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3 years ago

An aircraft has a liftoff speed of 33 m/s. What is the minimum constant acceleration an

Physics

1 answer:

JulsSmile [24]3 years ago

5 0

Answer: The minimum acceleration for the air plane is 2.269m/s2.

Explanation: To solve such problem the equation of motion are applicable.

The initial velocity is 0 since the airplane was initially standing. We are going to use this equation

V^2=U^2+2as

33^2=0+2a (240)

a= 2.269m/s2

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Unlike addition polymers, condensation polymers form from _____.

AysviL [449]

Both end of the chain

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3 years ago

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Un lote de construcción rectangular mide 100 metros por 150 metros. Calcula el área de este lote en yardas cuadradas.

antoniya [11.8K]

Answer: 17939.74 yards

Explanation:

Given , A rectangular measures 100 meters by 150 meters

To find : Area of rectangle.

Formula :

Area of rectangle = Length x width

Here, let length = 100 meters and width = 150 meters

Then, Area of rectangle = 100 meters x 150 meters = 15,000 square meters

Also , 1 meter = 1.09361 yards

Then, Area of rectangle = 15,000 x 1.09361 x 1.09361 square yards

= 17939.7424815 square yards

≈ 17939.74 yards

Hence, the area of rectangle is 17939.74 yards .

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3 years ago

Express 0.00000000062 kg in scientific notation. 6.2 x 1010kg 6.2 x 10-10kg 0.62 x 10-9kg

PilotLPTM [1.2K]

I believe the correct answer from the choices listed above is the second option. The scientific notation of the measurement 0.00000000062 kg would be <span>6.2 x 10^-10 kg. Scientific notation is used to express too large and too small values of numbers. Hope this helps. Have a nice day.</span>

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3 years ago

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A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coe

andrey2020 [161]

Answer:

The magnitude of force is 1593.4N

Explanation:

The sum of the horizontal components of the friction and the normal force will be equal to the centripetal force on the car. This can be represented as

fcostheta + Nsintheta = mv^2/r

Where F = force of friction

Theta = angle of banking

N = normal force

m = mass of car

v = velocity of car

r = radius of curve

The car has no motion in the vertical direction so the sum of forces = 0

The vertical component of the normal force acts upwards whereas the weight of the car and the vertical component friction acts downwards.

Taking the upward direction to be positive,rewrite the equation above to get:

Ncos thetha = mg - fsintheta =0

Ncistheta = mg + fain theta

N = mg/cos theta + sintheta/ costheta

fcostheta +[mg/costheta + ftan theta] sin theta = mv^2/r

Substituting gives:

f = (1/(costheta + tanthetasintheta) + mgtantheta = mv^2/r - mgtantheta)

Substituting given values into the above equation

f = 1/(cos25 + tan 25 )(sin25)[ 600×30/120 - (600×9.81)tan

f = 1593.4N25

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3 years ago

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The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increa

Nostrana [21]

Answer: The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increased by 2.0 kg, the period is found to be 3.0 s. Then the mass m is 0.625kg.

Explanation: To find the answer, we need to know more about the simple harmonic motion.

<h3>What is simple harmonic motion?</h3>

A particle is said to execute SHM, if it moves to and fro about the mean position under the action of restoring force.
We have the equation of time period of a SHM as,

$T=2\pi \sqrt{\frac{m}{k} }$