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Igoryamba
3 years ago
13

What forces were exerted during the air resistance experiment?

Physics
1 answer:
zepelin [54]3 years ago
6 0

The forces that were exerted during the air resistance was the gravity.

<u>Explanation:</u>

Air resistance is the frictional power air applies against a moving article. As an article moves, air opposition backs it off. The quicker the article's movement, the more prominent the air obstruction applied against it.

The force acting upon the objects dropped in the air was air resistance and gravity. The force acting upon the objects dropped in the vacuum was gravity only.

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Which option most likely describes scientist(s) working in pure science?
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Two scientists in a lab examining vials of urine they are analyzing for levels of excreted protein.

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This circuit has five bulbs (A, B, C, D, and E). Select which bulbs would turn off if bulb C was unscrewed.
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Answer:

C, A and E

Explanation:

In theory you always assume that the prostive is going to the negtive and since C, A and E are in series they all will turn off will turn off

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Tasting and smelling both involve the ability to sense chemicals. true or false
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3 years ago
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Two circus performers rehearse a trick in which a ball and a dart collide. Horatio stands on a platform 9.8 m above the ground a
xenn [34]

Answer:

Time = 0.55 s

Height = 8.3 m

Explanation:

The ball is dropped and therefore has an initial velocity of 0. Its acceleration, g, is directed downward in the same direction as its displacement, h_b.

The dart is thrown up in which case acceleration, g, acts downward in an opposite direction to its displacement, h_d. Both collide after travelling for a time period, t. Let the height of the dart from the ground at collision be h_d and the distance travelled by the ball measured from the top be h_b.

It follows that h_d+h_b=9.8.

Applying the equation of motion to each body (h = v_0t + 0.5at^2),

Ball:

h_b=0\times t + 0.5\times 9.8t^2 (since v_{b0} =0.)

h_b=4.9t^2

Dart:

h_d=17.8\times t - 0.5\times9.8t^2 (the acceleration is opposite to the displacement, hence the negative sign)

h_d=17.8\times t - 4.9t^2

But

h_b+h_d =9.8

17.8\times t - 4.9t^2+4.9t^2 =9.8

17.8\times t = 9.8

t = 0.55

The height of the collision is the height of the dart above the ground, h_d.

h_d=17.8\times t - 4.9t^2

h_d=17.8\times 0.55 - 4.9\times(0.55)^2

h_d=9.79 - 1.48225

h_d=8.3

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Answer:

What are you asking

Explanation:

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