Which statement is true about the Big Bang Theory?

A fully loaded cart with a mass of 2200 kg starts from the top of a 12-meter hill on a roller coaster.

Salsk061 [2.6K]

Answer:

A. potential energy is 258720 Joule

Explanation:

A.Gravitational potential energy is: PE = m × g × h

velocity = 15.33 m/s when the car reaches the bottom of the hill.

where, m = mass

g = acceleration due to gravity

h = height from the bottom of hill.

The potential energy is : m×g×h

=(2200×9.8×12)

=258720 Joule

B. at the bottom of the hill, the potential energy is converted into kinetic energy so PE at top = KE at bottom

kinetic energy= $\frac{1}{2}$ ( $m*v^{2}$ )

where v = velocity

m= mass

therefore, v= $\sqrt\frac{2*K.E}{m} {}$

or, v= $\sqrt{\frac{2*258720}{2200} }$

or, v=15.33 m/s

7 0

3 years ago

Mountains are part of which earth sphere?? Am I correct?

sammy [17]

You are correct. Mountains are part of the lithosphere.

8 0

3 years ago

Read 2 more answers

So u see what had happened was i need help again..

liubo4ka [24]

first off lemme just say this is really easy man, just look at the directions

Blank #1: -23

Blank #2: 23

8 0

3 years ago

When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how f

Y_Kistochka [10]

Answer:

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Explanation:

Geometrically speaking, the distance between the rocket and the observer ( $r$ ), measured in kilometers, can be represented by a right triangle:

$r = \sqrt{x^{2}+y^{2}}$ (1)

Where:

$x$ - Horizontal distance between the rocket and the observer, measured in kilometers.

$y$ - Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket ( $\theta$ ), measured in sexagesimal degrees, is defined by the following trigonometric relation:

$\tan \theta = \frac{y}{x}$ (2)

If we know that $x = 5\,km$ , then the expression is:

$\tan \theta = \frac{y}{5}$

And the rate of change of this angle is determined by derivatives:

$\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y$

$\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}$

$\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}$

$\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}$

Where:

$\dot \theta$ - Rate of change of the angle of elevation, measured in sexagesimal degrees.

$\dot y$ - Vertical speed of the rocket, measured in kilometers per hour.

If we know that $y = 4\,km$ and $\dot y = 400\,\frac{km}{h}$ , then the rate of change of the angle of elevation is:

$\dot \theta = 48.780\,\frac{\circ}{s}$

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

3 0

3 years ago

I need help with questions b and d, that’s all. Thank you.

Mkey [24]

b). The power depends on the RATE at which work is done.

Power = (Work or Energy) / (time)

So to calculate it, you have to know how much work is done AND how much time that takes.

In part (a), you calculated the amount of work it takes to lift the car from the ground to Point-A. But the question doesn't tell us anywhere how much time that takes. So there's NO WAY to calculate the power needed to do it.

The more power is used, the faster the car is lifted. The less power is used, the slower the car creeps up the first hill. If the people in the car have a lot of time to sit and wait, the car can be dragged from the ground up to Point-A with a very very very small power ... you could do it with a hamster on a treadmill. That would just take a long time, but it could be done if the power is small enough.

Without knowing the time, we can't calculate the power.

...

d). Kinetic energy = (1/2) · (mass) · (speed squared)

On the way up, the car stops when it reaches point-A.

On the way down, the car leaves point-A from "rest".

WHILE it's at point-A, it has no speed. So it has no (zero) kinetic energy.

7 0

3 years ago