Answer: hello options related to your question is missing attached below is the missing part of your question
answer: No charge of the length of the bonds expected because the rod did not touch the charge source ( option A )
Explanation:
When the Charge is first, Furthest away and second and closest to the source charge. <em>The spring like bonds can be said to have No charge of the length of the bonds expected because the rod did not touch the charge source </em><em>when Furthest away the bond with charge will be less effective </em>
<span>b) The force with a distance of 150 km is 889 N
c) The force with a distance of 50 km is 8000 N
This question looks like a mixture of a question and a critique of a previous answer. I'll attempt to address the original question.
Since the radius of the spherical objects isn't mentioned anywhere, I will assume that the distance from the center of each spherical object is what's being given. The gravitational force between two masses is given as
F = (G M1 M2)/r^2
where
F = Force
G = gravitational constant
M1 = Mass 1
M2 = Mass 2
r = distance between center of masses for the two masses.
So with a r value of 100 km, we have a force of 2000 Newtons. If we change the distance to 150 km, that increases the distance by a factor of 1.5 and since the force varies with the inverse square, we get the original force divided by 2.25. And 2000 / 2.25 = 888.88888.... when rounded to 3 digits gives us 889.
Looking at what looks like an answer of 890 in the question is explainable as someone rounding incorrectly to 2 significant digits.
If the distance is changed to 50 km from the original 100 km, then you have half the distance (50/100 = 0.5) and the squaring will give you a new divisor of 0.25, and 2000 / 0.25 = 8000. So the force increases to 8000 Newtons.</span>
Speed = (acceleration) x (time)
Velocity = (speed) in (direction of the speed)
Speed = (-3 m/s²) x (5 s) = 15 m/s
Velocity =
(15 m/s) in the direction opposite to the direction you call positive.
Displacement = (distance between start-point and end-point)
in the direction from start-point to end-point.
Distance = (1/2) (acceleration) (time)²
Distance = (1/2) (3 m/s²) (5 s)²
= (1/2) (3 m/s²) (25 s²) = 37.5 meters
Displacement =
37.5 meters in the direction opposite to the direction you call positive.
Answer:
d = 68.5 x 10⁻⁶ m = 68.5 μm
Explanation:
The complete question is as follows:
An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is 1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?
The answer can be given by using the formula derived from Young's Double Slit Experiment:
where,
d = slit separation = ?
λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m
L = distance from screen (detector) = 1.7 m
y = distance between bright fringes = 15.7 mm = 0.0157 m
Therefore,
<u>d = 68.5 x 10⁻⁶ m = 68.5 μm</u>
