Ask question

Ask question

All categories

3 years ago

7

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

nced by a pilot flying in a circle of constant radius at a constant speed of 475 m / s if the radius of the circle is 2310 m?

1 answer:

sweet-ann [11.9K]3 years ago

8 0

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.

You might be interested in

50 points !! I need help asap.......Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to t

r-ruslan [8.4K]

1) At the top of the building, the ball has more potential energy

2) When the ball is halfway through the fall, the potential energy and the kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

4) The potential energy at the top of the building is 784 J

5) The potential energy halfway through the fall is 392 J

6) The kinetic energy halfway through the fall is 392 J

7) The kinetic energy just before hitting the ground is 784 J

Explanation:

1)

The potential energy of an object is given by

$PE=mgh$

where

m is the mass

g is the acceleration of gravity

h is the height relative to the ground

While the kinetic energy is given by

$KE=\frac{1}{2}mv^2$

where v is the speed of the object

When the ball is sitting on the top of the building, we have

$h=40 m$ , therefore the potential energy is not zero
$v=0$ , since the ball is at rest, therefore the kinetic energy is zero

This means that the ball has more potential energy than kinetic energy.

2)

When the ball is halfway through the fall, the height is

$h=20 m$

So, half of its initial height. This also means that the potential energy is now half of the potential energy at the top (because potential energy is directly proportional to the height).

The total mechanical energy of the ball, which is conserved, is the sum of potential and kinetic energy:

$E=PE+KE=const.$

At the top of the building,

$E=PE_{top}$

While halfway through the fall,

$PE_{half}=\frac{PE_{top}}{2}=\frac{E}{2}$

And the mechanical energy is

$E=PE_{half} + KE_{half} = \frac{PE_{top}}{2}+KE_{half}=\frac{E}{2}+KE_{half}$

which means

$KE_{half}=\frac{E}{2}$

So, when the ball is halfway through the fall, the potential energy and the kinetic energy are equal, and they are both half of the total energy.

3)

Just before the ball hits the ground, the situation is the following:

The height of the ball relative to the ground is now zero: $h=0$ . This means that the potential energy of the ball is zero: $PE=0$

The kinetic energy, instead, is not zero: in fact, the ball has gained speed during the fall, so $v\neq 0$ , and therefore the kinetic energy is not zero

Therefore, just before the ball hits the ground, it has more kinetic energy than potential energy.

4)

The potential energy of the ball as it sits on top of the building is given by

$PE=mgh$

where:

m = 2 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity

h = 40 m is the height of the building, where the ball is located

Substituting the values, we find the potential energy of the ball at the top of the building:

$PE=(2)(9.8)(40)=784 J$

5)

The potential energy of the ball as it is halfway through the fall is given by

$PE=mgh$

where:

m = 2 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity

h = 20 m is the height of the ball relative to the ground

Substituting the values, we find the potential energy of the ball halfway through the fall:

$PE=(2)(9.8)(20)=392 J$

6)

The kinetic energy of the ball halfway through the fall is given by

$KE=\frac{1}{2}mv^2$

where

m = 2 kg is the mass of the ball

v = 19.8 m/s is the speed of the ball when it is halfway through the fall

Substituting the values into the equation, we find the kinetic energy of the ball when it is halfway through the fall:

$KE=\frac{1}{2}(2)(19.8)^2=392 J$

We notice that halfway through the fall, half of the initial potential energy has converted into kinetic energy.

7)

The kinetic energy of the ball just before hitting the ground is given by

$KE=\frac{1}{2}mv^2$

where:

m = 2 kg is the mass of the ball

v = 28 m/s is the speed of the ball just before hitting the ground

Substituting the values into the equation, we find the kinetic energy of the ball just before hitting the ground:

$KE=\frac{1}{2}(2)(28)^2=784 J$

We notice that when the ball is about to hit the ground, all the potential energy has converted into kinetic energy.

Learn more about kinetic and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

4 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Oduvanchick [21]

<u>Answer</u>

D. Base units

<u>Explanation</u>

Basic units are also called fundamental units. They are the standard units agreed internationally for measurements. Most of these measurements are taken from the ground and they are used to derive other units. They are seven in number. There are:

The metre (m)

The kilogram (kg)

The second (s)

The ampere (A)

The kelvin (K)

The candela (cd)

The mole (mol)

6 0

3 years ago

the roque requried to turn the crank on an ice cream maker is 4.50 N.m how much work does it take to turn the crank through 300

Alexus [3.1K]

Answer:

the work required to turn the crank at the given revolutions is 8,483.4 J

Explanation:

Given;

torque required to turn the crank, T = 4.50 N.m

number of revolutions, = 300 turns

The work required to turn the crank is given as;

W = 2πT

W = 2 x 3.142 x 4.5

W = 28.278 J

1 revolution = 28.278 J

300 revlotions = ?

= 300 x 28.278 J

= 8,483.4 J

Therefore, the work required to turn the crank at the given revolutions is 8,483.4 J

4 0

3 years ago

A resistor of 5 Ω is placed in a circuit. The voltage drop across the resistor is 12 V. What is the current through the resistor

Kaylis [27]

<span>Data:

</span>R = 5 Ω
U = 12 V
i = ?
<span>
Formula:

</span> $U = R*i$ → $i = \frac{U}{R}$ <span>

Solving:

</span> $i = \frac{U}{R}$
$i = \frac{12}{5}$
$\boxed{i = 2.4A}$

Answer:

<span>The current through the resistor is 2.4 Amperes</span><span>

</span>

5 0

3 years ago

Why football players boots have spikes their sole<br>Give short and sweet answer

Marizza181 [45]

Answer:

it helps with balance and speed.

"The football shoes have spikes or studs because the studs or spikes provides larger frictional force than normal shoes while running on the grass. The studs prevents player from slipping on the grass and help to run faster and change direction quickly without slipping"

3 0

3 years ago

Read 2 more answers