A ball is launched directly in the air at a speed of 40 feet per second from a platform located 10 feet in the air. The motion o
1 answer:
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Answer:
1)31/3Ω
2)18/31A
3)4.06V
Explanation:
According to the diagram and 10 Ω resistors are parallel to each other, parallel resistor can be calculated using the formula below
1/R= 1/R1 + 1/R2
But we know R1= 5 Ω and R2= 10 Ω
1/R= 1/5 + 1/10
R= 10/3 Ω
=3.33 Ω
Then if we follow the given figure, 10/3 Ω and 7Ω are now in series then
Req = 10/3 +7
Req= 31/3 Ω
Therefore, equivalent resistance = 31/3 Ω
According to ohms law we know that V= IR
Then I= V/R
Where I= current
R= resistance
V= voltage
I= 6/(31/3)
I= 18/31A
We can now calculate the voltage accross the resistor which is
V=(18/31)× 7
V=4.06V
Therefore, the voltage accross the 7 ohm resistor is 4.06V
CHECK THE FIQURE AT THE ATTACHMENT
Answer:
40 ms¯².
Explanation:
To solve this problem, we shall illustrate the question with a diagram.
The attached photo gives a better understanding of the question.
From the attached photo:
Velocity (v) = 160 ms¯¹
Time (t) = 4 secs.
Acceleration (a) =?
Acceleration (a) = Velocity (v) /time (t)
a = v/t
a = 160/4
a = 40 ms¯²
Therefore, the initial acceleration of the rocket is 40 ms¯².
