Which happens during the day at the beach?

Because of surface tension, it is possible, with care, to support an object heavier than water on the water surface. The maximum

boyakko [2]

Answer:

$\pi 1 = \frac{h}{l}$

$\pi 2 =$ б / $l^2gp$

$\frac{h}{l} =$ Ф ( б / $l^2gp$ )

Explanation:

<u>Develop a suitable set of dimensionless parameters for this problem</u>

The set of dimensionless parameters for this problem is :

$\pi 1 = \frac{h}{l}$

$\pi 2 =$ б / $l^2gp$

$\frac{h}{l} =$ Ф ( б / $l^2gp$ )

and they are using the pi theorem, MLT systems

attached below is a detailed solution

7 0

2 years ago

Boston Red Sox pitcher Roger Clemens could routinely throw a fastball at a horizontal

azamat

Answer: 0.145 seconds

Explanation:

Given that Roger Clemens could routinely throw a fastball at a horizontal speed of 119.7 m/s. How long did the ball take to reach home plate 17.3 m away

Since the speed is horizontal

Using the formula for speed which is

Speed = distance/time

Where

Speed = 119.7 m/s

Distance covered = 17.3 m

Time is what we are looking for

Substitute all the parameters into the formula

119.7 = 17.3/ time

Make time the subject of formula

Time = 17.3 / 119.7

Time = 0.145 seconds.

Therefore, it will take 0.145 seconds to reach the home plates

6 0

2 years ago

A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel

Dafna1 [17]

Answer:

$1.69\cdot 10^{10}J$

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

$E=-G\frac{mM}{2r}$

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

$M=5.98\cdot 10^{24}kg$ is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

$r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m$

So the initial total energy is

$E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J$

When the satellite hits the ground, it is now on Earth's surface, so

$r=R=6370 km=6.37\cdot 10^6 m$

so its gravitational potential energy is

$U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J$

And since it hits the ground with speed

$v=1.90 km/s = 1900 m/s$

it also has kinetic energy:

$K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J$

So the total energy when the satellite hits the ground is

$E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J$

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

$\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J$