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deff fn [24]
3 years ago
6

Stand on a bathroom scale on a level floor, and the reading on the scale shows the gravitational force on you, mg. If the floor

is slanted at an angle, the scale reading will be less than mg. Discuss why this is so, and why it is a good idea to measure your weight when the scale is on a horizontal surface.
Physics
1 answer:
kari74 [83]3 years ago
7 0

Answer:

<em>The angle introduces an error on the mesure of the weight</em>

Explanation:

<u><em>Weight and Normal Forces</em></u>

When an object is resting on a horizontal surface, its weight is directed downwards and the normal force has the same magnitude and opposite direction, i.e. directed upwards. When some angle α exists between the surface and the horizontal plane, the scale keeps 'feeling' the Normal force, but it's not equal to the weight anymore, but to the perpendicular component of the weight to the surface where the scale is placed. It can be found that the component of the weight is m.g.cosα

If, for example \alpha=10^o, the real measure is

m.g.cos 10^o=0.985(mg)

That is, 98.5% of the correct weight. So it's important to measure weight on horizontal surfaces

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Answer:

Conductors.

Explanation:

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In magazine car tests an important indicator of performance is the zero to 60 mph (0 to 96.6 km/h) acceleration time. A time bel
slavikrds [6]

Answer:

2.73414 seconds

467622.66798 J

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

v=60\times \dfrac{1609.34}{3600}=26.822\ m/s

mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{26.822^2}{2\times 9.81}\\\Rightarrow h=36.66766\ m

s=ut+\frac{1}{2}at^2\\\Rightarrow 36.66766=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{36.66766\times 2}{9.81}}\\\Rightarrow t=2.73414\ s

or

v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{26.822-0}{9.81}\\\Rightarrow t=2.73414\ s

The time taken is 2.73414 seconds

The potential energy is given by

U=mgh\\\Rightarrow U=1300\times 9.81\times 36.66766\\\Rightarrow U=467622.66798\ J

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Answer:

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Therefore, if you square the year of each planet and divide it by the distance that it is from the star, you will get the same number for all the other planets.

Thus, the units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.

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