Answer:
<h3>The 28 loops wound on the square armature</h3>
Explanation:
Peak output voltage 
V
Area of square armature 
Magnetic field 
T
Angular frequency 
According to the law of electromagnetic induction,
Where 
number of loops of wire.
≅ 28
Thus, 28 loops of wire should be wound on the square armature.
Answer:
11025 N / m²
Explanation:
Los siguientes datos se obtuvieron de la pregunta:
Área (A) = 400 cm²
Masa (m) = 45 Kg
Aceleración por gravedad (g) = 9,8 m / s²
Presión (P) =?
A continuación, determinaremos la fuerza aplicada. Esto se puede obtener de la siguiente manera:
Masa (m) = 45 Kg
Aceleración por gravedad (g) = 9,8 m / s²
Fuerza (F) =.?
F = m × g
F = 45 × 9,8
F = 441 N
A continuación, convertiremos 400 cm² a m². Esto se puede obtener de la siguiente manera:
1 cm² = 0,0001 m²
Por lo tanto,
400 cm² = 400 cm² × 0,0001 m² / 1 cm²
400 cm² = 0,04 m²
Por tanto, 400 cm² equivalen a 0,04 m².
Finalmente, determinaremos la presión ejercida de la siguiente manera:
Área (A) = 0.04 m².
Fuerza (F) = 441 N
Presión (P) =?
P = F / A
P = 441 / 0,04
P = 11025 N / m²
Por tanto, la presión ejercida es 11025 M / m²
Answer:
a) a = 3.06 10¹⁵ m / s
, b) F= 1.43 10⁻¹⁰ N, c) F_total = 14.32 10⁻²⁶ N
Explanation:
This exercise will average solve using the moment relationship.
a ) let's use the relationship between momentum and momentum
I = ∫ F dt = Δp
F t = m 
- m v₀
F = m (v_{f} -v₀o) / t
in the exercise indicates that the speed module is the same, but in the opposite direction
F = m (-2v) / t
if we use Newton's second law
F = m a
we substitute
- 2 mv / t = m a
a = - 2 v / t
let's calculate
a = - 2 4.59 10²/3 10⁻¹³
a = 3.06 10¹⁵ m / s
b) F= m a
F= 4.68 10⁻²⁶ 3.06 10¹⁵
F= 1.43 10⁻¹⁰ N
c) if we hit the wall for 1015 each exerts a force F
F_total = n F
F_total = n m a
F_total = 10¹⁵ 4.68 10⁻²⁶ 3.06 10¹⁵
F_total = 14.32 10⁻²⁶ N
Answer:
the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal
Explanation:
initial veetical speed V₀y=0
Horizontal speed Vx = Vx₀= 3.80m/s
Vertical drop height= 3.90m
Let Vy = vertical speed when it got to the water downward.
g= 9.81m/s² = acceleration due to gravity
From kinematics equation of motion for vertical drop
Vy²= V₀y² +2 gh
Vy²= 0 + ( 2× 9.8 × 3.90)
Vy= √76.518
Vy=8.747457
Then we can calculate the velocity of the fish relative to the water when it hits the water using Resultant speed formula below
V= √Vy² + Vx²
V=√3.80² + 8.747457²
V=9.537m/s
The angle can also be calculated as
θ=tan⁻¹(Vy/Vx)
tan⁻¹( 8.747457/3.80)
=66.52⁰
the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal
