The graph shows the federal budget from 1980 to 2010. In which period did the federal budget show the greatest deficit?

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.0 m/s at ground level.

BARSIC [14]

Answer:

There is an interval of 24.28s in which the rocket is above the ground.

Explanation:

$y_{i}=0m$

$v_i=80\frac{m}{s}$

$a=4\frac{m}{s^2}$

$y_{e}=1000m$

$g=9.8\frac{m}{s^2}$

From Kinematics, the position $y$ as a function of time when the engine still works will be:

$y(t)=v_it+\frac{1}{2}at^2$

At what time the altitud will be $y_{e}=1000m$ ?

$v_it+\frac{1}{2}at^2=y_{e}$ ⇒ $\frac{1}{2}at^2+v_it-y_{e}=0$

Using the quadratic formula: $t_1=10s$ .

How much time does it take for the rocket to touch the ground? No the function of position is:

$y(t)=y_{e}+v_et-\frac{1}{2}gt^2$

Where our new initial position is $y_{max}$ , the velocity when the engine breaks is $v_e=v_i+at=120\frac{m}{s}$ and the only acceleration comes from gravity (which points down).

Now, when the rocket tounches the ground:

$y_{e}+v_et-\frac{1}{2}gt^2=0$

Again, using the quadratic ecuation:

$t_2=24.49s$

Now, the total time from the moment it takes off and the moment it tounches the ground will be:

$t_T=t_1+t_2=34.49s$ .

6 0

3 years ago

Pls help I will mark brainliest

irina1246 [14]

B, a disurbance would make it unstable and affect the center of mass, which would then affect the equilibrium.

4 0

2 years ago

A total charge of 0.009 passes through a cross sectional area of a nichrome wire in 3.5 s. What is the current in the wire?

Marrrta [24]

Answer:0.003 amperes

Explanation:

Quantity of charge=0.009

Time=3.5seconds

current=quantity of charge ➗ time

Current=0.009 ➗ 3.5

Current=0.003 amperes

5 0

2 years ago

A seaplane flies horizontally over the ocean at 50 meters/second. It releases a buoy, which lands after 21 seconds. What's the v

pochemuha

The motion of the buoy is a composition of two independent motions:
- a uniform motion on the horizontal axis, with constant speed vx=50 m/s
- an uniformly accelerated motion on the vertical axis, with constant acceleration $g=9.81 m/s^2$

Since we want to find the vertical displacement, we are only interested in the vertical motion.
The law of motion on the vertical direction is given by:
$y(t)=h+v_{0y} t+ \frac{1}{2}gt^2$
where
h is the initial height of the buoy
$v_{0y}$ is the initial vertical velocity of the buoy, which is zero
t is the time

We know that the buoy lands after t=21 seconds, this means that the vertical position at t=21 s is y(21 s)=0. If we substitute these data into the equation, we can find the value of h, the initial height of the buoy:
$0=h+ \frac{1}{2}gt^2$
$h= -\frac{1}{2}gt^2= -\frac{1}{2}(9.81 m/s^2)(21 s)^2=-2163 m$
And this corresponds to the vertical displacement of the buoy.

7 0

3 years ago

Constants Part A Two small charged spheres are 9.20 cm apart. They are moved, and the force on each of them is found to have bee

jok3333 [9.3K]

Answer:

5.3 cm

Explanation:

Let the charges on the spheres be q and Q.

r = 9.20 cm

F = k Q q / (9.20)^2 ..... (1)

Let the new distance be r' and force F'

F' = 3 F

F' = k Q q / r'^2

3 F = k Q q / r'^2 ...... (2)

Divide equation (1) by (2)

F / 3 F = r'^2 / 84.64

1 / 3 = r'^2 / 84.64

r' = 5.3 cm

4 0

2 years ago