Answer:
a)y = 485 m
, v = 220 m / s
, b) y = 2954.39 m
, c) t_total = 51 s
,
d) v = 240.59 m / s
Explanation:
a) We can use vertical launch ratios for this exercise
the speed of the rocket the run out the fuel is
v = v₀ + a t
the rocket departs with initial velocity v₀ = 0
v = a t
v = 55 4
v = 220 m / s
the height at this point is
y = y₀ + v₀t + ½ a t²
y = y₀ + 1/2 a t²
y = 45 + ½ 55 4²
y = 485 m
b) the maximum height of the rocket is when its speed is zero
for this part we will use as the initial speed the speed at the end of the fuel (v₀´ = 220 m / s) and the height of y₀´ = 485 m
v² = v₀´² + 2 g (y-y₀´ )
0 = v₀´² + 2 g (y-y₀´ )
y = y₀´ + v₀´² / 2g
y = 485 + 220 2/2 9.8
y = 2954.39 m
c) the time that the rocket is in the air is the acceleration time t₁ = 4 s, plus the rise time (t₂) plus the time to reach the ground (t₃)
let's calculate the rise time
v = v₀´- g t
v = 0
t₂ = v₀´ / g
t₂ = 220 / 9.8
t₂ = 22.45 s
Now let's calculate the time it takes to get from this point (y₀´´ = 2954.39 m) to the floor
y = y₀´´ + v₀´´ t - ½ g t²
0 = y₀´´ - ½ g t²
t = √ (2 y₀´´ / g)
t = √ (2 2954.39 / 9.8)
t = 24.55 s
the total flight time is
t_total = t₁ + t₂ + t₃
t_total = 4 + 22.45 + 24.55
t_total = 51 s
d) the veloicda right now
v = vo + g t
v = 9.8 24.55
v = 240.59 m / s
Answer:
A. Kate’s location is at a lower latitude.
C. Kate’s location absorbs sunlight that travels a shorter distance through the atmosphere.
Explanation:
^
Answer:
curved
blue
light
matter
gravity
The type of stars in the areas give different colors to the galactic disk. The galactic disk appears blue because it has a large proportion of young, hot, main-sequence stars. The galactic bulge appears red because it contains many red giants and supergiants. The reddish hue of the galactic bulge indicates that new stars aren't being formed there.
Explanation:
The problem about describes a perfectly inelastic collision. We are tasked to find the initial velocity of an object having a mass of 6 kg moving due west. It is given in the problem that after collision the cart sticks together and it stops. Thus, the final mass is the sum of the two cart and the final velocity is zero. For a perfectly inelastic collision,
m1v1-m2v2=vf(m1+m2)
By Substitution,
3(4)-6(v2)=0
6v2=12
v2=2
Therefor, the initial velocity if a 6 kg cart is 2 m/s
Answer:
The acceleration towards the center will be
Explanation:
Given the running track is an oval shape, and the diameter of each half-circle is 74 meters.
Also, the runner took 1 minute and 40 seconds to complete 400 m one round of the track.
We need to find the acceleration towards the center.
First, we will find the speed.
Where is the speed.
is the distance covered by the rider that is 400 meters.
is the time taken by the rider to complete the lap which is 1 minute and 40 seconds. seconds.
So,
And
Where is the acceleration towards the center.
is the radius which will be the half of the diameter 74 meters.
Hence, the radius will be 37 meters.
So, the centripetal acceleration of the rider will be