and (3) the two balls swing up together just after the collision to their highest point. ..............conserved in parts (1) and (3) as the balls swing like pendulums. During the collision in part (2) ................. conserved as the collision is ................. Explain.Match the words in the left column to the appropriate blanks in the sentences on the rightboth energy and momentum areonly energy is only momentum is.........both energy and momentum are only energy is only momentum iselasticinelastic
Answer:
In parts 1 and 3 the energy
In part 2 moment. inelastic
conserved
Explanation:
In this exercise, we are asked to describe the conservation processes for each part of the exercise.
In parts 1 and 3 the energy is conserved since the bodies do not change
In part 2 the bodies change since they are united therefore the moment is conserved and part of the kinetic energy is converted into potential energy.
Energy
moment .inelastic
conserved
The two balls swing up together just after the collision to their highest point. energy is conserved.
What is the law of conservation of momentum?According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
According to the law of conservation of momentum
Momentum before collision =Momentum after collision
When the first ball is released and just before it hits the stationary ball, The two balls collide, The two balls swing up together just after the collision to their highest point. energy is conserved.
The balls swing like pendulums. During the collision in part (2) energy is conserved as the collision is inelastic.
We are requested to describe the conservation methods for each element of the activity in this exercise.
Because the bodies do not change in sections 1 and 3, energy is conserved.
Because the bodies change in part 2 is joined, the moment is conserved and some of the kinetic energy is transformed into potential energy.
Hence the two balls swing up together just after the collision to their highest point. energy is conserved.
To learn more about the law of conservation of momentum refer to;
https://brainly.com/question/1113396
two point charges two point charges are separated by 25 cm in the figure find The Net electric field these charges produced at point a and point b
The image is missing and so i have attached it.
Answer:
A) E = 8740 N/C
B) E = -6536 N/C
Explanation:
The formula for electric field is;
E = kq/r²
Where;
q is charge
k is a constant with value 8.99 x 10^(9) N•m²/C²
A) Now, to find the net electric field at point A, the formula would now be;
E = (kq1/(r1)²) - (kq2/(r2)²)
Where;
r1 is distance from charge q1 to point A
r2 is distance from charge q2 to point A.
q1 = -6.25 nC = -6.25 × 10^(-9) C
q2 = -12.5 nC = -12 5 × 10^(-9) C
From the attached image, r1 = 25 cm - 10 cm = 15 cm = 0.15 m
r2 = 10 cm = 0.1 m
Thus;
E = (8.99 x 10^(9)) × ((-6.25 × 10^(-9))/0.15^(2)) - ((-12.5 × 10^(-9))/0.1^(2))
E = 8740 N/C
B) similarly, electric field at point B;
E = (kq1/(r1)²) + (kq2/(r2)²)
Where;
r1 is distance from charge q1 to point B
r2 is distance from charge q2 to point B.
q1 = -6.25 nC = -6.25 × 10^(-9) C
q2 = -12.5 nC = -12 5 × 10^(-9) C
From the attached image, r1 = 10 cm = 0.1 m
r2 = 25cm + 10 cm = 35 cm = 0.35 m
Thus;
E = (8.99 x 10^(9)) × ((-6.25 × 10^(-9))/0.1^(2)) + ((-12.5 × 10^(-9))/0.35^(2))
E = -6536 N/C
A cannon and a supply of cannonballs are inside a sealed railroad car of length L, as in Fig. 7-33. The cannon fires to the right; the car recoils to the left. The cannonballs remain in the car after hitting the far wall. (a) After all the cannonballs have been fired, what is the greatest distance the car can have moved from its original position
Answer:
Initially let n cannonballs with a total mass of m be to the left of the center of mass at L /2 and the mass of the car at L/2
x1 = [-m / (m + M)] * L / 2 is the original position of the CM
x2 = (m (x + L/2) + M x) / (m + M) * L/2 final position of CM with all cannon balls to the right
[-m x - m L / 2 + m x - M x] / (M + m) * L/2
= - ( m L / 2 + M x) / (m + M) * L/2 = Xcm
Check the math, but maximum distance occurs when the cannonballs of mass m move from -L/2 to L/2 and the car of mass M moves from zero to -x
A uniform metre rule of mass 10g is balanced on a knife edge placed at 45cm mark. Calculate the distance of a mass 25g from the pivot
Answer:
The distance of a mass 25g from the pivot is 18cm
Explanation:
Given
[tex]m_1 = 10g[/tex]
[tex]d_1 = 45cm[/tex]
[tex]m_2 = 25g[/tex]
Required
Distance of m2 from the pivot
To do this, we make use of:
[tex]m_1 * d_1 = m_2 * d_2[/tex] --- moments of the masses
So, we have:
[tex]10 * 45= 25* d_2[/tex]
[tex]450= 25* d_2[/tex]
Divide both sides by 25
[tex]18= d_2[/tex]
Hence:
[tex]d_2 = 18[/tex]
A wave pulse travels along a stretched string at a speed of 200 cm/s. What will be the speed if:
a. The string's tension is doubled?
b. The string's mass is quadrupled (but its length is unchanged)?
c. The string's length is quadrupled (but its mass is unchanged)?
d. The string's mass and length are both quadrupled?
Answer:
a. 282.84 cm/s b. 100 cm/s c. 400 cm/s d. 200 cm/s
Explanation:
The speed of the wave v = √(T/μ) where T = tension and μ = mass per unit length = m/l where m = mass of string and l = length of string.
So, v = √(T/μ)
v = √(T/m/l)
v = √(Tl/m)
a. The string's tension is doubled?
If the tension is doubled, T' = 2T the new speed is
v' = √(T'l/m)
v' = √(2Tl/m)
v' = √2(√Tl/m)
v' = √2v
v' = √2 × 200 cm/s
v' = 282.84 cm/s
b. The string's mass is quadrupled (but its length is unchanged)?
If the mass is quadrupled, m' = 4m the new speed is
v' = √(Tl/m')
v' = √(Tl/4m)
v' = (1/√4)(√Tl/m)
v' = v/2
v' = 200/2 cm/s
v' = 100 cm/s
c. The string's length is quadrupled (but its mass is unchanged)?
If the length is quadrupled, l' = 4l the new speed is
v' = √(Tl'/m)
v' = √(T(4l)/m)
v' = √4)(√Tl/m)
v' = 2v
v' = 200 × 2 cm/s
v' = 400 cm/s
d. The string's mass and length are both quadrupled?
If the length is quadrupled, l' = 4l and mass quadrupled, m' = 4m, the new speed is
v' = √(Tl'/m')
v' = √(T(4l)/4m)
v' = √(Tl/m)
v' = v
v' = 200 cm/s
A wheel rotates about a fixed axis with an initial angular velocity of 13 rad/s. During a 8-s interval the angular velocity increases to 57 rad/s. Assume that the angular acceleration was constant during this time interval. How many revolutions does the wheel turn through during this time interval
Answer:
The number of revolutions is 44.6.
Explanation:
We can find the revolutions of the wheel with the following equation:
[tex]\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}[/tex]
Where:
[tex]\omega_{0}[/tex]: is the initial angular velocity = 13 rad/s
t: is the time = 8 s
α: is the angular acceleration
We can find the angular acceleration with the initial and final angular velocities:
[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]
Where:
[tex] \omega_{f} [/tex]: is the final angular velocity = 57 rad/s
[tex] \alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2} [/tex]
Hence, the number of revolutions is:
[tex] \theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev [/tex]
Therefore, the number of revolutions is 44.6.
I hope it helps you!
An electron is pushed into an electric field where it acquires a 1-V electrical potential. Suppose instead that two electrons are pushed the same distance into the same electric field (but far enough apart that they don't effect eachother). What is the electrical potential of one of the electrons now?
Answer:
0.5 V
Explanation:
The electric potential distance between different locations in an electric field area is unaffected by the charge that is transferred between them. It is solely dependent on the distance. Thus, for two electrons pushed together at the same distance into the same field, the electric potential will remain at 1 V. However, the electric potential of one of the two electrons will be half the value of the electric potential for the two electrons.