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3 years ago

Examples of increase in pressure due to increase in applied force

Physics

1 answer:

Nina [5.8K]3 years ago

5 0

Answer:

injecting

Explanation:

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if mass of an object is decreased to half and acting force is reduced by quarter the acceleration of its motion

11111nata11111 [884]

Answer:

See explanations below

Explanation:

According to Newtons second law of motion

F = mass * acceleration

F = ma

If mass of an object is decreased to half, then m₂ = 1/2 m

If acting force is reduced by quarter, then F₂ = 3/4 F

F₂ = m₂a₂

3/4F = 1/2m a₂

Divide both expressions

(3/4F)/F = (1/2m)a₂/ma

3/4 = 1/2a₂/a

3/4 = a₂/2a

4a₂ = 6a

2a₂ = 3a

a₂ = 3/2 a

Hence the acceleration of its motion will be one and a half of its original acceleration.

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3 years ago

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3 years ago

A pilot can withstand an acceleration of up to 9g, which is about 88 m/s2, before blacking out. What is the acceleration experie

sweet-ann [11.9K]

Answer:

Acceleration, $a=97.67\ m/s^2$

Explanation:

Given that,

Speed of pilot, v = 475 m/s

Radius of the circle, r = 2310 m

If the pilot is moving in a circular path, it will experience a centripetal acceleration. It is given by the formula as :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(475\ m/s)^2}{2310\ m}$

$a=97.67\ m/s^2$

So, the acceleration experienced by a pilot flying in a circle is $97.67\ m/s^2$ . Hence, this is the required solution.

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3 years ago

Consider a ship of mass 2 x 10^6 kg with propeller blades of radius 7 m and it can push water backwards to a speed of 2 m/s. Giv

PolarNik [594]

Answer:

$a = 0.308 m/s^2$

Explanation:

As we know that the force due to propeller when it will push the water backwards is given as

$F = \frac{dP}{dt}$

now we know that

$P = mv$

so we have

$F = v\frac{dm}{dt}$

$F = v \rhoA\frac{dx}{dt}$

$F = \rho A v^2$

here we know that

$A = \pi r^2$

$A = \pi(7^2)$

$A = 154 m^2$

now the force is given as

$F = (1000)(154)(2^2)$

$F = 616000 N$

now the acceleration of the ship is given by Newton's II law

$a = \frac{F}{m}$

$a = \frac{616000}{2 \times 10^6}$

$a = 0.308 m/s^2$

3 0

3 years ago

Examples of increase in pressure due to increase in applied force​

Examples of increase in pressure due to increase in applied force