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cluponka [151]
2 years ago
11

A stunt pilot weighing 0.70 kN performs a vertical circular dive of radius 0.80 km.At the bottom of the dive, the pilot has a sp

eed of 0.20 km/s which at that instant is notchanging. What force does the plane exert on the pilot?
Physics
1 answer:
Yuliya22 [10]2 years ago
3 0

Answer:

 Force plane exert on pilot  = 4270 N

Explanation:

first convert radius and speed to ms

using formula from force we know that

mass = weight/ gravity =  700 N/ 9.8N/kg=  71.4 kg

Fc=  N-mg

N= Fc+ mg                  As Fc = mv²/R

N= mv²/R + mg

taking m common

N= m( v²/R +g)

   = 71.4( (200)²/ 800 + 9.8 )

 Force  = 4270 N

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Based on the time measurements in the table, what can be said about the speed of the car on the lower track as compared to the h
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1.a

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B ) Ascend using my buddy alternative air source / make an emergency Ascent

Explanation:

From the description it can be seen his buddy is close by of which he can easily use the alternative air source. Also we can see that he is closer to the water surface than his buddy, of which controlled emergency swimming ascent is highly favourable in this condition.

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a 10.0 kg block on a smooth horizontal surface is acted upon by two forces: a horizontal force of 30 N acting to the right and a
Eddi Din [679]

Answer:

a=-3\ m/s^2

Explanation:

<u>Second Newton's Law</u>

It allows to compute the acceleration of an object of mass m subject to a net force Fn. The relation is given by

F_n=m.a

The net force is the sum of all vector forces applied to the object. The block has two horizontal forces applied (in absence of friction): The 30 N force acting to the right and the 60 N force to the left. The positive horizontal direction is assumed to the right, so the net force is

F_n=30\ N-60\ N=-30\ N

Thus, the acceleration can be computed by

\displaystyle a=\frac{F_n}{m}=\frac{-30}{10}=-3\ m/s^2

\boxed{\displaystyle a=-3\ m/s^2}

The negative sign indicates the block is accelerated to the left

7 0
2 years ago
A truck is traveling down a road with a 4-percent grade at a speed of 75 mi/h when its brakes are applied to slow it down to 22.
kvasek [131]

Answer:

3.964 s

Explanation:

Metric unit conversion:

1 miles = 1.6 km = 1600 m.

1 hour = 60 minutes = 3600 seconds

75 mph = 75 * 1600 / 3600 = 33.3 m/s

22.5 mph = 22.5 * 1600/3600 = 10 m/s

Let g = 9.81 m/s2

Friction is the product of coefficient and normal force, which equals to the gravity

F_f = \mu N = \mu mg

The deceleration caused by friction is friction divided by mass according to Newton 2nd law.

a = F_f / m = \mu mg / m = \mu g = 0.6 *9.81 = 5.886 m/s^2

So the time required to decelerate from 33.3 m/s to 10 m/s so the wheels don't slide, with the rate of 5.886 m/s2 is

t = \frac{\Delta v}{a} = \frac{33.3 - 10}{5.886} = 3.964 s

3 0
1 year ago
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