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3 years ago

When flying in an airplane, you are most likely in which layer of the atmosphere? mesosphere thermosphere stratosphere trosphere

Physics

2 answers:

Sergeeva-Olga [200]3 years ago

6 0

Lower stratosphere, this is to avoid turbulence

Oliga [24]3 years ago

3 0

Answer:

its troposphere!!

Explanation:

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Gases have two specific heat values .....?

kipiarov [429]

A At one constant temp and another at a constant pressure

6 0

3 years ago

A circular disc of mass 20kg and radius 15cm is mounted in an horizontal cylindrical axle of radius

disa [49]

Using the concepts of energy, rotational Newton's second law and rotational kinematics we can find the kinematic energy of the system formed by the disk and the cylindrical axis

KE = 0.23 J

given parameters

Disk radius R = 15 cm = 0.15 m
Cylinder radius r = 1.5 cm = 0.0015 m
Disk mass M = 20 kg
Time t = 1.2 s
Force F = 12 N

to find

Kinetic energy (KE)

This exercise must be solved in parts:

1st part. Endowment kinetic energy is the energy due to the circular motion of an object and is described by the equation

KE = ½ I w²

Where KE is the kinetic energy, I the moment of inertia and w the angular velocity

The moment of inertia is a magnitude that measures the inertia for rotational movement, it is a scalar quantity, therefore it is additive. In this system it is composed of two bodies, the disk and the cylindrical axis, for which the total moment of inertia it is

I_{ total} = I_{ disk} + I_{ cylinder}

the moments of inertia with respect to an axis passing through the center of mass are tabulated

disk I_{disk} = ½ M R²

cylinder I_{cylinder} = ½ m r²

where M and m are the masses of the disk and cylinder respectively, R and r their radii

I_{total} = ½ (M R² + m r²) = ½ M R² ( $1 + \frac{m}{M} \ (\frac{r}{R})^2$ )

I_{total} = ½ M R² ( $1+ \frac{m}{20} (\frac{0.015}{0.15} )^2$ ) = $\frac{1}{2}$ M R² (1 + 0.005 m)

As the shaft mass is much lighter than the disk mass , the last term is very small, which is why we despise it.

I_{total} = ½ M R²

2nd part. Let's use Newton's second law for endowment motion

τ = I α

α = $\frac{\tau }{I_{total}}$ l

τ = F R

α = $\frac{F \ R}{I_{total}}$

With the rotational kinematics expressions, we assume that the system starts from rest (w₀ = 0)

w = w₀ + α t

where w is the angular velocity, alpha is the angular acceleration and t is the time

w = 0 + $\frac{\tau }{I_{total}} \ t$

we substitute in the kinetic energy equation

KE = ½ I_{total} ( $\frac{ \tau }{I_{total}} \ t$ )²

KE = ½ $\frac{ \tau^2 }{I_{total}} \ t^2$

let's substitute

KE = $\frac{F^2 \ R^4}{M \ R^2 } \ t^2$

KE = F² R² t² / M

let's calculate

KE = 12² 0.15² 1.2² / 20

KE = 0.23 J

With the concepts of energy and rotational kinematics we can find the kinetic energy of the system is

KE = 0.23 j

learn more about rotational kinetic energy here:

brainly.com/question/20261989

4 0

3 years ago

Four identical particles of mass 0.980 kg each are placed at the vertices of a 4.14 m x 4.14 m square and held there by four mas

Zielflug [23.3K]

Answer:

a) The rotational inertia when it passes through the midpoints of opposite sides and lies in the plane of the square is 16.8 kg m²

b) I = 50.39 kg m²

c) I = 16.8 kg m²

Explanation:

a) Given data:

m = 0.98 kg

a = 4.14 * 4.14

The moment of inertia is:

$I=mr^{2} \\r=\frac{a}{2} \\I=m(a/2)^{2} \\I=\frac{ma^{2} }{4}$

For 4 particles:

$I=4(\frac{ma^{2} }{4} )=ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}$

b) Distance from top left mass = x = a/2

Distance from bottom left mass = x = a/2

Distance from top right mass = x = √5 (a/2)

The total moment of inertia is:

$I=m(\frac{a}{2} )^{2} +m(\frac{a}{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}=\frac{12ma^{2} }{4} =3ma^{2} =3*0.98*(4.14)^{2} =50.39kgm^{2}$