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Tju [1.3M]
3 years ago
15

What are the phenomena that physics is applied in?

Physics
2 answers:
Ratling [72]3 years ago
5 0

Answer:

For empiricists like van Fraassen, the phenomena of physics are the appearances observed or perceived by sensory experience. Constructivists, however, regard the phenomena of physics as artificial structures generated by experimental and mathematical methods.

omeli [17]3 years ago
5 0

Answer:

Physics is applied every where? For example if you talking with you friends, that mean the sound that you produce oscillate through your friends' ear so he/ she can respond to you.

Explanation:

Physics is the fundamental unit about understanding the world around us.

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For the different values given for the radius of curvature RRR and speed vvv, rank the magnitude of the force of the roller-coas
nirvana33 [79]

Explanation:

The force of the roller-coaster track on the cart at the bottom is given by :

F=\dfrac{mv^2}{R}, m is mass of roller coaster

Case 1.

R = 60 m v = 16 m/s

F=\dfrac{(16)^2m}{60}=4.26m\ N

Case 2.

R = 15 m v = 8 m/s

F=\dfrac{(8)^2m}{15}=4.26m\ N

Case 3.

R = 30 m v = 4 m/s

F=\dfrac{(4)^2m}{30}=0.54m\ N

Case 4.

R = 45 m v = 4 m/s

F=\dfrac{(4)^2m}{45}=0.36m\ N

Case 5.

R = 30 m v = 16 m/s

F=\dfrac{(16)^2m}{30}=8.54m\ N

Case 6.

R = 15 m v =12 m/s

F=\dfrac{(12)^2m}{15}=9.6m\ N

Ranking from largest to smallest is given by :

F>E>A=B>C>D

5 0
3 years ago
A cylindrical tube sustains standing waves at the following frequencies: 600 Hz, 800 Hz, and 1000 Hz. The tube does not sustain
Thepotemich [5.8K]

Answer:

Explanation:

In  closed organ pipe notes with odd harmonics are produced and in open organ pipe notes with all odd and even harmonics are produced. notes with frequencies 600, 800 and 1000 Hz are produced. These are 3 , 4 and 5 times 200 Hz. ie both odd and even times of 200 . So fundamental frequency appears to be 200 Hz. There is no note available between 800 and 1000. It also indicates that 200 Hz is the fundamental frequency and the pipe is open at both ends.

3 0
3 years ago
On a hypothetical scale X The ice point is 40° and steam point is 120°.
arlik [135]

Answer:

The reading of Y is -10°.

Explanation:

For scale X, the ice point is 40° and steam point is 120°.

Difference between the two extremes for scales X = 120 - 40 = 80

For scale X, the ice point and steam points are -30° and 130° respectively.

Difference between the two extremes for scales X = 130 - (-30) = 160

Comparing both scales:

One unit of scale X = x

One unit of scale Y = y

Scale X has 80 divisions while scale Y has 160

80x = 160y

x = 2y

50° in scale X = 10x + ice point in X scale

10 divisions in Y scale = 20y

Reading of Y scale = ice point of Y + 20y

= -30° + 20°

= -10°

7 0
3 years ago
How much energy (in joules) is required to evaporate 0.0005 kg of liquid ammonia to vapor at the same temperature?
Lana71 [14]

Answer:

685.6 J

Explanation:

The latent heat of vaporization of ammonia is

L = 1371.2 kJ/kg

mass of ammonia, m = 0.0005 Kg

Heat = mass x latent heat of vaporization

H = 0.0005 x 1371.2

H = 0.686 kJ

H = 685.6 J

Thus, the amount of heat required to vaporize the ammonia is 685.6 J.

6 0
3 years ago
Given a wind turbine with blades that sweep out a 10 m diameter circle, and a wind speed of 2 m/s, approximately what is the max
Mars2501 [29]

Answer:

223.55 W

Explanation:

v = Velocity of wind = 10 m/s

r = Radius of circle = 5 m

S = Swept area

\rho = Density of air = 1.2 kg/m³

C_p = Power coefficient = 0.593

P=C_p\frac {1}{2}}\rho Sv^{3}\\\Rightarrow P=0.593\times \frac{1}{2}1.2 \pi \times 5^2\times 2^{3}\\\Rightarrow P=223.55\ W

The maximum possible power that can be produced by the turbine is 223.55 W

8 0
4 years ago
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