Answer:
a. 2 Hz b. 0.5 cycles c . 0 V
Explanation:
a. What is period of armature?
Since it takes the armature 30 seconds to complete 60 cycles, and frequency f = number of cycles/ time = 60 cycles/ 30 s = 2 cycles/ s = 2 Hz
b. How many cycles are completed in T/2 sec?
The period, T = 1/f = 1/2 Hz = 0.5 s.
So, it takes 0.5 s to complete 1 cycles. At t = T/2 = 0.5/2 = 0.25 s,
Since it takes 0.5 s to complete 1 cycle, then the number of cycles it completes in 0.25 s is 0.25/0.5 = 0.5 cycles.
c. What is the maximum emf produced when the armature completes 180° rotation?
Since the emf E = E₀sinθ and when θ = 180°, sinθ = sin180° = 0
E = E₀ × 0 = 0
E = 0
So, at 180° rotation, the maximum emf produced is 0 V.
On the list of choices that you provided, there is no such statement.
Answer:
R = 5.28 103 km
Explanation:
The definition of density is
ρ = m / V
V = m /ρ
Where m is the mass and V the volume of the body
The volume of a sphere is
V = 4/3 π r³
Let's replace
4/3 π r³ = m / ρ
R =∛ ¾ m / ρ π
The mass of the planet is
M = 5.5 Me
R = ∛ ¾ 5.5 Me /ρ π
Let's reduce the density to SI units
ρ = 1.76 g / cm³ (1 kg / 10³ g) (10² cm / 1 m)³
ρ = 1.76 10³ kg / m³
Let's calculate
R = ∛ ¾ 5.5 5.97 10²⁴ / (1.76 10³ pi)
R = ∛ 0.14723 10²¹
R = 0.528 10⁷ m
R = 0.528 104 km
R = 5.28 103 km
Answer:
2) c) give-way vessel
3) a) With one short blast
Explanation:
2) A vessel that is required to take early substantial action to ensure avoiding collision called Give way vessel
In overtaking, the vessel intending to overtake is the Give-Way Vessel the vessel that is going to be overtaken is the Stand-On Vessel
Therefore, the correct option is c) give-way vessel
3) When vessels use sound signals in a meeting head on situation both vessel are Give-Way vessels and both vessel pass the each other by turning to the starboard side therefore they intend to pass each other on their port side requiring one short blast
Therefore, the correct option is a) With one short blast.
The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N
the slow masses that must be quicker are the pulley, ring, and the rolling sphere.
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
