What is 1960 in scientific notation

A large asteroid of mass 98700 kg is at rest far away from any planets or stars. A much smaller asteroid, of mass 780 kg, is in

Citrus2011 [14]

Answer:

1.81 x 10^-4 m/s

Explanation:

M = 98700 kg

m = 780 kg

d = 201 m

Let the speed of second asteroid is v.

The gravitational force between the two asteroids is balanced by the centripetal force on the second asteroid.

$\frac{GMm}{d^{2}}=\frac{mv^2}{d}$

$v=\sqrt{\frac{GM}{d}}$

Where, G be the universal gravitational constant.

G = 6.67 x 10^-11 Nm^2/kg^2

$v=\sqrt{\frac{6.67 \times 10^{-11}\times 98700}{201}}$

v = 1.81 x 10^-4 m/s

7 0

3 years ago

Calculate the amount of heat (in kJ) required to convert 97.6 g of water to steam at 100° C. (The molar heat of vaporization of

Ratling [72]

Answer:

221.17 kJ

Explanation: Note the heat of vaporization is in kJ/mol,then to determine the number of moles of water: divide the mass by 18. Then multiply the number of moles by the molar heat of vaporization of water.

N = 97.6 ÷ 18

Q=molar heat *moles

Q = (40.79) * (97.6 ÷ 18)

This is approximately 221.17 kJ

4 0

2 years ago

Now repeat the procedure for line 2. Count a steady beat, and time the motion of your finger so it crosses a dot at the end of e

Doss [256]

Answer:

The finger moves at a changing rate along Line 2.

Explanation: Hope this helped! :p

3 0

2 years ago

Read 2 more answers

The erg is a unit of work in units of centimeters (cm), grams (g), and seconds (s), and 1 erg=1 g⋅cm^2/s^2 . Recall that the SI

OverLord2011 [107]

For the given problem, the amount of work done expressed in ergs is 3200 ergs.

Answer: Option A

<u>Explanation: </u>

The work done on an objects are the force acting on it to move the object to a particular distance. So, work done on the object will be directly proportional to the force acting on it and the displacement.

Here, the force acting on the object is given as 0.010 N and the displacement of the object is 0.032 m. So, the work done on the object is

$\text { Work done }=\text { Force } \times \text { displacement }$

$\text { Work done }=0.010 \mathrm{N} \times 0.032 \mathrm{m}=0.00032 \mathrm{Nm}$

It is known that $1 N=1 \mathrm{kg} \mathrm{ms}^{-2}$

So, the work done can be expressed in $k g m s^{-2}$ as,

$\text { Work done }=0.00032 \mathrm{kgm}^{2} \mathrm{s}^{-2}$

It is known that $1 \mathrm{erg}=1 \mathrm{g} \mathrm{cm}^{2} / \mathrm{s}^{2}$ , so the conversion of units from Nm to erg will be done as follows:

$\text { Work done }=0.00032 \mathrm{kgm}^{2} \mathrm{s}^{-2} \times \frac{1000 \mathrm{g}}{1 \mathrm{kg}} \times \frac{100 * 100 \mathrm{cm}^{2}}{m^{2}}=3200 \mathrm{g} \mathrm{cm}^{2} \mathrm{s}^{-2}$

Thus, work done in ergs is 3200 ergs.

6 0

3 years ago

Suppose that the potential energy of a particle constrained

denpristay [2]

The graph of the potential energy pass through zero origin, cuts across positive x-axis and negative x-axis.

<h3>Slope of the graph</h3>

The slope of the graph when the potential energy of a particle constrained in on x-axis is determined as follows;

F = -dU/dx

-dU = Fdx

dU = -Fdx

U = -∫(kx + ax³)dx

U = kx²/2 - ax⁴/4

When, x = 0, U = 0

When U = 0;

$x = \pm \sqrt{\frac{2k}{a} }$

The graph of the potential energy pass through zero origin, cuts across positive x-axis and negative x-axis.

Complete question is below:

Suppose that the potential energy of a particle constrainedon x-axis and subjected to a force in the same direction, at a fuction of F(x) = -kx + ax³.

Find the functional form of the potential energy.

Learn more about potential energy here: brainly.com/question/1242059

#SPJ1

4 0

2 years ago