The second ball traveled a greater distance when compared to the first ball because the second ball spent more time in motion.
The given parameters;
- time of fall of the first ball, t = 1 s
- time of fall of the second ball, t = 3 s
The distance traveled by each ball is calculated using the second equation of motion as shown below.
The distance traveled by the first ball is calculated as follows;

The distance traveled by the second ball is calculated as follows;

Thus, the second ball traveled a greater distance because it spent more time in motion.
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Answer:
The awnser is d
Explanation:
i know cause i took the test
Answer:
1. 11 A
2. 240 V
3. 8 Ω
4. 60 C
5. 14400 C
Explanation:
1. Determination of the current.
Voltage (V) = 110 V
Resistance (R) = 10 Ω
Current (I) =?
V = IR
110 = I × 10
Divide both side by 10
I = 110 / 10
I = 11 A
2. Determination of the voltage
Current (I) = 3 A
Resistance (R) = 80 Ω
Voltage (V) =?
V = IR
V = 3 × 80
V = 240 V
3. Determination of the resistance.
Current (I) = 0.5 A
Voltage (V) = 4 V
Resistance (R) =?
V = IR
4 = 0.5 × R
Divide both side by 0.5
R = 4 / 0.5
R = 8 Ω
4. Determination of the charge
Current (I) = 2 A
Time (t) = 30 s
Charge (Q) =?
Q = it
Q = 2 × 30
Q = 60 C
5. Determination of the charge.
We'll begin by converting 20 mins to seconds. This can be obtained as follow:
1 min = 60 s
Therefore,
20 mins = 20 × 60
20 mins = 1200 s
Finally, we shall determine the charge as follow:
Current (I) = 12 A
Time (t) = 1200 s
Charge (Q) =?
Q = it
Q = 12 × 1200
Q = 14400 C
Answer:
C. The voltage drop across the resistor is 2.1V and nothing about the current through the resistor.
Explanation:
When connected in parallel, voltage across the resistances are the same. So if 2.1V was dropped across the LED then 2.1V was also dropped across the resistor. However, this tells us nothing about the current through the resistor. We can find the current across the resistor if we know the resistance of the resistor, but that's about it.
If it were a series connection, then the current would have been the same, but the voltage drop were another story.