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VashaNatasha [74]

2 years ago

10

One of the largest planes ever to fly, and the largest to fly frequently, is the Ukrainian-built Antonov An-124 Ruslan. The grav

itational potential energy associated with the plane at an altitude of 1450 m is 3,360,000,000 J. What is the airplane's mass?

1 answer:

Lerok [7]2 years ago

4 0

Answer:

m = 236212 [kg]

Explanation:

The potential energy can be determined by means of the product of mass by gravity by height. In this way, we have the following equation.

$P=m*g*h\\$

where:

P = potential energy = 3360000000 [J]

m = mass [kg]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = 1450 [m]

Now, we can clear the mass from the equation above:

$3360000000=m*9.81*1450\\m = 236212 [kg]$

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What is the difference between internal and external force

AnnZ [28]

Answer:External forces are forces caused by external agent outside of the system. Internal forces are forces exchanged by the objects in the system.

Explanation:or our purposes, we will simply say that external forces include the applied force, normal force, tension force, friction force, and air resistance force. And for our purposes, the internal forces include the gravity forces, magnetic force, electrical force, and spring force.HOPE THIS HELPS!!! ^w^

7 0

2 years ago

Two flywheels of negligible mass and different radii are bonded together and rotate about a common axis (see below). The smaller

jekas [21]

Answer:

$F_2 = 29.54 N$

Explanation:

As we know that the combination is maintained at rest position

So we will take net torque on the system to be ZERO

so we know that

$\tau = \vec r \times \vec F$

here we will have

$\vec r_1 \times F_1 = \vec r_2 \times F_2$

so we have

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so we have

$F_2 = \frac{13 \times 50}{22}$

$F_2 = 29.54 N$

8 0

2 years ago

A rod of length Lo moves iwth a speed v along the horizontal direction. The rod makes an angle of (θ)0 with respect to the x' ax

Colt1911 [192]

Answer:

From the question we are told that

The length of the rod is $L_o$

The speed is v

The angle made by the rod is $\theta$

Generally the x-component of the rod's length is

$L_x = L_o cos (\theta )$

Generally the length of the rod along the x-axis as seen by the observer, is mathematically defined by the theory of relativity as

$L_xo = L_x \sqrt{1 - \frac{v^2}{c^2} }$

=> $L_xo = [L_o cos (\theta )] \sqrt{1 - \frac{v^2}{c^2} }$

Generally the y-component of the rods length is mathematically represented as

$L_y = L_o sin (\theta)$

Generally the length of the rod along the y-axis as seen by the observer, is also equivalent to the actual length of the rod along the y-axis i.e $L_y$

Generally the resultant length of the rod as seen by the observer is mathematically represented as

$L_r = \sqrt{ L_{xo} ^2 + L_y^2}$

=> $L_r = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}$

=> $L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}$

=> $L_r = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}$

=> $L_r = \sqrt{L_o^2 * cos^2(\theta) [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}$

=> $L_r = \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }$

=> $L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Hence the length of the rod as measured by a stationary observer is

$L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Generally the angle made is mathematically represented

$tan(\theta) = \frac{L_y}{L_x}$

=> $tan {\theta } = \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }$

=> $tan(\theta ) = \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }$

Explanation:

7 0

2 years ago

What is your speed (in m/s) if you walk 1000m in 20 minutes? (Hint: how many seconds are in a minute)

labwork [276]

Answer:

20min = 20 × 60 = 1200sec.

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V = 1000/1200

V = 0.833m per sec.

Explanation:

7 0

2 years ago

A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processe

Allushta [10]

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzman law that is responsible for calculating radioactive energy.

Mathematically this expression can be given as

$P = \sigma Ae\Delta T^4$

Where

A = Surface area of the Object

$\sigma =$ Stefan-Boltzmann constant

e = Emissivity

T = Temperature (Kelvin)

Our values are given as

$A = 1.36m^2$

$\Delta T^4 = T_2^4 -T_1^4 = 307^4-T_1^4$

$\sigma = 5.67*10^{-8} J/(s m^2 K^4)$

$P = 122J/s$

$e = 0.7$

Replacing at our equation and solving to find the temperature 1 we have,

$P = \sigma Ae\Delta T^4$

$P = \sigma Ae (T_2^4 -T_1^4)$

$122 = (5.67*10^{-8})(1.36)(0.7)(307^4-T_1^4)$

$T_1 = 285.272K = 12.122\°C$

Therefore the the temperature of the coldest room in which this person could stand and not experience a drop in body temperature is 12°C

8 0

2 years ago