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Otrada [13]
3 years ago
13

Explain experimental physics. ​

Physics
1 answer:
insens350 [35]3 years ago
4 0

Answer:

This is a category of physics dealing with the observable phenomena, and experimental aspects by investigation, and scientific testing to measure the properties of matter, and energy.

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A person rolls a barrel up an inclined plane with a length of 6m and a height of 3m/ How much force is needed to roll up the bar
snow_lady [41]

Answer:

Length (l) = 6 m

height (h) = 3 m

Load(L) = 500 N

Effort (E) = ?

we know the principal that

E * l = L * h

6 E = 500 * 3

6E = 1500

E = 250

therefore 250 N work is done on the barrel.

8 0
4 years ago
How much work is accomplished when a force of 300N pushes a box across the floor for a distance of 100 meters?
Nesterboy [21]

So the correct ans is B.

hope it helps u.

5 0
4 years ago
Read 2 more answers
A 10cm tall object is placed perpendicular to
wlad13 [49]

My answer is in the picture

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4 0
3 years ago
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
To test a slide at an amusement park, a block of wood with mass 3.00 kgkg is released at the top of the slide and slides down to
dem82 [27]

Answer:

511.1 J

Explanation:

We are given that

Mass of wood block=m=3 kg

Vertical distance,h=23 m

Horizontal distance =x=30 m

Distance traveled in downward direction y=40 m

Initial velocity,u=0

y=ut+\frac{1}{2} gt^2

Where g=9.8 m/s^2

40=0+\frac{1}{2}(9.8)t^2=4.9t^2

t^2=\frac{40}{4.9}

t=\sqrt{\frac{40}{4.9}}=2.86 s

x=v_x\times t

30=v_x(2.86)

v=v_x=\frac{30}{2.86}=10.49 m/s

By work energy theorem

Change in kinetic energy=Work done= mgh-W

\frac{1}{2}mv^2-\frac{1}{2}mu^2=3\times 9.8\times 23-W

\frac{1}{2}(3)(10.49)^2-0=676.2-W

W=676.2-\frac{1}{2}(3)(10.49)^2=511.1 J

Hence, the work done due to friction on the block as it slides down the ramp=511.1 J

6 0
3 years ago
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