All categories

2 years ago

Set local ground level to 700 ft, and record the inches of mercury.

Physics

1 answer:

pashok25 [27]2 years ago

4 0

The altimeter reading is 29.17 from the Kollsman window is 29.17 in Hg

<h3>How to determine the altimeter reading?</h3>

The given parameters are:

Ground level = 700 ft i.e. the field elevation
Pressure altitude = 1450 ft

The formula to calculate the pressure altitude is:

Altitude = (29.92 – Altimeter reading) * 1,000 + Ground level

Substitute the known values

1450 = (29.92 - Altimeter reading) * 1000 + 700

Evaluate the like terms

750 = (29.92 - Altimeter reading) * 1000

Divide both sides by 1000

0.75 = 29.92 - Altimeter reading

Evaluate the like terms

Altimeter reading = 29.17

Hence, the altimeter reading is 29.17 from the Kollsman window is 29.17 in Hg

You might be interested in

Egocentrism, animism, and artificialism are characteristic of which of Jean Piaget's stages of cognitive development?

lara31 [8.8K]

Answer:

preoperational

Explanation:

Please mark me Brainliest!!!

3 0

2 years ago

f the magnitude of the acceleration of a propeller blade's tip exceeds a certain value amaxamax, the blade tip will fracture. If

Luden [163]

Answer:

The angular velocity is $w= \sqrt[4]{\frac{a_{max}^2}{r^2} - \alpha ^2}$

Explanation:

Generally the acceleration experienced by the propeller blade's is broken down into

The Radial acceleration which is mathematically represented as

$a_r = \frac{v^2}{r} = w^2r$

And the Tangential acceleration which is mathematically represented as

$a_r = \alpha r$

The net acceleration is evaluated as

$a = \sqrt{a_r^2 + a_t^2}$

Now since angular speed varies directly with angular acceleration so when acceleration is maximum the angular velocity is maximum also and this point if the propeller blade's tip exceeds it the blade would fracture

So at maximum angular acceleration we a have

$a_{max} = \sqrt{a_r^2 + a_t^2}$

$a_{max}^2 = a_r^2 + a_t^2$

$a_{max}^2 = (w^2r)^2 + (\alpha r)^2$

$a_{max}^2 = r^2 w^4 + r^2 \alpha ^2$

$a_{max}^2 = r^2 (w^4 + \alpha^2 )$

$w^4 +\alpha ^2 = \frac{a_{max}^2}{r^2}$

$w^4 = \frac{a_{max}^2}{r^2} - \alpha ^2$

$w= \sqrt[4]{\frac{a_{max}^2}{r^2} - \alpha ^2}$

3 0

3 years ago

If you fired a rifle straight upwards at 1000 m/s, how far up will the bullet get?

nadya68 [22]

Answer:

h = 51020.40 meters

Explanation:

Speed of the rifle, v = 1000 m/s

Let h is the height gained by the bullet. It can be calculated using the conservation of energy as :

$\dfrac{1}{2}mv^2=mgh$

$h=\dfrac{v^2}{2g}$

$h=\dfrac{(1000\ m/s)^2}{2\times 9.8\ m/s^2}$

h = 51020.40 meters

So, the bullet will get up to a height of 51020.40 meters. Hence, this is the required solution.

4 0

2 years ago

10. A 25 kg apple cart is being pushed with an applied force of 115 N. The coefficient of friction between the ground and the ca

ziro4ka [17]

Answer:

1.1 m/s²

Explanation:

From the question,

F -mgμ = ma.................... Equation 1

Where F = applied force, m = mass of the apple cart, g = acceleration due to gravity, μ = coefficient of friction., a = acceleration of the apple cart.

Given: F = 115 N, m = 25 kg, μ = 0.35

Constant: g = 10 m/s²

Substitute these values into equation 2

115-(25×10×0.35) = 25×a

115-87.5 = 25a

25a = 27.5

a = 27.5/25

a = 1.1 m/s²

8 0

3 years ago

If the man on the left pulls on the object with a force of 500 N and the man on the right pulls on the object with a force of 75

Igoryamba

Force is a vector quantity
so pulling from opposite side will be negative
so
750+(-500)= 250N
C is the right answer
becauseause the man on the right applies greater force.

3 0

3 years ago