An object is falling downward at a rate of 25 m/s. Two seconds later, what is its acceleration?

At the Earth's surface, a projectile is launched straight up at a speed of 10.0km / s. To what height will it rise? Ignore air r

icang [17]

At h = 5199.28 km it will rise. Kinetic energy is a type of energy that is present in a particle or object in motion.

<h3>What is kinetic energy?</h3>

Kinetic energy is a type of energy that is present in a particle or object in motion. When work, which entails the transfer of energy, is done on an object by applying a net force, that object acquires kinetic energy.

There are five types of kinetic energy: mechanical, electrical, acoustic, thermal, and radiant. Kinetic energy (KE), which is the energy of a body in motion, is essentially the energy of all moving objects. It is one of the two main types of energy together with potential energy, which is the stored energy existing in objects at rest.

<h3>To what height will it rise?</h3>

Equating the initial kinetic energy to the final potential energy

At the highest point, the body has only potential energy

initial kinetic energy = .5 * m * v2

velocity = 10.1 km/s

= 10.1 * 10100 m/s

= 10100 m/s

KE = .5 * m * (10100)2 = 51.005 x 106 x m J

potential energy = m x g x h

max height reached = m x 9.81 x h

m x 9.81 x h = 51.005 x 106 x m J

h = 5199.28 km

To learn more about kinetic energy, refer to:

brainly.com/question/25959744

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6 0

1 year ago

(b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 1510 m/s? (c) With wh

makvit [3.9K]

Answers:

(a) $2509.98 m/s$

(b) 397042.215 m

Explanation:

The question is incomplete, please remember to write the whole question :) However, part (a) is written below:

(a) What is the escape speed on a spherical asteroid whose radius is $700 km$ and whose gravitational acceleration at the surface is $a_{g}=4.5 m/s^{2}$

Knowing this, let's begin:

a) In this part we need to find the escape speed $V_{e}$ on the asteroid:

$V_{e}=\sqrt{\frac{2GM}{R}}$ (1)

Where:

$G$ is the universal gravitational constant

$M$ is the mass of the asteroid

$R=700 km=700(10)^{3} m$ is the radius of the asteroid

On the other hand we know the gravitational acceleration is $a_{g}=4.5 m/s^{2}$ , which is given by:

$a_{g}=\frac{GM}{R^{2}}$ (2)

Isolating $GM$ :

$GM=a_{g}R^{2}$ (3)

Substituting (3) in (1):

$V_{e}=\sqrt{\frac{2a_{g}R^{2}}{R}}=\sqrt{2a_{g}R}$ (4)

$V_{e}=\sqrt{2(4.5 m/s^{2})(700(10)^{3} m)}$ (5)

$V_{e}=2509.98 m/s$ (6) This is the escape velocity

b) In this part we will use the Conservation of mechanical energy principle:

$E_{o}=E_{f}$ (7)

Being:

$E_{o}=K_{o}+U_{o}=\frac{1}{2}m V^{2} - \frac{GMm}{R}$ (8)

$E_{f}=K_{f}+U_{f}=0 - \frac{GMm}{R+h}$ (9)

Where:

$E_{o}$ is the initial mechanical energy

$E_{f}$ is the final mechanical energy

$K_{o}$ is the initial kinetic energy

$K_{f}=0$ is the final kinetic energy

$U_{o}$ is the initial gravitational potential energy

$U_{f}$ is the final gravitational potential energy

$m$ is the mass of the object

$V=1510 m/s$ is the radial speed of the object

$h$ is the distance above the surface of the object

Then:

$\frac{1}{2}m V^{2} - \frac{GMm}{R}=- \frac{GMm}{R+h}$ (10)

Isolating $h$ :

$h=\frac{2 a_{g} R^{2}}{2a_{g}R-V^{2}}-R$ (11)

$h=\frac{2 (4.5 m/s^{2}) (700(10)^{3} m)^{2}}{2(4.5 m/s^{2})(700(10)^{3} m)-(1510 m/s)^{2}}-700(10)^{3} m$ (11)

$h=397042.215 m$ (12) This is the distance above the asteroid's surface

c) We will use the Conservation of mechanical energy principle again, but now the condition is that the object is dropped at a distance $h=981.8 km=981.8(10)^{3} m$ . This means that at the begining the object only has gravitational potential energy and then it has kinetic energy and gravitational potential energy: