The shaded boxes contain the first half of four statements. The unshaded boxes

Artists can create emphasis in a competition using

nirvana33 [79]

Answer:

Microphone and speaker

5 0

2 years ago

Calculate the Reynolds number for a person swimming through maple syrup. The density of syrup is about 1400 kg/m^3 and the visco

Tema [17]

Answer:

The Reynolds number is 5600.

Explanation:

Given that,

Density = 1400 kg/m³

Viscosity = 0.5 Pa's

Length = 2 m

Speed = 1 m/s

We need to calculate the Reynolds number

Using formula of Reynolds number

$R_{e}=\dfrac{\rho V\times L}{\mu}$

Where, $\rho$ = density of fluid

v = speed of syrup

l = length of a person

$\mu$ =Viscosity

Put the all value into the formula

$R_{e}=\dfrac{1400\times1\times2}{0.5}$

$R_{e}=5600$

Hence, The Reynolds number is 5600.

5 0

3 years ago

14)

Mama L [17]

Watt is a unit for power that is also equal to J/s. We therefore need to convert the minutes to seconds first before answering. Every minute is comprised of 60 seconds. Therefore, 3 minutes are composed of 180 seconds. Multiplying the number of seconds to the given power will give us,
Work = Power x time
= (1500 J/s) x (180 s)
= 270,000 J
Therefore, the answer is letter D.

6 0

3 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:

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