Answer:both diagrams are attached.
Explanation:
Answer:
The rocket has to be launched 8 m from the hoop
Explanation:
Let's analyze this problem, the rocket is on a car that moves horizontally, so the rocket also has the same speed as the car; The initial horizontal rocket speed is (v₀ₓ = 3.0 m/s).
On the other hand, when starting the engines we have a vertical force, which creates an acceleration in the vertical axis, let's use Newton's second law to find this vertical acceleration
F -W = m a
a = (F-mg) / m
a = F/m -g
a = 7.0/0.500 - 9.8
a = 4.2 m/s²
We see that we have a positive acceleration and that is what we are going to use in the parabolic motion equations
Let's look for the time it takes for the rocket to reach the height (y = 15m) of the hoop, when the rocket fires its initial vertical velocity is zero (I'm going = 0)
y =
t + ½ a t²
y = 0 + ½ a t²
t = √ 2y/a
t = √( 2 15 / 4.2)
t = 2.67 s
This time is also the one that takes in the horizontal movement, let's calculate how far it travels
x = v₀ₓ t
x = 3 2.67
x = 8 m
The rocket has to be launched 8 m from the hoop
Answer:
rpm= 916.7436 rev/min
Explanation:
First determine the perimeter of the wheel, to know the horizontal distance it travels in a revolution:
perimeter= π×diameter= π × 22 inches × 0.0254(m/inche)= 1.7555m
Time we divide the speed of the car, which is the distance traveled horizontally over time unit, by the perimeter of the wheel that is the horizontal distance traveled in a revolution, this dates us the revolutions over the time unit:
revolutions per time= velocity/perimeter
velocity= (60 mi/hr) × (1609.34m/mi) = 96560m/h
revolutions per time= (96560.6m/h) / (1.7555m)= 55004.614 rev/hr
rpm= (55004.614 rev/hr) × (hr/60min)= 916.7436 rev/min
It's B.
The current passing through each bulb is the same. As you add more light bulbs, the current will decrease in each, but it will always be the same current passing through each one.
Answer:
The average power dissipated is 72 W.
Explanation:
Given;
peak voltage of the AC circuit, V₀ = 120 V
resistance of the resistor, R = 100 -ohm
The average power dissipated by the resistor is given by;

where;
is the root-mean-square-voltage

The average power dissipated by the resistor is calculated as;

Therefore, the average power dissipated is 72 W.