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3 years ago

5

As its speed increases, what happens to its energy of motion?

1 answer:

irina [24]3 years ago

5 0

Answer:

The energy of motion increases

Explanation:

The reason the energy of motion increases is because as the speed increases there is more motion and more energy

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An unknown object is placed inside of a spherical container and dropped from an airplane. When empty, the spherical container ha

hichkok12 [17]

Answer:

The mass of unknown object is 8.62Kg

Explanation:

To develop this problem it is necessary to apply the equations related to the Drag force and the Force of Gravity.

For the given point, that is, the moment at which the terminal velocity is reached, the two forces equalize, that is,

$F_D =F_g$

By definition we know that the Drag force is defined as

$F_D= \frac{1}{2} C_d \rho A V^2$

Where,

$C_d =$ Drag coefficient

$\rho =$ Density

A =Cross-sectional Area

V = Velocity

In the other hand we have,

$F_g = (m_1 +m_2) g$

Where,

$m_1 =$ Mass of sphere

$m_2 =$ Mass of unknown object

Equating the two equations we have to

$(m_1 +m_2) g=\frac{1}{2} C_d \rho A V^2$

Re-arrange for m_2,

$m_2 = \frac{1}{2g} C_d \rho A V^2 -m_1$

Our values are given by,

$C_d = 0.5$

$\rho = 1.22Kg/m^3$

$V = 66.7m/s$

$m_1 = 3Kg$

$d= 32.7*10^{-2}m$

$r = 16.35*10^{-2}m$

Replacing in the equation we have,

$m_2 = \frac{1}{2(9.8)}(0.5) (1.22) (\pi*(16.35*10^{-2})^2)*66.7^2 -3$

$m_2 = 8.62Kg$

<em>Therefore the mass of unknown object is 8.62Kg</em>

4 0

3 years ago

A 480N sphere 40.0cm in radius rolls without slipping 1200cm down a ramp that is inclined at 53 0 with the horizontal. What is t

IgorC [24]

As we know that sphere roll without slipping so there is no loss of energy in this case

so here we can say that total energy is conserved

Initial Kinetic energy + initial potential energy = final kinetic energy + final potential energy

$\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2+ mgh'$

as we know that ball start from rest

$v_i = 0$

height of the ball initially is given as

$h = Lsin\theta$

$h = 1200sin53 = 960 cm$

also we know that

$I = \frac{2}{5}mR^2$

also for pure rolling

$v = r\omega$

also we know that

$480 = m*9.8$

$m = 49 kg$

now plug in all data in above equation

$480*9.60 + 0 = \frac{1}{2}*49*(0.40*\omega)^2 + \frac{1}{2}*\frac{2}{5}*49*(0.40)^2\omega^2 + 0$

$4608 = 3.92\omega^2 + 1.568\omega^2$

$\omega^2 = 839.65$

$\omega = 29 rad/s$

So speed at the bottom of the inclined plane will be 29 rad/s

7 0

3 years ago

Temperature is a measure of _____.

nevsk [136]

<span>I think it would kinetic energy which is the same as potentail energy</span>

8 0

3 years ago

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Making payments to a victim in order to repay them for losses they incurred during or as a result of the crime is called

FrozenT [24]

<span>the answer is victim restitution</span>

3 0

3 years ago

A train traveled 85 mph for 7 hours how far did it travel?

Sergio [31]

Answer:

12.1429 miles per hour

Explanation:

4 0

3 years ago

Read 2 more answers