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Ivanshal [37]
3 years ago
11

on aircraft carriers, catapults are used to accelerate jet air craft to flight speeds in a short distance. One such catapult tak

es 18,000-kg jet from 0 to 70 m/s in 2.5s. What is the acceleration of the jet?
Physics
1 answer:
sineoko [7]3 years ago
8 0

Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) - (speed at the beginning)

The jet's change in speed = (70 m/s) - (zero) = 70 m/s

So acceleration = (70 m/s) / (2.5 s)

Acceleration = (70 / 2.5) m/s²

<em>Acceleration = 28 m/s²</em>  

That's about 2.9 G's .  Jet pilots can endure a lot more than that, but maybe the catapult or the hook on the airplane can't.  Let's look a little closer:

F = m A (Newton #2)

The force on the airplane = (18,000 kg) x (28 m/s²)

Force on the airplane = 504,000 Newtons

That's about 113,000 pounds !  Maybe the part of the airplane that the catapult pushes on can't handle any more force than that.  Or maybe that's the most force the catapult can deliver.

Also, the REACTION force on the catapult is the same 113,000 pounds.  Maybe the hooks or the chains or the struts on the catapult can't handle any more force than that.

That's almost 57 tons for gosh sakes !  Maybe the DECK of the carrier can't handle more force than that, and that's why they can't launch the airplane with acceleration of more than 2.9 G's .

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You create a ramp using two text books and a 0.50m board. Using a timer you determine that a cart can roll down the ramp in 0.55
ahrayia [7]

Answer:

The velocity of the cart at the bottom of the ramp is 1.81m/s, and the acceleration would be 3.30m/s^2.

Explanation:

Assuming the initial velocity to be zero, we can obtain the velocity at the bottom of the ramp using the kinematics equations:

v=v_0+at\\\\v^2=v_0^2+2ad

Dividing the second equation by the first one, we obtain:

v=\frac{v_0^2+2ad}{v_0+at}

And, since v_0=0, then:

v=\frac{2ad}{at}\\\\v=\frac{2d}{t}\\\\v=\frac{2(0.50m)}{0.55s}\\\\v=1.81m/s

It means that the velocity at the bottom of the ramp is 1.81m/s.

We could use this data, plus any of the two initial equations, to determine the acceleration:

v=v_0+at\\\\\implies a=\frac{v}{t}\\\\a=\frac{1.81m/s}{0.55s}\\\\a=3.30m/s^2

So the acceleration is 3.30m/s^2.

7 0
4 years ago
Write any two importance of gravitational force​
saul85 [17]

Answer:

plz mark me as brainliest plz

Explanation:

The gravitational force of the earth keeps us bound to the earth. Gravitational force between earth and sun makes the earth move around the sun. Gravitational force between moon and earth makes the moon go around the earth.

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3 years ago
Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an
ki77a [65]

Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega

Power factor is given by

F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802

The power factor is 0.28802

The average power to the circuit is given by

P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W

The average power to the circuit is 2.57162 W

Power to resistor

P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W

Power to resistor is 14.28 W

Power to inductor

P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W

Power to the inductor is 53.55 W

Power to the capacitor

P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W

The power to the capacitor is 6.07142 W

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Answer:

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Explanation:

fevwygafjdbhsaukchavbgaekh

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What is one Pascal pressure? What is its unit?​
BabaBlast [244]

Explanation:

the force acting perpendicularly on unit area of surface

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