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3 years ago

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LW Hydraulic engineers in the United States often use, as a unit of volume of water, the acre-foot, defined as the volume of wat

er that will cover 1 acre of land to a depth of 1 ft. A severe thunderstorm dumped 2.0 in. of rain in 30 min on a town of area 26 km2. What volume of water, in acre-feet, fell on the town?

1 answer:

ruslelena [56]3 years ago

4 0

Answer:

volume of water is 1066.50 acre feet

Explanation:

given data

dia = 2 in = 0.166 feet

area = 26 km² = 26 × 247.105 acre = 6424.73 acre

to find out

volume of water

solution

we know volume of water is express as

volume = area × dia ....................1

put here value in equation 1

volume = 6424.73 acre × 0.166 feet

volume = 1066.50

so volume of water is 1066.50 acre feet

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$F(x)=-3 N - (3N)x^2$

Explanation:

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If we use the potential energy function given in this problem:

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$F(x)=-\frac{d}{dx}(3x+x^3)=-3-3x^2$

So, the force is

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On a playground, there is a merry‑go‑round. In order to get it moving, Bonnie applies a force of 31 N31 N . The merry‑go‑round m

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Answer:

The magnitude of the torque is 263.5 N.

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3 years ago

Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center

kiruha [24]

Answer:

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Explanation:

From the question we are told that

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The initial angular velocity is $w_1 = 0.5 \ rad /s$

The distance between the two astronauts after the rope is pulled is $d_f = 4 \ m$

Generally the radius is mathematically represented as $r_f = \frac{d_f}{2} = \frac{4}{2} = 2\ m$

Generally from the law of angular momentum conservation we have that

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Here $I_{k_1 }$ is the initial moment of inertia of the first astronauts which is equal to $I_{p_1}$ the initial moment of inertia of the second astronauts So

$I_{k_1} = I_{p_1 } = m * r_i^2$

Also $w_{k_1 }$ is the initial angular velocity of the first astronauts which is equal to $w_{p_1}$ the initial angular velocity of the second astronauts So

$w_{k_1} =w_{p_1 } = w_1$

Here $I_{k_2 }$ is the final moment of inertia of the first astronauts which is equal to $I_{p_2}$ the final moment of inertia of the second astronauts So

$I_{k_2} = I_{p_2} = m * r_f^2$

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$w_{k_2} =w_{p_2 } = w_2$

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=> $w_f = \frac{3.5^2 * 0.5}{ 2^2 }$

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Metal sphere A has a charge of − Q . −Q. An identical metal sphere B has a charge of + 2 Q . +2Q. The magnitude of the electric

Goshia [24]

Complete Question:

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A. 2F

B. F/4

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D. F

E. 4F

Answer:

D.

Explanation:

If both spheres can be treated as point charges, they must obey the Coulomb's law, that can be written as follows (in magnitude):

$F =\frac{kQ*2Q}{r^{2} }$

As it can be seen, this force is proportional to the product of the charges, so it must be the same for both charges.

As this force obeys also the Newton's 3rd Law, we conclude that the magnitude of the electric force on sphere A due to sphere B, must be equal to the the magnitude of the force on the sphere B due to the sphere A, i.e., just F.

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