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andrew11 [14]
3 years ago
13

Look at the picture and answer what the time of day the earth is and what caused it to look like it.

Physics
2 answers:
katrin2010 [14]3 years ago
6 0

Answer:

The top picture is day and its bc the sun is facing that side of the Earth. The bottom pic is night time and that side of earth is not facing the sun :/

Explanation:

idunooo

oksano4ka [1.4K]3 years ago
4 0

First Square is Like 12:00pm The Next is Like 12:00am

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A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the
djverab [1.8K]

Complete Question:

A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns.

They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) i + (4.00 m/s2 ) j .At that instant and in unit-vector notation, what is the acceleration of the wallet

Answer:

aw = 3 i + 6 j m/s2

Explanation:

  • Since both objects travel in uniform circular motion, the only acceleration that they suffer is the centripetal one, that keeps them rotating.
  • It can be showed that the centripetal acceleration is directly proportional to the square of the angular velocity, as follows:

       a_{c} = \omega^{2} * r (1)

  • Since both objects are located on the same radial line, and they travel in uniform circular motion, by definition of angular velocity, both have the same angular velocity ω.

       ∴ ωp = ωw (2)

           ⇒ a_{p} = \omega_{p} ^{2} * r_{p} (3)

               a_{w} = \omega_{w}^{2} * r_{w} (4)

  • Dividing (4) by (3), from (2), we have:

        \frac{a_{w} }{a_{p}} = \frac{r_{w} }{r_{p}}

  • Solving for aw, we get:

        a_{w} = a_{p} *\frac{r_{w} }{r_{p} } = (2.0 i + 4.0 j) m/s2 * 1.5 = 3 i +6j m/s2

7 0
3 years ago
Holding onto a tow rope moving parallel to a frictionless ski slope, a 61.8 kg skier is pulled up the slope, which is at an angl
nata0808 [166]

Answer:

a) Frope= 71.7 N

b) Frope=6.7 N

Explanation:

In the figure the skier is simulated as an object, "a box".

a) At constant velocity we can say that the object is in equilibrium, so we apply the Newton's first law:

∑F=0

Frope=w*sen6.8°

Frope=71.71N

Take into account that w is the weight that is calculated as mass per gravitiy constant:

w=m*g

w=61.8Kg*9.8\frac{m}{s^{2} }

w=605.64N

b) In this case the system has an acceleration of 0.109m/s2.  Then, we apply Newton's second law of motion:

F=m*a

F=61.8Kg*0.109m/s2

Frope=6.73N

8 0
3 years ago
Develop expression for equation for uniform motion in a straight line
frozen [14]
According to cuneiform tablets in the ancient world, straight lines cannot cross, and no motion in the world is not relative. Btw...I KNOW!!! GOT MILK???
4 0
3 years ago
What is the resultant of the two vectors shown?
BARSIC [14]

Answer:

B is the right answer of the following statement

5 0
3 years ago
An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in
Gwar [14]

The final temperature is 83 K.

<u>Explanation</u>:

For an adiabatic process,

T {V}^{\gamma - 1} = \text{constant}

\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}

Given:-

{T}_{1} = 275 \; K  

{T}_{2} = T \left( \text{say} \right)

{V}_{1}  = V

{V}_{2} = 6V

\gamma = \cfrac{5}{3} \;    (the gas is monoatomic)

\therefore \cfrac{T}{275} = {\left( \cfrac{V}{6V} \right)}^{\frac{5}{3} - 1}

 

\Rightarrow \cfrac{T}{275} = {\left( \cfrac{1}{6} \right)}^{\frac{2}{3}}  

T  =  275 \times 0.30

T  =  83 K.

3 0
4 years ago
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