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Nonamiya [84]
3 years ago
9

Andrew is pushing this block to the right across on the floor. Which force in the free body diagram is pointing in the wrong dir

ection?
Question 11 options:

the applied force


friction


the normal force


the weight

Physics
2 answers:
soldier1979 [14.2K]3 years ago
8 0

Answer:

the normal force

Explanation:

The free-body diagram represents all the forces acting on an object. In this example, there are four forces acting on the box: an applied force, the friction (which always act opposite to the applied force), the weight of the box (which is always downward), and the normal force.

The normal force is the reaction force exerted by the surface on which the box is moving on the box, and this reaction force is always opposite to the force exerted by the box on the surface. Since the latter is downward, it means that the normal force must be upward, so in the diagram it is wrong.

Deffense [45]3 years ago
6 0

Normal force since it's perpendicular to the contact force and should be going up.

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Well, they're not quite the way Newton expressed it, but out of all this mess of statements, there are two that are correct AND come from Newton's 2nd Law of Motion:

<em>-- The smaller the mass of an object, the greater the acceleration of that object when a force is applied. </em>

<em>-- The greater the force applied, the greater the acceleration.</em>

For the <u><em>other </em></u>statements in the question:

-- <em>Every reaction is equal to the force applied.</em>  True; comes from Newton's <u><em>3rd</em></u> law of motion.

-- <em>Forces are balanced when they are equal and opposite.</em>  True; kind of a definition, not from Newton's laws of motion.

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The sound from a loud speaker has an intensity level of 80 db at a distance of 2.0m. Consider the speaker to be a point source,
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Answer:

2.83m

Explanation:

The information that we have is

Intensity at 2.0 m: I=80dB and r_{1}=2m

we need an intensity level of: I_{2}=40dB

thus, we are looking for the distance r_{2}.

which we can find with the law for intensity and distance:

(\frac{r_{2}}{r_{1}} )^2=\frac{I_{1}}{I_{2}}

we solve for r_{2}:

\frac{r_{2}}{r_{1}}=\sqrt{\frac{I_{1}}{I_{2}} }\\\\r_{2}=r_{1}\sqrt{\frac{I_{1}}{I_{2}} }

and we substitute the known values:

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