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VMariaS [17]
2 years ago
9

Does the mass of a parachute affect terminal velocity?

Physics
1 answer:
vovangra [49]2 years ago
6 0

Answer:

The greater weight increases the terminal velocity by acting as an extra force against gravity and air resistance.

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Gravity can be described as..?
Tju [1.3M]

Answer:

D

Explanation:

Gravity is the force of attraction between two objects.

Each object creates a gravitational field in wich every other object is affected by it.

6 0
4 years ago
What are generated at transform fault plate boundaries?
tankabanditka [31]

Answer: The Earth's layer, which has the covering and layer, is made of a progression of things, or structural plates, that creep after some time. Along these lines, at intersecting limits, mainland outside is made and maritime covering is devastated. 2 plates slippy past each other structures a redesign plate limit.

5 0
3 years ago
Formular for coefficient of friction<br>​
Bumek [7]

Answer:

hope you like it mark as brainliest

Explanation:

Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There are several types of friction: Dry friction is a force that opposes the relative lateral motion of two solid surfaces in contact.

The coefficient of friction (fr) is a number that is the ratio of the resistive force of friction (Fr) divided by the normal or perpendicular force (N) pushing the objects together. It is represented by the equation: fr = Fr/N.

7 0
3 years ago
Which two ways dose inertia affect the monitors of the planets
dedylja [7]

Answer:

the inertia provides the tendency to maintain speed and keep moving

7 0
2 years ago
Determine the critical crack length for a through crack contained within a thick plate of 7150-T651 aluminum alloy that is in un
Mila [183]

Explanation:

Formula to determine the critical crack is as follows.

          K_{IC} = \gamma \sigma_{f} \sqrt{\pi \times a}

  \gamma = 1,     K_{IC} = 24.1

  [/tex]\sigma_{y}[/tex] = 570

and,   \sigma_{f} = 570 \times \frac{3}{4}

                       = 427.5

Hence, we will calculate the critical crack length as follows.

      a = \frac{1}{\pi} \times (\frac{K_{IC}}{\sigma_{f}})^{2}

        = \frac{1}{3.14} \times (\frac{24.1}{427.5})^{2}

       = 10.13 \times 10^{-4}

Therefore, largest size is as follows.

            Largest size = 2a

                                 = 2 \times 10.13 \times 10^{-4}

                                 = 20.26 \times 10^{-4}

Thus, we can conclude that the critical crack length for a through crack contained within the given plate is 20.26 \times 10^{-4}.

4 0
3 years ago
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