Answer:
Electric dipole
Explanation:
the axis of a dipole makes an angle with the electric field. depending on the direction (clockwise/anticlockwise) we can get the torque (positive/negative).
hope this helps :D
What is meant civilized?
Answer:
at an advanced stage of social and cultural development. "a civilized society"
Explanation:
polite and well-mannered "I went to talk to them and we had a very civilized conversation" hope this helps you :)
You walk into an elevator, step onto a scale, and push the "down" button to go directly from the tenth floor to the first floor. You also recall that your normal weight is w= 635 N. If the elevator has an initial acceleration of magnitude 2.45 m/s2, what does the scale read? Express your answer in newtons.
Answer: 479. 425 N
Explanation: the calculation of a body in an elevator obeys Newton law. When it is accelerating upward, the scale reading is greater than the true weight of the person.
It is given by N= m(g+a)
When it is accelerating downward, the scale reading is less than the true weight.
It so given by N = m(g-a)
The answer to the above questions is in the attached photo
Answer:
the scale will read 476.414 N
Explanation:
Weight = 635 N
mass = (weight) ÷ (acceleration due to gravity 9.81 m/^2)
mass m = 635 ÷ 9.81 = 64.729 kg
initial acceleration of the elevator a = 2.45 m/s^2
the force produced by the acceleration of the elevator downwards = ma
your body inertia force try to counteract this force, by a force equal and opposite to the direction of this force, leading to an apparent weight loss
apparent weight = weight - ma
apparent weight = 635 - (64.729 x 2.45)
apparent weight = 635 - 158.586 = 476.414 N
in a _system supply and demand forces affect the production and consumption decisions. There is little to no _control in such a system
Answer:
in a free market system supply and demand forces affect the production and consumption decisions. There is little to no government control in such a system .
Explanation:
A free market is an economic system in which prices are based on competition between private actors and are not affected by other factors besides supply and demand, that is, where there are no external variables that condition the market.
Free market economy systems are characterized by limited government intervention, which characterizes democratic, liberal states and the modern global economy, in which the market in its private face makes most of the economic decisions, leaving the government a minimum amount of necessary regulations.
A 30 W engine generates 3600 J of energy. How long did it run for?
Answer:
so the time taken will be 120 seconds
Explanation:
power=30W
work done=3600J
time=?
as we know that
[tex]power=\frac{work done}{time taken}[/tex]
evaluating the formula
power×time taken=work done
[tex]time taken=\frac{work done}{power}[/tex]
[tex]time taken=\frac{3600J}{30W}[/tex]
[tex]Time taken=120seconds[/tex]
i hope this will help you :)
How do you use these muscles in your everyday life? What daily activities do you complete that mimic the movements of these exercises
Answer:
If ur talking abkut hamstrings then it would be running that mimics them xplanation:
This was on a gym class quizz and I got it wrong but turned out this was the right answer
Answer:
In this activity, I exercised my hips, thighs, knees, calves, ankles, and legs. In some exercises, I specifically worked on only one type of muscle or on a combination of muscles. For example, the lunges mainly exercised the muscles of the inner thighs while the dead lift worked the muscles of the leg as well as the back and shoulders. I haven't consciously exercised my leg muscles before, but I have often noticed their tightening during my daily body movements, like when I climb the stairs or run to catch the school bus.
Explanation:
Hope this helped:)
To prevent damage to floors (and to increase friction) a crutch will often have a rubber tip attached to its end. If the end of the crutch is a circle of radius 0.95 cm without the tip, and the tip is a circle of radius 2.0cm, by what factor does the tip reduce the pressure exerted by the crutch
Answer:
By a factor of about 0.23
Explanation:
Pressure is force over an area: P=F/A
Let's call the pressure without the tip P₁ and the pressure with the rubber piece P₂.
-P₁ = F/A₁= F/(πr₁²)=F/(π0.95²)
-P₂=F/A₂=F/(πr₂²)=F/(π2²)
When they ask "by what factor" it signals that we should find a ratio between the two pressures. To do this, let's divide P₁ by P₂ (I'm going to mathematical step here):
P₁/P₂=[F/(π0.95²)]x[(π2²)/F]= 2²/0.95² = 4/0.9025
So with that we can say:
P₁=(4/0.9025)P₂=4.4P₂ or
P₂=(0.9025/4)P₁=0.23P₁
What this means is that the rubber tip reduced the pressure by almost one quarter, 0.25, of what it would have been without it. Note that because we took a ratio between the two pressures that the units reduce; meaning the ratio is unitless.
By a factor of about 0.23 the tip reduces the pressure exerted by the crutch.
PressureFriction exists as the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There exist several types of friction: Dry friction is a force that disagrees with the relative lateral motion of two solid surfaces in contact.
Pressure exists as force over an area: P=F/A
Let's name the pressure without the tip P₁ and the pressure with the rubber piece P₂.
-P₁ = F/A₁= F/(πr₁²)=F/(π0.95²)
-P₂=F/A₂=F/(πr₂²)=F/(π2²)
let's divide P₁ by P₂
P₁/P₂=[F/(π0.95²)]x[(π2²)/F]= 2²/0.95² = 4/0.9025
So with that, we can say:
P₁=(4/0.9025)P₂=4.4P₂ or
P₂=(0.9025/4)P₁=0.23P₁
Hence, By a factor of about 0.23 the tip reduces the pressure exerted by the crutch,
To learn more about Pressure refer to:
https://brainly.com/question/912155
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An accelerating voltage of 2.25 103 V is applied to an electron gun, producing a beam of electrons originally traveling horizontally north in vacuum toward the center of a viewing screen 36.4 cm away. (a) What is the magnitude of the deflection on the screen caused by the Earth's gravitational field
Answer:
s= 8.28×10⁻¹⁶m
Explanation:
given
V= 2.25×10³V
from conservation of energy
mv²/2=qΔV
v=√(2qΔV/m)
v= √(2×1.6×10⁻¹⁹×2.25×10³/9.1×10⁻³¹)
=√7.9×10¹⁴m/s
=2.8×10⁷m/s
the deflection of electron beam is
S= gt²/2
recall t= d/v
s=g([tex]\frac{d}{v}[/tex])²/2
s= [tex]\frac{1}{2}[/tex]×9.8×(0.364/2.8×10⁷)²
s= 8.28×10⁻¹⁶m
A block is released from the top of a frictionless incline plane as pictured above. If the total distance travelled by the block is 1.2 m to get to the bottom, calculate how fast it is moving at the bottom using Conservation of Energy.
Complete Question
The diagram for this question is showed on the first uploaded image (reference homework solutions )
Answer:
The velocity at the bottom is [tex]v = 11.76 \ m/ s[/tex]
Explanation:
From the question we are told that
The total distance traveled is [tex]d = 1.2 \ m[/tex]
The mass of the block is [tex]m_b = 0.3 \ kg[/tex]
The height of the block from the ground is h = 0.60 m
According the law of energy
[tex]PE = KE[/tex]
Where PE is the potential energy which is mathematically represented as
[tex]PE = m * g * h[/tex]
substituting values
[tex]PE = 3 * 9.8 * 0.60[/tex]
[tex]PE = 17.64 \ J[/tex]
So
KE is the kinetic energy at the bottom which is mathematically represented as
[tex]KE = \frac{1}{2} * m v^2[/tex]
So
[tex]\frac{1}{2} * m* v ^2 = PE[/tex]
substituting values
=> [tex]\frac{1}{2} * 3 * v ^2 = 17.64[/tex]
=> [tex]v = \sqrt{ \frac{ 17.64}{ 0.5 * 3 } }[/tex]
=> [tex]v = 11.76 \ m/ s[/tex]
Two charged particles are projected into a region where a magnetic field is directed perpendicular to their velocities. If the charges are deflected in opposite directions, what are the possible relative charges and directions? (Select all that apply.)
Answer:
*If the particles are deflected in opposite directions, it implies that their charges must be opposite
*the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.
Explanation:
When a charged particle enters a magnetic field, it is subjected to a force given by
F = q v x B
where bold letters indicate vectors
this expression can be written in the form of a module
F = qv B sin θ
and the direction of the force is given by the right-hand rule.
In our case the magnetic field is perpendicular to the speed, therefore the angle is 90º and the sin 90 = 1
If the particles are deflected in opposite directions, it implies that their charges must be opposite, one positive and the other negative.
Furthermore, the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.
A sunbather stands waist deep in the ocean and observes that six crests of periodic surface waves pass each minute. The crests are 16.00 meters apart. What is the wavelength, frequency, period, and speed of the waves
Answer:
Wavelength = 16 m
Frequency = 0.1 Hz
Period = 10 s^-1
speed of the wave = 1.6 m/s
Explanation:
The crests of the wave is 16.00 m apart
Also, 6 crests pass per minute
The wavelength of this wave is the distance between consecutive corresponding troughs or crests. This means that the wavelength λ is 16 m
Frequency is defined as a number of cycles per seconds.
A minute has 60 sec, therefore, the frequency of this wave is
==> f = 6/60 = 0.1 Hz
Period is the inverse of the frequency, therefore period of the wave is
==> T = 1/0.1 = 10 s^-1
Speed of the wave is the frequency times the wavelength
v = λf = 16 x 0.1 = 1.6 m/s
3. Which is a general chemical equation for an endothermic, double-replacement reaction?
✓
O AB+ CD + energy-> AD+ CB
O AB+ CD AD + CB + energy
O AB+ C + energy - A+ CB
O AB+C- A+ CB + energy
The correct answer is A. AB+ CD + energy-> AD+ CB
Explanation:
In chemistry, a reaction is endothermic if the reaction involves absorption of heat or energy and this is necessary for the reaction to start. In terms of the chemical equation, this implies energy is part of the reactants or initial substances. Besides this, if the reaction is a double-replacement reaction this means two ions of the original substances are swapped or replaced, which means new substances in the products.
According to this, option A is the correct chemical equation because energy is part of the reactants, which shows the reaction is endothermic and the reactants AB + CD lead to the products AD + CB which shows two ions of the compounds were replaced (double replacement).
Answer:
The proper Answer is A) AB+CD + Energy --> AD + CB
Explanation:
The pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 0.600 g is supported by a needle, the tip of which is a circle 0.240 mm in radius, what pressure is exerted on the record in N/m2?
Answer:
[tex]P=3.25x10^{4}\frac{N}{m^2}[/tex]
Explanation:
Hello,
In this case, since pressure is defined as the force applied over a surface:
[tex]P=\frac{F}{A}[/tex]
We can associate the force with the weight of the needle computed by using the acceleration of the gravity:
[tex]F=0.600g*\frac{1kg}{1000g}*9.8\frac{m}{s^2} =5.88x10^{-3}N[/tex]
And the area of the the tip (circle) in meters:
[tex]A=\pi r^2=\pi (0.240mm)^2=\pi (0.240mm*\frac{1m}{1000mm} )^2\\\\A=1.81x10^{-7}m^2[/tex]
Thus, the pressure exerted on the record turns out:
[tex]P=\frac{5.88x10^{-3}N}{1.81x10^{-7}m^2} \\\\P=3.25x10^{4}\frac{N}{m^2}[/tex]
Which is truly a large value due to the tiny area on which the pressure is exerted.
Best regards.
The electric field strength is 1.70 × 104 N/C inside a parallel-plate capacitor with a 0.800 m spacing. An electron is released from rest at the negative plate. What is the electron's speed when it reaches the positive plate?
Answer:
Here, "v" is the velocity of electron and "V" is the potential.
Light from a helium-neon laser (? = 633 nm) is used to illuminate two narrow slits. The interference pattern is observed on a screen 3.2m behind the slits. Eleven bright fringes are seen, spanning a distance of 60mm .
What is the spacing (in mm) between the slits?
Answer:
0.3376 mm
Explanation:
The computation of the spacing in mm between the slits is shown below:
As we know that
[tex]d = \frac{m\lambda L}{\Delta y}[/tex]
where,
[tex]\lambda[/tex] = wavelength
L = distance from the scrren
[tex]\Delta y[/tex] = spanning distance
As there are 11 bright fingers seen so m would be
= 11 - 1
= 10
Now placing these values to the above formula
So, the spacing is
[tex]= \frac{(10)(633 \times 10^{-9})(3.2m)}{60 \times 10^{-3}}[/tex]
= 0.3376 mm
We simply applied the above formula.
Answer:
Explanation:
Maximum occurs when the path difference is an integral multiple of wavelength
Here [tex]\lambda[/tex] - Wavelength, [tex]d-[/tex] slit separation and [tex]m-[/tex] Order of pattern
Rearrange the equation for
[tex]\begin{aligned}d &=\frac{m \lambda}{\sin \theta} \\
\text { Here, } \sin \theta &=\frac{y}{L} \quad\left(\begin{array}{l}
\text { Here, } L-\text { separation between slit and screen } \\
y-\text { Distance between respective fringe from center on screen }\end{array}\right)[/tex]
[tex]d=\frac{m \lambda}{\left(\frac{y}{L}\right)} \\
&=\frac{m \lambda L}{y}[/tex]
Here, order
Due to the fact that there are 11 bright fringes seen, you take [tex]11-1=10[/tex]
since starts from 0,1,2,3
Substitute given values
[tex]\begin{aligned}d &=\frac{(10)\left(633 \times 10^{-9} \mathrm{m}\right)(3.2 \mathrm{m})}{60 \times 10^{-3} \mathrm{m}} \\&=\left(3.376 \times 10^{-4} \mathrm{m}\right)\left(\frac{1 \mathrm{mm}}{10^{-3} \mathrm{m}}\right) \\&=0.3376 \mathrm{mm}\end{aligned}[/tex]
A 25.0 kg block is initially at rest on a horizontal surface. A horizontal force of 75 N is required to set the block in motion, after which a horizontal force of 60 N is required to keep the block in moving with constant speed. Find the coefficient of static and kinetic friction between the block and the surface.
Answer:
μs = 0.30
μk = 0.24
Explanation:
In order to calculate the kinetic friction and static friction between the block and the surface, you take into account that the kinetic friction is important when the block is moving and the static friction when the block is at rest.
You use the following formula to find the coefficient of static friction:
[tex]F_1=\mu_s Mg[/tex] (1)
F1 = 75N
μs: coefficient of static friction = ?
M: mass of the block = 25.0kg
g: gravitational acceleration = 9.8m/s^2
You solve for μs in the equation (1):
[tex]\mu_s=\frac{F_1}{Mg}=\frac{75N}{(25.0kg)(9.8m/s^2)}=0.30[/tex]
For the coefficient of kinetic friction you have:
[tex]F_2=\mu_k Mg[/tex] (2)
F2 = 60N
μk: coefficient of kinetic friction = ?
You solve for μk in the equation (2):
[tex]\mu_k=\frac{F_2}{Mg}=\frac{60N}{(25.0kg)(9.8m/s^2)}=0.24[/tex]
Then, you have:
coefficient of static friction = 0.30
coefficient of kinetic friction = 0.24
When the charges in the rod are in equilibrium, what is the magnitude of the electric field within the rod?
Answer: If we have equilibrium, the magnitude must be zero.
Explanation:
If the charges are in equilibrium, this means that the total charge is equal to zero.
And as the charges must be homogeneously distributed in the rod, we can conclude that the electric field within the rod must be zero, so the magnitude of the electric field must be zero
Two carts connected by a 0.05 m spring hit a wall, compressing the spring to 0.02 m. The spring constant k is
N
100
m
What is the elastic potential energy stored from the spring's compression?
Choose 1 answer:
-3.0J
-0.045 J
0.090 J
0.045 J
Answer:
0.045 J
Explanation:
From the question,
The elastic potential energy stored in a spring is given as,
E = 1/2ke²...................... Equation 1
Where E = elastic potential energy, k = spring constant, e = compression.
Given: k = 100 N/m, e = 0.05-0.02 = 0.03 m
Substitute these values into equation 1
E = 1/2(100)(0.03²)
E = 50(9×10⁻⁴)
E = 0.045 J
Hence the right option is 0.045 J
Two carts connected by a 0.05 m spring hit a wall, compressing the spring to 0.02 m.The spring constant k is 100 N/m.
What is the elastic potential energy stored from the spring’s compression?
Answer: 0.045 J
A 1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 5000 N . Rolling friction can be neglected. You may want to review (Pages 165 - 168) . Part A What is the magnitude of the force of the car on the truck
Answer:
a) 3344 N
b) 3344 N
Explanation:
This is the complete question
1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 5000 N. Rolling friction can be neglected. A. What is the magnitude of the force of the car on the truck? Express your answer to two significant figures and include the appropriate units. B. What is the magnitude of the force of the truck on the car?
Mass of the car = 1100 kg
Mass of the truck = 2200 kg
Force exerted on the ground by the car = 5000 N
The total mass in the system = 1100 + 2200 = 3300 Kg
Total force in the system = 5000 N
Recall that the force in the system = mass x acceleration
therefore,
5000 = 3300 x a
Total acceleration in the system = 5000/3300 = 1.52 m/s^2
The force on the truck individually fro the car, will be the product of this acceleration and its mass
Force on the truck = 2200 x 1.52 = 3344 N
b) Force on the car From the truck will be equal to this force but will act in the opposite direction.
Force on the car from the truck is 3344 N
Bromine, a liquid at room temperature, has a boiling point
Yes it does ! The so-called "boiling point" is the temperature at which Bromine liquid can change state and become Bromine vapor, if enough additional thermal energy is provided. The boiling point is higher than room temperature.
Stress is a factor that contributes to heart disease risk.true or false
A 25.0 kg block is initially at rest on a horizontal surface. A horizontal force of 75.0 N is required to set the block in motion, after which a horizontal force of 60.0 N is required to keep the block moving with constant speed. Find
(a) the coefficient of static friction.
(b) the coefficient of kinetic friction between the block and the surface.
Answer:
(a) 0.31
(b) 0.245
Explanation:
(a)
F' = μ'mg.................... Equation 1
Where F' = Horizontal Force required to set the block in motion, μ' = coefficient of static friction, m = mass of the block, g = acceleration due to gravity.
make μ' the subject of the equation above
μ' = F'/mg............. Equation 2
Given: F' = 75 N, m = 25 kg
constant: g = 9.8 m/s²
Substitute these values into equation 2
μ' = 75/(25×9.8)
μ' = 75/245
μ' = 0.31.
(b) Similarly,
F = μmg.................. Equation 3
Where F = Horizontal force that is required to keep the block moving with constant speed, μ = coefficient of kinetic friction.
make μ the subject of the equation
μ = F/mg.............. Equation 4
Given: F = 60 N, m = 25 kg, g = 9.8 m/s²
Substitute these values into equation 4
μ = 60/(25×9.8)
μ = 60/245
μ = 0.245
Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by a distance of 40.0 mm, and the potential difference between them is 370 V
A. What is the magnitude of the electric field (assumed to be uniform) in the region between the plates?
B. What is the magnitude of the force this field exerts on a particle with a charge of 2.40 nC ?
C. Use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower.
D. Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.
Answer:
Explanation:
A )
electric field E = V / d where V is potential difference between plates separated by distance d .
putting the given values
E = 370 / .040 V / m
= 9250 V / m
B )
Force on charged particle of charge q in electric field E
F = q E
F = 2.4 x 10⁻⁹ x 9250
= 22200 x 10⁻⁹
= 222 x 10⁻⁷ N .
C ) since field is uniform , force will be constant
work done by electric field putting up this force
= force x displacement
= 222 x 10⁻⁷ x 40 x 10⁻³
= 888 x 10⁻⁹ J
D )
change in potential energy
= q ( V₁ - V₂ )
= 2.40 X 10⁻⁹ x 370
= 888 x 10⁻⁹ J .
(a) The magnitude of electric field in the region between the plates is 9,250 V/m.
(b) The magnitude of the force the field exerts on a particle with the given charge is 2.22 x 10⁻⁵ N.
(c) The work done by the field on the particle as it moves from the higher potential plate to the lower is [tex]8.88 \times 10^{-7} \ J[/tex].
(d) the change of the potential energy is [tex]8.88 \times 10^{-7} \ J[/tex].
The given parameters;
distance between the two metal plates, d = 40 mmpotential difference between the plates, V = 370 V(a) The magnitude of electric field in the region between the plates is calculated as;
[tex]E = \frac{V}{d} \\\\E = \frac{370 }{40 \times 10^{-3} } \\\\E = 9,250 \ V/m[/tex]
(b) The magnitude of the force the field exerts on a particle with the given charge is calculated as follows;
F = Eq
F = 9,250 x 2.4 x 10⁻⁹
F = 2.22 x 10⁻⁵ N
(c) The work done by the field on the particle as it moves from the higher potential plate to the lower is calculated as follows;
[tex]W = Fd\\\\W = 2.22 \times 10^{-5} \times 40\times 10^{-3} \\\\W =8.88 \times 10^{-7} \ J[/tex]
(d) the change of the potential energy is calculated as;
[tex]\Delta U = q \Delta V\\\\\Delta U = q(V_1 - V_2)\\\\\\Delta U = 2.4 \times 10^{-9}(370)\\\\\Delta U = 8.88 \times 10^{-7} \ J[/tex]
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Check Your UnderstandingSuppose the radius of the loop-the-loop inExample 7.9is 15 cm and thetoy car starts from rest at a height of 45 cm above the bottom. What is its speed at the top of the loop
Answer:
v = 1.7 m/s
Explanation:
By applying conservation of energy principle in this situation, we know that:
Loss in Potential Energy of Car = Gain in Kinetic Energy of Car
mgΔh = (1/2)mv²
2gΔh = v²
v = √(2gΔh)
where,
v = velocity of car at top of the loop = ?
g = 9.8 m/s²
Δh = change in height = 45 cm - Diameter of Loop
Δh = 45 cm - 30 cm = 15 cm = 0.15 m
Therefore,
v = √(2)(9.8 m/s²)(0.15 m)
v = 1.7 m/s
A block is attached to a horizontal spring and it slides back and forth in simple harmonic motion on a frictionless horizontal surface. At one extreme end of the oscillation cycle, where the block comes to a momentary halt before reversing the direction of its motion, another block is placed on top of the first block without changing its zero velocity. The simple harmonic motion then continues. What happens to the amplitude and the angular frequency of the ensuing motion of the two-block system
Answer:
A = A₀ , w = w₀/√2
Explanation:
This is a problem that we must solve with Newton's second law. They indicate that at the end of the initial movement where the speed is zero, add a mass to the block, we assume that it has the same mass, therefore the total mass is m_total = 2 m. Let's write Newton's second law at this point
[tex]F_{e}[/tex] = m_total a
the elastic force is
F_{e} = - k x
acceleration is
a = d²x / dt²
we substitute
- k x = m_total d²x / dt²
d²x / dt² + (k / m_total) x = 0
we substitute
d²x / dt² + (k /2m) x = 0
the solution to this differential equation is
x = A cos (wt + Ф)
where
w = √ (k / 2m)
to find the constant Ф we use the velocity
v = dx / dt = - Aw sin (wt + Ф)
At the most extreme point and when the new movement begins (t = 0) they indicate that v = 0
0 = - A w sin Ф
for this expression to be zero the sine must be zero therefore Ф = 0
when replacing
x = A cos (wt)
w = 1 /√2 √ (k / m)
if we want to relate to the initial movement (before placing the block)
w₀ = √ (k / m)
w = w₀ /√ 2
The amplitude of the movement is the distance from the equilibrium point to where the movement begins, in this case it is the same as in the initial movement
A = A₀
the subscript is used to refer to the oscillations before placing the second block
we substitute to have the final equation
x = A₀ cos (w₀ t /√2)
A = A₀
w = w₀/√2
a. Using the information below, calculate the cyclotron period of an electron that is launched into a magnetic field of strength 1 Gauss with a speed 200 m/s.
Electron Mass = 9.11 x 10^-31 kg
Proton Mass = 1.67 x 10^-27 kg
Elementary Charge = 1.602 x10^9 Nm/C
b. Using the same information from above, calculate the net work done on the charged particle by the magnetic field as it makes one full rotation.
Answer:
Explanation:
In cyclotron charged particle moves in a circular path in a magnetic field .
for rotation
mv² / R = Bqv where m is mass and q be charge of the particle which moves on circular path of radius R with velocity v .
v = BqR / m
Time period of rotation
T = 2πR / v
= 2πR m / BqR
= 2π m / Bq
For electron
T = 2π x 9.1 x 10⁻³¹ / (1 x 10⁻⁴ x 1.602 x 10⁻¹⁹)
= 35.67 x 10⁻⁸ s
b )
work done on the charged particle will be zero because force on charged particle is perpendicular to its movement so work done will be zero
A piece of tape is pulled from a spool and lowered toward a 100-mg scrap of paper. Only when the tape comes within 8.0 mm is the electric force magnitude great enough to overcome the gravitational force exerted by Earth on the scrap and lift it.
Requried:
Determine the magnitude and direction of the electric force exerted by the tape on the paper at this distance.
Answer:
The magnitude of the electric force is [tex]F_e = 0.00098 \ N[/tex]
Explanation:
From the question we are told that
The mass of the paper is [tex]m= 100 mg = 100 *10^{-6} \ kg[/tex]
The position is [tex]d = 8.0\ mm = 0.008 \ m[/tex]
Generally the magnitude of the electric force at the point of equilibrium between the electric force and the gravitational force is mathematically represented as
[tex]F_e = F_g = mg[/tex]
Where [tex]F_g[/tex] is gravitational force
substituting values
[tex]F_e = 100 *10^{-6} * 9.8[/tex]
[tex]F_e = 0.00098 \ N[/tex]
Now generally the gravitational force acts downward (negative y axis ) hence the reason the electric force is same magnitude but opposite in direction (upward + y - axis )
On Apollo missions to the Moon, the command module orbited at an altitude of 160 km above the lunar surface. How long did it take for the command module to complete one orbit?
Answer:
T = 2.06h
Explanation:
In order to calculate the time that the Apollo takes to complete an orbit around the moon, you use the following formula, which is one of the Kepler's law:
[tex]T=\frac{2\pi r^{3/2}}{\sqrt{GM_m}}[/tex] (1)
T: time for a complete orbit = ?
r: radius of the orbit
G: Cavendish's constant = 6.674*10^-11 m^3.kg^-1.s^-2
Mm: mass of the moon = 7.34*10^22 kg
The radius of the orbit is equal to the radius of the moon plus the distance from the surface to the Apollo:
[tex]r=R_m+160km\\\\[/tex]
Rm: radius of the moon = 1737.1 km
[tex]r=1737.1km+160km=1897.1km=1897.1*10^3 m[/tex]
Then, you replace all values of the parameters in the equation (1):
[tex]T=\frac{2\pi (1897.1*10^3m)^{3/2}}{\sqrt{(6.674*10^{-11}m^3/kgs^2)(7.34*10^22kg)}}\\\\T=7417.78s[/tex]
In hours you obtain:
[tex]T=7417.78s*\frac{1h}{3600s}=2.06h[/tex]
The time that the Apollo takes to complete an orbit around the moon is 2.06h
What is the length of a contention slot in CSMA/CD for (a) a 2-km twin-lead cable (signal propagation speed is 82% of the signal propagation speed in vacuum)
Answer:
1.99*10-4sec
Explanation:
Signal propagation speed=0.82∗2.46∗108m/s
d=2000 m
Tp=20000/0.82∗2.46∗108 sec
ContentionPeriod=2Tp=2∗20000/0.82∗2.46∗10^8
= 1.99* 10^-4seconds
A student has made the statement that the electric flux through one half of a Gaussian surface is always equal and opposite to the flux through the other half of the Gaussian surface. This is:_______.
a. never true.
b. never false.
c. true whenever enclosed charge is symmetrically located at a center point, or on a center line or centrally placed plane
d. true whenever no charge is enclosed within the Gaussian surface.
e. true only when no charge is enclosed within the Gaussian surface.
Answer:
E.true only when no charge is enclosed within the Gaussian surface.
Explanation:
Because Gauss’s law states that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge.
Passengers in a carnival ride move at constant speed in a circle of radius 5.0 m, making a complete revolution in 4.0 s. As they spin, they feel their backs pressing against the wall holding them in the ride. A. What is the direction of the passengers' acceleration? a. No direction (zero acceleration) b. Directed towards center c. Directed away from center d. Directed tangentially B. What is the passengers' linear speed in m/s? C. What is the magnitude of their acceleration in m/s^2? D. What is their angular speed in rad/s?
Answer:
A. b) Directed towards center
B. [tex]v = 7.854\ m/s[/tex]
C. [tex]a_c = 12.337\ m/s^2[/tex]
D. [tex]w = 1.57\ rad/s[/tex]
Explanation:
The "force" that they feel pressing their backs against the wall is because the reaction to the centripetal acceleration .
A.
This acceleration has its direction towards the center of the circle. (option b)
B.
Their linear speed can be calculated with the equation:
[tex]v = (\theta/t)*r[/tex]
Where [tex]\theta[/tex] is the total angular position moved in radians ([tex]1\ rev = 2\pi\ radians[/tex]), 't' is the time elapsed for the angular position moved and 'r' is the radius. So we have that:
[tex]v = (2\pi/4)*5 = 7.854\ m/s[/tex]
C.
The centripetal acceleration is given by the equation:
[tex]a_c = v^2/r[/tex]
[tex]a_c = 7.854^2/5[/tex]
[tex]a_c = 12.337\ m/s^2[/tex]
D.
Their angular speed is given by the equation:
[tex]w = \theta/t = 2\pi/4 = \pi/2 = 1.57 \ rad/s[/tex]