if a toaster produces 8 ohms of resistance in a 120 v circuit, what is the amount of current in the circuit
1 answer:
You might be interested in
Answer:
(a)
(b) 5220 j
(c) 1740 watt
(d) 3446.66 watt
Explanation:
We have given mass m = 290 kg
Initial velocity u = 0 m/sec
Final velocity v = 6 m/sec
Time t = 3 sec
From first equation of motion
v = u+at
So
(a) We know that force is given by
F = ma
So force will be
(b) From second equation of motion we know that
We know that work done is given by
W = F s = 580×9 =5220 j
(c) Time is given as t = 3 sec
We know that power is given as
(d) Time t = 1.5 sec
So
Refer to the diagram shown below.
We want to find V, the initial launch velocity of a projectile, without knowing θ, the launch angle.
To find V, we need to perform an experiment to determine:
d, the horizontal distance traveled,
t, the time taken to travel the horizontal distance.
Assume that aerodynamic resistance is negligible.
The horizontal component of the velocity is
Vx = V cosθ
Because the horizontal distance, d, is traveled in time, t, therefore
(V cosθ)*t = d
Vcosθ = d/t (1)
Assume that ground level has height zero.
Note that g, the acceleration due to gravity is known.
The vertical travel between the time of launch and return to ground level obeys the equation
Vsinθ*t - (g/2)*t² = 0
Therefore
t[Vsinθ - (gt)/2] = 0
Obtain
t = 0, which corresponds to launch
or
t = (2Vsinθ)/g
That is,
Vsinθ = (gt)/2 (2)
From (1) and (2), obtain
(Vcosθ)² + (Vsinθ)² = (d/t)² + (g²t²)/4
V²(cos² + sin² ) = (d/t)² + (g²t²)/4
Because g,t and d are known, V can be calculated.
We want to find V, the initial launch velocity of a projectile, without knowing θ, the launch angle.
To find V, we need to perform an experiment to determine:
d, the horizontal distance traveled,
t, the time taken to travel the horizontal distance.
Assume that aerodynamic resistance is negligible.
The horizontal component of the velocity is
Vx = V cosθ
Because the horizontal distance, d, is traveled in time, t, therefore
(V cosθ)*t = d
Vcosθ = d/t (1)
Assume that ground level has height zero.
Note that g, the acceleration due to gravity is known.
The vertical travel between the time of launch and return to ground level obeys the equation
Vsinθ*t - (g/2)*t² = 0
Therefore
t[Vsinθ - (gt)/2] = 0
Obtain
t = 0, which corresponds to launch
or
t = (2Vsinθ)/g
That is,
Vsinθ = (gt)/2 (2)
From (1) and (2), obtain
(Vcosθ)² + (Vsinθ)² = (d/t)² + (g²t²)/4
V²(cos² + sin² ) = (d/t)² + (g²t²)/4
Because g,t and d are known, V can be calculated.