PlzHELP

1 answer:

8 0

Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion.Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.

To calculate displacement, simply draw a vector from your starting point to your final position and solve for the length of this line. If your starting and ending position are the same, like your circular 5K route, then your displacement is 0. In physics, displacement is represented by Δs.

For me to solve this I would need to know the time, but I can give you a handy displacement calculator I used that helped me.

https://www.easycalculation.com/physics/classical-physics/constant-acc-displacement.php

Hope I helped.

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a stone is dropped from the top of 50 m high tower simultaneously another stone is thrown upward with a speed of 20 m/s . calcul

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'H' = height at any time
'T' = time after both actions
'G' = acceleration of gravity
'S' = speed at the beginning of time
Let's call 'up' the positive direction.
Let's assume that the tossed stone is tossed from the ground, not from the tower.

For the stone dropped from the 50m tower:

H = +50 - (1/2) G T²

For the stone tossed upward from the ground:

H = +20T - (1/2) G T²

When the stones' paths cross, their Heights are equal.

50 - (1/2) G T² = 20T - (1/2) G T²

Wow ! Look at that ! Add (1/2) G T² to each side of that equation,
and all we have left is:

50 = 20T Isn't that incredible ? ! ?

Divide each side by 20 :

2.5 = T

The stones meet in the air 2.5 seconds after the drop/toss.

I want to see something:
What is their height, and what is the tossed stone doing, when they meet ?

Their height is +50 - (1/2) G T² = 19.375 meters

The speed of the tossed stone is +20 - (1/2) G T = +7.75 m/s ... still moving up.
I wanted to see whether the tossed stone had reached the peak of the toss,
and was falling when the dropped stone overtook it. The answer is no ... the
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This doesn’t make any sense

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Answer:

Explanation:

Force needed to apply start the box is greater than the force needed to keep it moving because static friction is greater than the kinetic friction .

A threshold force is needed to move the box and when box started to move kinetic friction comes into play.

Friction force is directly related to the weight of the box as the friction force is

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And Normal reaction is equal to the weight of box if no force is applied.

$f_r=\mu N$

$N=mg$

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