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3 years ago

PLZ EXPLAIN IM SO CONFUSED AND THIS IS DUE TONIGHT. I WILL GIVE 50 POINTS!

1 answer:

8 0

When you first pull back on the pendulum, and when you pull it back really high the Potential Energy is high and the Kinetic Energy is low, But when up let go, and it gets right around the middle, that's when the Potential energy transfers to Kinetic, at that point the kinetic Energy is high and the potential Energy is low. But when it comes back up at the end. The same thing will happen, the Potential Energy is high, and the Kinetic Energy is low. Through all of that the Mechanical Energy stays the same.

I hope this helps. :)

Brainliest?

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HELP ME ASAP PLS, ( zoom in on the picture )

il63 [147K]

Answer:

the direction of the velocity is downward and the acceleration decreases throughout the motion

Explanation:

since the gradient is negative it is decelerating

6 0

2 years ago

A car is to be hoisted by elevator to the fourth floor of a parking garage, which is 48 ft above the ground. If the elevator can

belka [17]

Answer: 21.91 s

Explanation:

Given that,

Maximum height of the car, h = 48 ft

Acceleration of the elevator, a = 0.6 ft/s²

Deceleration of the elevator, -a = 0.3 ft/s²

Maximum speed of the elevator, v = 8 ft/s

Initial speed of the elevator, u = 0

If when the elevator accelerate from 0 to maximum velocity, v.

Let s be the vertical distance traveled during acceleration.

v² = u² - 2as

s = (v² - u²) / 2a

s = (8² - 0) / 2*0.6

s = 64 / 1.2

s = 53.33 ft

If when the elevator decelerates from maximum velocity, v to zero.

Let S be the vertical distance traveled during deceleration

u² = v² + 2aS

S = (u² - v²) / 2a

S = (0 - 8²) / 2 * 0.3

S = -64 / 0.6

S = 106.67 ft

Since he sum of s and S (i.e s + S) is greater than 48 ft, then the elevator will switch from acceleration to deceleration

without reaching the maximum velocity. Below, the switching point is labeled y.

v² = u² + 2ay

y = v²/2a

Inserting this into the earlier deceleration equation, we have

-v²/2 = d * [48 - (v²/2a)], where

d = deceleration

a = acceleration

Therefore, v = [4.√6. a √-(a.b/a)] / b

Where b = acceleration - deceleration

v = 4.382 ft/s

Using this newly found v, we proceed to find our s

s = (u² + v²)/2a

s = 19.2 / 1.2

s = 16 ft

The transport times for each segment are found from

v = u + a*t, thus upward t1

4.382 = 0 + 0.6 * t

t = 4.382/0.6

t = 7.303 s

Also,

4.382 = 0 + 0.3 * T

T = 4.382/0.3

T = 14.607 s

The total travel time is then t + T =

7.303 + 14.607

Total time of travel is 21.91 s

5 0

3 years ago

Look at the variables for this lab. Which variable would we have to change in order to change the amount of current flowing thro

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Where are the variables from the lab?

3 0

2 years ago

Difference between freefall and weightlessness.

Margaret [11]

Answer:

Differences between freefall and weightlessness are as follows:

<h3><u>Freefall</u></h3>

When a body falls only under the influence of gravity, it is called free fall.
Freefall is not possible in absence of gravity.
A body falling in a vacuum is an example of free fall.

<h3><u>Weightlessness</u></h3>

Weightlessness is a condition at which the apparent weight of body becomes zero.
Weightlessness is possible in absence of gravity.
A man in a free falling lift is an example of weightlessness.

Hope this helps....

Good luck on your assignment....

6 0

3 years ago

The "Giant Swing" at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper e

Mashcka [7]

Answer:

Explanation:

When the central shaft rotates , the seat along with passenger also rotates . Their rotation requires a centripetal force of mw²R where m is mass of the passenger and w is the angular velocity and R is radius of the circle in which the passenger rotates.

This force is provided by a component of T , the tension in the rope from which the passenger hangs . If θ be the angle the rope makes with horizontal ,

T cos θ will provide the centripetal force . So

Tcosθ = mw²R

Tsinθ component will balance the weight .

Tsinθ = mg

Dividing the two equation

Tanθ = $\frac{g}{\omega^2R}$

Hence for a given w , θ depends upon g or weight .

8 0

3 years ago

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