1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
spin [16.1K]
3 years ago
8

Calculate the constant acceleration a in g’s which the catapult of an aircraft carrier must provide to produce a launch velocity

of 188 mi/hr in a distance of 299 ft. Assume that the carrier is at anchor.
Physics
1 answer:
goblinko [34]3 years ago
5 0
The solution to the problem is as follows:

<span>First, I'd convert 188 mi/hr to ft/s. You should end up with about ~275.7 ft/s.
 
So now write down all the values you know:

Vfinal = 275.7 ft/s

Vinitial = 0 ft/s

distance = 299ft
</span>
<span>Now just plug in Vf, Vi and d to solve
</span>
<span>Vf^2 = Vi^2 + 2 a d 

</span><span>BTW: That will give you the acceleration in ft/s^2. You can convert that to "g"s by dividing it by 32 since 1 g is 32 ft/s^2.</span>
You might be interested in
A dad takes his kids to their school just 8.0 miles down the road but with traffic it takes him 30 minutes and the fastest he ca
VARVARA [1.3K]

Answer:C 24 mi/hr

Explanation:

8 0
4 years ago
Suppose you buy some inflated party balloons that are at room temperature (about 20°C). What will happen to those balloons if yo
Zina [86]
The balloons will begin to shrink or deflate.
7 0
3 years ago
Ahsoka pulls a wounded soldier (mass 64 kg) across the ground with a force 18 Newtons. If they have an acceleration of 1.5 m/s2
Kryger [21]

Answer:

0.124

Explanation:

18-(9.81*64)*k=64*1.5

4 0
3 years ago
Read 2 more answers
What is Newton's Law
r-ruslan [8.4K]
I hope you are referring to Newton's Laws of Motion by "Newton's Law".

There are three laws quoted by the great physicist of all time Sir Isaac Newton.

These laws are the building-blocks of the field of the Physics known as Classical Physics or Classical Mechanics.

Law 1. If no force applied externally, then a body in rest will always be in rest and a body in motion will continue to move in a straight line with a uniform velocity.

Law 2. The rate of change of momentum is directly proportional to the force applied and the direction of motion is always in the direction of the force applied.

Law 3. Every action has an equal and opposite reaction.

The Law 1 is also called the law of Inertia.

From the 2nd law, we can derive the equation of force, that is F =m.a (m = mass; a = acceleration)
5 0
4 years ago
In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope
zysi [14]

Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude g.

If we have the initial velocity v_o and its angle \theta, we can obtain the vertical component of the velocity v_{oy} using trigonometry:

v_{oy}=v_osin\theta

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}

Finally, from the equation of horizontal motion with constant speed, we have that:

x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s}  }{9.8\frac{m}{s^{2} } } =7.49m

In words, the projectile travels 7.49m horizontally before it lands.

8 0
4 years ago
Other questions:
  • What relationship was uncovered by Ørsted’s observation?
    11·2 answers
  • You pull your little sister across a flat snowy field on a sled. your sister plus the sled have a mass of 20 kg.the rope is at a
    5·1 answer
  • When it is burned, 1 gallon of gasoline produces 1.3Ã108j of energy. 1500 kg car accelerates from rest to 37 m/s in 10 s. the en
    9·1 answer
  • I need help setting this problem up. Please help! A mother pushes a stroller up a hill, with the center of the stroller 1 meter
    8·1 answer
  • What is the volume of gold with the mass of 1500g if the density of gold is 19.32g ml?
    5·1 answer
  • If you want to create an electric current, which device should you use? a generator a transformer a motor a magnet
    7·2 answers
  • What is projectile motion is caused by?
    7·2 answers
  • Calculated in kilowatt hour , then in joule the energy consumed by a electrical dry towels of puissance 800 W.
    10·1 answer
  • Technician A says that outer tie rod ends should be replaced in pairs, even if only one is worn. Technician B says that inner ti
    14·1 answer
  • What would happen to a ray of light when it enters: - a) glass to water b) water to air?​
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!