Well you need to have lots of heat
Answer:
the number of turns in the primary coil is 13000
Explanation:
Given the data in the question;
V₁ = 13000 V
V₂ = 120 V
N₁ = ?
N₂ = 120 turns
the relation between the voltages and the number of turns in the primary and secondary coils can be expressed as;
V₁/V₂ = N₁/N₂
V₁N₂ = V₂N₁
N₁ = V₁N₂ / V₂
so we substitute
N₁ = (13000 V × 120 turns) / 120 V
N₁ = 1560000 V-turns / 120 V
N₁ = 13000 turns
Therefore, the number of turns in the primary coil is 13000
Answer:
(a) 0.42 m
(b) 20.16 N/m
(c) - 0.42 m
(d) - 0.21 m
(e) 17.3 s
Solution:
As per the question:
Mass, m = 0.56 kg
x(t) = (0.42 m)cos[cos(6 rad/s)t]
Now,
The general eqn is:
where
A = Amplitude
= angular frequency
t = time
Now, on comparing the given eqn with the general eqn:
(a) The amplitude of oscillation:
A = 0.42 m
(b) Spring constant k is given by:
Thus
(c) Position after one half period:
(d) After one third of the period:
(e) Time taken to get at x = - 0.10 m:
t = 17.3 s
Its potential energy decreases and kinetic energy.
Answer:
20.1 m
Explanation:
First, find the time it takes for the cannonball to travel the horizontal distance of 50.0 m.
Given (in the x direction):
Δx = 50.0 m
v₀ = 68 cos 25° m/s
a = 0 m/s²
Find: t
Δx = v₀ t + ½ at²
(50.0 m) = (68 cos 25° m/s) t + ½ (0 m/s²) t²
t = 0.811 s
Now find the vertical displacement after that time.
Given (in the y direction):
v₀ = 68 sin 25° m/s
a = -9.8 m/s²
t = 0.811 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (68 sin 25°) (0.811 s) + ½ (-9.8 m/s²) (0.811 s)²
Δy = 20.1 m
