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3 years ago

Which formula can be used to find the x-component of the resultant vector?

1 answer:

5 0

If we are dealing with the usual situation where we have a vector in terms of length and an angle, we could find the x-component using the formula:

x = r*cos(theta)

where r is the length of the vector, and theta is the angle.

We might need more information to say what the exact formula would be.

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Determine the maximum shearing stress in a solid shaft of 1.5-in. diameter as it transmits 75 hp at a speed of 1800 rpm. (Round

erma4kov [3.2K]

Answer:

$\tau=3.96\ ksi$

Explanation:

Given that

d= 1.5 in ( 1 in = 0.0254 m)

d= 0.0381 m

P= 75 hp ( 1 hp = 745.7 W)

P= 55927.5 W

N= 1800 rpm

We know that power P is given as

$P=\dfrac{2\pi N\ T}{60}$

T=Torque

N=Speed

$55927.5=\dfrac{2\times \pi \times 1800\ T}{60}$

T=296.85 N.m

The maximum shear stress is given as

$\tau=\dfrac{16 T}{\pi d^3}$

$\tau=\dfrac{16\times 296.85}{\pi \times 0.0381^3}$

$\tau=27.35\ MPa$

We know that 1 MPa =0.145 ksi

$\tau=3.96\ ksi$

3 0

3 years ago

Suppose the acceleration of the particle moving gin a circle of radius "r" with the uniform speed "v" is proportional to some po

garik1379 [7]

So i say the power or the radius cortex in my equinox when i look at this V is the determination of V

5 0

2 years ago

Consider this situation: A rope is used to lift an actor upward off the stage. Of the forces listed, identify which act upon the

Katarina [22]

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6 0

3 years ago

Read 2 more answers

If an atom has an atomic number of 2, it will be stable with 2 electrons in its valence shell. Group of answer choices True Fals

torisob [31]

Answer: True

Explanation:

Atomic number is defined as the number of protons or the number of electrons that are present in an electrically neutral atom.

Atomic number = Number of protons = number of electrons = 2

Electronic configuration represents the total number of electrons that a neutral element contains. We add all the superscripts to know the number of electrons in an atom.

The electronic configuration will be $1s^2$

As its duplet is already complete and it has noble gas configuration , it is stable with 2 valence electrons.

3 0

3 years ago

Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted

galina1969 [7]

Answer:

1) When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

$\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}$

$Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}$

Where:

f = Focal length of the mirror = R/2

$d_{i}$ = Image distance from the mirror

$d_{o}$ = Object distance from the mirror

$h_{i}$ = Image height

$h_{o}$ = Object height

$d_{o}$ is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

$d_{i}$ is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

5 0

3 years ago