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stira [4]
3 years ago
8

A box fails to slide down a ramp at a warehouse. what happens if you put the box on a cart that has wheels?

Physics
2 answers:
My name is Ann [436]3 years ago
5 0

Well the Wheels would help to move the box but you need to have in mind that if the box is heavy you must make sure that it does not role-down because of how heavy the box is.

Explanation:

Wheels would help to move the box down the ramp; yet, if the box is large you must make sure that it does not role-down supporting its own weight.

The ramp on an auto transport truck is handled to lift cars high into the air. An oblique plane makes it easier to raise something heavy, like a rock. Rather of lifting the rock straight up, you can raise it from its initial location with less force by shouldering it up a ramp.


amm18123 years ago
4 0
<span>Well the Wheels would help to move the box but you need to have in mind that if the box is heavy you must make sure that it does not role-down because of how heavy the box is</span>
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How is electron movement related to the bonding in sodium chloride? A) The atomic orbitals overlap and share electrons causing a
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Answer: Option (C) is the correct answer.

Explanation:

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Two rockets are flying in the same direction and are side by side at the instant their retrorockets fire. Rocket A has an initia
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Answer:

a_2\ =\ -33.65\ m/s^2

Explanation:

Given,

For the first rocket,

  • Initial velocity of the first rocket A = u_1\ =\ 4600\ m/s.
  • Acceleration of the first rocket = a_1\ =\ -18\ m/s^2

For the second rocket,

  • Initial velocity of the second rocket B = u_2\ =\ 8200 m/s.
  • Displacement of both the rockets A and B = s = 0 m

Fro the first rocket,

Let 't' be the time taken by the first rocket A for whole the displacement

\therefore s\ =\ u_1t\ +\ \dfrac{1}{2}a_1t^2\\\Rightarrow 0\ =\ 4600t\ -\ 0.5\times 18t^2\\\Rightarrow t\ =\ \dfrac{4600}{0.5\times 18}\\\Rightarrow t\ =\ 511.11 sec

Let a_2 be the acceleration of the second rocket B for the same time interval

from the kinematics,

\therefore s\ =\ ut\ +\ \dfrac{1}{2}at^2\\\Rightarrow s\ =\ u_2t\ +\ \dfrac{1}{2}a_2t^2\\\Rightarrow a_2\ =\ \dfrac{2s\ -\ 2u_2t}{t^2}\\\Rightarrow a_2\ =\ \dfrac{0\ -\ 2u_2t}{t^2}\\

\Rightarrow a_2\ =\ -\dfrac{2u_2}{t}\\\Rightarrow a_2\ =\ -\dfrac{2\times 8600}{511.11}\\\Rightarrow a_2\ =\ -33.65\ m/s^2

Hence the acceleration of the second rocket B is -33.65\ m/s^2.

6 0
3 years ago
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