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shtirl [24]
3 years ago
11

What is a satellite?

Physics
1 answer:
Bumek [7]3 years ago
8 0

Answer:

c

Explanation:

answer is c I hope it is correct

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At its widest point, the diameter of a bottlenose dolphin is 0.50 m. Bottlenose dolphins are particularly sleek, having a drag c
fiasKO [112]

Answer:

497.00977 N

3742514.97005

Explanation:

\rho = Density of water = 1000 kg/m³

C = Drag coefficient = 0.09

v = Velocity of dolphin = 7.5 m/s

r = Radius of bottlenose dolphin = 0.5/2 = 0.25 m

A = Area

Drag force

F_d=\frac{1}{2}\rho CAv^2\\\Rightarrow F_d=\frac{1}{2}\times 1000 \times 0.09(\pi 0.25^2)7.5^2\\\Rightarrow F_d=497.00977\ N

The drag force on the dolphin's nose is 497.00977 N

at 20°C

\mu = Dynamic viscosity = 1.002\times 10^{-3}\ Pas

Reynold's Number

Re=\frac{\rho vd}{\mu}\\\Rightarrow Re=\frac{1000\times 7.5\times 0.5}{1.002\times 10^{-3}}\\\Rightarrow Re=3742514.97005

The Reynolds number is 3742514.97005

8 0
3 years ago
If the light wave has a wavelength of 10m what would be its velocity
s2008m [1.1K]

If this case could ever happen, the speed would follow from this formula:

v = f \cdot \lambda

with f the frequency and lambda the wavelength. We are give a wavelength of 10m. The frequencies of the visible light can range between 400 to about 790 Terahertz, so let us pick a middle point of 600 THz ("green-ish") as a "representative."

v = 600THz\cdot 10m = 6\cdot 10^{14} \frac{1}{s}\cdot 10 m = 6\cdot10^{15}\frac{m}{s}

The speed of such a wave would have to be 6e+15 m/s (which would be 7 orders of magnitude higher than the universal speed of light constant)

7 0
3 years ago
Read 2 more answers
Plzz helpppp meee!!!!
spin [16.1K]
A=f/m
A=900/425
A=2.18

To determine acceleration you divide the force by the mass.
4 0
2 years ago
Read 2 more answers
Visible electrical discharge from a storm cloud is called?
SOVA2 [1]

id have to say its thunder

7 0
3 years ago
A bowling ball (mass = 7.2 kg, radius = 0.11 m) and a billiard ball (mass = 0.38 kg, radius = 0.028 m) may each be treated as un
Semenov [28]

Answer:

Explanation:

Given that

Mass of bowling ball M1=7.2kg

The radius of bowling ball r1=0.11m

Mass of billiard ball M2=0.38kg

The radius of the Billiard ball r2=0.028m

Gravitational constant

G=6.67×10^-11Nm²/kg²

The magnitude of their distance apart is given as

r=r1+r2

r=0.028+0.11

r=0.138m

Then, gravitational force is given as

F=GM1M2/r²

F=6.67×10^-11×7.2×0.38/0.138²

F=9.58×10^-9N

The force of attraction between the two balls is

F=9.58×10^-9N

3 0
3 years ago
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