A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the grou
nd. What one of the following statements is true if air resistance is neglected? The acceleration of the rock is zero when it is at the highest point. The speed of the rock is negative while it falls toward the ground. As the rock rises, its acceleration vector points upward. At the highest point the velocity is zero, the acceleration is directed downward. The velocity and acceleration of the rock always point in the same direction.
1 answer:
Answer:
At the highest point the velocity is zero, the acceleration is directed downward.
Explanation:
This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.
I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.
At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.
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<h2>Answer:</h2>
(a) 3.18Ω
(b) 3.18Ω
<h2>Explanation:</h2>Let the two bulbs be A and B
Given;
= Resistance in bulb A = 3.0Ω
= Resistance in bulb B = 2.5Ω
Since the two bulbs are connected in parallel;
i. their effective resistance () is given by
=
+
---------------(i)
Substitute the values of and
into equation (i)
=> =
+
Solve for
= 1.36Ω
ii. voltage (potential difference), V, across them is the same;
Therefore we can get the total current (I) that will flow through them if the voltage to be supplied is 2.4V.
Use the Ohm's law;
V = I x R -----------------(ii)
Where;
V = voltage across them = 2.4V
I = total current flowing through them
R = their effective resistance = = 1.36Ω
Substitute these values into equation (ii) as follows;
2.4 = I x 1.36
I = 2.4 / 1.36
I = 1.76A
(a) Now get the value of R
Since the voltage across the two bulbs is 2.4V out of the 8.0V supplied by the source, then the remaining (8.0 - 2.4 = 5.6)V will pass across the resistor R.
Also, since the two bulbs make a series connection with the resistor R, the same total current (I = 1.76A) that flows through these bulbs will flow through the resistor R.
Therefore, to get the value of R, we use the relation
V = I x R ------------------------------(iii)
Where;
V = voltage across the resistor = 5.6V
I = current through the resistor = 1.76A
<em>Substitute these values into equation (iii)</em>
=> 5.6 = 1.76 x R
=> R = 5.6 / 1.76
=> R = 3.18Ω
Therefore, the value of R to be chosen in order to supply each bulb with a voltage of 2.4V is 3.18Ω
(b) The potential difference and voltage across refer to the same thing. Therefore, the value of R that would make the potential difference across each of the bulbs be 2.4V is the same as the one calculated in (a) which is 3.18Ω