Answer:
I = 0.2 A
Explanation:
Lamp is rated at 300 mA
I_lamp = 0.3 A
Voltage is; V = 3V
Thus; Resistance is given by;
R = V/I
R = 3/0.3
R = 10 ohms
Now, since the ammeter of 5 ohms is connected in series with the lamp. Thus equivalent resistance;
R_eq = 10 + 5
R_eq = 15 ohms
Ammeter current will be;
I = V/R_eq
I = 3/15
I = 0.2 A
Answer: energy source, path, and load
Explanation:
Answer:
No!
Explanation:
The correct answer is x = 9.
To start with solving this
problem, let us assume a launch angle of 45 degrees since that gives out the
maximum range for given initial speed. Also assuming that it was launched at
ground level since no initial height was given. Using g = 9.8 m/s^2, the
initial velocity is calculated using the formula:
(v sinθ)^2 = (v0 sinθ)^2
– 2 g d
where v is final
velocity = 0 at the peak, v0 is the initial velocity, d is distance = 11 m
Rearranging to find for
v0: <span>
v0 = sqrt (d * g/ sin(2 θ)) </span>
<span>v0 = 10.383 m/s</span>
