Using clay electric rates, how much does it cost to watch Netflix for 4 hours on a smart tv?

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

A car travels around a level, circular track that is 750m across. What coefficient of friction is required to ensure the car can

Crank

The coefficient of friction must be 0.196

Explanation:

For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:

$\mu mg = m\frac{v^2}{r}$

where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:

$\mu$ is the coefficient of friction between the tires and the road

m is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

v is the speed of the car

r is the radius of the curve

In this problem,

r = 750 m is the radius

$v=85 mph \cdot \frac{1609}{3600}=38.0 m/s$ is the speed

And solving for $\mu$ , we find the coefficient of friction required to keep the car in circular motion:

$\mu = \frac{v^2}{rg}=\frac{38.0^2}{(750)(9.8)}=0.196$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

8 0

3 years ago

Two wires A and B with circular cross-section are made of the same metal and have equal lengths, but the resistance of wire A is

Nesterboy [21]

Answer:

r₁/r₂ = 1/2 = 0.5

Explanation:

The resistance of a wire is given by the following formula:

R = ρL/A

where,

R = Resistance of wire

ρ = resistivity of the material of wire

L = Length of wire

A = Cross-sectional area of wire = πr²

r = radius of wire

Therefore,

R = ρL/πr²

FOR WIRE A:

R₁ = ρ₁L₁/πr₁² -------- equation 1

FOR WIRE B:

R₂ = ρ₂L₂/πr₂² -------- equation 2

It is given that resistance of wire A is four times greater than the resistance of wire B.

R₁ = 4 R₂

using values from equation 1 and equation 2:

ρ₁L₁/πr₁² = 4ρ₂L₂/πr₂²

since, the material and length of both wires are same.

ρ₁ = ρ₂ = ρ

L₁ = L₂ = L

Therefore,

ρL/πr₁² = 4ρL/πr₂²

1/r₁² = 4/r₂²

r₁²/r₂² = 1/4

taking square root on both sides:

r₁/r₂ = 1/2 = 0.5

6 0

3 years ago

A soap bubble, when illuminated at normal incidence with light of 463 nm, appears to be especially reflective. If the index of r

Sedbober [7]

Answer:

the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm

Explanation:

Given the data in the question;

wavelength of light; λ = 463 nm = 463 × 10⁻⁹ m

Index of refraction; n = 1.35

Now, the thinnest thickness of the soap film can be determined from the following expression;

$t_{min$ = ( λ / 4n )

so we simply substitute in our given values;

$t_{min$ = ( 463 × 10⁻⁹ m ) / 4(1.35)

$t_{min$ = ( 463 × 10⁻⁹ m ) / 5.4

$t_{min$ = ( 463 × 10⁻⁹ m ) / 4(1.35)

$t_{min$ = 8.574 × 10⁻⁸ m

$t_{min$ = 85.74 × 10⁻⁹ m

$t_{min$ = 85.74 nm

Therefore, the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm

5 0

3 years ago

The critical angle for a certain air-liquid surface is 47.7°. What is the index of refraction of the liquid? Round to the neares

KengaRu [80]

Answer:

The index of refraction of the liquid is 1.35.

Explanation:

It is given that,

Critical angle for a certain air-liquid surface, $\theta_1=47.7^{\circ}$

Let n₁ is the refractive index of liquid and n₂ is the refractive index of air, n₂ =1

Using Snell's law for air liquid interface as :

$n_1\ sin\theta_1=n_2\ sin(90)$

$n_1\ sin(47.7)=1$

$n_1=\dfrac{1}{sin(47.7)}$

$n_1=1.35$

So, the index of refraction of the liquid is 1.35. Hence, this is the required solution.

5 0

3 years ago