What is the momentum of a 0.5 kg ball which is thrown at 20 m/s?

Match each term with the appropriate definition.

Ratling [72]

Answer:

mass- the amount of matter in an object

balance- tool used to measure mass

scale- a tool used to measure weight

weight- the downward pull on an object due to gravity

8 0

2 years ago

An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 23.0 days on average to co

Gelneren [198K]

Answer:

See it in the pic

Explanation:

See it in the pic

7 0

3 years ago

Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w

gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

$U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}$

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

At x = 1.00 m

$u(1) = \frac{-2}{1} + \frac{4}{1^2}$

= 4J

Kinetic energy = (1/2)mv²

$= \frac{1}{2} * 4(3)^2$

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

$u(5) = \frac{-2}{5} + \frac{4}{5^2}$

$= \frac{4-10}{25} = \frac{-6}{25} J$

= -0.24J

Kinetic energy =

$\frac{1}{2} * 4Vf^2$

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

$Vf = \sqrt{frac{22.4}{2}$

= 3.33 m/s

b) magnitude of force when x = 5.0m

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

$\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}$

$= \frac{2}{x^2} - \frac{8}{x^3}$

At x = 5.0 m

$\frac{2}{5^2} - \frac{8}{5^3}$

$F = \frac{2}{25} - \frac{8}{125}$

= 0.016N

8 0

3 years ago

A 49.3 g ball of copper has a net charge of 2.0 µc. what fraction of the copper's electrons has been removed? (each copper atom

Serggg [28]

First, find how many copper atoms make up the ball:
moles of atoms = (49.3 g) / (63.5 g per mol of atoms) = 0.77638mol
 # of atoms = (0.77638 mol) (6.02 × 10^23 atoms per mol) = 4.6738*10^23 atoms 

 There is normally one electron for every proton in copper. This means there are normally 29 electrons per atom:
 normal # electrons = (4.6738 × 10^23 atoms) (29 electrons per atom) = 1.3554× 10^25 electrons

 Currently, the charge in the ball is 2.0 µC, which means -2.0 µC worth of electrons have been removed.
 # removed electrons = (-2.0 µC) / (1.602 × 10^-13 µC per electron) = 1.2484 × 10^13 electrons removed

 # removed electrons / normal # electrons = 
(1.2484 × 10^13 electrons removed) / (1.3554 × 10^25 electrons) = 9.21 × 10^-13 

 That's 1 / 9.21 × 10^13

7 0

2 years ago

How do defy gravity?

kaheart [24]

Exert force upward.
Like when you pick something up from the floor, or walk up the stairs.

6 0

3 years ago