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barxatty [35]
3 years ago
11

What is the center of gravity

Physics
2 answers:
Gala2k [10]3 years ago
7 0
The center of gravity is a SIKKE bro you really thought I’d give the answer of something so simple
ivanzaharov [21]3 years ago
3 0

Answer:

The center of gravity is the average location of the weight of an object

Explanation:

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If the mean velocity adjacent to the top of a wing of 1.8 m chord is 40 m/s and that adjacent to the bottom of the wing is 31 m/
Cloud [144]

Answer:

lift per meter of span = 702 N/m

Explanation:

See attached pictures.

7 0
3 years ago
A 123-turn circular coil of radius 2.41 cm is immersed in a uniform magnetic field that is perpendicular to the plane of the coi
klasskru [66]

Answer:

67.44 V

Explanation:

Number of turns N =123

Radius = 2.41 cm =0.0241 m

The magnetic field strength increases from 50.9 T to 96.3 T so change in magnetic field dB = 96.3-50.9=45.4 T

Time interval dt = 0.151 sec

We know that the induced emf e=-NA\frac{dB}{dt}

Area A=\pi r^2=3.14\times 0.0241^2=1.8237\times 10^{-3}m^2

Putting all these values in emf equation e=-123\times 1.8237\times 10^{-3}\times \frac{45.4}{0.151}=-67.44\ V here negative sign indicates that it opposes the cause due to which it is produced

8 0
3 years ago
What is a solvent and a solute
artcher [175]

Answer:

solvent: the substance in which a solute dissolves to produce a homogeneous mixture. solute: the substance that dissolves in a solvent to produce a homogeneous mixture.

Explanation:

eSo basically  thats  aswer

7 0
3 years ago
A 9.6 cm diameter circular loop of wire is in a 1.10 T magnetic field perpendicular to the plane of the loop. The loop is remove
Natali5045456 [20]

Answer:

Thus induced emf is 0.0531 V

Solution:

As per the question:

Diameter of the loop, d = 9.6\ cm = 0.096\ m

Thus the radius of the loop, R = 0.048 m

Time in which the loop is removed, t = 0.15 s

Magnetic field, B = 1.10 T

Now,

The average induced emf, e is given by Lenz Law:

e = - \frac{\Delta \phi_{B}}{\Delta t}

e = - \frac{\Delta \phi_{B}}{\Delta t}

where

\phi_{B} = magnetic flux = A\Delta B

where

A = cross sectional area

Also, we know that:

e = - \frac{A\Delta B}{\Delta t}

e = - \frac{\pi r^{2}\times (0 - 1.10)}{0.15}

e = - \frac{\pi \times 0.048^{2}\times (0 - 1.10)}{0.15}

e = 0.0531 V

The sketch is shown in the figure, where I indicates the direction of the induced current.

3 0
3 years ago
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Which is brighter in our sky, a star with apparent magnitude 5 or a star with apparent magnitude 10 ?
andrew-mc [135]
The answer to that would be a magnitude of 5

4 0
3 years ago
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