What are earths two main motions called

If the result of your calculation of a quantity has si units kg•m^2/s^2•C, that quantity could be

Novosadov [1.4K]

It would be Joules.
Workdone is measured in Joules.
Workdone = Force * distance
Force = mass * acceleration
= kg * ms⁻²
= kgms⁻²

Distance = m

So, Force * distance
kgms⁻² * m

Apply laws of indices that says
x² * x³ = x²⁺³ = x⁵

Therefore, It would be kgm²s⁻²
m¹ * m¹ = m¹⁺¹ = m²
s⁻² is also = s / 2

4 0

3 years ago

Read 2 more answers

What form of matter is "vapor"?

elena55 [62]

Vapor is usually made with water and it would be made with gas.

4 0

3 years ago

Which of the following statements are true?

Likurg_2 [28]

Answer:

Explanation:

(A) True: It is true.

In junction law, the current entering at a junction is equal to teh current leaving at the junction.

(B) False: It is false.

The kirchhoff's junction law is based on the conservation of charge.

(C) True: It is true.

Energy is used in the circuit.

(D) True: It is true.

It is based on the conservation of charge.

4 0

2 years ago

What is the eulerian description of fluid motion how does it differ from the lagrangian description?

Alex_Xolod [135]

Kinematics : Study of motion

Fluid kinematics : study of how fluid flows and how to describe its motion.

There are two ways to describe fluid motion

one is Eulerian, where the variations are described at all fixed stations as a function of time.

the other is Lagrangian, in which one follows all fluid particles and describes the variations around each fluid particle along its trajectory.

DIFFRENCE BETWEEN LAGRANGIAN AND EULERIAN:

1.Both Lagrangian and Eulerian describes time variation.

2. Eulerian describes the rate of change in one point of space

Lagrangian descries rate of change of a property of material system.

To know more about the Lagrangian and Eulerian :\brainly.com/question/14944792

#SPJ4

3 0

1 year ago

You drop a rock down a well that is 5.4 m deep. How long does it take the rock to hit the bottom of the well?

Natali5045456 [20]

Equation of motion:

$y_{f}=y_{o}+v_{o}t+ \frac{1}{2} at^{2}$

Since initial velocity is zero, the second term goes away:

$y_{f}=y_{o}+0+ \frac{1}{2} at^{2}$

$y_{f}=y_{o}+\frac{1}{2} at^{2}$

$y_{f}-y_{o}= \frac{1}{2} at^{2}$

$y_{f}-y_{o}=5.4m$

$5.4m= \frac{1}{2} at^{2}$

$\frac{2(5.4m) }{a} = t^{2}$

$a = g = 9.81 \frac{m}{ s^{2}}$

$\frac{2(5.4m) }{9.81 \frac{m}{ s^{2} } } = t^{2}$

$1.1 s^{2} = t^{2}$ $\sqrt{1.1 s^{2}} = \sqrt{t^{2}}$

 $t = 1.05s$ 

4 0

3 years ago

Read 2 more answers