Answer:
Explanation:
side of the square loop, a = 7 cm
distance of the nearest side from long wire, r = 2 cm = 0.02 m
di/dt = 9 A/s
Integrate on both the sides

i = 9t
(a) The magnetic field due to the current carrying wire at a distance r is given by


(b)
Magnetic flux,





(c)
R = 3 ohm

magnitude of voltage is
e = 1.89 x 10^-7 V
induced current, i = e / R = (1.89 x 10^-7) / 3
i = 6.3 x 10^-8 A
I think the answer would be: The G-note's wavelength is longer
Here are the formula to calculate wavelength
Wavelength = Wave speed/Frequency
Which indicates that the wavelength will become larger as the frequency became smaller.
No, not exactly. They jiggle and tremble and vibrate a lot, but
they always basically stay in very nearly the same place.
It's like if you're allowed to go anywhere you want in your jail cell,
you wouldn't exactly call that "moving about freely".
<u>First Symbol </u>: Cobalt (Co)
Its Group Number - 9
Its Period Number - 4
Its Family Name - Transition Metal
<u>Second Symbol</u> : Silicon (Si)
Its Group Number - 14
Its Period Number - 2
Its Family Name - Semiconductor
<u>Third Symbol</u> : Astatine (At)
Its Group Number - 17
Its Period Number - 6
Its Family Name - Halogen
<u>Fourth Symbol </u>: Magnesium (Mg)
Its Group Number - 2
Its Period Number - 3
Its Family Name - Alkaline Earth Metal
<u>Fifth Symbol</u> : Xenon (Xe)
Its Group Number - 18
Its Period Number - 5
Its Family Name - Noble Gas
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y =
t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45
)
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² =
- 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct