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Diano4ka-milaya [45]
3 years ago
7

The rate which light flows through a given area of space is referred to as ita

Physics
1 answer:
Brilliant_brown [7]3 years ago
7 0
Hey there! 

The correct answer to your question is: Intensity

The rate which light flows through a given area of space is referred to as its intensity. I<span>ntensity and wavelength are two factors which contribute to light energy.</span>
Thank you!
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Two hockey pucks, labeled A and B, are initially at rest on asmooth ice surface and are separated by a distance of18.0 m. Simult
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Answer:

8.505 m

Explanation:

Let V1 and V2 be velocities of puck A and B respectively

Since A and B move in the same direction, so the relative velocity will be V1+V2=3.5+3.9=7.4m/s

Or

Vr=7.4 m/s

Distance=S= 18 m

Time =t=?

S=Vr×t

==> t=S/Vr

==> t= 18/7.4=2.43 sec

At this time both will strike together

<em><u>Distance by puck A</u></em>

<em>V1=3.5 m/s</em>

Time=t= 2.43 sec

Distance covered=d=?

d=V1×t=3.5×2.43=8.505 m

So, puck A will cover 8.505 meters before collision

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3 years ago
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Which of these would have the lowest albedo?
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John pushes Hector on a plastic toboggan.The free-body diagram is shown. A free body diagram with 4 force vectors. The first vec
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4 years ago
The pressure at the bottom of a cylindrical container with a cross-sectional area of 45.5 cm2 and holding a fluid of density 420
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Answer:

3.33 m

Explanation:

Pressure is the distributed force applied to the surface of an object per unit area. The force is applied perpendicular to the surface of the object. The SI unit of pressure is N/m² or Pa.

Hydrostatic pressure is the pressure that a fluid exerts at a point due to the force of gravity.

The relationship between  pressure on the bottom of the container,  atmospheric pressure and the pressure due to the depth of the fluid is given by:

P_{bottom}-P_{atm}=P_{depth}\\\\where\ P_{bottom}=pressure\ at\ the \ fluid\ bottom,\ P_{atm}=atmospheric\ pressure\\P_{depth}=pressure\ due\ to\ fluid\ depth=\rho gh. \ Hence:\\\\P_{bottom}-P_{atm}=\rho gh\\\\Given \ that\ P_{bottom}=115\ kPa=115*10^3\ Pa, let\ us\ assume\ P_{atm}=101\ kPa=101*10^3\ Pa,\rho=420\ kg/m^3,g=acceleration\ due\ to \ gravity=10\ m/s^2.\\\\Therefore:\\\\115*10^3-101*10^3=420*10*h\\\\14*10^3=4200h\\\\h=3.33\ m\\\\

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4 years ago
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