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3 years ago

Calculate the acceleration of a car that is maintaining a constant velocity of 1m/s

Physics

2 answers:

lora16 [44]3 years ago

5 0

"Constant velocity" is another way of saying "zero acceleration".

Papessa [141]3 years ago

3 0

Answer:

Acceleration, a =0

Explanation:

If the velocity is constant, it means the acceleration of the object is equal to zero. Mathematically, the acceleration of an object is given by :

$a=\dfrac{dv}{dt}$

It is also defined as the change in speed per unit time. Its SI unit is $m/s^2$ . It is a vector quantity.

So, it is concluded that if a car is maintaining a constant velocity of 1 m/s, its acceleration is equal to zero. Hence, this is the required solution.

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What happens to the strength of the magnetic field as you come closer to the current carrying wire

Leokris [45]

Form concentric circles around the wire

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3 years ago

A net force of 60 N north acts on an object with a mass of 30 kg. Use Newton's second law of

earnstyle [38]

Answer:

Explanation:

F = ma. For us, this looks like

60 = 30a and

a = 2 m/s/s

If the force goes up to, say, 90, then

90 = 30a and

a = 3...if the force goes up, the acceleration also goes up.

If the mass goes up to say, 60, and the force stays the same, then

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2 years ago

An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below

grin007 [14]

<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let $P$ be the pressure at a point.

Let $p$ be the density fluid at a point.

Let $v$ be the velocity of fluid at a point.

Bernoulli's equation states that $P+\frac{1}{2}pv^{2}+pgh=constant$ for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let $p_{up}$ be the pressure of a point just above the wing.

Let $p_{do}$ be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

$\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}$

So, $p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa$

Force is given by the product of pressure difference and area.

Given that area is $27ms^{2}$ .

So,lifting force is $16398.48\times 27=442758.96N$

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Answer:

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2 years ago