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3 years ago

At what points in the model do the cars have maximum and minimum gravitational potential energy?

Physics

1 answer:

Leviafan [203]3 years ago

8 0

Explanation:

The gravitational potential energy is given by :

E = mgh

Here, mg is force acting on the car

h is height at which it is placed

It is directly proportional to the height or the position of the car. It means if the height is less then it will have minimum potential energy while if the height is more then it will have maximum potential energy.

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You are riding a broomstick and see a golden ball hovering motionless 30 meters directly in front

Alisiya [41]

You should fly at an angle of $14.0^{\circ}$

Explanation:

The motion of the ball is a free fall motion, so the vertical displacement of the ball (in negative directions) is given by

$y=\frac{1}{2}gt^2$

where

$g=9.8 m/s^2$ is the acceleration due to gravity

t is the time

The horizontal distance you have to cover to catch the ball is

$x = 30 m=d$

The vertical and horizontal displacement form a right triangle, so that we can write

$tan \theta = \frac{y}{x}$

where $\theta$ is the direction of your motion. Substituting,

$tan \theta = \frac{gt^2}{2d}$ (1)

We also know that the horizontal position of the person at time t is

$x(t) = v cos \theta t$

And so, for the time t at which the person catches the ball,

$d=v cos \theta t\\t=\frac{d}{v cos \theta}\\t^2=\frac{d^2}{v^2 cos^2 \theta}$

Substituting this expression into (1),

$tan \theta = \frac{g}{2d} \frac{d^2}{v^2 cos^2 \theta}\\2 tan \theta cos^2 \theta = \frac{gd}{2v^2}$

And using

$tan \theta = \frac{sin \theta}{cos \theta}, 2sin \theta cos \theta = sin(2\theta)$

we find:

$sin (2\theta) = \frac{gd}{v^2}=\frac{(9.8)(30)}{25^2}=0.470\\2\theta = sin^{-1}(0.470)=28.0^{\circ}\\\theta=\frac{28}{2}=14.0^{\circ}$

Learn more about free fall motion:

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#LearnwithBrainly

4 0

3 years ago

When the Moon is in the position shown, how would the Moon look from Earth?

Vlad [161]

Answer:When the Moon is in the position shown below, how would the Moon look to an observer on the North Pole? The moon's rotational period is equal to its revolutional period around Earth. This statement is false because at night we see one side and during the day we see the opposite.

Explanation:

8 0

3 years ago

A car starts from rest, speeds up with constant acceleration, and travels 400 meters in 10 seconds. What was the acceleration of

PIT_PIT [208]

Answer:

Acceleration of the car will be $a=8m/sec^2$

Explanation:

We have given that car starts from rest so initial velocity of the car u = 0 m/sec

And car traveled 400 m in 10 sec

So distance traveled by car s = 400 m

Time taken to compete this distance t = 10 sec

We have to find the acceleration of the car

From second equation of motion we know that $s=ut+\frac{1}{2}at^2$

So $400=0\times 10+\frac{1}{2}\times a\times 10^2$

$a=8m/sec^2$

So acceleration of the car will be $a=8m/sec^2$

7 0

3 years ago

Four forces act at a point in the image above. calculate the resultant.

levacccp [35]

Answer:

50° and 60°

Explanation:

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3 0

3 years ago

A car that increases its speed from 20 km/h to 100 km/h undergoes

Kamila [148]

Answer:

accelaration

Explanation:

3 0

3 years ago