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Serhud [2]
3 years ago
11

Keisha is making a diagram of a simple machine in the body.

Physics
2 answers:
Amiraneli [1.4K]3 years ago
3 0

Answer:

Sorry for being late. It is...

A.) X: Load, Y: Fulcrum, Z: Lever

aleksandrvk [35]3 years ago
3 0

Answer:

A

Explanation:

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Hydraulic engineers often use, as a unit of volume of water, the "acre-foot", defined as the volume of water that will cover 1 a
Alex17521 [72]

Answer:

Volume = 1,015 acre-feet (Approx)

Explanation:

Given:

Rain = 1.7 in

Time = 30 min

Area = 29 km²

Find:

Volume in acre-feet

Computation:

1 km = 1,000 m

1 m = 3.28 feet

1 km² = 247.105 acre

d = 1.7 in = 1.7 / 12 = 0.14167 ft

Area = 29 × 247.105 = 7,166.045 acre

Volume = 7,166.045 acre × 0.14167 ft

Volume = 1,015 acre-feet (Approx)

7 0
3 years ago
Study the image of the moving car.
gayaneshka [121]

Answer:

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Explanation:

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6 0
2 years ago
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A wire of radius R has a current I uniformly distributed across its cross-sectional area. Ampere's law is used with a concentric
MrMuchimi

Answer:

Please refer to the figure.

Explanation:

The crucial point here is to calculate the enclosed current. If the current I is flowing through the whole cross-sectional area of the wire, the current density is

J = \frac{I}{\pi R^2}

The current density is constant for different parts of the wire. This idea is similar to that of the density of a glass of water is equal to the density of a whole bucket of water.

So,

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This enclosed current is now to be used in Ampere’s Law.

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Here, \int \, dl represents the circular path of radius r. So we can replace the integral with the circumference of the path, 2\pi r.

As a result, the magnetic field is

B = \frac{\mu_0}{2\pi}\frac{Ir}{R^2}

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Which best describes the history of scientific knowledge?
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A path of inferences guided to be cherry picked as for which ones were reasonable and which ones had no ability in the real world to sustain in scientific law
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A ball is thrown straight up. At the top of its path its acceleration is
timurjin [86]
C. Acceleration is the rate of change of velocity. So at the top of the path, while the velocity is zero, the CONSTANT GRAVITATIONAL ACCELERATION is about 10 m/s^2 (9.8)
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