The strength of the gravitational forces between two masses depends on
-- the product of the masses,
-- the distance between their centers of mass.
Answer:
B) A planet's speed as it moves around the sun will not be the same in six months.
Explanation:
A planet's speed as it moves around the sun will not be the same in six months, is a statement that CANNOT be supported by Kepler's laws of planetary motion.
Answer:
F = 0i (in the x-direction), 0j (in the y-direction),-8.59*10^-4 N k (In the z-direction)
Explanation:
The force given by charged particles moving in a magnetic field is given below (cross is cross product, they don't have that format in the equation tool):
Now we can perform the cross product between v and B
d{array}\right][/tex]
Now multiply by Q (charge) to get the force
F = -8.59*10^-4 N k
F = 0i, 0j, (-8.59*10^-4)k
Well we know the correct answer cannot be "a" bcause velocity is tangent to the circlular path of an object experienting centripical motion. Velocity DOES NOT point inward in centripical motion.
we know the correct answer cannot be "b" because "t" stands for "time" which cannot point in any direction. so, time cannot point toward the center of a circle and therefore this answer must be incorrect.
I would choose answer choice "c" because both force and centripical acceleration point toward the center of the circle.
I do not think answer choice "d" can be correct because the velocity of the mass moves tangent to the circle. velocity = (change in position) / time. Therefore, by definition the mass is moving in the direction of the velocity which does not point to the center of the circle.
does this make sense? any questions?
In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
(1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit
In our problem,
while the distance between the first and the fifth minima is
(2)
If we use the formula to rewrite
, eq.(2) becomes
Which we can solve to find a, the width of the slit: