3 years ago

What happens to gravitational potential energy as a roller coaster moves down a hill?

Physics

1 answer:

Kaylis [27]3 years ago

5 0

It is converted to kinetic energy.

You might be interested in

For an object to slow down and stop due to friction while sliding across the floor which of the following would be true.

EastWind [94]

Answer:10-4

Explanation:

4 0

2 years ago

two engines are turned on for 763 s at a moment when the velocity of the craft has x and y components of v0x = 6380 m/s and v0y

svet-max [94.6K]

Answer:

Explanation:

Given

initial velocity component of engines is

$v_0_x=6380 m/s$

$v_0_y=6770 m/s$

time period of engine running=763 s

Displacement in $x=4.50\times 10^6$

$y=7.27\times 10^6$

Using $s=ut+\frac{at^2}{2}$ in x and y direction

$x=v_0_x\times t+\frac{at^2}{2}$

$4.50\times 10^6=6380\times 763+\frac{a\times 763^2}{2}$

$4.50\times 10^6-4.86\times 10^6=\frac{a\times 763^2}{2}$

$a=-1.23 m/s^2$

In y direction

$y=v_0_y\times t+\frac{a't^2}{2}$

$7.27\times 10^6=6770\times 763+\frac{a\times 763^2}{2}$

$7.27\times 10^6-5.16\times 10^6=\frac{a\times 763^2}{2}$

$a=7.24 m/s^2$

x component $=-1.23 m/s^2$

y component $=7.24 m/s^2$

3 0

2 years ago

Read 2 more answers

Vibrations that move through the ground carrying the energy released during an earthquake are called

Alekssandra [29.7K]

Answer is Seismic Wave

4 0

2 years ago

Choose the appropriate descriptor for the term cubic centimeter .

Arada [10]

Answer:volume

Explanation:

4 0

3 years ago

Read 2 more answers

A copper wire has a radius of 3.5 mm. When forces of a certain equal magnitude but opposite directions are applied to the ends o

denpristay [2]

Answer:

The tensile stress on the wire is 550 MPa.

Explanation:

Given;

Radius of copper wire, R = 3.5 mm

extension of the copper wire, e = 5.0×10⁻³ L

L is the original length of the copper wire,

Young's modulus for copper, Y = 11×10¹⁰Pa.

Young's modulus, Y is given as the ratio of tensile stress to tensile strain, measured in the same unit as Young's modulus.

$Y =\frac{Tensile \ stress}{Tensile \ strain} \\\\Tensile \ stress = Y*Tensile \ strain\\\\But, Tensile \ strain = \frac{extension}{original \ Length} = \frac{5.0*10^{-3} L}{L} = 5.0*10^{-3}\\\\Tensile \ stress = 11*10^{10} *5.0*10^{-3} \ = 550*10^6 \ Pa$

Therefore, the tensile stress on the wire is 550 MPa.

8 0

2 years ago