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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
Refer to the diagram shown.
There are twelve 5-minute divisions.
Each 5-minute division is equal to 360°/12 = 30°.
By convention, angles are measured counterclockwise from the positive x-axis.
The angular position of the minute hand at 2:55 is
θ = 90° + 30° = 120°
Because 360° = 2π radians, therefore
θ = (120/360)*2π = (2π)/3 radians = 2.0944 radians
Answer: (2π)/3 radians ofr 2.0944 radians.
There are twelve 5-minute divisions.
Each 5-minute division is equal to 360°/12 = 30°.
By convention, angles are measured counterclockwise from the positive x-axis.
The angular position of the minute hand at 2:55 is
θ = 90° + 30° = 120°
Because 360° = 2π radians, therefore
θ = (120/360)*2π = (2π)/3 radians = 2.0944 radians
Answer: (2π)/3 radians ofr 2.0944 radians.