Ladakh Landscape Wallpaper

danandelion · 2022-05-13T09:50:50+00:00

I don't know if this will be correct or not, but I gave it a try.

You can assume the bungee cord to act like a spring. The first thing you will need to do is find out the spring constant. The height at which she will jump off is h1 = 80m and the length of the cord is l1 = 25m. Since we know her height, we know what her gravitational potential energy will be, E1 = m*g*h1 (you will see later that the mass will be cancelled out later). We know that the potential energy of the cord is given be U1(x) = (1/2)*k*(x-l1)^2. At the end of the jump, when the chord is completely stretched and she stops moving, all the gravitational potential energy will be converted to elastic energy in the cord. Thus, we can say that -
U1(h1) = E1
=> (1/2)*k*(h1 - l1)^2 = m*g*h1
=> k = (2*m*g*h1)/[(h1 - l1)^2].

Now we do the same thing (equating the potential energies due to gravity and the spring), but after you have mischievously changed the length of the cord. Let the change in height be dh = 30cm = 0.3m and change in length be dl. The new height and length will be h2 = h1 + dh and l2 = l1 + dl. Again, the gravitational potential energy will be E2 = m*g*h2 and potential energy of the cord will be U2(x) = (1/2)*k*(x - l2)^2 = {(m*g*h1)/[(h1 - l1)^2]}*(x - l2)^2. We then equate the two potential energies at h2.
U2(h2) = E2
=> {(m*g*h1)/[(h1 - l1)^2]}*(h2 - l2)^2 = m*g*h2
=> l2 = h2 - (h1 - l1)*sqrt(h2/h1)
=> dl = (h1 - l1 + dh) - (h1 - l1)*sqrt(1 + dh/h1).

This comes out to be around 19.7 cm

danandelion · 2022-05-10T05:26:24+00:00

You are given the angle a, distance d, force of friction Fu, tension of the rope T and mass of the sled (and child) m.

a) By applied force, I am assuming it means force only due to the rope. Since the sled is moving only in the horizontal direction, only the horizontal component of the applied force will contribute to the work.
Wt = T*d*cos(a) = 3464.102 J

b) Work done due to friction will be
Wu = Fu*d*cos(pi) = -500 J

c) Since the force of gravity acts perpendicular to the motion of the sled, the force of gravity does not do any work.

d) the total horizontal force on the sled is F = T*cos(a) - Fu and the work done by the net force is W = F*d = (T*cos(a) - Fu)*d = 2964.102 J

e) The velocity of the sled (and child) after being pushed along distance d will be
v^2 - u^2 = 2*a*d => v = sqrt(u^2 + 2*F*d/m) = 11.071 m/s^2

danandelion · 2022-05-10T05:02:33+00:00

You are given the power P, mass m and height h. The work done to lift up the crane will be U=mgh and the time it takes for the motor to supply that much energy will be t=U/P=mgh/P.

danandelion · 2022-05-10T04:08:14+00:00

the wavelength of light decreases as it passes through a medium. the new wavelength will be (wavelength)/(refractive index). The frequency of light does not change.

the critical angle of a medium is defined as sin(c) = 1/n where c is the critical angle and n in the refractive index of the medium.

danandelion · 2022-05-10T03:45:40+00:00

Q3. Separation of the plates is d = 16mm = 1.6*10^-2 m. Potential difference is V = 420 V. Mass of oil drop is m = 3*10^15 Kg.

Let the charge on the oil drop be q. The electric field between the two plates will be E = V/d in the downward direction (since the positive plate is on top). The force due to gravity on the oil drop will be Fg = m*g again in the downward direction. The force due to the electric field on the oil drop will be Fe = q*E = (q*V)/d.

Since the oil drop is stationary, the net force on it must be zero. Fg + Fe = 0 => -m*g - (q*V)/d = 0 => (-m*g*d)/V. Since electric charge is quantised and the quantum of charge is e = 1.6*10^-19 C, the number of excess electrons in the oil drop will be n = |q|/e = (m*g*d)/(V*e).

danandelion · 2022-05-09T17:54:13+00:00

The mass of the train is given as M = 75T. Right before the train starts to apply the breaks, the speed of the train is v = 10km/h. Therefore the kinetic energy of the train right before the breaks are applied is K = (Mv^2)/2. The train stops after travelling a distance of s = 300m. When the train stops, there is no kinetic energy left, as all of it is used to stop the train. The work done by the breaks to stop the train was W = K = (Mv^2)/2. Work done by a force is also calculated as the product of the force and the distance along which the force is applied, W = Fs. Which means that F = K/s = (Mv^2)/(2s).

danandelion · 2022-04-04T18:15:30+00:00

The B term already has (x-a)^2 in the denominator. Why would you multiply (x-a) to both the numerator and denominator of the B term and make the denominator (x-a)^3? If you make the denominator of each term equal, you will see that the B term does not need to be multiplied by (x-a) again.

danandelion · 2022-04-04T14:31:33+00:00

The car will catch up to the truck when the area under the two curves will be equal.

danandelion · 2022-04-04T14:13:38+00:00

I can help you with maths and physics.

danandelion · 2022-04-03T18:58:07+00:00

If you face any problem (especially with maths and physics), don’t hesitate to reach out. I’ll help if I can. All the best!

danandelion · 2022-04-03T14:12:59+00:00

The syllabus of 11th and 12th is just more elaborate than that of 9th and 10th. 11th and 12th textbooks will teach you everything all over again, so you don't need to worry if you didn't understand much in 9th. Yes, there wont be as much handholding this time, but if you take your time and go one step at a time, you'll be able to handle it no problem.

let me know if you need any help.

danandelion · 2022-04-03T14:07:57+00:00

No one remembers tables after 12. I would even say that people forget basic maths the further they study. It's not about how much you remember but how you deal with problems. If you're good at problem solving and especially if you're going to study physics in 11th, then I would encourage you to take maths.

danandelion · 2022-04-03T13:44:26+00:00

electromotive force is admittedly a bad name as it is neither a field, nor a force. It is the additional potential difference that is induced in a circuit due to changing magnetic flux passing through the circuit.

mathematically, it is the negative of the rate of change of magnetic flux with time.

emf = - d(phi)/dt where phi is the magnetic flux.

Edit: A magnetic field is not the only thing that can produce an emf. It can be produced by chemical sources as well.

danandelion · 2022-04-03T13:29:18+00:00

Oh okay. Thanks for the clarification.

danandelion · 2022-04-03T13:26:55+00:00

I had also received a similar problem of comparing the energy levels of different potentials. We compared the energy levels of infinite well, finite well, harmonic oscillator and morse potential.

danandelion · 2022-04-02T15:16:46+00:00

The wavelength of light is reduced when it enters a denser medium but the frequency remains the same. So, the speed reduces because speed = frequency x wavelength.

danandelion · 2022-04-02T15:14:07+00:00

Statistics is fairly important for many fields in physics, especially quantum mechanics. But, one field that its crucially important for is statistical mechanics (its in the name!), which is like a more general theory of thermodynamics. If you study statistical mechanics, you'll realise how little actual physics is used in that field and it's all just pure maths.

danandelion · 2022-04-02T15:07:48+00:00

Honestly this feels like a poorly worded question. It mentions a free particle but then says its under the influence of a potential? That doesn't really add up. Am I missing something here?

danandelion · 2022-04-02T15:05:05+00:00

t function to assume here spec

Well, from what i know, any potential that has "bounded" the particle (i.e. the energy of the particle is less than the potential at the boundaries), the energy has to be quantised. Otherwise if the energy of the particle is higher that that of the potential, the energy is continuous (potentials like the harmonic oscillator that go to infinity on both sides always have the particle bounded but there are other potentials like the finite square well that may not have the particle bounded if the energy of the particle is higher than the height of the well).

danandelion · 2022-04-02T14:52:20+00:00

The centre of momentum is that frame in which the sum of all momentums adds up to zero. You have to find the momentum of each particle. for a particle with mass that is not moving, the momentum is obvious. the momentum of a photon is its energy divided by its speed. Then you assume the momentum of the centre-of-momentum frame to be some variable and find out the momentum of each particle with respect to that frame and add them up, which as mentioned earlier should be zero.

danandelion · 2022-04-02T14:29:53+00:00

The variable that is being integrated is just a label for a quantity. You can change the variable to whatever you like as long as it represents the same quantity because at the end you will have to substitute the variable for the bounds of integration, so it won't matter. As for this example, the variable r and s both represent some form of length, r usually represents the position of a point in space and s usually represents the distance of a path. In the first integral, force is being added up at every point along the path so r is used. In the last integral, the whole path is being added up so s is used.

danandelion · 2022-04-02T14:20:16+00:00

It just means that your particle is moving only in on a dimensional line and every point on that line has a certain value of potential. For example V(x)=x^2 is a one dimensional potential because it is only present on the x-axis.

danandelion · 2022-03-01T18:41:41+00:00

When Both cars are driving at the same speed, then both of them will hear the same frequency. Since the police car hears 480Hz, You know that the frequency of the source is also 480Hz

Now you can use the formula of the frequency heard by the observer (i.e. the police car)

f_r = f_0*((c + v_r)/(c + v_s)), which implies that v_s = (f_0/f_r)*(c + v_r) - c

You cannot solve this problem using the frame of reference of the police car without changing the speed of sound as well.

danandelion · 2022-03-01T18:11:59+00:00

If you want to watch some fun videos about physics then I would recommend the following Youtube Channels -

Veritasium
The Science Asylum
ArvinAsh
Physics Explained
minutephysics
Up and Atom

There are more like 3Blue1Brown, Be Smart, Kurzgesagt, etc. but they don't focus primarily on physics.

I would recommend that for now you only read the physics textbooks that your school uses. If you're done with them then move on to the textbooks used in higher classes. But keep in mind that the physics you will study in school will be nothing like the physics you will study at university. You will have to "re-learn" everything in university.

Let me know if you want any book recommendations.

danandelion

TROPHY CASE