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Montano1993 [528]

2 years ago

5

A Formula 1 car acelerations from 0 to 60 mph (26.8 m/s) in 2.1 seconds

1 answer:

gladu [14]2 years ago

4 0

Answer:

¿Y CUÁL ES LA PREGUNTA?

Explanation:

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The parallel axis theorem relates Icm, the moment of inertia of an object about an axis passing through its center of mass, to I

erma4kov [3.2K]

Answer:

Part a)

$I_{end} = \frac{mL^2}{3}$

Part b)

$I_{edge} = \frac{2ma^2}{3}$

Explanation:

As we know that by parallel axis theorem we will have

$I_p = I_{cm} + Md^2$

Part a)

here we know that for a stick the moment of inertia for an axis passing through its COM is given as

$I = \frac{mL^2}{12}$

now if we need to find the inertia from its end then we will have

$I_{end} = I_{cm} + Md^2$

$I_{end} = \frac{mL^2}{12} + m(\frac{L}{2})^2$

$I_{end} = \frac{mL^2}{3}$

Part b)

here we know that for a cube the moment of inertia for an axis passing through its COM is given as

$I = \frac{ma^2}{6}$

now if we need to find the inertia about an axis passing through its edge

$I_{edge} = I_{cm} + Md^2$

$I_{edge} = \frac{ma^2}{6} + m(\frac{a}{\sqrt2})^2$

$I_{edge} = \frac{2ma^2}{3}$

7 0

3 years ago

A solid of density 8000 kgm.. weighs 0.8 kgf in air. When it is completely

ad-work [718]

Answer:<em> 0.5kg</em>

Explanation:

5 0

2 years ago

An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temper

dimulka [17.4K]

Answer:

The power output of this engine is $P = 17.5 W$

The the maximum (Carnot) efficiency is $\eta_c = 0.7424$

The actual efficiency of this engine is $\eta _a = 0.46$

Explanation:

From the question we are told that

The temperature of the hot reservoir is $T_h = 1250 \ K$

The temperature of the cold reservoir is $T_c = 322 \ K$

The energy absorbed from the hot reservoir is $E_h = 1.37 *10^{5} \ J$

The energy exhausts into cold reservoir is $E_c = 7.4 *10^{4} J$

The power output is mathematically represented as

$P = \frac{W}{t}$

Where t is the time taken which we will assume to be 1 hour = 3600 s

W is the workdone which is mathematically represented as

$W = E_h -E_c$

substituting values

$W = 63000 J$

So

$P = \frac{63000}{3600}$

$P = 17.5 W$

The Carnot efficiency is mathematically represented as

$\eta_c = 1 - \frac{T_c}{T_h}$

$\eta_c = 1 - \frac{322}{1250}$

$\eta_c = 0.7424$

The actual efficiency is mathematically represented as

$\eta _a = \frac{W}{E_h}$

substituting values

$\eta _a = \frac{63000}{1.37*10^{5}}$

$\eta _a = 0.46$

7 0

3 years ago

In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.

Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular displacement of the object

t is the time elapsed

$\omega$ is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

$\theta=2\pi$ rad

And the time taken is

$t=95 min \cdot 60 =5700 s$

Therefore, the angular velocity of the telescope is

$\omega=\frac{2\pi}{5700}=0.0011 rad/s$

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

$v=\omega r$

where

v is the linear velocity

$\omega$ is the angular velocity

r is the radius of the circular orbit

In this problem:

$\omega=0.0011 rad/s$ is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

$r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m$

Therefore, the linear velocity of the telescope is:

$v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s$

4 0

3 years ago

Two particles with charges are initially very far apart (effectively an infinite distance apart). They are then fixed at positio

ddd [48]

Answer:

potential energy increases.

Explanation:

The potential energy between the two charged particles is given by

U = k Q q / r

If they are very far apart then r tends to infinity and the potential energy is zero.

If they come closer then the potential energy between the two charged particles increases.

Thus, the potential energy increases.

3 0

3 years ago