Answer:
a) 0.0130 m
b') w' = =6.46*10^{-3] m
Explanation:
given data:
\lambda of light = 633 nm
width of siit a =0.360 mm
distance from screen = 3.75 m
a) the first minima is located at

=


with of central fringe = 2y_1 = 2*6.54 *10^{-3} = 0.0130 m
b)
width of the first bright fringe on either side of the central one = 
calculation for y_2

= 



w' = =6.46*10^{-3] m
Answer:
Half
Explanation:
Given that:
- radial distance of satellite from the earth,

Now, if the satellite is moved to a distance 
<u>We have the mathematical expression for the potential energy fue to gravitational field as:</u>
...................(1)
where:

M = mass of earth
m = mass of satellite
R = radial distance of satellite
<u>Now from eq. (1) initially we have:</u>

<u>after the satellite is moved, we have:</u>



which is half of the initial condition.
The black squirrel has zero kinetic energy (if it's not moving) and lower gravitational potential energy than the red squirrel or zero gravitational potential energy if the ground is assumed to be zero gravitational potential line.
There are two particular cases, the first is when Object A is attracted to the neutral wall. This would indicate that the object is not neutral, as there is an attraction.
At the same time we know that Object A is attracted to an object B. And therefore, the load of A must be opposite to that of B. Remember that opposite charges attract each other. If the charge of object B is positive, then the charge of object A will be negative.
Option B is correct: It has a negative charge.
Answer:
B. The "Learner" was working with Milgram.
Explanation:
just took the test
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