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2 years ago

Normal force is exerted _____to the surface of an object.

Physics

2 answers:

denis23 [38]2 years ago

7 0

<h2>Answer:</h2>

<u>The correct option is </u><u>A. perpendicular</u>

<h2>Explanation:</h2>

In Physics, the normal force is the component of a contact force that is perpendicular to the surface that an object contacts. It is the support force exerted upon an object that is in contact with another stable object. Normal Force will be always acting opposite to the force falling on the surface. For example, if a book is resting upon a surface, then the surface is exerting an upward force upon the book in order to support the weight of the book.

Elanso [62]2 years ago

3 0

The normal force is always (underline, bold) is always perpendicular to the surface an object is sitting on. If the object is on an inclined plane, then the normal will not be vertical but it will be perpendicular to the angle of the incline.

The diagram below (left) shows a normal force (GH) that is not vertical, but it is perpendicular to the surface. The object on the right is the more usual normal a mass on a table top.

The vertical line on the right is the normal and it points up.

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Atmospheric pressure is due to the weight _______

Kryger [21]

Of the gravitational pull and other things like mass. Planet earth it,s self as you said sir.

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3 years ago

Friction is necessary when you are on a bike to stay

zzz [600]

Answer:

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2 years ago

Please help with this

sertanlavr [38]

Answer:I have to say 56

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2 years ago

A spring of spring constant k is attached to a support at the bottom of a ramp that makes an angle θ with the horizontal. A bloc

Nikitich [7]

Answer:

$x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

Explanation:

From the law of conservation of energy

Energy lost by the spring, W=Kinetic energy gained, KE+Potential energy gained, PE+Work done by friction, Fr

$0.5kd^{2}=0.5mv^{2}+mgLsin\theta+\mu_{k}mgcos\theta x$

$x(mgsin\theta+\mu_{k}mgcos\theta)=0.5kd^{2}$

$x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

The required distance from A to B is $x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

5 0

3 years ago

You know your mass is 62 kg but when you stand on a bathroom scale in an elevator it says your mass is 77 kg what is the acceler

lbvjy [14]

In a stationary situation, the weight of person is
$W=mg=(62 kg)(9.81 m/s^2) = 608.2 N$
This is the weight "felt" by the scale, which is basically the normal reaction applied by the scale on the person, and which uses the value of g (9.81) as reference to convert the weight (602.8 N) into a mass (62 kg).

When the person is in the elevator, the scale says 77 kg. The scale is still using the same value of conversion (9.81), so the apparent weight "felt" by the scale is
$W' = m'g=(77 kg)(9.81 m/s^2)=755.4 N$
This is the normal reaction applied by the scale on the person, and which is directed upward. Besides this force, there is still the weight W of the person, acting downward. So, if we use Newton's second law:
$\sum F = ma$
$W-W'=ma$
where a is the acceleration of the elevator. If we solve for a, we find
$a= \frac{W-W'}{m}= \frac{608.2N-755.4N}{62 kg}=-2.37 m/s^2$
The negative sign means the acceleration is in the opposite direction of g (which we take positive), so it means the elevator is going upward.

4 0

3 years ago