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fgiga [73]
3 years ago
10

A quelle distance le soletl se trouve-t-il de la terre?

Physics
1 answer:
Mnenie [13.5K]3 years ago
5 0

The explanation for the following answer is explained below.

Explanation:

The sun is at an average distance of about 93,000,000 miles(150 million kilometers) away from the earth.It is so far away that light from the Sun,travelling at a speed of 186,000 miles (300,000 kilometers) per second, takes about 8 minutes to reach the earth.Earth does not travel around the Sun in a perfect circle.Instead its orbit is elliptical,like a stretched circle,with the sun just off the center of the orbit. At its closest,the Sun is 91.4 million miles (147.1 million kilometers ) away us.At its farthest ,the Sun is 94.5 million miles (152.1 million km) away.The Earth is closest to the Sun during winter in the northern hemisphere

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John drives a distance of 90 Km to the right at a steady speed of V1=25.00 m/s. Then he stops at the gas station for 10 minutes.
aleksley [76]

a. Speed is defined as rate of change of distance per unit time whereas velocity is defined as rate of change of displacement per unit time.

b. t=6000\ s is the total time taken in the trip

c. d=126000\ m is the total distance

d. s=54000\ m towards right from the starting point.

e. v_a=21\ m.s^{-1}

f. \vec v_a=9\ m.s^{-1} towards right.

Explanation:

a.

Speed is a scalar quantity while velocity is a vector quantity.

Speed is defined as rate of change of distance per unit time whereas velocity is defined as rate of change of displacement per unit time.

Speed is a directionless quantity while velocity constitutes direction.

b.

<em>Total time of round trip when we're given:</em>

  • distance travelled to the right, d_r=90000\ m
  • speed while travelling to the right, v_r=25\ m.s^{-1}
  • time spent at gas station, t_g=600\ s
  • time spent while travelling back towards the left, t_l=30\times 60=1800\ s
  • speed while travelling to the left, v_{_l}=20\ m.s^{-1}

<em>Now time taken for travelling towards right:</em>

t_r=\frac{d_r}{v_r}

t_r=\frac{90000}{25}

t_r=3600\ s

<u>Therefore total time taken in the round trip:</u>

t=t_r+t_l+t_g

t=3600+600+1800

t=6000\ s

c.

<em>Now, distance travelled towards left:</em>

d_l=v_{_l}\times t_l

d_l=20\times1800

d_l=36000\ m

<u>Therefore total distance:</u>

d=d_l+d_r

d=36000+90000

d=126000\ m

d.

Now, total displacement:

s=d_r-d_l

s=90000-36000

s=54000\ m towards right from the starting point.

e.

<u>Average speed:</u>

v_a=\frac{d}{t}

v_a=\frac{126000}{6000}

v_a=21\ m.s^{-1}

f.

<u>Average velocity:</u>

\vec v_a=\frac{s}{t}

\vec v_a=\frac{54000}{6000}

\vec v_a=9\ m.s^{-1} towards right.

4 0
3 years ago
What is the momentum of a 5.6 kg ball going 22m/s ?
maks197457 [2]
<h2>momentum(p) = mass(m)×acceleration (a)</h2>

<h3>p = 5.6 × 22</h3><h3>p = 123.2 kg m/s</h3>

6 0
3 years ago
Plz Help
Angelina_Jolie [31]

Answer: There is greater genetic variety in offspring.

Explanation: This is because with asexual reproduction it is faster, but the offspring is identical to the parent. While sexual reproduction is slower yes, but there is a variety in genetics.

Hope this helps!

7 0
3 years ago
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Where is there potential energy in this system?
mario62 [17]
In the air and the metal
4 0
3 years ago
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Captain Hook is ghting Peter Pan, and they are about to step onto a tightrope strung horizontally between two maststhat are 16 m
andreyandreev [35.5K]

Complete Question

Captain Hook is fighting  Peter Pan, and they are about to step onto a tightrope strung horizontally between two masts that are 16 m apart. When Pan and Hook are standing exactly halfway between the masts, the rope makes a 3 anglewith the horizontal. The rope has a diameter of 0.02 m and a Youngs modulus of 35 GPa.

What is the combined mass,in kilograms, of Peter Pan and Captain Hook?

Answer:

Their combined mass is m= 161.2kg

Explanation:

A sketch that describes the question is shown on the first uploaded image  

  From the question we are told that

          The distance apart is d_A = 16m

          The angle the rope makes is \theta = 3^o

          The diameter of the rope is d = 0.02m

          The Young modulus is  Y = 35Pa

From the diagram we see that the elongation of the rope can be  mathematically evaluated as

         \Delta L = 2x - 16

And applying  SOHCATOH rule    x = \frac{8}{cos \theta}

   Substituting values

                              x = \frac{8}{cos (3)}

                                = 8.01m

      And   \Delta L = \frac{16}{cos 3}  -16

                     \Delta L = 0.02196m

The Tension on the rope can be mathematically represented as

               T = Y A * \frac{\Delta L}{L}

Where A is the area and is mathematically represented as

              A = \frac{\pi}{4} d^2

 Substituting values

            A = \frac{\pi}{4} (0.02)^2

Now Substituting values into the formula for the tension on the rope

          T = (35*10^9) * \frac{\pi}{4} (0.02)^2 * \frac{(0.02196)}{16}

             =15093.4 N

From the diagram we can mathematically evaluate the the weight of peter and hook as

              W = 2T sin \theta

Where W = mg

Now substituting this into the equation and making m the subject

                   m = \frac{2Tsin \theta}{g}

Substituting values

                m = \frac{2* 15093.4 sin(3)}{9.8}

                    m= 161.2kg

Note  SOHCATOH is

                         Sin \theta = \frac{opposite}{hypotenuse}\\ Cos \theta = \frac{adjacent }{hypotenuse} \\Tan \theta = \frac{opposite}{adjacent}      

8 0
3 years ago
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