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3 years ago

12) Does the label “centripetal force” ever appear in an FBD?

1 answer:

6 0

"Though a force, or a resulting sum of forces, may be centripetal, there is not an actual "Centripetal Force" that would appear on an FBD. Since an FBD is a diagram of physical forces at work upon a body, and a centripetal force is not the reaction of some physical object upon another, it would not be proper to show it. You might have a tension force, a weight force, a normal force, electric forces, magnetic forces, or any number of reactions, but actually labeling any of them "centripetal" would be unconventional".
-Yahoo

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A 1.0-cm-diameter pipe widens to 2.0, then narrows to 0.5. Liquid flows through the first segment at a speed of 4.0. What is the

uysha [10]

Answer:

3.14 × 10⁻⁴ m³ /s

Explanation:

The flow rate (Q) of a fluid is passing through different cross-sections remains of pipe always remains the same.

Q = Area x velocity

Given:

Diameters of 3 sections of the pipe are given as

d1 = 1.0 cm, d2 = 2.0 cm and d3 = 0.5 cm.

Speed in the first segment of the pipe is

v1 = 4 m/s.

From the equation of continuity the flow rate through different cross-sections remains the same.

Flow rate = Q = A1 v1 = A2 v2 = A3 v3.

Q = A1v1

=π/4 d²1 v1 = π/4 * 0.01² ×4.0 m³/s = 3.14 × 10⁻⁴ m³ /s

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2 years ago

A vertical spring (ignore its mass), whose spring constant is 1070 N/m, is attached to a table and is compressed 0.100 m.

harina [27]

I can not solve the problem if I do not have the mass.

3 0

2 years ago

A force is applied to a block sliding along a surface (Figure 2). The magnitude of the force is 15 N, and the horizontal compone

Lemur [1.5K]

Answer:

Fy = 14.3 [N]

Explanation:

To be able to solve this problem we must know that the force is a vector and has magnitude and direction, therefore it can be decomposed into the force in the X & y components:

When we have the components on the horizontal and vertical axes we must use the Pythagorean theorem.

$F = \sqrt{F_{x}^{2} +F_{y}^{2} }$

where:

F = 15 [N]

Fx = horizontal component = 4.5 [N]

Fy = vertical component [N]

$15=\sqrt{4.5^{2}+F_{y}^{2}}\\ 15^{2}= (\sqrt{4.5^{2}+F_{y}^{2}})^{2} \\225 = 4.5^{2}+F_{y} ^{2}\\ F_{y}^{2} =225 -4.5^{2}\\ F_{y}^{2}=204.75\\F_{y}=\sqrt{204.75}\\ F_{y}=14.3 [N]$

7 0

2 years ago

(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th

julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of r from a point charge, Q, is given by

$V = -\dfrac{kQ}{r}$

where $k = 9\times 10^9 \text{ F/m}$

Difference in potential between the points is

$kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}$