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Zina [86]
4 years ago
9

Pleasee help me helppppp

Physics
1 answer:
jeka944 years ago
8 0

Answer:

B

Explanation:

If the light is pointing down ,and then hits a surface of a 60 degree angle, then the direction will be away from the light since the light is touching a surface that is flat of a 60 degree angle then it will reflect off of it Perpendicular.

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When a wire is made smaller, the resistance increases. Which happens to the electric current?
Lelu [443]

When a wire is made shorter, the resistance decreases.

When a wire is made thinner, the resistance increases.

8 0
4 years ago
A rubber ball rolls across a level surface.The ball will eventually stop rolling due to
Illusion [34]
C. Friction is the answer
7 0
4 years ago
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A pressure vessel that has a volume of 10m3 is used to store high-pressure air for operating a supersonic wind tunnel. If the ai
Anna71 [15]

Answer:

The mass of air stored in the vessel is 235.34 kilograms.

Explanation:

Let supossed that air inside pressure vessel is an ideal gas, The density of the air (\rho), measured in kilograms per cubic meter, is defined by following equation:

\rho = \frac{P\cdot M}{R_{u}\cdot T} (1)

Where:

P - Pressure, measured in kilopascals.

M - Molar mass, measured in kilomoles per kilogram.

R_{u} - Ideal gas constant, measured in kilopascal-cubic meters per kilomole-Kelvin.

T - Temperature, measured in Kelvin.

If we know that P = 2026.5\,kPa, M = 28.965\,\frac{kg}{kmol}, R_{u} = 8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} and T = 300\,K, then the density of air is:

\rho = \frac{(2026.5\,kPa)\cdot \left(28.965\,\frac{kg}{kmol} \right)}{\left(8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} \right)\cdot (300\,K)}

\rho = 23.534\,\frac{kg}{m^{3}}

The mass of air stored in the vessel is derived from definition of density. That is:

m = \rho \cdot V (2)

Where m is the mass, measured in kilograms.

If we know that \rho = 23.534\,\frac{kg}{m^{3}} and V = 10\,m^{3}, then the mass of air stored in the vessel is:

m = \left(23.534\,\frac{kg}{m^{3}} \right)\cdot (10\,m^{3})

m = 235.34\,kg

The mass of air stored in the vessel is 235.34 kilograms.

7 0
3 years ago
At a certain elevation, the pilot of a balloon has a mass of 120 lb and a weight of 119 lbf. What is the local acceleration of g
Strike441 [17]

Answer:

31.905 ft/s²

Explanation:

Given that

Mass of the pilot, m = 120 lb

Weight of the pilot, w = 119 lbf

Acceleration due to gravity, g = 32.05 ft/s²

Local acceleration of gravity of found by using the relation

Weight in lbf = Mass in lb * (local acceleration/32.174 lbft/s²)

119 = 120 * a/32. 174

119 * 32.174 = 120a

a = 3828.706 / 120

a = 31.905 ft/s²

Therefore, the local acceleration due to gravity at that elevation is 31.905 ft/s²

3 0
3 years ago
A car undergoing uniform acceleration travels 100 meters from a standing start in a given period of time. If the time were incre
iogann1982 [59]

Answer:

d = 1600 m

Explanation:

If car start from rest and accelerates uniformly then the displacement of the car is given as

d = v_i t + \frac{1}{2}at^2

now we know

d = 0 + \frac{1}{2}at^2

now if the same car will accelerate from rest for four times of the previous time interval

then we will have

d = 0 + \frac{1}{2}a(4t)^2

d = 16(\frac{1}{2}at^2)

so the distance will be 16 times more

d = 1600 m

5 0
4 years ago
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