How much ancient plant debris is necessary to form one foot of coal

The period of rotation of Mars is 1 day and 37 minutes. Determine its frequency of rotation in Hertz.

Sholpan [36]

The frequency of rotation of Mars is 0.0000113 Hertz.

Given the following data:

Period = 1 day and 37 minutes.

To find the frequency of rotation in Hertz:

First of all, we would convert the the value of period in days and minutes to seconds because the period of oscillation of a physical object is measured in seconds.

Conversion:

1 day = 24 hours

24 hours to minutes = $60$ × $24$ = $1440$ minutes

$1440 + 37 = 1477 \; minutes$

1 minute = 60 seconds

1477 minute = X seconds

Cross-multiplying, we have:

$X = 60$ × $1477$

X = 88620 seconds

Now, we can find the frequency of rotation of Mars by using the formula:

$Frequency = \frac{1}{Period}\\\\Frequency = \frac{1}{88620}$

Frequency of rotation = 0.0000113 Hertz

Therefore, the frequency of rotation of Mars is 0.0000113 Hertz.

8 0

2 years ago

What systems of the body are involved in preparing and eating a sandwich

jek_recluse [69]

The systems of the body involved in preparing and eating a sandwich are :
The skeletal, muscular, and digestive system. They are all directly involved in eating and preparing a sandwich.

5 0

3 years ago

A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 152 om and makes an angl

aliya0001 [1]

Answer:

D₂= 167,21 cm : Magnitude of the second displacement

β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement

Explanation:

We find the x-y components for the given vectors:

i: unit vector in x direction

j:unit vector in y direction

D₁: Displacement Vector 1

D₂: Displacement Vector 2

R= resulta displacement vector

D₁= 152*cos110°(i)+152*sin110°(j)=-51.99i+142.83j

D₂= -D₂(i)-D₂(j)

R= 131*cos38°(i)+ 131*sin38°(j) = 103.23i+80.65j

We propose the vector equation for sum of vectors:

D₁+ D₂= R

-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j

-51.99i+D₂x(i)=103.23i

D₂x=103.23+51.99=155.22 cm

+142.83j-D₂y(j) =+80.65j

D₂y=142.83-80.65=62.18 cm

Magnitude and direction of the second displacement

$D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2} }$

$D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2} }$

D₂= 167.21 cm

Direction of the second displacement

$\beta = tan^{-1} \frac{D_{y}}{D_{x} }$

$\beta = tan^{-1} \frac{62.18}{155.22 }$

β= 21.8°

D₂= 167,21 cm : Magnitude of the second displacement

β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

6 0

3 years ago

Can we use momentum to see how fast the earth is going?

Kisachek [45]

Yes, if we know the Earth's mass

Explanation:

The momentum of an object is a vector quantity given by the equation

$p=mv$

where

m is the mass of the object

v is its velocity

In this case, we are asked if we can find the velocity of the Earth by starting from its momentum. Indeed, we can. In fact, we can rewrite the equation above as

$v=\frac{p}{m}$