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valentinak56 [21]
1 year ago
11

A 375-pound concrete cylinder has a base area of 144 square inches. with the cylinder resting on its base, the pressure exerted

by the concrete cylinder is greater than sea level pressure (i.e., 14.7 psi). the formula for pressure is:
Physics
1 answer:
Akimi4 [234]1 year ago
6 0

The pressure exerted by the concrete cylinder is 2.60 pound/in².

We need to know about the pressure to solve this problem. Pressure is a unit that describes how much force is applied to a surface area. It can be determined as

P = F / A

where P is pressure, F is force and A is area.

From the question above, we know that

F = 375 pound

A = 144 in²

By substituting the given parameters, we can calculate the pressure

P = F / A

P = 375 / 144

P = 2.60 pound/in²

Thus, the pressure should be 2.60 pound/in².

Find more on pressure at: brainly.com/question/25965960

#SPJ4

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A is just the most logical. Sagittarius doesn't mean anything here.

Light does not originate anywhere. DNA is not formed in a black hole.

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The speed of an object and the direction in which it moves constitute a vector quantity known as the velocity. an ostrich is run
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6 0
4 years ago
Calculate the power if the current in a circuit is 3A and the resistance of the circuit is 50hms
Tresset [83]

Answer:

450W

Explanation:

V= I*R

V=3A*50 Ohms

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3 0
3 years ago
A positive point charge Q is located at x=a and a negative point charge −Q is at x=−a. A positive charge q can be placed anywher
Scilla [17]

Answer:

\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x

Explanation:

The Coulomb's Law gives the force by the charges:

\vec{F} = K\frac{q_1q_2}{r^2}\^r

Let us denote the positon of the charge q on the y-axis as 'y'.

The force between 'Q' and'q' is

F_1 = K\frac{Qq}{x^2 + y^2}\\F_1_x = F_1\cos(\theta)

where Θ is the angle between F_1 and x-axis.

F_1_x = K\frac{Qq}{x^2 + y^2}(\frac{x}{\sqrt{x^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}

whereas

F_2_x = K\frac{-Qq}{a^2 + y^2}(-\frac{a}{\sqrt{a^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}

Finally, the x-component of the net force is

\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x

8 0
4 years ago
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