Speed of wave is given as
![v = 5 m/s](https://tex.z-dn.net/?f=v%20%3D%205%20m%2Fs)
Wavelength of the wave is given as
![\lambda = 20 m](https://tex.z-dn.net/?f=%5Clambda%20%3D%2020%20m)
now from the formula of wave time period we can say
![speed = \frac{wavelength}{time period}](https://tex.z-dn.net/?f=speed%20%3D%20%5Cfrac%7Bwavelength%7D%7Btime%20period%7D)
![5 = \frac{20}{T}](https://tex.z-dn.net/?f=5%20%3D%20%5Cfrac%7B20%7D%7BT%7D)
![T = \frac{20}{5}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B20%7D%7B5%7D)
![T = 4 s](https://tex.z-dn.net/?f=T%20%3D%204%20s)
so it will have time period of T = 4 s
Answer
given,
resistance = 0.05 Ω
internal resistance of battery = 0.01 Ω
electromotive force = 12 V
a) ohm's law
V = IR
and volage
now,
![I(R+r) = \epsilon](https://tex.z-dn.net/?f=I%28R%2Br%29%20%3D%20%5Cepsilon)
inserting the values
I = 200 A
b) Voltage
V = I R
V = 200 x 0.05
V = 10 V
c) Power
P = I V
P = 200 x 10 = 2000 W
d) total resistance = 0.05 + 0.09 = 0.14 Ω
I = 80 A
V = 80 x 0.05 = 4 V
P = 4 x 80 = 320 W
Answer:
Since the area of the perfect square is 11650, and all of a squares sides ar equal, we just need to find the square root.
The square root of 11650 is 107.935166.
One side of the square is 107.935166
107.935166 x 107.935166 = 11650
(っ◔◡◔)っ ♥ Hope It Helps ♥
Answer:
F₁ = 4 F₀
Explanation:
The force applied on the string by the ball attached to it, while in circular motion will be equal to the centripetal force. Therefore, at time t₀, the force on ball F₀ is given as:
F₀ = mv₀²/r --------------- equation (1)
where,
F₀ = Force on string at t₀
m = mass of ball
v₀ = speed of ball at t₀
r = radius of circular path
Now, at time t₁:
v₁ = 2v₀
F₁ = mv₁²/r
F₁ = m(2v₀)²/r
F₁ = 4 mv₀²/r
using equation (1):
<u>F₁ = 4 F₀</u>
Since my givens are x = .550m [Vsub0] = unknown
[Asubx] = =9.80
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0])
Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in
0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)
0 = [Vsub0x]^2 -10.78
10.78 = [Vsub0x]^2
Sqrt(10.78) = 3.28 m/s