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Aleks04 [339]
3 years ago
8

An aircraft is loaded 110 pounds over maximum certificated gross weight. If fuel (gasoline) is drained to bring the aircraft wei

ght within limits, how much fuel should be drained?
Physics
1 answer:
lorasvet [3.4K]3 years ago
5 0

Answer:

242 lts de fuel

Explanation:

We know density of fuel (is variable) but we can say that is close to 1 Kg/lts.

If we have 110 pounds over maximun certificated gross weight, we need to get rid of that 110 pounds  or 110 * 2.2  = 242 Kgs

Now    by rule of three

If     1 lt        weight      1   kg.

        x                          242 kgs

x  = 242 lts de fuel

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The dimension line is connected to the part being measured by _______________. A. Hidden lines B. Extension lines C. Visible lin
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B: Extension Lines! You could have just searched this up on google
3 0
4 years ago
Read 2 more answers
A 2:2 kg toy train is con ned to roll along a straight, frictionless track parallel to the x-axis. The train starts at the origi
Liula [17]

Answer:

a) 10.51 J

b) 3.48 m/s

Explanation:

Given data :

mass of train ( M ) = 2.2 kg

Given initial velocity ( u ) = 1.6 m/s

<u>a) calculating work done by the force over the journey of the train</u>

F = mx + b  ------ ( 1 )

m = slope  = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m

x = distance travelled on the x axis by the train = 7.5 m

F = force experienced by the train = 2.8 N

x = 0

∴ b = 2.8

hence equation 1 can be written as

F = ( -0.373) x + 2.8   ----- ( 2 )

hence to determine the work done by the force

W   = \int\limits^7_0 { ( -0.373) x + 2.8  )} \, dx     Note:  the limits are actually 7.5 and 0

∴ W ( work done ) = -10.49 + 21 = 10.51 J

<u>b) calculate the speed of the train at the end of its journey</u>

we will apply the work energy theorem

W = 1/2 m*v^2  -  1/2 m*u^2

∴ V^2 = 2 / M ( W + 1/2 M*u^2 )  ( input values into equation )

 V^2 = 12.11

hence V = 3.48 m/s

6 0
3 years ago
Durante un intento juego de croquet, una bola de 0.52 kg en reposo sobre el cesped es golpeada por un mazo con una fuerza promed
Romashka [77]

Answer: 2.63 m/s

Explanation:

The question in english is written below:

During an attemp of croquet game, a 0.52 kg ball at rest on the grass is hit by a mallet with an average force of 190 N, if the mallet is in contact with the ball for 7.2(10)^{-3} s,  Whay is the speed of the ball just after being hit?

According to Newton's second law of motion, the force F exerted on an object is directly proportional to its mass m and acceleration a:

F=m.a (1)

On the other hand, acceleration is defined as the variation of velocity V in time t:

a=\frac{V}{t} (2)

Substituting (2) in (1):

F=m\frac{V}{t} (3)

Where:

F=190 N

m=0.52 kg

t=7.2(10)^{-3} s

Isolating the velocity V from (3):

V=\frac{Ft}{m} (4)

V=\frac{(190 N)(7.2(10)^{-3} s)}{0.52 kg}

Finally:

V=2.63 m/s

6 0
4 years ago
When you abbreviate very large or very small numbers, you are using
andrew-mc [135]

Answer:

A

Explanation:

its scientific notation

3 0
3 years ago
A proton moves perpendicular to a uniform magnetic field B at a speed of 2.40 107 m/s and experiences an acceleration of 1.90 10
jonny [76]

Answer:

Magnitude of the magnetic field = B = 0.00827 T = 8.27 m T

Direction = B is directed in the negative y-direction.

Explanation:

Given

Charge moving through the magnetic field is a proton, Hence,

Charge = (1.602 × 10⁻¹⁹) C

Mass of the proton = (1.673 × 10⁻²⁷) kg

Magnitude of velocity = v = (2.40 × 10⁷) m/s

Acceleration = (1.90 × 10¹³) m/s²

Angle between the proton's motion and the magnetic field = 90°

First of, we find the magnitude of the magnetic field.

F = qvB sin θ

where

F = magnitude of the force experienced by the proton = ma = (mass of a proton) × (acceleration of the proton)

F = ma = (1.673 × 10⁻²⁷ × 1.90 × 10¹³)

F = (3.179 × 10⁻¹⁴) N

(3.179 × 10⁻¹⁴) = (1.602 × 10⁻¹⁹ × 2.40 × 10⁷ × B × sin 90°)

B = 0.00827 T = 8.27 m T

To get the direction of the magnetic field, we can use the right hand rule, or the vector multiplication method.

The right hand rule expresses the directions of the velocity of proton, force on the proton and the magnetic field using the first 3 fingers on the right hand (the thumb, the pointing finger and the middle finger). It states that whenthose first 3 fingers are pointed in a way that they are at right angles to one another, the direction of the magnetic force on a moving charge points in the direction of the middle finger, the thumb points in the direction of the velocity, the pointing finger is in the direction of B.

Practicalizing this, with the directions already stated in the question (velocity is in the positive x direction, the force is in the direction of the acceleration in the positive z-direction), the middle finger is left to point in the negative y-direction.

Using vector notation,

F = qV × B

F = Fx î

qV = qV(z) k

B = Bxî + Byj + Bzk

Doing the cross product,

Fx î = (-By)(qVz)

Since Fx, q, Vz are all positive quantities, By has to be a negative quantity to turn the minus into a plus.

Hence, the magnetic field is directed in the negative y-direction.

Hope this Helps!!!

5 0
4 years ago
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