The distance between slit and the screen is 1.214m.
To find the answer, we have to know about the width of the central maximum.
<h3>How to find the distance between slit and the screen?</h3>
- It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.
- We have the expression for slit width w as,
where, d is the distance between slit and the screen, and a is the slit width.
- Thus, distance between slit and the screen is,
Thus, we can conclude that, the distance between slit and the screen is 1.214m.
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Answer:
-4.0 N
Explanation:
Since the force of friction is the only force acting on the box, according to Newton's second law its magnitude must be equal to the product between mass (m) and acceleration (a):
(1)
We can find the mass of the box from its weight: in fact, since the weight is W = 50.0 N, its mass will be
And we can fidn the acceleration by using the formula:
where
v = 0 is the final velocity
u = 1.75 m/s is the initial velocity
t = 2.25 s is the time the box needs to stop
Substituting, we find
(the acceleration is negative since it is opposite to the motion, so it is a deceleration)
Therefore, substituting into eq.(1) we find the force of friction:
Where the negative sign means the direction of the force is opposite to the motion of the box.
Answer:
Explanation:
gravitational acceleration of meteoroid
= GM / R²
M is mass of planet , R is radius of orbit of meteoroid from the Centre of the planet .
R = (.9 x 6370 + 600 )x 10³ m
= 6333 x 10³ m
M , mass of the planet = 5.97 x 10²⁴ kg .
gravitational acceleration of meteoroid
= GM / R²
= (6.67 x 10⁻¹¹ x 5.97 x 10²⁴ kg / (6333 x 10³ m)²
9.92m/s²
Answer: Hot spot
The volcanic hot spot is an area in the mantle from which heat rises as a thermal plume from deep in the Earth.
Answer:
<em>B) 1.0 × 10^5 V</em>
Explanation:
<u>Electric Potential Due To Point Charges
</u>
The electric potential produced from a point charge Q at a distance r from the charge is
The total electric potential for a system of point charges is equal to the sum of their individual potentials. This is a scalar sum, so direction is not relevant.
We must compute the total electric potential in the center of the square. We need to know the distance from all the corners to the center. The diagonal of the square is
where a is the length of the side.
The distance from any corner to the center is half the diagonal, thus
The total potential is
Where V1 and V2 are produced by the +4\mu C charges and V3 and V4 are produced by the two opposite charges of 
. Since all the distances are equal, and the charges producing V3 and V4 are opposite, V3 and V4 cancel each other. We only need to compute V1 or V2, since they are equal, but they won't cancel.
The total potential is
