We want to calculate the distance covered by the drag racer. Recall, the formula for calculating distance is expressed as
Distance = speed x time
From the information given,
speed = 320 m/s
time = 4.5 s
By substituting these values into the formula, we have
Distance = 320 m/s x 4.5s
s cancels out. We are left with m. Thus,
Distance = 1440m
Answer:
The two most common types of orbit are "geostationary" and "polar."
Answer:
The population will increase because grasshoppers are the prey.
Explanation:
If there has been a significant decrease in the praying mantis population in recent years. This most likely affect the grasshopper population over time by the population will increase because grasshoppers are the prey. If there is less praying mantis then grasshoppers won't be eaten as much so the population of the grasshoppers will thrive
Answer:
a) a = 34.375 m / s², b) v_f = 550 m / s
Explanation:
This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.
a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed
v_f =
we substitute the values
v_f =
The initial part of the movement is carried out with acceleration
v_f = v₀ + a t
x₁ = x₀ + v₀ t + ½ a t²
the rocket starts from rest v₀ = 0 with an initial height x₀ = 0
x₁ = ½ a t²
v_f = a t
we substitute the values
x₁ = 1/2 a 16²
x₁ = 128 a
v_f = 16 a
let's write our system of equations
v_f =
x₁ = 128 a
v_f = 16 a
we substitute in the first equation
16 a =
16 4 a = 6600 - 128 a
a (64 + 128) = 6600
a = 6600/192
a = 34.375 m / s²
b) let's find the time to reach this height
x = ½ to t²
t² = 2y / a
t² = 2 5100 / 34.375
t² = 296.72
t = 17.2 s
We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part
v_f = 16 a
v_f = 16 34.375
v_f = 550 m / s