Answer:
Δt = 5.85 s
Explanation:
For this exercise let's use Faraday's Law
emf = 
- d fi / dt
= B. A
\phi = B A cos θ
The bold are vectors. It indicates that the area of the body is A = 0.046 m², the magnetic field B = 1.4 T, also iindicate that the normal to the area is parallel to the field, therefore the angle θ = 0 and cos 0 =1.
suppose a linear change of the magnetic field
emf = - A 
Dt = - A 
the final field before a fault is zero
let's calculate
Δt = - 0.046 (0- 1.4) / 0.011
Δt = 5.85 s
Answer:
Any choices for the blank?
Answer:
H = 45 m
Explanation:
First we find the launch velocity of the ball by using the following formula:
v₀ = √(v₀ₓ² + v₀y²)
where,
v₀ = launching velocity = ?
v₀ₓ = Horizontal Component of Launch Velocity = 15 m/s
v₀y = Vertical Component of Launch Velocity = 30 m/s
Therefore,
v₀ = √[(15 m/s)² + (30 m/s)²]
v₀ = 33.54 m/s
Now, we find the launch angle of the ball by using the following formula:
θ = tan⁻¹ (v₀y/v₀ₓ)
θ = tan⁻¹ (30/15)
θ = tan⁻¹ (2)
θ = 63.43°
Now, the maximum height attained by the ball is given by the formula:
H = (v₀² Sin² θ)/2g
H = (33.54 m/s)² (Sin² 63.43°)/2(10 m/s²)
<u>H = 45 m</u>
Answer:
ΔL = 0.66 m
Explanation:
The change in length on an object due to rise in temperature is given by the following equation of linear thermal expansion:
ΔL = αLΔT
where,
ΔL = Change in Length of the bridge = ?
α = Coefficient of linear thermal expansion = 11 x 10⁻⁶ °C⁻¹
L = Original Length of the Bridge = 1000 m
ΔT = Change in Temperature = Final Temperature - Initial Temperature
ΔT = 40°C - (-20°C) = 60°C
Therefore,
ΔL = (11 x 10⁻⁶ °C⁻¹)(1000 m)(60°C)
<u>ΔL = 0.66 m</u>
