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Ray Of Light [21]

2 years ago

5

A package rests on the floor of an elevator that is rising with constant speed. the elevator exerts an upward normal force on th

e package, and hence does positive work on it.why doesn't the kinetic energy of the package increase?

1 answer:

Tomtit [17]2 years ago

6 0

Because the actual package isn't moving so the gravitational potential energy stays the same. kinetic energy is gained through motion of the object not the motion of the surroundings of the object so
unless the package is falling out of a window its losing gravitational potential energy and gaining kinetic energy. hope this helps :-)

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Can anyone help with these questions please

Svet_ta [14]

Here... I've finished this course already

6 0

3 years ago

A thin-walled cylindrical pressure vessel is subjected to an internal gauge pressure, p=75 psip=75 psi. It had a wall thickness

Mekhanik [1.2K]

To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as

$\sigma_h = \frac{Pd}{2t}$

Here,

P = Pressure

d = Diameter

t = Thickness

At the same time the longitudinal stress is given as,

$\sigma_l = \frac{Pd}{4t}$

The letters have the same meaning as before.

Then he hoop stress would be,

$\sigma_h = \frac{Pd}{2t}$

$\sigma_h = \frac{75 \times 8}{2\times 0.25}$

$\sigma_h = 1200psi$

And the longitudinal stress would be

$\sigma_l = \frac{Pd}{4t}$

$\sigma_l = \frac{75\times 8}{4\times 0.25}$

$\sigma_l = 600Psi$

The Mohr's circle is attached in a image to find the maximum shear stress, which is given as

$\tau_{max} = \frac{\sigma_h}{2}$

$\tau_{max} = \frac{1200}{2}$

$\tau_{max} = 600Psi$

Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi

6 0

3 years ago

The lever allows Jeff 150 pounds to lift a much greater weight 600 pounds because A) the larger force is applied over a longer d

kvasek [131]

Jeff uses all of his weight to lift the 600 lbs. C is your answer i do believe

4 0

2 years ago

If the charges attracting each other in the preceding problem have equal magnitudes,show that the magnitude of each charge is 2.

Schach [20]

Answer:

The magnitude of each charge is $2.82\times10^{-6}\ C$

Explanation:

Suppose the two point charges are separated by 6 cm. The attractive force between them is 20 N.

We need to calculate the magnitude of each charge

Using formula of force

$F=\dfrac{kq^2}{r^2}$

Where, q = charge

r = separation

Put the value into the formula

$20=\dfrac{9\times10^{9}\times q^2}{(6\times10^{-2})^2}$

$q^2=\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}$

$q=\sqrt{\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}}$

$q=2.82\times10^{-6}\ C$

Hence, The magnitude of each charge is $2.82\times10^{-6}\ C$

6 0

3 years ago

A housefly walking across a clean surface can accumulate a significant positive or negative charge. In one experiment, the large

Lisa [10]

Answer:

$4.56*10^8 \text{ electrons.}$

Explanation:

Since the fly accumulated a positive charge of +73pC, it must have lost an equal number of negative charge of -73pC to the surface (because the housefly was neutral to begin with).

Therefore, to answer our question we have to ask ourselves <em>how many electrons combine to make -73pC of charge? </em>

The answer is since one electron carries a charge of $-1.6*10^{-19}C$ , the number $n$ of electrons that make up -73pC $(-73*10^{-12}C)$ are

$n= \dfrac{-73*10^{-12}C}{-1.6*10^{-19}C/e}$

$\boxed{n= 4.56*10^8e.}$

Thus, the housefly lost about 456 million electrons to the surface!

4 0

3 years ago