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# Tutor profile: Daniel S.

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Daniel S.
Graduate Physics Student with three years of tutoring experience
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## Questions

### Subject:Physics (Newtonian Mechanics)

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Question:

Two carts, cart A and cart B, start at rest on a table. Cart B is twice as heavy as cart A. The two carts are pushed across the table with the same constant force for same distance(you can't assume they are pushed for the same amount of time). Assume there is no friction between the carts and the table. Which cart is going faster after the carts have been pushed? What is the ratio of the speed of cart A to the speed of cart B after the carts have been pushed?

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Daniel S.

When a force is applied to an object, work is done on the object. If the force applied to the object is constant, the work done is equal to the displacement of the object times the magnitude of the force, $(W=sF$)where $$W$$ is the work done, $$s$$ is the displacement, and $$F$$ is the magnitude of the force. The work done on the object is also equal to the change in energy of the object. Both carts were pushed with the same force for the same distance, this means that the same amount of work was done on the carts, so the carts have the same final energy. Kinetic energy is the only form of energy that the carts have ( at least as far as classical mechanics is concerned), so the carts have the same kinetic energy. $(KE_A=KE_B$)$(\frac{1}{2}m_Av_A^2=\frac{1}{2}m_Bv_B^2$) plugging in $$2m_A=m_B$$, $(\frac{1}{2}m_Av_A^2=m_Av_B^2$)$(\frac{1}{2}v_A^2=v_B^2$)$(2=\frac{v_A^2}{v_B^2}$)$(\sqrt{2}=\frac{v_A}{v_B}$) so cart A is going $$\sqrt2$$ times as fast as cart B after the carts are pushed.

### Subject:Physics (Electricity and Magnetism)

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Question:

Electrons are fired into a box, the electrons initially travel in the $$\hat{y}$$ direction. Inside the box there is a constant magnetic field with magnitude B in the $$\hat{x}$$ direction, and a constant electric field with magnitude E in the $$\hat{z}$$ direction. At what speed should the electrons be fired for them to not be deflected as they travel through the box?

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Daniel S.

Charged particles in an electromagnetic field experience a Lorentz force. The magnitude and direction of the force can be calculated using, $(\textbf{F}=q(\textbf{E}+\textbf{v}\times\textbf{B})$)Where $$\textbf{F}$$ is the force on the charged particle, $$q$$ is the charge of the particle, $$\textbf{E}$$ is the electric field, $$\textbf{B}$$ is the magnetic field, and $$\textbf{v}$$ is the velocity of the particle. In this case $$q$$ is the charge of the electron, $$q=-e$$. We need to find the velocity where the force on the electron is zero, with no force on the particle it won't be deflected. Setting $$\textbf{F}=0$$ and plugging in $$-e$$ for $$q$$ we get, $(0=-e(\textbf{E}+\textbf{v}\times\textbf{B})$)$(\textbf{E}=-\textbf{v}\times\textbf{B}$)plugging in the directions of the fields and velocity we get,$(E\hat{z}=-(v\hat{y})\times(B\hat{x})$)$(E\hat{z}=-vB(\hat{y}\times\hat{x})$)using the right hand rule $$\hat{y}\times\hat{x}=-\hat{z}$$. Plugging this is we get, $(E\hat{z}=vB\hat{z}$)$(v=\frac{E}{B}$)

### Subject:Calculus

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Question:

Find $$\frac{\partial^2 f(x,y)}{\partial x\partial y}$$, where $$f(x,y)=ye^{yx}+y$$.

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Daniel S.

The partial derivatives can be taken in either order, $( \frac{\partial^2 f(x,y)}{\partial x\partial y} =\frac{\partial}{\partial x}(\frac{\partial f(x,y)}{\partial y})= \frac{\partial}{\partial y}(\frac{\partial f(x,y)}{\partial x})$). When taking the partial x derivative y is treated as a constant so, $( \frac{\partial}{\partial y}(\frac{\partial f(x,y)}{\partial x})= \frac{\partial}{\partial y}\frac{\partial}{\partial x}(ye^{yx}+y)= \frac{\partial}{\partial y}(y^2e^{yx}+0)$). Using the chain rule, $(\frac{\partial}{\partial y}(y^2e^{yx})=e^{yx}\frac{\partial}{\partial y}(y^2)+y^2\frac{\partial}{\partial y}(e^{yx})=e^{yx}(2y)+y^2(xe^{yx})$) so, $( \frac{\partial^2 f(x,y)}{\partial x\partial y}=e^{yx}(2y+y^2x)$)

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