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3 years ago

The aircraft has a mass of 6x 104kg. Two identical jet engines provide a thrusting

Physics

1 answer:

astraxan [27]3 years ago

7 0

Explanation:

I assume the acceleration calculated in part (b) is the 3.33 m/s² from your other question.

Use Newton's second law to find the total force:

F = ma

F = (60,000 kg) (3.33 m/s²)

F = 200,000 N

Since there are 2 engines, the thrust from each is half of this:

F = 100,000 N

In reality, there are forces other than thrust. There are also drag forces (rolling friction and air resistance).

From Newton's second law, if we increase the mass and keep the force the same, the acceleration decreases. So it would take longer to reach the take-off speed.

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What is the electrical consumption in KVA of a motor powered by a 3-phase, 60 Hz, 460 VAC supply that continuously draws 17 A

MrRa [10]

Answer:

15.34 kVA

Explanation:

A motor is a device that converts electrical energy into mechanical energy. It takes in electrical energy at the input and produce torque (motion) at the output.

The power consumption for a three phase motor is the product of voltage and current and √3. The √3 is because it is a three phase supply.

Hence Power (P) =√3 × voltage (V) × current (I)

P = √3 × V × I

Given that voltage (V) = 460 V, current (I) = 17 A. Hence:

P = √3 × V × I = √3 × 460 × 17 = 13544.64 VA

But 1000 VA = 1 kVA. Hence:

$P=13544.64\ VA*\frac{1\ kVA}{1000\ VA}=13.54\ kVA$

8 0

3 years ago

(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup

KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

$W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx$

Where

$x_{o}$ , $x_{f}$ - Initial and final position, respectively, measured in meters.

$F(x)$ - Force as a function of position, measured in newtons.

Given that $F = k\cdot x$ and the fact that $F = 25\,N$ when $x = 0.3\,m - 0.2\,m$ , the spring constant ( $k$ ), measured in newtons per meter, is:

$k = \frac{F}{x}$

$k = \frac{25\,N}{0.3\,m-0.2\,m}$

$k = 250\,\frac{N}{m}$

Now, the work function is obtained:

$W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx$

$W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]$

$W = 0.313\,J$

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be $r(\theta) = 2\cdot \sin 5\theta$ . The area of the region enclosed by one loop of the curve is given by the following integral:

$A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta$

$A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta$

By using trigonometrical identities, the integral is further simplified:

$A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta$

$A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta$

$A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta$

$A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)$

$A = 4\pi$

The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

5 0

3 years ago

What direction does current flow from a battery in a series circuit?

telo118 [61]

What's now called "Conventional current" is thought of as the flow of positive charge, from the battery's positive terminal to its negative one.

But it turns out that positive charges don't flow. The physical flow of charge is the flow of electrons. They come out of the battery's negative terminal, and carry negative charge around the circuit to the battery's positive one.

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3 years ago

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Why does the moon's gravity have a greater effect on the earth's ocean than the sun.

White raven [17]

Answer:

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garri49 [273]

Answer:

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3 years ago