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Snezhnost [94]
2 years ago
12

If an object reflects all of the wavelengths of visible light you would see what color

Physics
1 answer:
Ahat [919]2 years ago
8 0

Answer:

white

An object has the color of the wavelengths it reflects. A material that reflects all wavelengths of visible light appears white. A material that absorbs all wavelengths of visible light appears black. A green lime absorbs most wavelengths but reflects green, so the lime looks green, as shown below.

Explanation:

hope that helps.

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Imagine a rock is dropped from the top of a tall building. After 2 seconds of falling, the rock’s instantaneous speed is approxi
shepuryov [24]
True
It's in free fall( 9.8 m/s), 2 seconds have passed.
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Tornadoes are most frequent from _____. April to June October to December January to March July to August
jenyasd209 [6]

Answer:

April to June

Explanation:

Not really an explanation, I just know it.

8 0
2 years ago
What Do You Already Know about Density? Material Design. Number each material and sort the items in order from lowest (1) to hig
Elena-2011 [213]

Answer:

1. Dry Beans - 591.75  kg/m^3

2. Flour - 593  kg/m^3

3. Wax - 900  kg/m^3

4. Wet sand - 2039 kg/m^3

5. Chalk - 2499 kg/m^3

6. Talcum Powder - 2776 kg/m^3

7. Copper - 8960  kg/m^3

Explanation:

Make sure your units are the same

7 0
3 years ago
The Earth revolves around the Sun once a year at an average distance of 1.50×1011m. Find the orbital radius that corresponds to
DedPeter [7]

Answer:

9.4\cdot 10^{10} m

Explanation:

We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write

\frac{r_a^3}{T_a^2}=\frac{r_e^3}{T_e^2}

where

r_o is the distance of the new object from the sun (orbital radius)

T_o=180 d is the orbital period of the object

r_e = 1.50\cdot 10^{11} m is the orbital radius of the Earth

T_e=365 d is the orbital period the Earth

Solving the equation for r_o, we find

r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m

3 0
3 years ago
g The electric power needs of a community are to be met by windmills with 40-m-diameter rotors. The windmills are to be located
Ksenya-84 [330]

Answer:

Explanation:

Given Data

The diameter of the wind mills is d = 40m

Velocity of the air is V = 6 m / s

Required power output is:  P ₀ = 2100 k W

Expression to calculate the exergy of the air is

E = V ² / 2

Substitute the value in above expression

E = ( 6 m / s ) ² / 2

E = 18 m ² / s ² x (1kJ/kg / 1000m²/s²)

E = 0.018 k J / k g

Expression to calculate the density of the air is

P v =m R T

m /v = P  /RT ⋯ ⋯( I )

Here  

m  is the mass of the air,  

v  is the volume of the air,  

P  is the atmospheric pressure,  

T  is the standard temperature at the atmospheric pressure and  

R  is the gas constant

As the density is

ρ = m /V

Substitute the value in expression (I)

ρ = 101  kP a /( 0.287 k J / k g ⋅ K ) ( 298 K )

ρ = 1.180 k g / m ²

Expression to calculate the mass flow rate is

m = ρ A V ⋯ ⋯ ( I I )

Here  A  is the area of the windmill

Expression to calculate the  A  is

A = π /4  d ²

Substitute the value in above expression

A = π /4 ( 40 m ²)

A = 1256.63 m ²

Substitute the value in expression (II)

m = ( 1.180 k g / m ³) ( 1256.63 m ²) ( 6 m / s )

m = 8896.94  k g / s

Expression to calculate the maximum power available to the windmill is

P w = m ( V ² /2 )

Substitute the value in above expression

P w = 8896.94  k g / s ( (6m/s)²/2 )

P w = 160144.92 W  × ( 1 W /1000 k W )

P w = 160.144 k W

Expression to calculate the number of windmills required is

n = P o /P w

Substitute the value in above expression

n=2100kw/160.144kw

n=13.11

8 0
3 years ago
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