If earth were a ping pong what size ball would jupiter be

Reil [10]

Answer:

Yes

Explanation:

When an object has more mass it takes more gravity to keep it down therefore producing friction which in return reduces the amount of kinetic energy created. A change in an object's speed has an greater effect on its kinetic energy. than a change in its mass has, because kinetic energy is proportional to.

3 0

3 years ago

A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou

Setler [38]

Answer:

a. wavelength of the sound, $\vartheta = 1.315\vartheta_{o}$

b. observed frequecy, $\lambda = 0.7604\lambda_{o}$

Given:

speed of sound source, $v_{s}$ = 80 m/s

speed of sound in air or vacuum, $v_{a}$ = 343 m/s

speed of sound observed, $v_{o}$ = 0 m/s

Solution:

From the relation:

v = $\vartheta \lambda$ (1)

where

v = velocity of sound

$\vartheta$ = observed frequency of sound

$\lambda$ = wavelength

(a) The wavelength of the sound between source and the listener is given by:

$\lambda = \frac{v_{a}}{\vartheta }$ (2)

(b) The observed frequency is given by:

$\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}$

$\vartheta = \frac{334}{334 - 80}\vartheta_{o}$

$\vartheta = 1.315\vartheta_{o}$ (3)

Using eqn (2) and (3):

$\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}$

$\lambda = 0.7604\lambda_{o}$

4 0

3 years ago

A 10 kg migratory swan cruises at 20m/s. A calculation that takes into ac-count the necessary forces shows that this motion requ

IgorC [24]

Answer:

Part A:

Distance=864000 m=864 km

Part B:

Energy Used=ΔE=8638000 Joules

Part C:

$\frac{\triangle m}{m}=0.004998=0.49985\%$

Explanation:

Given Data:

v=20m/s

Time =t=12 hours

In Secs:

Time=12*60*60=43200 secs

Solution:

Part A:

Distance = Speed**Time

Distance=v*t

Distance= 20*43200

Distance=864000 m=864 km

Part B:

Energy Used=ΔE= Energy Required-Kinetic Energy of swans

Energy Required to move= Power Required*time

Energy Required to move=200*43200=8640000 Joules

Kinetic Energy= $\frac{1}{2}mv^2$

$K.E\ of\ Swans=\frac{1}{2} *10*(20)^2=2000\ Joules$

Energy Used=ΔE=8640000 -2000

Energy Used=ΔE=8638000 Joules

Part C:

Fraction of Mass used=Δm/m

For This first calculate fraction of energy used:

Fraction of energy=ΔE/Energy required to move

ΔE is calculated in part B

Fraction of energy=8638000/8640000

Fraction of energy=0.99977

Kinetic Energy= $\frac{1}{2}mv^2$

Now, the relation between energies ratio and masses is:

$\frac{\triangle E}{E}=\frac{\triangle m}{2m}v^2$

$\frac{\triangle m}{m}=\frac{2}{v^2} *\frac{\triangle E}{E}\\\frac{\triangle m}{m}=\frac{2}{20^2} *0.99977$

$\frac{\triangle m}{m}=0.004998=0.49985\%$

3 0

3 years ago

Why does ice melt faster in water than in oil when both liquids are at the same temperature

vekshin1

Water has a high specific heat capacity. Oil has a smaller heat capacity.

7 0

3 years ago

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Assume that a satellite orbits mars 150km above its surface. Given that the mass of mars is 6.485 X 10^23kg, and the radius of m

Kisachek [45]

<span>3598 seconds The orbital period of a satellite is u=GM p = sqrt((4*pi/u)*a^3) Where p = period u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits. a = semi-major axis of orbit. Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2 The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So 150000 m + 3.396x10^6 m = 3.546x10^6 m Substitute the known values into the equation for the period. So p = sqrt((4 * pi / u) * a^3) p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3) p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3) p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3) p = sqrt(1.2945785x10^7 s^2) p = 3598.025212 s Rounding to 4 significant figures, gives us 3598 seconds.</span>

8 0

3 years ago