A jet on an aircraft carrier can be launched from rest to 40 m/s in 2 seconds. What is the acceleration of the aircraft? Show steps please​

Question:

A small rocket is launched vertically upward from the edge of a cliff 80 ft off the ground at a speed of 69 ft/s. Its height (in feet) above the ground is given by  [tex]h(t) = - 25t^2 + 200t + 400[/tex] where t represents time measured in seconds.

Assuming the rocket is launched at t=0, what is an appropriate domain for h?

Answer:

[tex]0 \le t \le 9.66[/tex]

Explanation:

Given

[tex]h(t) = - 25t^2 + 200t + 400[/tex]

Required

Determine the domain of h

The initial value of t is 0.

i.e.

[tex]t = 0[/tex] --- This is given in the question

To determine the final value, we simply solve for t in [tex]h(t) = - 25t^2 + 200t + 400[/tex]

Rewrite the expression:

[tex]- 25t^2 + 200t + 400=0[/tex]

Divide through by -25

[tex]\frac{-25t^2}{-25} + \frac{200t}{-25} + \frac{400}{25}=\frac{0}{-25}[/tex]

[tex]t^2 - 8t - 16 = 0[/tex]

Using quadratic formula:

[tex]t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex]

Where

[tex]a= 1[/tex]    [tex]b = -8[/tex]  and [tex]c = -16[/tex]

The expression becomes:

[tex]t = \frac{-(-8)\±\sqrt{(-8)^2 - 4*1*(-16)}}{2*1}[/tex]

[tex]t = \frac{8\±\sqrt{64+ 64}}{2}[/tex]

[tex]t = \frac{8\±\sqrt{128}}{2}[/tex]

[tex]\sqrt{128} = 11.31[/tex]

So, we have:

[tex]t = \frac{8\±11.31}{2}[/tex]

Split

[tex]t = \frac{8+11.31}{2}\ or\ t = \frac{8-11.31}{2}[/tex]

[tex]t = \frac{19.31}{2}\ or\ t = \frac{-3.31}{2}[/tex]

[tex]t = 9.66\ or\ t = -1.66[/tex]

But time can't be negative.

So:

[tex]t = 9.66[/tex]

Hence, the domain is:

[tex]0 \le t \le 9.66[/tex]

Answer:

46 degrees

Explanation:

90+44+x=180

44+x=90

x=46

Answer:well isn't it 90 like it said on the image?

Explanation?????

Answer:

θ = 83.6º

Explanation:

Once the shell is in the air, the only influence acting on it (assuming that the resistance of the air is negligible) is gravity.Gravity acts on the shell accelerating it in the vertical direction with a=g=-9.8 m/s2 (taking the upward direction as the positive one).Since movements in perpendicular directions are independent each other, we conclude that in the horizontal direction, speed must be constant.So, we can find the speed in the horizontal direction (the projection of the velocity vector along the horizontal axis) applying simply the definition of average velocity, as follows:

       [tex]v_{avg} =\frac{\Delta x}{\Delta t} (1)[/tex]

Taking the launch site as the origin, and making the horizontal direction coincident with the x-axis (positive to the right), we know that the horizontal displacement is 81.7 m, and the time at the explosion, 8.6 s, so we can write the horizontal component of the velocity vector (its projection along the x-axis) as follows:

       [tex]v_{x} = \frac{81.7m}{8.6s} = v_{o} * cos \theta = 9.5 m/s (2)[/tex]

In the vertical direction, since we know that the shell will explode when it reaches to the maximum height, at this point just before exploding, the vertical component of the velocity is just zero, even though it continues being accelerated downward.Applying the definition of acceleration, replacing a by the value of g, we can write the following expression:

        [tex]v_{fy} = v_{oy} - g*t = v_{o} * sin \theta -g*t = 0 (3)[/tex]

⇒      v₀* sin θ = g*t = 9.8m/s2* 8.6s = 84.3 m/s (4)

Dividing (4) by (2), we can find tg θ (the angle of the initial velocity with the x-axis), as follows:

       [tex]\frac{v_{o}* sin \theta}{v_{o}* cos \theta} = tg \theta= \frac{84.3m/s}{9.5m/s} = 8.87 (5)[/tex]

⇒    θ = tg⁻¹ (8.87) = 83.6º

Answer:

35.2Ns

Explanation:

The change in momentum of a body is also its impulse;

Given:

Force of kicking  = 1600N

Time of action  = 0.022 s

Unknown:

Impulse  = ?

Solution:

The impulse on the body is;

      Impulse  = Force x time

Now;

      Impulse  = 1600 x 0.022  = 35.2Ns

Answer:

24.89degree angle

Explanation:

Thats what i think because its my opinion it took me a while to answer your question but i wish you the best in luck with your degree angles

Answer:

the factor that contributed to the fact that he was able to survive this fall is due to the fact that his body is made up of bone which are strong and his skeleton also protected all his vitals internal organs. His body went limp after he was knocked in the head when he fell, all of these made room allowing his muscles to relax and this gave his bones the opportunity to absorb the impact of the fall

If a man was dropped by the tornado from a distance of a quarter away from home. As per the authorities, he was dropped least at a speed of 30 mph. The man suffered no main injuries and survived.

The factors that would have contributed to the man's body able to withstand the drop were the air current that were so strong that he may have landed in a safe place.  The fact that he fell from lesser height or the man might have been fat.

Learn more about the up-by a tornado and dropped a quarter-mile away.

brainly.com/question/1492400.

Answer:

Workdone = 147Nm

Explanation:

Given the following data;

Mass = 5kg

Height = 3m

We know that acceleration due to gravity is 9.8m/s²

First of all, we would find the force applied;

[tex] Force = mass * acceleration [/tex]

Force = 5*9.8

Force = 49N

Now, to find the workdone;

[tex] Workdone = force * distance [/tex]

Substituting into the equation, we have;

Workdone = 49 * 3

Workdone = 147Nm

–0.2 m/s2 I did it on edge 2020

Answer:

the answer is C

Explanation:

i got it wrong the first time because the answer above says -0.2 but its actually 0.2

i hope this helps