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3 years ago

9

A fighter jet is launched from an aircraft carrier with the aid of its own engines and a steam-powered catapult. The thrust of i

ts engines is 2.3 x 10^5 N. In being launched from rest it moves a distance of 87 m and has a kinetic energy of 4.5 x 10^7 J at lift-off. What is the work done on the jet by the catapult?

1 answer:

kirza4 [7]3 years ago

8 0

Answer:

2.5 x 10⁷ J

Explanation:

F = thrust of the engine = 2.3 x 10⁵ N

d = distance traveled = 87 m

Work done by the engine is given as

W = F d = (2.3 x 10⁵) (87) = 200.1 x 10⁵ J

W' = Net work done

W'' = work done by catapult

KE₀ = initial kinetic energy = 0 J

KE = final kinetic energy = 4.5 x 10⁷ J

Net work done is given as

W' = KE - KE₀

W' = 4.5 x 10⁷ J

We know that

W' = W + W''

4.5 x 10⁷ = 2.001 x 10⁷ + W''

W'' = 2.5 x 10⁷ J

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The first PCS technology used a form of time division multiplexing called ____. a. Time Division Multiple Access (TDMA) b. Gl

salantis [7]

Answer:

A. Time Division multiple Access (TDMA)

Explanation:

Time-division multiple access (TDMA) is a channel access method for shared-medium networks

It allows several users to share the same frequency channel by dividing the signal into different time slots. The users transmit in rapid succession, one after the other, each using its own time slot.

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2 years ago

A block (mass = 61.2 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1/2MR2 kg · m2, wh

kolezko [41]

Answer:

The angular velocity is $w = 53.35 \ rounds /minute$

Explanation:

From the question we are told that

The mass of the block is $m = 61.2kg$

The of the pulley is $M = 14.2 kg$

The radius of the pulley is $R = 1.5m$

The radius of the cord around the pulley is $r = 1.5 m$

The distance of the block to the floor is $d = 8.0 m$

From the question we are told that the moment of inertia of the pulley is

$I = \frac{1}{2} MR^2 kg \cdot m^2$

Substituting value

$I = \frac{1}{2} * 14.2 * (1.5)^2$

$I = 15.975 kg \cdot m^2$

Using the Newtons law we can express the force acting on the vertical axis as

$ma = mg -T$

=> $T = mg -ma$

Now when the pulley is rotated that torque generated on the massless cord as a r result of the tension T and the radius of the cord around the pulley is mathematically represented as

$\tau = I \alpha$

Here $\alpha$ is the angular acceleration

Here $\tau$ is the torque which can be equivalent to

$\tau = T r$

Substituting this above

$Tr = I \alpha$

Substituting for T

$(mg - ma ) r = I\ r \alpha$

Here $a$ is the linear acceleration which is mathematically represented as

$a = r\alpha$

$(mg - m(r\alpha ) ) r = I\ r \alpha$

$mgr = I\alpha + m(r\alpha ) r$

$mgr = \alpha [ I + mr^2]$

making $\alpha$ the subject

$\alpha = \frac{mgr}{I -mr ^2}$

Substituting values

$\alpha = \frac{61.2 * 1.5 * 9.8}{15.975 + (61.2 ) * (1.5)^2}$

$\alpha =5.854 rad /s^2$

Now substituting into the equation above to obtain the acceleration

$a = 5.854 * 1.5$

$a=8.78 m/s^2$

This acceleration is $a = \frac{v}{t}$

and v is the linear velocity with is mathematically represented as

$v = \frac{d}{t}$

Substituting this into the formula acceleration

$a = \frac{d}{t^2}$

making t the subject

$t = \sqrt{\frac{d}{a} }$

substituting value

$t = \sqrt{\frac{8}{8.78}}$

$t = 0.9545 \ s$

Now the linear velocity is

$v = \frac{8}{0.9545}$

$v = 8.38 m/s$

The angular velocity is

$w = \frac{v}{r}$

So

$w = \frac{8.38}{1.5}$

$w = 5.59 rad/s$

Generally 1 radian is equal to 0.159155 rounds or turns

So 5.59 radian is equal to x

Now x is mathematically obtained as

$x = \frac{5.59 * 0.159155}{1}$

$= 0.8892 \ rounds$

Also

60 second = 1 minute

So 1 second = z

Now z is mathematically obtained as

$z = \frac{ 1}{60}$

z $= 0.01667 \ minute$

Therefore

$w = \frac{0.8892}{0.01667}$

$w = 53.35 \ rounds /minute$

8 0

3 years ago

Potassium is a crucial element for the healthy operation of the human

Degger [83]

Answer:

1

The mass of the Potassium-40 is $m_{40}} = 2.88*10^{-6} kg$

2

The Dose per year in Sieverts is $Dose_s = 26.4 *10^{-10}$

Explanation:

From the question we are told that

The isotopes of potassium in the body are Potassium-39, Potassium-40, and Potassium- 41

Their abundance is 93.26%, 0.012% and 6.728%

The mass of potassium contained in human body is $m = 3.0 g = \frac{3}{1000} = 0.0003 \ kg$ per kg of the body

The mass of the first body is $m_1 = 80 \ kg$

Now the mass of potassium in this body is mathematically evaluated as

$m_p = m * m_1$

substituting value

$m_p = 80 * 0.0003$

$m_p =0.024 kg$

The amount of Potassium-40 present is mathematically evaluated as

$m_{40}} =$ 0.012% * 0.024

$m_{40}} = \frac{0.012}{100} * 0.024$

$m_{40}} = 2.88*10^{-6} kg$

The dose of energy absorbed per year is mathematically represented as

$Dose = \frac{E}{m_1}$

Where E is the energy absorbed which is given as $E = 1.10 MeV = 1.10 * 10^6 * 1.602*10^{-19}$

Substituting value

$Dose = \frac{ 1.10 * 10^6 * 1.602*10^{-19}}{80}$

$Dose = 22*10^{-10} J/kg$

The Dose in Sieverts is evaluated as

$Dose_s = REB * Dose$

$Dose_s = 1.2 * 22*10^{-10}$

$Dose_s = 26.4 *10^{-10}$

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2 years ago

A small child has a wagon with a mass of 10 kilograms. The child pulls on the wagon with a force of

Sholpan [36]

$F=ma$
where:
F - force
m - mass
a - acceleration

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$a= \frac{F}{m}$
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3 years ago

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serious [3.7K]

Answer:

B

Explanation:

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2 years ago