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g100num [7]
3 years ago
13

Does water pressure depend on the total amount of water present? (Will pick brainliest)

Physics
1 answer:
yaroslaw [1]3 years ago
3 0

Answer:

No, because pressure is determined by force and the area over which that force acts.

Explanation:

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A spring with spring-constant of 100 n/m is compressed 0.10 m. what is its maximum stored elastic potential energy
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Elastic potential energy stored in a spring is

(1/2) · (spring constant) · (stretch or compress)² .

PE = (1/2) · (100 N/m) · (0.1 m)²

PE = (50 N/m) · (0.01 m²)

PE = (50 · 0.01) (N · m / m²)

PE = 0.5 N · m

PE = 0.5 Joule

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3 years ago
Electricity that comes into our homes from the grid is known as:
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2 years ago
Four equal masses m are so small they can be treated as points, and they are equallyspaced along a long, stiff mass less wire. T
gavmur [86]

The moment of inertia of a point mass about an arbitrary point is given by:

I = mr²

I is the moment of inertia

m is the mass

r is the distance between the arbitrary point and the point mass

The center of mass of the system is located halfway between the 2 inner masses, therefore two masses lie ℓ/2 away from the center and the outer two masses lie 3ℓ/2 away from the center.

The total moment of inertia of the system is the sum of the moments of each mass, i.e.

I = ∑mr²

The moment of inertia of each of the two inner masses is

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The moment of inertia of each of the two outer masses is

I = m(3ℓ/2)² = 9mℓ²/4

The total moment of inertia of the system is

I = 2[mℓ²/4]+2[9mℓ²/4]

I = mℓ²/2+9mℓ²/2

I = 10mℓ²/2

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3 years ago
If we assume that, in the African savanna, the predator population remains constant, what other, nonmobile, factor would influen
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On usa test prep the answer is c
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3 years ago
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New 5G networks utilize millimeter-wave radiation. Millimeter-wave radiation refers to electromagnetic waves with frequencies in
seraphim [82]

Answer:

It corresponds to 1mm-10 mm range.

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        v = \lambda * f  (1)

  • Replacing v= c=3*10⁸ m/s, and the extreme values of f (which are givens), in (1) and solving for λ, we can get the free-space wavelengths that correspond to the 30-300 GHz range, as follows:

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      \lambda_{high} = \frac{c}{f_{low}}  = \frac{3e8m/s}{30e9Hz} = 10 mm (3)

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2 years ago
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