Answer:
a) battery-->electrical current-->copper wire rotor -->magnet--> mechanical energy
Explanation:
We are given information:
m = 0.0450 kg
Δv = 25.2 m/s
Δt = 1.95 ms = 0.00195s
To find force we use formula:
F = m * a
a is acceleration. To find it we use formula:
a = Δv / Δt
a = 25.2 / 0.00195
a = 12923.1 m/s^2
Now we can find force:
F = 0.0450 * 12923.1
F = 581.5 N
To check the effect of the ball's weight on this movement we need to calculate it and then compare it to this force.
W = m * g
W = 0.0450 * 9.81
W = 0.44145 N
We can see that weight is much smaller than the applied force so it's influence in negligible.
Answer:
<h2> 145km</h2>
Explanation:
The displacement is a vector quantity, it tells how far away from a point a distance or a destination is
given that the distance covered are
50. km, 30. km, and 65 km
the displacement is expressed as
= 50+30+65
=145km
We actually performed straight addition because in all the movement the antarctic explorers did not record any deviation from the initial direction, hence they maintained a linear movement from the beginning to the end
Answer:
If R₂=25.78 ohm, then R₁=10.58 ohm
If R₂=10.57 then R₁=25.79 ohm
Explanation:
R₁ = Resistance of first resistor
R₂ = Resistance of second resistor
V = Voltage of battery = 12 V
I = Current = 0.33 A (series)
I = Current = 1.6 A (parallel)
In series
![\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2](https://tex.z-dn.net/?f=%5Ctext%7BEquivalent%20resistance%7D%3DR_%7Beq%7D%3DR_1%2BR_2%5C%5C%5Ctext%20%7BFrom%20Ohm%27s%20law%7D%5C%5CV%3DIR_%7Beq%7D%5C%5C%5CRightarrow%20R_%7Beq%7D%3D%5Cfrac%7B12%7D%7B0.33%7D%5C%5C%5CRightarrow%20R_1%2BR_2%3D36.36%5C%5C%20Also%5C%20R_1%3D36.36-R_2)
In parallel
![\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}](https://tex.z-dn.net/?f=%5Ctext%7BEquivalent%20resistance%7D%3D%5Cfrac%7B1%7D%7BR_%7Beq%7D%7D%3D%5Cfrac%7B1%7D%7BR_1%7D%2B%5Cfrac%7B1%7D%7BR_2%7D%5C%5C%5CRightarrow%20%7BR_%7Beq%7D%3D%5Cfrac%7BR_1R_2%7D%7BR_1%2BR_2%7D)
![\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0](https://tex.z-dn.net/?f=%5Ctext%20%7BFrom%20Ohm%27s%20law%7D%5C%5CV%3DIR_%7Beq%7D%5C%5C%5CRightarrow%20R_%7Beq%7D%3D%5Cfrac%7B12%7D%7B1.6%7D%5C%5C%5CRightarrow%20%5Cfrac%7BR_1R_2%7D%7BR_1%2BR_2%7D%3D7.5%5C%5C%5CRightarrow%20%5Cfrac%7BR_1R_2%7D%7B36.36%7D%3D7.5%5C%5C%5CRightarrow%20R_1R_2%3D272.72%5C%5C%5CRightarrow%2836.36-R_2%29R_2%3D272.72%5C%5C%5CRightarrow%20R_2%5E2-36.36R_2%2B272.72%3D0)
Solving the above quadratic equation
![\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}](https://tex.z-dn.net/?f=%5CRightarrow%20R_2%3D%5Cfrac%7B36.36%5Cpm%20%5Csqrt%7B36.36%5E2-4%5Ctimes%20272.72%7D%7D%7B2%7D)
![\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega](https://tex.z-dn.net/?f=%5CRightarrow%20R_2%3D25.78%5C%20or%5C%2010.57%5C%5C%20If%5C%20R_2%3D25.78%5C%20then%5C%20R_1%3D36.36-25.78%3D10.58%5C%20%5COmega%5C%5C%20If%5C%20R_2%3D10.57%5C%20then%5C%20R_1%3D36.36-10.57%3D25.79%5COmega)
∴ If R₂=25.78 ohm, then R₁=10.58 ohm
If R₂=10.57 then R₁=25.79 ohm