Answer is B. ABAB. Hope it helped you, and have a great day.
-Charlie
(A) P(v) = 0.135v
(B) P(h) = 0.234v
<u>Explanation:</u>
Given-
Mass of the ball, m = 0.27kg
Force, F = 125N
angle of projection, θ = 30°
Let v be the velocity of the ball.
A) vertical component of the momentum of the volleyball
We know,
P(vertical) = mvsinθ
P(V) = 0.27 X v X sin 30°
P(V) = 0.27 X v X 0.5
P(V) = 0.135v
B) horizontal component of the momentum of the volleyball
We know,
P(Horizontal) = mvcosθ
P(h) = 0.27 X v X cos 30°
P(h) = 0.27 X v X 0.866
P(h) = 0.234v
Answer with Explanation:
We are given that
Area of loop=![(20\times 20) cm^2=400\times 10^{-4} m^2](https://tex.z-dn.net/?f=%2820%5Ctimes%2020%29%20cm%5E2%3D400%5Ctimes%2010%5E%7B-4%7D%20m%5E2)
![1 cm^2=10^{-4} m^2](https://tex.z-dn.net/?f=1%20cm%5E2%3D10%5E%7B-4%7D%20m%5E2)
Resistance, R=![0.1\Omega](https://tex.z-dn.net/?f=0.1%5COmega)
B=
We know that magnetic flux
![\phi=BA](https://tex.z-dn.net/?f=%5Cphi%3DBA)
Emf ,![E=\mid \frac{d\phi}{dt}\mid =\mid\frac{d(BA}{dt}\mid =\mid A\frac{dB}{dt}=400\times 10^{-4}\times \frac{4t-2t^2}{dt}\mid =\mid400\times 10^{-4}\times(4-4t)\mid](https://tex.z-dn.net/?f=E%3D%5Cmid%20%5Cfrac%7Bd%5Cphi%7D%7Bdt%7D%5Cmid%20%3D%5Cmid%5Cfrac%7Bd%28BA%7D%7Bdt%7D%5Cmid%20%3D%5Cmid%20A%5Cfrac%7BdB%7D%7Bdt%7D%3D400%5Ctimes%2010%5E%7B-4%7D%5Ctimes%20%5Cfrac%7B4t-2t%5E2%7D%7Bdt%7D%5Cmid%20%3D%5Cmid400%5Ctimes%2010%5E%7B-4%7D%5Ctimes%284-4t%29%5Cmid)
Current, ![I=\frac{E}{R}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7BE%7D%7BR%7D)
Current, ![I=\frac{\mid 400\times 10^{-4}(4-4t)\mid }{0.1}=1.6\mid (1-t)\mid](https://tex.z-dn.net/?f=I%3D%5Cfrac%7B%5Cmid%20400%5Ctimes%2010%5E%7B-4%7D%284-4t%29%5Cmid%20%7D%7B0.1%7D%3D1.6%5Cmid%20%281-t%29%5Cmid)
Substitute t=0 s
Then, I=
=1.6 A
Substitute t=1 s
Then, I=
=0
Substitute
t=2 s
Current, I=
=1.6 A
<span>Here is an example, the allele for blue eyes and the allele for brown eyes are different versions of the gene for eye color. Alleles are located at the same genetic locus . </span>
Answer:
=20 turns
Explanation:
The given case is a step down transformer as we need to reduce 120 V to 6 V.
number of turns on primary coil N_{P}= 400
current delivered by secondary coil I_{S}= 500 mA
output voltage = 6 V (rms)
we know that
![I_{p}=\frac{V_{out}}{V_{in}\times I_{s}}](https://tex.z-dn.net/?f=I_%7Bp%7D%3D%5Cfrac%7BV_%7Bout%7D%7D%7BV_%7Bin%7D%5Ctimes%20I_%7Bs%7D%7D)
putting values we get
![I_{p}=\frac{6}{120\times 0.5}](https://tex.z-dn.net/?f=I_%7Bp%7D%3D%5Cfrac%7B6%7D%7B120%5Ctimes%200.5%7D)
![I_{p}= 0.1 A](https://tex.z-dn.net/?f=I_%7Bp%7D%3D%200.1%20A)
to calculate number of turns in secondary
![\frac{N_{2}}{400} =\frac{6}{120}](https://tex.z-dn.net/?f=%5Cfrac%7BN_%7B2%7D%7D%7B400%7D%20%3D%5Cfrac%7B6%7D%7B120%7D)
therefore,
=20 turns