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3 years ago

If the sign of work is negative,

Physics

1 answer:

aivan3 [116]3 years ago

5 0

If the sign of work is negative, that means the force and the motion are in opposite directions.

Let's say you see something roll off of the shelf. You catch it, and you let it down slowly and gently.

Gravity exerted down-force on it and it moved down. Gravity did positive work on it.

YOU exerted UP-force on it and it moved down. YOU did negative work on it.

(Also, the falling object exerted down-force on your hand, and your hand moved down. The falling object did positive work on your hand ! Where did THAT energy come from ? It came from the potential energy that the object had while it was on the shelf. Your hand absorbed that energy on the way down, doing negative work. So the object didn't have any kinetic energy when it reached the floor, and it did NOT splinter the floor or shatter in smithereens. It had barely enough energy left to make a sound when it hit the floor.)

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3 years ago

Read 2 more answers

If two normal distributions each have the same mean, but one has a standard deviation of 1 and the other has a standard deviatio

likoan [24]

In comparison to a distribution with a standard deviation of 5, one has a curve with a higher peak. The variability will be greater the larger the standard deviation. It denotes increased variability in a distribution with a standard deviation of 5.

<h3>What do you mean by the term standard deviation?</h3>

The term "standard deviation" (or "") refers to a measurement of the data's dispersion from the mean. A low standard deviation implies that the data are grouped around the mean, whereas a large standard deviation shows that the data are more dispersed. In contrast, a high or low standard deviation indicates that the data points are, respectively, above or below the mean. A standard deviation that is close to zero implies that the data points are close to the mean. the curve at the top is more dispersed and has a greater standard deviation than the curve at the bottom, which is more concentrated around the mean and has a lower standard deviation.

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1 year ago

A straight segment of a current-carrying wire has a current element IL where I = 2.70 A and L = 3.20 cm i + 4.30 cm j. The segme

myrzilka [38]

The component of the force in negative z-direction is -0.144 N.

The given parameters;

<em>current in the wire, I = 2.7 A</em>
<em>length of the wire, L = (3.2 i + 4.3j) cm</em>
<em>magnetic filed, B = 1.24 i</em>

The force on the segment of the wire is calculated as follows;

$F = ILBsin(\theta)$

where;

<em>θ is the angle wire and magnetic field</em>

<em />

The force on the wire segment will be perpendicular in negative z-direction (applying right hand rule), so there won't be any x and y component of the force.

The angle between the wire and the magnetic field is calculated as follows;

$\theta = tan^{-1} (\frac{y}{x} )\\\\\theta = tan^{-1} (\frac{4.3}{3.2} )\\\\\theta = 53.3 \ ^0$

The magnitude of the wire length is calculated as follows;

$|l | = \sqrt{3.2^2 + 4.3^2} = 5.36 \ cm = 0.0536 \ m$

The component of the force in negative z-direction is calculated as;

$F_z = -ILB sin(\theta)\\\\F_z = -2.7 \times 0.0536 \times 1.24 \times sin(53.3)\\\\F_z = -0.144 \ N$

Thus, the component of the force in negative z-direction is -0.144 N.

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3 years ago