The maximum potential difference between points A and B is 100 V, and the maximum energy stored in the three-capacitor arrangement is determined by the capacitor with the smallest capacitance.
The maximum potential difference (voltage) between points A and B in a series combination of capacitors is limited by the capacitor with the lowest breakdown voltage. Given C2 = 20.0 μF and C3 = 25.0 μF, and no capacitor can exceed 100 V, the maximum potential difference that can exist between points A and B is 100 V.
The maximum energy that can be stored in the three-capacitor arrangement is determined by the capacitor with the least energy storage capability. In this case, the capacitors are in series, so the total energy stored in the arrangement is the energy stored by the capacitor with the smallest capacitance, which is 20 μF, at the maximum allowed voltage of 100 V.
You are driving home from school steadily at 95 km/h for 180km. It then begins to rain and you slow to 65 km/h. You arrive home after driving 4.5 h. (a) How far is your hometown from school? (b) What was your average speed?
Answer:
the total trajectory length is 349.39 km
Explanation:
for the first trajectory
Time taken in first trajectory = First trajectory length /velocity = 180 km/95 km/h = 1.894 hours
therefore since the total time is 4.5 hours
Time taken in second trajectory = Second trajectory length /velocity
4.5 hours- 1.894 hours = Second trajectory length / 65 km/h
Second trajectory length = 169.39 km
therefore the total trajectory length is 180 km + 169.39 km = 349.39 km
Explanation:
Below is an attachment containing the solution.
A large truck collides head-on with a small car. The car is severely damaged as a result of the collision. According to Newton's third law, how do the forces acting between the truck and car compare during the collision
Answer: Force on the truck is equal to force on the car.
Explanation: According to the Newton's third law of motion which states that; For every action, there is an equal and opposite reaction. These pair of forces are regarded as action - reaction forces. These size or magnitude of the forces on the colliding objects are equal or the same, while the direction of the colliding objects are opposite.
In the scenario above, both truck and carry have the same mass, however, the damage suffered by the car is based on its smaller mass which makes it unable to withstand the acceleration resulting from the collision.
ma = m(-a)
m= mass, a= acceleration
A car of mass m = 1030 kg is traveling down a θ = 13-degree incline. When the car's speed is v0 = 14 m/s, a mechanical failure causes all four of its brakes to lock. The coefficient of kinetic friction between the tires and road is μk = 0.45.
a. Write an expression for the magnitude of the force of kinetic friction
b. Write an expression for the magnitude of the change in the car's height, h, along the y-direction, assuming it travels a distance L down the incline.
c. Calculate the distance the car travels down the hill 1 in meters until it comes to a stop at the end
Answer:
a. [tex]F_f = \mu mg cos\theta[/tex]
b. h = Lsinθ
c. 22.78 m
Explanation:
a. The kinetic friction is the product of kinetic coefficient and normal force N, which is the gravity force in the direction normal to the incline
[tex]F_f = \mu N = \mu mg cos\theta[/tex]
b. As the car travels a distance L down the incline of θ degrees, vertically speaking it would have traveled a distance of:
h = Lsinθ
As we can treat L and h in a right triangle where L is the hypotenuse and h is a side length in opposite of incline angle θ
c. Let g = 9.81 m/s2. the acceleration caused by kinetic friction according to Newton's 2nd law is
[tex]a = F_f/m = \mu g cos\theta = 0.45*9.81*cos13^o = 4.3 m/s^2[/tex]
We can use the following equation of motion to find out the distance traveled by the car:
[tex]v^2 - v_0^2 = 2a\Delta s[/tex]
where v = 0 m/s is the final velocity of the car when it stops, [tex]v_0[/tex] = 14m/s is the initial velocity of the car when it starts braking, a = -4.3 m/s2 is the deceleration of the car, and [tex]\Delta s[/tex] is the distance traveled, which we care looking for:
[tex]0^2 - 14^2 = 2(-4.3)\Delta s[/tex]
[tex]\Delta s = 14^2 / (2*4.3) = 22.78 m[/tex]
a. An expression for the magnitude of the force of kinetic friction is [tex]\(f_{\text{friction}} = \mu_k \cdot N\).[/tex]
b. An expression for the magnitude of the change in the car's height, h, along the y-direction, assuming it travels a distance L down the incline is: [tex]\(h = L \cdot \sin(\theta)\).[/tex]
c. The car will travel approximately 94.69 meters down the hill before coming to a stop due to the locked brakes.
The detailed explanation is as follows:
a. The magnitude of the force of kinetic friction can be calculated using the formula:
[tex]\(f_{\text{friction}} = \mu_k \cdot N\),[/tex]
Where:
[tex]\(f_{\text{friction}}\)[/tex] is the force of kinetic friction,
[tex]\(\mu_k\)[/tex] is the coefficient of kinetic friction (given as 0.45),
[tex]\(N\)[/tex] is the normal force.
The normal force can be calculated using the equation:
[tex]\(N = m \cdot g \cdot \cos(\theta)\),[/tex]
Where:
[tex]\(m\)[/tex] is the mass of the car (1030 kg),
[tex]\(g\)[/tex] is the acceleration due to gravity (approximately 9.81 m/s²),
[tex]\(\theta\)[/tex] is the angle of the incline (13 degrees converted to radians).
b. The magnitude of the change in the car's height [tex](\(h\))[/tex] along the y-direction can be found using trigonometry. When the car travels a distance [tex]\(L\)[/tex] down the incline, the vertical displacement [tex](\(h\))[/tex] can be calculated as:
[tex]\(h = L \cdot \sin(\theta)\).[/tex]
c. To calculate the distance the car travels down the hill until it comes to a stop, you can use the work-energy theorem. The work done by the force of kinetic friction will be equal to the initial kinetic energy of the car. The work-energy theorem is given as:
[tex]\(W = \Delta KE\),[/tex]
Where:
[tex]\(W\)[/tex] is the work done by friction (negative, as it opposes motion),
[tex]\(\Delta KE\)[/tex] is the change in kinetic energy.
The initial kinetic energy is:
[tex]\(KE_0 = \frac{1}{2} m v_0^2\).[/tex]
The final kinetic energy is zero because the car comes to a stop.
So, the work done by friction is:
[tex]\(W = -\frac{1}{2} m v_0^2\).[/tex]
Now, you can use the work-energy theorem to find the distance \(L\) down the incline:
[tex]\(W = -\frac{1}{2} m v_0^2 = \Delta KE = KE_f - KE_0\),[/tex]
Where [tex]\(KE_f = 0\)[/tex] (final kinetic energy).
Solve for [tex]\(L\):[/tex]
[tex]\(-\frac{1}{2} m v_0^2 = -\mu_k m g L \cos(\theta)\).[/tex]
Now, solve for [tex]\(L\):[/tex]
[tex]\[L = \frac{v_0^2}{2 \mu_k g \cos(\theta)}.\][/tex]
Substitute the known values:
[tex]\[L = \frac{(14 m/s)^2}{2 \cdot 0.45 \cdot 9.81 m/s^2 \cdot \cos(13^\circ)} \approx 94.69 \, \text{meters}.\][/tex]
So, the car will travel approximately 94.69 meters down the hill before coming to a stop due to the locked brakes.
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Calculate the work done (in J) by a 90.0 kg man who pushes a crate 4.25 m up along a ramp that makes an angle of 20.0° with the horizontal (see below). He exerts a force of 535 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate and on his body to get up the ramp.
Answer:
W = 3.4x0³ J.
Explanation:
The work done by the man is given by the following equation:
[tex] W = F_{t}\cdot d [/tex] (1)
where W: is the work, Ft is the total force and d: is the displacement = 4.25 m.
We need to find first the total force Ft, which is:
[tex] Ft = Fm + W [/tex]
where Fm: is the force exerted by the man = 535 N, W: is the weight = m*g*sin(θ), m: is the mass of the man, g: is the gravitational acceleration = 9.81 m/s², and θ: is the angle = 20.0°.
[tex] F_{t} = Fm + W = 500 N + 90.0 kg*9.81 m/s^{2} * sin(20.0) = 802.0 N [/tex]
Hence, the work is:
[tex] W = 802.0 N \cdot 4.25 m = 3.4 \cdot 10 ^{3} J [/tex]
Therefore, the work done by the man is 3.4x10³ J.
I hope it helps you!
10. A satellites is in a circular orbit around the earth at a height of 360 km above the earth’s surface. What is its time period? What is its orbital speed?
Answer:
Orbital speed=8102.39m/s
Time period=2935.98seconds
Explanation:
For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g(R2(R+h)2)
V2R+h=g(R2(R+h)2)
V=√g(R2R+h)
V= sqrt(9.8 × (6371000)^2/(6371000+360000)
V= sqrt(9.8× (4.059×10^13/6731000)
V=sqrt(65648789.18)
V= 8102.39m/s
Time period ,T= sqrt(4× pi×R^3)/(G× Mcentral)
T= sqrt(4×3.142×(6.47×10^6)^3/(6.673×10^-11)×(5.98×10^24)
T=sqrt(3.40×10^21)/ (3.99×10^14)
T= sqrt(0.862×10^7)
T= 2935.98seconds
an object of unknown mass oscillates on the end of a spring wit period 8 s. A 10kg object is attached to the first object, changing te period to 12 s. What is the mass of the first object
Answer:
Explanation:
Let the first object have a mass of M
And a period of T1=8sec
The second object has a mass 10kg and a period of T2=12 sec
It is know that,
The period of a spring-mass system is proportional to the square root of the mass and inversely proportional to the square root of the spring constant.
T=2π√(m/k)
Then the constant in this equation is the spring constant (k) and 2π, which does not change for the same material.
Then, make k subject of formulas
T²=4π²(m/k)
T²k=4π²m
Then, k/4π²=m/T²
So the k is directly proportional to m and inversely proportional to T²
M1/T1²=M2/T2²
Since, M1 is unknown, M2=10kg, T1=8sec and T2=12
Then,
M1/T1²=M2/T2²
M1/8²=10/12²
M1/64=0.06944
M1=0.06944×64
M1=4.444kg
The mass of the first object is 4.44kg
Using a good pair of binoculars, you observe a section of the sky where there are stars of many different apparent brightnesses. You find one star that appears especially dim. This star looks dim because it is:_______.
Answer:
Farther,
Because the stars are far from one another
Explanation:
The star look dim because a star's brightness also depends on its proximity to us. The more distant an object is, the dimmer it appears.
The sun appears very bright to us because it is closer to us, the sun distance from the earth is one light year which is around 92,955,807 miles. Now the closest star to the earth is 4.22 light-years, which is four times that of the sun and so it slowly spread out over time.
Therefore, if two objects have the same level of brightness, but one is farther away, the closer star will appear brighter than the more distant star - even though they are equally bright!
The same applies to star.
The free throw line in basketball is 4.570 m (15 ft) from the basket, which is 3.050 m (10 ft) above the floor. A player standing on the free throw line throws the ball with an initial speed of 7.157 m/s, releasing it at a height of 2.440 m above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket?
Answer:
[tex]\theta = 86.491^{\textdegree}[/tex]
Explanation:
The equations for the horizontal and vertical position of the ball are, respectivelly:
[tex]4.570\,m = [(7.157\,\frac{m}{s})\cdot\cos \theta]\cdot t\\3.050\,m = 2.440\,m +[(7.157\,\frac{m}{s})\cdot \sin \theta]\cdot t - \frac{1}{2}\cdot (9.807\,\frac{m}{s^{2}} )\cdot t^{2}[/tex]
By isolating each trigonometric component and summing each equation:
[tex]20.885\,m^{2} = [51.223\,\frac{m^{2}}{s^{2}}\cdot \cos^{2} \theta]\cdot t^{2}[/tex]
[tex][0.61\,m + \frac{1}{2}\cdot (9.807\,\frac{m}{s^{2}} )\cdot t^{2}]^{2} = [51.223\,\frac{m^{2}}{s^{2}}\cdot \sin^{2} \theta]\cdot t^{2}[/tex]
[tex]21.257\,m^{2} + (5.982\,\frac{m^{2}}{s^{2}})\cdot t^{2}+(24.044\,\frac{m^{2}}{s^{4}} )\cdot t^{4} = (2623.796\,\frac{m^{2}}{s^{2}})\cdot t^{2}[/tex]
[tex]21.257\,m^{2} - (2617.814\,\frac{m^{2}}{s^{2}})\cdot t^{2}+(24.044\,\frac{m^{2}}{s^{4}} )\cdot t^{4} = 0[/tex]
The positive real roots are:
[tex]t_{1} = 10.434\,s,t_{2} = 0.09\,s[/tex]
The needed angle is:
[tex]\theta = \cos^{-1} [\frac{4.570\,m}{(7.157\,\frac{m}{s} )\cdot t} ]\\\theta_{1} = 86.491^{\textdegree}\\\theta_{2} = NaN[/tex]
Two small balls, A and B, attract each other gravitationally with a force of magnitude F. If we now double both masses and the separation of the balls, what will now be the magnitude of the attractive force on each one?A) 16F
B) 8F
C) 4F
D) F
E) F/4
Answer:
D) F
Explanation:
Let m and M be the mass of the balls A and B respectively and r be the distance between the two balls. The magnitude of attractive gravitational force experienced by the balls due to each other is given by the relation :
[tex]F=\frac{GMm}{r^{2} }[/tex] ......(1)
Now, if the masses of both the balls gets doubled as well as there separation distance also gets doubled, then let F₁ be the new gravitational force acting on them.
Since, New mass of ball A = 2M
New mass of ball b = 2m
Distance between the two balls = 2r
Substitute these values in equation (1).
[tex]F_{1} =\frac{G(2M)(2m)}{(2r)^{2} }[/tex]
[tex]F_{1} =\frac{4GMm}{4r^{2} }=\frac{GMm}{r^{2} }[/tex]
Using equation (1) in the above equation.
F₁ = F
In a particular television picture tube, the measured beam current is 23.3 µA . How many electrons strike the tube screen every 28 s ? The fundamental charge is 1.602 × 10−19 C. Answer in units of electrons.
Answer:
Explanation:
Given that
Beam current (i)=23.3µA
And the time to strike(t)=28s
Also, a fundamental charge e=1.602×10^-19C
Then, the charge quantity is given as,
q=it
Then, q=23.3×10^-6×28
q=6.524×10^-4C
Also, the number of electron N is given as
q=Ne
Therefore, N=q/e
So, N=6.524×10-4/1.602×10^-19
N=4.072×10^15
There are 4.072×10^15 electrons strike the tube screen every 28 s.
When The fundamental charge is 1.602 × 10−19 C So, There are [tex]4.072\times 10\wedge 15[/tex] electrons that strike the tube screen every 28 s.
Calculate of Units of Electrons
Given that information as per the question
Beam current (i) is =23.3µA
And also the time to strike(t)=28s
Also, a fundamental charge e=[tex]1.602\times10\wedge -19C[/tex]
Then, the charge quantity is given as,
q is =it
Then, q=[tex]23.3\times 10\wedge-6×28[/tex]
q is =[tex]6.524\times10\wedge -4C[/tex]
Also, When the number of electron N is given as
Now, q=Ne
Therefore, N is =q/e
So, N is = [tex]6.524\times10-4/1.602\times10\wedge-19[/tex]
N is = [tex]4.072\times 10\wedge 15[/tex]
Therefore, There are [tex]4.072\times 10\wedge 15[/tex] electrons that strike the tube screen every 28 s.
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A boulder is raised above the ground, so that its potential energy relative to the ground is 200 J. Then it is dropped. Estimate what its kinetic energy will be just before hitting the ground.
Answer:
200 J
Explanation:
In this problem, I assume there is no air resistance, so the system is isolated (=no external forces).
For an isolated system, the total mechanical energy is constant, and it is given by:
[tex]E=KE+PE[/tex]
where
KE is the kinetic energy
PE is the potential energy
The kinetic energy is the energy due to the motion of the object, while the potential energy is the energy due to the position of the object relative to the ground.
At the beginning, when the boulder is raised above the ground, its height above the ground is maximum, while its speed is zero; it means that all its mechanical energy is just potential energy, and it is:
[tex]E=PE_{max}=200 J[/tex]
As the boulder falls down, its altitude decreases, so its potential energy decreases, while the speed increases, and the kinetic energy increases. Therefore, potential energy is converted into kinetic energy.
Eventually, just before the boulder hits the ground, the height of the object is zero, and the speed is maximum; this means that all the energy has now converted into kinetic energy, and we have
[tex]E=KE_{max}=200 J[/tex]
Therefore, the kinetic energy just before hitting the ground is 200 J.
The area of each plate of a parallel plate capacitor is 0.021 m2. The plates are 2.75 mm apart with a dielectric material (κ = 3.0) between them. The maximum possible electric field between the plates is 3.25 ✕ 105 V/m.What is the maximum energy that can be stored in the capacitor?
Explanation:
The given data is as follows.
Dielectric constant, K = 3.0
Area of the plates (A) = 0.021 [tex]m^{2}[/tex]
Distance between plates (d) = [tex]2.75 \times 10^{-3} m[/tex]
Maximum electric field (E) = [tex]3.25 \times 10^{5} V/m[/tex]
Now, we will calculate the capacitance as follows.
C = [tex]\frac{k \epsilon_{o} \times A}{d}[/tex]
= [tex]\frac{3.0 \times 8.85 \times 10^{-12} \times 0.021}{2.75 \times 10^{-3}}[/tex]
= [tex]\frac{0.55755 \times 10^{-12}}{2.75 \times 10^{-3}}[/tex]
= [tex]0.203 \times 10^{-9}[/tex] F
Formula to calculate electric charge is as follows.
E = [tex]\frac{\sigma}{k \epsilon_{o}}[/tex]
or, Q = [tex]E \times k \times \epsilon_{o}A[/tex] (as [tex]\frac{\sigma}{\epsilon_{o}} = \frac{Q}{A}[/tex])
= [tex]3.25 \times 10^{5} \times 3.0 \times 8.85 \times 10^{-12} \times 0.021[/tex]
= [tex]181.2 \times 10^{-9} C[/tex]
Formula to calculate the energy is as follows.
U = [tex]\frac{1 \times Q^{2}}{2 \times C}[/tex]
= [tex]\frac{(181.2 \times 10^{-9} C)^{2}}{2 \times 1.6691 \times 10^{-9}}[/tex]
= [tex]\frac{32833.44 \times 10^{-18}}{3.3382 \times 10^{-9}}[/tex]
= [tex]9835.67 \times 10^{-9}[/tex]
or, = [tex]98.35 \times 10^{-7} J[/tex]
Thus, we can conclude that the maximum energy that can be stored in the capacitor is [tex]98.35 \times 10^{-7} J[/tex].
A cosmic ray (an electron or nucleus moving ar speeds close to the speed of light) travels across the Milky Way at a speed of 0.98 times the spee of light. If the Milky way is 30,000 pc across, how long does this journey take in our (essentially stationary) frame of reference and the cosmic ray's frame of reference, in years?
Answer:
Cosmic ray's frame of reference: 99,875 years
Stationary frame of reference: 501,891 years
Explanation:
First of all, we convert the distance from parsec into metres:
[tex]d=30,000 pc =9.26\cdot 10^{20} m[/tex]
The speed of the cosmic ray is
[tex]v=0.98 c[/tex]
where
[tex]c=3.0 \cdot 10^8 m/s[/tex] is the speed of light. Substituting,
[tex]v=(0.98)(3.0\cdot 10^8)=2.94\cdot 10^8 m/s[/tex]
And so, the time taken to complete the journey in the cosmic's ray frame of reference (called proper time) is:
[tex]T_0 = \frac{d}{v}=\frac{9.26\cdot 10^{20}}{2.94\cdot 10^8}=3.15\cdot 10^{12} s[/tex]
Converting into years,
[tex]T_0 = \frac{3.15\cdot 10^{12}}{(365\cdot 24\cdot 60 \cdot 60}=99,875 years[/tex]
Instead, the time elapsed in the stationary frame of reference is given by Lorentz transformation:
[tex]T=\frac{T_0}{\sqrt{1-(\frac{v}{c^2})^2}}[/tex]
And substituting v = 0.98c, we find:
[tex]T=\frac{99,875}{\sqrt{1-(\frac{0.98c}{c})^2}}=501,891 years[/tex]
Two cars having different weights are traveling on a level surface at different constant velocities. Within the same time interval, greater force will always be required to stop the car that has greater
Answer:
The answer to the question is
Momentum
Explanation:
Momentum in physics refer to the attribute that a body has by virtue of its mass and velocity. It is found by multiplying the mass of the moving object and the velocity, hence
Momentum = Mass × Velocity = m·v
Newton first law of motion states that the force acting on an object is proportional to the rate of change of momentum produced
Therefore when we find the momentum of the two cars, the one that has the greater momentum will require the most force to stop it.
Momentum is a physics term; it refers to the quantity of motion that an object has. A sports team that is on the move has the momentum. If an object is in motion (on the move) then it has momentum
Momentum
On your first trip to Planet X you happen to take along a 290g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You're curious about the free-fall acceleration on Planet X, where ordinary tasks seem easier than on earth, but you can't find this information in your Visitor's Guide. One night you suspend the spring from the ceiling in your room and hang the mass from it. You find that the mass stretches the spring by 21.1cm . You then pull the mass down 11.2cm and release it. With the stopwatch you find that 11 oscillations take 18.2sCan you now satisfy your curiosity?what is the new g?
Answer:
Explanation:
11 oscillations in 18.2 s
Time period is defined as the time taken to complete one oscillation.
T = 18.2 / 11 = 1.655 s
mass, m = 290 g = 0.29 kg
Δx = 21.1 cm = 0.211 m
ω = 2π / T = (2 x 3.14) / 1.655 = 3.796 rad/s
[tex]\omega =\sqrt{\frac{K}{m}}[/tex]
Where, K is the spring constant
K = ω² m = 3.796 x 3.796 x 0.29 = 4.18 N/m
Now, mg = K Δx
0.29 x g = 4.18 x 0.211
g = 3.04 m/s²
A 9-year-old has just fallen off the monkey bars. She has obvious deformity to her mid forearm. How should the radiographer care for her when taking x-rays?
Answer:
He will need to support the joint above and below the deformity, this will allow the xray to be taken without generating more damage to the child.
I hope you find this information useful and interesting! Good luck!
A plastic pipe carries deionized water in a microelectronics clean room, and one end of it is capped. the water pressure is p0 = 60 psi, and the cap is attached to the end of the pipe by an adhesive. calculate the shear stress present in the adhesive.
Answer:
Where the height of the cap is less than or equal to the radius of the bore of the plastic pipe then the ⇵⇵shear stress is less than o equal to 60 psi.
Explanation:
Shear stress = F/A
where F = Applied force
A = Cross sectional area of the member experiencing the force
If the cap covers a section of the pipe of height, h then the area = 2πrh
Where the 60 psi pressure is acting on the pipe bore, we have P₀ = 60 psi = F₀/A₀ where A₀ = area of pipe bore ≈ πr². Therefore if F₀ = F then we have
F₀/A₀ = F₀/πr² and F/A = F₀/2πrh = where h greater than or equal to 0.5×r then the shear stress is less than or equal to 60 psi
To calculate the shear stress in the adhesive, use the formula for shear stress and the given pressure and area.
Explanation:To calculate the shear stress present in the adhesive, we can use the formula for shear stress:
Shear stress = Force / Area
In this case, the force exerted by the water pressure on the capped end of the pipe is equal to the pressure multiplied by the area of the capped end. Since the water pressure is given as 60 psi, we need to convert it to a consistent unit, such as Pascals. 1 psi is equal to approximately 6895 Pascals. The area of the capped end can be calculated using the formula for the area of a circle: Area = pi * radius^2.
Once we have the force and area, we can substitute them into the formula for shear stress to find the value.
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(a) Use the de Broglie relation λ = h/p to nd the wavelength of a raindrop with mass 1 mg and speed 1 cm/s. Is there a way to set up a raindrop-diraction experiment and detect the wave-like properties of rain?
Question:
a). Use the de Broglie relation λ=h/p to find the wavelength of a raindrop with mass m=1 mg and speed 1cm/s.
ii). Does it seem likely that the wave properties of a raindrop could be easily detected?
b). Find the wavelength of electrons with KE = 500 eV.
c). If a neutron has the same wavelength as blue light (λ=450 nm) what is it's KE?
ii). What if it's an electron?
Answer:
The answers to the question are
a). The wavelength of the raindrop is 6.626*10⁻²⁶ m
The properties of the rain drop will be hardly detected
b). The wavelength of the electrons is 5.491×10⁻¹¹ m
c). The KE of the neutron is 5.242510⁻²⁸ J
ii). For an electron it will increase to be KE (electron) = 9.6392639×10⁻²⁵ J
Explanation:
Using de Broglie relations, we have
p = h/λ and E = h·f also E = 1/2·m·v²
a). λ= h/p, E= p²/2·m, p = √(2·m·E), λ = h/√(2·m·E)
Where
λ=wavelength
E = energy
p = momentum
m = mass
The kinetic energy of the rain drop is [tex]\frac{1}{2}[/tex]×m×v² = 0.5×(1×10⁻⁶)(0.01)2
= 5× 10⁻¹¹ J
λ = h/√(2·m·E) = 6.626*10-34 Js/√(2×1×10⁻⁶×5× 10⁻¹¹)
= 6.626*10⁻²⁶ m
The properties of the rain drop will not be easily detected
b). The electron energy is equivalent to 500 eV ⇒500 eV × 1.6×10⁻¹⁹ J/eV
= 8×10⁻¹⁷ J
λ = h/√(2·m·E) = 6.626×10⁻³⁴ Js/√(2*×9.1×10⁻³¹×8×10⁻¹⁷)
= 5.491×10⁻¹¹ m
c). λ = h/√(2·m·E) then √(2·m·E) = h/ λ or E = (h/λ)²/(2·m)
= (6.626×10⁻³⁴/5.0×10⁻⁷)²/(2×1.674927471×10⁻²⁷)
E = 5.242510⁻²⁸ J
ii). For an electron, we have m = 9.10938356 × 10⁻³¹ kg
λ = (6.626×10⁻³⁴/5.0×10⁻⁷)²/(2×9.10938356×10⁻³¹) = 9.6392639×10⁻²⁵ J
A proton (????p = +????, mp = 1.0 u; where u = unified mass unit ≃ 1.66 × 10−27kg), a deuteron (???????? = +????, m???? = 2.0 u) and an alpha particle (???????? = +2????, m???? = 4.0 u) are accelerated from rest through the same potential difference ????, and then enter the same region of uniform magnetic field ????⃗⃗ , moving perpendicularly to the direction of the magnetic field.
a. What is the ratio of the proton’s kinetic energy Kp to the alpha particle’s kinetic energy K?????
b. What is the ratio of the deuteron’s kinetic energy K???? to the alpha particle’s kinetic energy K?????
c. If the radius of the proton’s circular orbit ????p = 10 cm, what is the radius of the deuteron’s orbit ?????????
d. What is the radius of the alpha particle’s orbit ?????????
Answer:
a. 1/2 b. 1/2 c, 20 cm d. 40 cm
Explanation:
Here is the complete question
A proton ( = +, = 1.0 u; where u = unified mass unit ≃ 1.66 × 10−27kg), a deuteron ( = +, = 2.0 u) and an alpha particle ( = +2, = 4.0 u) are accelerated from rest through the same potential difference , and then enter the same region of uniform magnetic field ⃗⃗ , moving perpendicularly to the direction of the magnetic field.
A) What is the ratio of the proton’s kinetic energy to the alpha particle’s kinetic energy ?
B) What is the ratio of the deuteron’s kinetic energy to the alpha particle’s kinetic energy ?
C) If the radius of the proton’s circular orbit = 10 cm, what is the radius of the deuteron’s orbit ?
D) What is the radius of the alpha particle’s orbit ?
Solution
a. For both particles, kinetic energy = electric potential energy
For proton K.E= K₁ = 1/2m₁v₁² = +eV , for alpha particle K.E = K₂ = 1/2m₂v₂²= +2eV
where m₁, m₂ and v₁, v₂ are the respective masses and velocities of the proton and alpha particle. So, the ratio of their kinetic energies is
1/2m₁v₁²/1/2m₂v₂² = +eV/+2eV
m₁v₁²/m₂v₂² = 1/2.
So the ratio K₁/K₂ = 1/2
b. For both particles, kinetic energy = electric potential energy
For deuteron K₁ = 1/2m₁v₁² = +eV , for alpha particle K₂ = 1/2m₂v₂²= +2eV
where m₁, m₂ and v₁, v₂ are the respective masses and velocities of the deuteron and alpha particle. So, the ratio of their kinetic energies is
1/2m₁v₁²/1/2m₂v₂² = +eV/+2eV
m₁v₁²/m₂v₂² = 1/2.
So the ratio K₁/K₂ = 1/2
c. The radius of the proton's circular is gotten from the centripetal force which equal the magnetic force. So,
mv²/r = Bev
r₁ = mv/Be
Since mass of deuteron m₂ equals twice mass of proton m₁, m₂ = 2m₁
So, radius of deuteron's circular orbit equals
r₂ = m₂v/Be = 2m₁v/Be = 2r₁ = 2 × 10 cm = 20 cm
d. The radius of the alpha particle is given by r₃ = m₃v/Be. Since mass of alpha particle equal four times mass of proton, m₃ = 4m₁.
So, radius of alpha particle's circular orbit equals
r₃ = m₃v/Be = 4m₁v/Be = 4r₁ = 4 × 10 cm = 40 cm
A very long, solid insulating cylinder has radius R; bored along its entire length is a cylindrical hole with radius a. The axis of the hole is a distance b from the axis of the cylinder, where a 6 b 6 R (Fig. P22.58). The solid material of the cylinder has a uniform volume charge density r. Find the magnitude and direction of the electric field E S inside the hole, and show that E S is uniform over the entire hole
The electric field inside the hollowed insulator can be calculated considering the missing charge in the cylindrical hole. This electric field is uniform and its magnitude is E = ρ.b/ϵ0; where ρ is the uniform charge density and ϵ0 is the permittivity constant. The direction of the electric field is from the axis of the cylinder toward the hole for positive ρ and opposite for negative ρ.
Explanation:The question refers to the electric field inside a hollowed cylindrical insulator. This is a physics problem related to the study of electrostatics. The concept of electric field refers to the influence exerted in the space around a charged object, which can result in forces on other charged objects.
We know the electric field inside a uniformly charged insulator is zero because charges move to the surface in an insulator. In this case, though, the insulator is not uniform; it has a cylindrical hole. So, we must calculate the contribution from the missing charge in the hole.
The electric field generated by the missing volume, dV, with charge density ρ is given by Coulomb's law, where the electric field corresponds to the charge divided by the square of the distance: E = k.Q/R^2.
Because the volume dV, at radius r from the axis of the cylinder has a charge equal to the volume of the cylindrical disk multiplied by ρ, we have dQ = ρ.dV and thus E = k.ρ.dV/R^2. Integrating over the total volume of the cylindrical hole, we find that the electric field is uniform and its magnitude, E, is given by E = ρ.b/ϵ0, where ϵ0 is the permittivity constant.
The direction of the E is from the axis of the cylinder towards the hole if ρ is positive, because positive charges repel, and vice versa if ρ is negative.
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To find the electric field inside the hole of the insulating cylinder, we use the superposition principle and Gauss's law, resulting in a uniform electric field due to the subtracted cylindrical charge distribution.
The question is asking for the magnitude and direction of the electric field inside a cylindrical hole bored inside another charged cylindrical insulator. To determine the electric field (ES) inside the hole, we apply the principle of superposition, which involves considering the charge distribution that would have been present had there been no hole and subtracting the charge distribution of a cylinder of radius a with charge density -ρ, placed such that it would carve out the hole.
Using Gauss's Law, the electric field due to the entire charged cylinder without the hole would be zero inside the cylinder. Therefore, the electric field inside the hole is solely due to the cylindrical charge distribution we subtracted. Since the cylindrical charge distribution is symmetric, the electric field ES within the hole will also be uniform and directed radially outward from the axis of the imaginary cylinder we subtracted. The magnitude of ES can be calculated using Gauss's Law and considering the symmetry of the problem.
If the raindrops hit at 8.3 m/s , estimate the magnitude of the force on the bottom of a 1.0-m2 pan due to the impacting rain which we assume does not rebound.
Answer:
F = 0.1153 N
Explanation:
Given:
- Rain fall rate = 5 cm/hr
- Velocity before impact vi = 8.3 m/s
- The Area of the pan A = 1.0 m^2
- The density of water ρ = 1000 kg/m^3
Find:
Estimate the magnitude of the force
Solution:
- Consider rain drops impacting a surface looses all its momentum. So the change in its momentum is just the momentum with which it impacted.
- We do not know the size of each drop. But the rain fall rate allows us to calculate the rate of change of momentum.
- Total Force: The total force experience by the surface due to the momentum transfer from the impacting rain drops is:
F = m(Δv) / t = ρ*v*A*Δh/Δt
Where
Δh/Δt = rain-fall rate.
F = 1000*8.3*1*0.05 / 3600
F = 0.1153 N
The magnitude of the force is 0.1153N
Given-
The velocity of the raindrop=8.3m/sec
Let when the raindrop hits the surface it loses its momentum
Now the magnitude of the force on the bottom of the given area can be calculated as
[tex]F=\dfrac{m\bigtriangleup v}{t}[/tex]
[tex]F= \rho vA \frac{\bigtriangleup h}{\bigtriangleup t}[/tex]
here rainfall rate
=[tex]\dfrac{\bigtriangleup h}{\bigtriangleup t}[/tex]
therefore,
[tex]F=1000\times 8.3\times 1 \times \dfrac{0.05}{3600}[/tex]
[tex]F=0.1153N[/tex]
Hence the magnitude of the force is 0.1153N
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You have been hired to help improve the material movement system at a manufacturing plant. Boxes containing 16 kg of tomato sauce in glass jars must slide from rest down a frictionless roller ramp to the loading dock, but they must not accelerate at a rate that exceeds 2.6 m/s2 because of safety concerns.a. What is the maximum angle of inclination of the ramp?b. If the vertical distance the ramp must span is 1.4 m, with what speed will the boxes exit the bottom of the ramp?c. What is the normal force on a box as it moves down the ramp?
a) [tex]15.4^{\circ}[/tex]
b) 5.2 m/s
c) 151.2 N
Explanation:
a)
When the box is on the frictionless ramp, there is only one force acting in the direction along the ramp: the component of the forc of gravity parallel to the ramp, which is given by
[tex]mg sin \theta[/tex]
where
m =16 kg is the mass of the box
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
[tex]\theta[/tex] is the angle of the ramp
According to Newton's second law of motion, the net force on the box is equal to the product of mass and acceleration, so:
[tex]F=ma\\mgsin \theta = ma[/tex]
where a is the acceleration.
From the equation above we get
[tex]a=g sin \theta[/tex]
And we are told that the acceleration must not exceed
[tex]a=2.6 m/s^2[/tex]
Substituting this value and solving for [tex]\theta[/tex], we find the maximum angle of the ramp:
[tex]\theta=sin^{-1}(\frac{a}{g})=sin^{-1}(\frac{2.6}{9.8})=15.4^{\circ}[/tex]
b)
Here we are told that the vertical distance of the ramp is
[tex]h=1.4 m[/tex]
Since there are no frictional forces acting on the box, the total mechanical energy of the box is conserved: this means that the initial gravitational potential energy of the box at the top must be equal to the kinetic energy of the box at the bottom of the ramp.
So we have:
[tex]GPE=KE\\mgh=\frac{1}{2}mv^2[/tex]
where:
m = 16 kg is the mass of the box
[tex]g=9.8 m/s^2[/tex]
h = 1.4 m height of the ramp
v = final speed of the box at the bottom of the ramp
Solving for v,
[tex]v=\sqrt{2gh}=\sqrt{2(9.8)(1.4)}=5.2 m/s[/tex]
c)
There are two forces acting on the box in the direction perpendicular to the ramp:
- The normal force, N, upward
- The component of the weight perpendicular to the ramp, downward, of magnitude
[tex]mg cos \theta[/tex]
Since the box is in equilibrium along the perpendicular direction, the net force is zero, so we can write:
[tex]N-mg cos \theta[/tex]
and by substituting:
m = 16 kg
[tex]g=9.8 m/s^2[/tex]
[tex]\theta=15.4^{\circ}[/tex]
We can find the normal force:
[tex]N=mg cos \theta=(16)(9.8)cos(15.4^{\circ})=151.2 N[/tex]
The maximum angle of inclination of the ramp and the final velocity of the boxes are calculated based on the given acceleration and distance. The normal force on a box moving down the ramp remains unaffected by the angle of inclination in this case as the ramp is devoid of friction. It's an application of physics concepts in real-world situations.
Explanation:This problem is a practical application of the concepts of physics, specifically mechanics. Let's break it down.
a. The maximum angle of inclination of the ramp can be found by utilizing the relationship between the acceleration, the gravitational constant, and the angle of inclination. The formula is as follow:
sin(θ) = acceleration / g
Substitute the given acceleration (2.6 m/s²) and the gravitational constant g (9.8 m/s²) and solve for θ.
b. To find the final velocity of the boxes, we can apply the equations of motion. Using the formula v² = u² + 2gs (where u is the initial velocity, g is the gravitational constant, s is the distance) and substituting the given values (u=0, g=2.6 m/s², s=1.4m), we find the final velocity.
c. The normal force on a box moving down the ramp can be found from the formula: Normal force = mg cos (θ) (where m is the mass, g is gravitational constant and θ is the angle of inclination). Here, θ does not affect the normal force because the box is moving down a frictionless ramp.
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This image shows a stream of positively charged particles being directed at gold foil. The positively charged particles are called "alpha particles” and each one is like a nucleus without any electrons.
What is the best explanation for why a particle is striking point X?
Alpha particles are mostly empty space, so they move in random directions, and one of these alpha particles can strike point X.
The gold atoms contain negative electrons, so when alpha particles strike the gold they keep moving but in various directions.
If an alpha particle hits the gold foil, a gold nucleus splits and a particle from it flies out at point X.
When the dense, positive alpha particle passes close to a positive nucleus of gold, the alpha particle repels and hits the screen at point X.
Answer:
D. When the dense, positive alpha particle passes close to a positive nucleus of gold, the alpha particle repels and hits the screen at point X.
Explanation:
D)When the dense, positive alpha particle passes close to a positive nucleus of gold, the alpha particle repels and hits the screen at point X.
What happens to alpha particles that pass close to the nucleus of a gold atom?The gold nucleus and alpha particle are both definitely charged therefore there is a repulsive pressure between the (gold) nucleus and the alpha particle. This causes the alpha particle to be deflected through a massive angle.
Maximum alpha debris surpassed instantly through the gold foil, which implied that atoms are ordinarily composed of open space. a few alpha particles had been deflected barely, suggesting interactions with different definitely charged particles in the atom.
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The electron current in a horizontal metal wire is 7.4 × 1018 electrons/s, and the electrons are moving to the left. What are the magnitude and direction of the conventional current?
Answer:
The magnitude and direction of the conventional current is then 1.186 Amps moving to the left direction
Explanation:
To answer the question, it should be noted that the direction of conventional current is in the opposite direction of the flow of electrons. Therefore, the direction of flow of conventional current will be to the right
The magnitude of the electric current is equal to the rate of flow of the electrons or the time it takes for the electrons to flow past a section of the wire. Therefore the magnitude is that of the 7.4 × 1018 electrons/s
However the unit of the electricity = ampere which is = coulombs/seconds
The 7.4 × 1018 electrons carry
7.4 × 10¹⁸×1.60217662 × 10⁻¹⁹ coulombs = 1.1856106988 coulombs
Therefore the magnitude of electric current = 1.186 coulombs/Seconds
= 1.186 Amps
Explanation:
Below is an attachment containing the solution.
A 92.6 kg weight-watcher wishes to climb a mountain to work off the equivalent of a large piece of chocolate cake rated at 735 (food) Calories. How high must the person climb? The acceleration due to gravity is 9.8 m/s 2 and 1 food Calorie is 103 calories. Answer in units of km.
Answer:
349 m
Explanation:
Parameters given:
Mass of climber, m = 92.6 kg
Amount of food calories = 735
1 food calorie = 103 calories
735 food calories = 75705 calories
1 joule is equal to 0.239 calories. Therefore, 75705 calories will be 316749.72 joules.
Hence, this is the amount of work the climber must do work off the food he ate.
Work done is given as:
W = Force * distance
W = m * g * h
h = W/(m * g)
h = 316749.72/(92.6 * 9.8)
h = 349 m
A comet is approaching Earth at a known velocity. Indicate the proper sequence that describes the wavelength of light you measure as it first approaches, then passes, and finally recedes from Earth.
Answer:
Explanation:
When it was approaching, the wavelength is blue-shifted and the rest wavelength when it was receding the wavelength is redshifted.
Any planet is an extremely faint light source compared to its parent star. Comets are considered as the most primitive objects in the Solar System.
Which of the following represents energy in its most disordered form? Group of answer choices Chemical-bond energy Electromagnetic (light) energy Heat energy Potential energy The kinetic energy of a moving object
Answer: Heat Energy
Explanation:
Heat is energy in its most disordered form. heat energy is the random jostling of molecules and is therefore not organized. As cells perform the chemical reactions that generate order within, some energy is inevitably lost in the form of heat. Because the cell is not an isolated system, the heat energy produced by the cell is quickly dispersed into the cell's surroundings where it increases the intensity of the thermal motions of nearby molecules. This increases the entropy of the cell's environment and keeps the cell from violating the second law of thermodynamics.
A generator uses a coil that has 100 turns and a 0.50-T magnetic field. The frequency of this generator is 60.0 Hz, and its emf has an rms value of 120 V. Assuming that each turn of the coil is a square (an approximation), determine the length of the wire from which the coil is made
Answer:
38 m
Explanation:
Number of turns=N=100
Magnetic field=B=0.50 T
Frequency of the generator=f=60 Hz
Rms value of emf=[tex]E_{rms}=120 V[/tex]
We have to find the length of the wire from which the coil is made.
Peak value of emf=[tex]E_0=E_{rms}\sqrt 2=120\times \sqrt 2=169.7 V[/tex]
Length of wire=[tex]4\sqrt{\frac{NE_0}{2\pi fB}}[/tex]
Substitute the values
Length of wire=[tex]4\times \sqrt{\frac{169.7\times 100}{0.50\times 2\pi\times 60}}[/tex]
Length of wire=38 m
Hence, the length of wire from which the coil is made=38 m
To determine the length of wire in a generator coil with 100 turns in a 0.50-T field and 120 V rms at 60 Hz, one can use the formula for rms emf of a generator and solve for the area of one turn to find the length per turn and multiply by the number of turns.
Explanation:The question is asking to determine the length of wire used to make a coil in a generator. The generator has 100 turns of wire, operates with a 0.50-T magnetic field, and has an rms value of the emf of 120 V with a 60.0 Hz frequency. Assuming the turns are squares, we can use the formula for the rms value of the emf (Erms) for a generator, which is Erms = NABωrms, where N is the number of turns, A is the area of the turn, B is the magnetic field, and ωrms is the rms angular velocity. The rms angular velocity ωrms is related to the frequency (f) by the equation ωrms = 2πf/√2.
To find the side length (L) of the square turns, we rearrange the formula to solve for A and then take the square root. Once we have L, we multiply by 4 to get the perimeter of one turn and then by 100 to find the total length of wire needed for all turns.
What is the angular velocity of the second hand on a clock? (Hint: It takes the second hand 6060 seconds to rotate 2π2π radians. Divide the number of radians by the number of seconds so that your answer has units of radians/second.)
Explanation:
Below is an attachment containing the solution.
A bird sits on a high-voltage power line with its feet 5.000 cm apart. The wire is made from aluminum, is 2.00 cm in diameter, and carries a current of 170.0 A. What is the potential difference between the bird's feet?
The potential difference between the bird's feet is 0.71 mV
Explanation:
Given-
Distance between the feet, x = 5 cm = 0.05 m
Diameter of the wire = 2 cm
Radius, r = 2/2 = 1 cm = 0.01 m
Current, I = 170 A
Potential difference, ΔV = ?
We know,
ΔV = IR
Where,
ΔV is the potential difference
I is the current
R is the resistance of the wire
And also
R = ρ L/ A
where,
ρ is the resistivity of aluminium wire (ρ = 2.65 X 10⁻⁸ Ωm)
L is the length
A is the area
Equating both the equations,
ΔV = I * ρL/A (The value of R is replaced in the first equation)
ΔV =
[tex]170 * \frac{(2.65 X 10^-^8) * 0.05}{\pi (0.01)^2} \\\\\frac{22.525 X 10^-^8}{0.000314} \\\\\frac{22.525 X 10^6}{314 X 10^8} \\\\0.0717 X 10^-^2\\\\7.17 X 10^-^4[/tex]
Therefore, the potential difference between the bird's feet is 0.71 mV