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:
480
Explanation:
resistance equals to potential difference divide by electric current
120÷0.25
=480
Answer:
Record your measured values of displacement and velocity for times t = 8.0 seconds and t = 10.0 seconds in the columns below.
Next, use the measured displacement and velocity values at t = 7.0 seconds and t = 9.0 seconds to interpolate the values of displacement and velocity at t = 8.0 seconds.
Use the following formula to interpolate and extrapolate. Remember, x and y here represent values on the x and y axes of the graph. The x values will really be time and the y values will be either displacement (x) or velocity (vx).
Explanation:
Record your measured values of displacement and velocity for times t = 8.0 seconds and t = 10.0 seconds in the columns below.
Next, use the measured displacement and velocity values at t = 7.0 seconds and t = 9.0 seconds to interpolate the values of displacement and velocity at t = 8.0 seconds.
Use the following formula to interpolate and extrapolate. Remember, x and y here represent values on the x and y axes of the graph. The x values will really be time and the y values will be either displacement (x) or velocity (vx).
This is the answer
Answer:
v₂ = 17.98 m/s
Explanation:
given,
mass of ball = m = 4.6 Kg
length of string = L = 6.6 m
force acting toward the center is equal to the force exerted by centripetal acceleration
now, calculating the speed of ball at the bottom of the circlr
work done by the gravity = change in kinetic energy
v₂ = 17.98 m/s
Use round wheels to roll the heap, help the weight by evacuating appended or contained protests, or place a smoother surface betwen the question and the floor. Water or different liquids can be utilized to diminish the protection, yet just on the off chance that they stay between the question and the floor.