A) It would be doubled.
Why?
To answer the question, we just need to calculate the momentum of the basketball using the following formula:
Now, we have calculated the momentum and the result is 12 kg.m/s, what would happen to the velocity if we double the momentum? Let's calculate it!
Hence, we can see that if the momentum is doubled, the velocity will be doubled too.
Have a nice day!
(a) <u>i₁ = 0.03818 A = 38.18 mA</u>
(b) downward
(c) <u>i₂ = 0.01091 A = 10.91 mA</u>
(d) rightward
(e) <u>i₃ = 0.02727 A = 27.27 mA</u>
(f) leftward
(g) <u>Eₐ = 3.818 Volts</u>
<h3>Question:</h3>The complete question is stated below and the figure is provided in the attachment:
In Fig. 27-47, E 1 = 6.00 V, E 2 = 12.0 V, R1 = 100 Ω,
R2 = 200 Ω, and R3 = 300 Ω. One point of the circuit is grounded
(V = 0). What are the (a) size and (b) direction (up or down) of the
current through resistance 1, the (c) size and (d) direction
(left or right) of the current through resistance 2, and the
(e) size and (f) direction of the current through resistance 3?
(g) What is the electric potential at point A?
<h3></h3><h3>Explanation:</h3>Applying Kirchoff's voltage law in the loops of botg E₁ and E₂, in the clockwise and anti clockwise direction:
E₁ - i₂R₂ - i₁R₁ = 0
E₂ - i₃R₃ - i₁R₁ = 0
If, we apply Kirchhoff's current law at junction A, we get:
i₁ = i₂ + i₃
Using these relations in loop equations, and re-arranging:
E₁ - i₂R₂ - (i₂ + i₃) R₁ = 0 ___________ eqn (1)
E₂ - i₃R₃ - (i₂ + i₃) R₁ = 0 ___________ eqn (2)
Eqn (1) implies:
6 - 200 i₂ - 100 i₂ - 100 i₃ = 0
i₂ = (6 - 100i₃)/300
Eqn (2) implies:
12 - 300 i₃ - 100 i₂ - 100 i₃ = 0
12 - 400 i₃ = 100 i₂
using value of i₃ from eqn (1)
12 - 400 i₃ = (1/3)(6 - 100 i₃)
36 - 1200 i₃ = 6 - 100 i₃
1100 i₃ = 30
<u>i₃ = 0.02727 A</u>
using this value in eqn of i₂:
i₂ = [6 - 100(0.02727)]/300
i₂ = (6 - 2.727)/300
<u>i₂ = 0.01091 A</u>
Since:
i₁ = i₂ + i₃
i₁ = 0.01091 A + 0.02727 A
<u>i₁ = 0.03818 A</u>
<u></u>
(a)
<u>i₁ = 0.03818 A = 38.18 mA</u>
(b)
Since, the value of current is positive, thus it will have the direction that was assumed.
Therefore, its direction will <u>downward</u>
(c)
<u>i₂ = 0.01091 A = 10.91 mA</u>
(d)
Since, the value of current is positive, thus it will have the direction that was assumed.
Therefore, its direction will <u>rightward</u>
(e)
<u>i₃ = 0.02727 A = 27.27 mA</u>
(f)
Since, the value of current is positive, thus it will have the direction that was assumed.
Therefore, its direction will <u>leftward</u>
<u>(g)</u>
With respect to the grounded portion, the potential drop at the resistance 1 will be equal to the potential at A Eₐ.
Therefore,
Eₐ = i₁R₁
Eₐ = (0.03818 A)(100 Ω)
<u>Eₐ = 3.818 Volts</u>
Answer:
(a) check attachment
(b)5 m/s²
Explanation:
Given: radius = 2.00mm: density = 2500kg/m³: viscosity of glycerin = 1.5pa: decity of glycerin = 1250kg/m³: g = 10N/kg = 10m/s²: Fdrag = 6πnrv
(a) for answer check attachment.
(b) For the magnitude of the balls initial acceleration:
Initial net force(f) = mg - upthrust
=
acceleration (a) =
c.) fromthe force diagram in the attachment; when the ball attains terminal velocity the net force will be zero(0)
d.) For the magnitude of terminal velocity:
e.) when the ball reaches terminal velocity, the acceleration is zero (0)
