Answer:
A) E = 0N/C
B) 0i + 0^^j
C) F = 0N
D) 0^i + 0^j
Explanation:
You assume that the rings are in the zy plane but in different positions.
Furthermore, you can consider that the origin of coordinates is at the midway between the rings.
A) In order to calculate the magnitude of the electric field at the middle of the rings, you take into account that the electric field produced by each ring at the origin is opposite to each other and parallel to the x axis.
You use the following formula for the electric field produced by a charge ring at a perpendicular distance of r:
(1)
k: Coulomb's constant = 8.98*10^9Nm^2/C
Q: charge of the ring
r: perpendicular distance to the center of the ring
R: radius of the ring
You use the equation (1) to calculate the net electric field at the midpoint between the rings:
The electric field produced by each ring has the same magnitude but opposite direction, then, the net electric field is zero.
B) The direction of the electric field is 0^i + 0^j
C) The magnitude of the force on a proton at the midpoint between the rings is:
D) The direction of the force is 0^i + 0^j
-- We know that the y-component of acceleration is the derivative of the
y-component of velocity.
-- We know that the y-component of velocity is the derivative of the
y-component of position.
-- We're given the y-component of position as a function of time.
So, finding the velocity and acceleration is simply a matter of differentiating
the position function ... twice.
Now, the position function may look big and ugly in the picture. But with the
exception of 't' , everything else in the formula is constants, so we don't even
need any fancy processes of differentiation. The toughest part of this is going
to be trying to write it out, given the text-formatting capabilities of the wonderful
envelope-pushing website we're working on here.
From the picture . . . . . y (t) = (1/2) (a₀ - g) t² - (a₀ / 30t₀⁴ ) t⁶
First derivative . . . y' (t) = (a₀ - g) t - 6 (a₀ / 30t₀⁴ ) t⁵ = (a₀ - g) t - (a₀ / 5t₀⁴ ) t⁵
There's your velocity . . . /\ .
Second derivative . . . y'' (t) = (a₀ - g) - 5 (a₀ / 5t₀⁴ ) t⁴ = (a₀ - g) - (a₀ /t₀⁴ ) t⁴
and there's your acceleration . . . /\ .
That's the one you're supposed to graph.
a₀ is the acceleration due to the model rocket engine thrust
combined with the mass of the model rocket
'g' is the acceleration of gravity ... 9.8 m/s² or 32.2 ft/sec²
t₀ is how long the model rocket engine burns
Pick, or look up, some reasonable figures for a₀ and t₀
and you're in business.
The big name in model rocketry is Estes. Their website will give you
all the real numbers for thrust and burn-time of their engines, if you
want to follow it that far.
Here are the parts of the comet:
1. NUCLEUS: This is the frozen part of the comet. It is also known as the core. It is made up of ice and dust which are completely covered by organic matter. The nucleus usually consist of frozen water but other materials that are in frozen forms can be found in it. Comet nuclei are usually less than 16 kilometer in diameter.
2. COMA: The atmosphere of dust and gases formed when the nucleus vaporize. The coma refers to the envelope of gases that surround the comet's nucleus. The coma plus the nucleus forms the head of the comet. The coma is about a million kilometer in diameter and is made up of gases and dust which sublime from the comet's nucleus.
3. ION TAIL: Tail made of ions that appear to point away from the comet's orbit. The charged solar particles convert the gases found in the comet to ions thus forming an ion tail. The ion tail can measure over 100 million kilometer long and it accelerate much faster than the dust tail.
4. DUST TAIL: Tail made up of small solid dust particles. It is formed by radiation from the sun, which forces dust particles away from the coma. It usually point away from the sun because the tail are shaped by the solar wind. As the distance from the sun increases, the dust tail usually become faint and diminished.
Answer:
The acceleration of the elevator machine is, a = 3 m/s²
Explanation:
Given data,
The mass on the one end, m = 5 kg
The mass on the other end, M = 10 kg
According to the Atwood's machine
Ma = Mg - T
ma = T - mg
Adding those equations,
a (M + m) = g ( M - m)
a = (M + m) / ( M - m)
Substituting the values,
a = (10 + 5) / (10 - 5)
= 3 m/s²
The acceleration of the elevator machine is, a = 3 m/s²
Answer: it is rocket 2
Explanation: I just took the test
Hope it's the right answer for you
