Two physics students are arguing about superconductors and their discovery, Jeffe says that he can use a

(f) What do you mean by: (i) 1 J energy (ii) 1 W power

max2010maxim [7]

Explanation:

Watts are defined as 1 Watt = 1 Joule per second (1W = 1 J/s)

which means that 1 kW = 1000 J/s. A Watt is the amount of energy (in Joules) that an electrical device (such as a light) is burning per second that it's running.

3 0

3 years ago

How big is the image produced by the periscope compared to the size of the object

ipn [44]

As long as both mirrors are set at 45% and the same size then you see the same as is reflected in the upper mirror 

Put a lens in the middle of the tube 

? 

We use mirrors when we drive cars ect 

Normally they are set across from a concealed entrance or one that is hard to see both ways like the inside of a hairpin bend. Sometimes only to help in one direction. 

Sonar which is sound waves that are sent out at a set rate then reflected by objects. The longer the gap between the two the further away it is, They still use periscopes to target boats though. 

The periscope can only reflect what is outside so if you could see it because there is enough light then Yes. If you could not see it because it is dark then No unless you get into Info-Red light or Image Intensifying systems as well

8 0

3 years ago

Two particles are fixed to an x axis: particle 1 of charge q1 = 2.78 × 10-8 c at x = 15.0 cm and particle 2 of charge q2 = -3.24

Oksi-84 [34.3K]

Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
$E=\frac{1}{4\pi\varepsilon_0}\frac{q}{(x-x')^2}$
Now At xp we have:
$\frac{1}{4\pi\varepsilon_0}\frac{q_1}{(x_p-x_1)^2}-\frac{1}{4\pi\varepsilon_0}\frac{3.29q_1}{(x_p-x_2)^2}=0$
$\implies (x_p-x_1)^2=\frac{(x_p-x_2)^2}{3.29}\\
\implies(1-\frac{1}{3.29})x_p^2+2(\frac{x_2}{3.29}-x_1)x_p+x_1^2-\frac{x_2^2}{3.29}=0$
Which is a second order equation, using the quadratic formula to solve for xp would give us:
$xp=\frac{-(\frac{x_2}{3.29}-x_1)-\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
or
$xp=\frac{-(\frac{x_2}{3.29}-x_1)+\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that BOTH are correct. This is simply explained by considring the following.

Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.