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ohaa [14]
2 years ago
12

If someone weighs 80n on Mars and comes back on earth will they weigh the same

Physics
2 answers:
Alexandra [31]2 years ago
8 0

Answer:

No

Explanation:

Your answer will be no because Mars has less gravity then Earth does but you would still have the same amount of mass therefor If someone goes to Mars and goes back to Earth they would not weigh the same as they did on earth.Hope it helps you!

xxMikexx [17]2 years ago
3 0
No i disagree cus its mars
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Why is it important to study electromagnetic waves?
labwork [276]
The study of EM is essential to understanding the properties of light, its propagation through tissue, scattering and absorption effects, and changes in the state of polarization. ... Since light travels much faster than sound, detection of the reflected EM radiation is performed with interferometry.
3 0
3 years ago
What is the internal energy of 2.00 mol of diatomic hydrogen gas (H2) at 35°C?
djyliett [7]
As you mentioned, we will use <span>Equipartition Theorem.
</span><span>H2 has 5 degrees of freedom; 3 translations and 2 rotation
</span>Therefore:
Internal energy = (5/2) nRT
You just substitute in the equation with the values of R and T and calculate the internal energy as follows:
Internal energy = (5/2) x 2 x <span>8.314 x 308 = 32.0089 x 10^3 J</span>
4 0
3 years ago
3. A certain wire, 3 m long, stretches by 1.2 mm when under tension of 200 N. By how much does
nikitadnepr [17]

Answer:

The extension of the second wire is   e_2 = 0.0024 \  m =  2.4 mm

Explanation:

From the question we are told that

    The length of the wire is L  = 3 \ m

     The elongation of the wire is  e =  1.2mm =  \frac{1.2}{1000} =  0.0012 m

        The tension is F  =  200 \ N

       The length of the second wire is  L_2   =  6 \ m

     

Generally the Young's modulus(Y) of this material is  

        Y  = \frac{stress}{strain }

Where stress =  \frac{F}{A}

    Where A is the area which is evaluated as  

           A = \pi r^2

  and   strain = \frac{extention}{length} =  \frac{e}{L}

   So

        Y  = \frac{\frac{F}{\pi r^2 } }{ \frac{e}{L}  }

Since the wire are of the same material Young's modulus(Y)  is constant

So we have  

              \frac{F * L }{r^2 e}  =  \pi * Y = constant

              F * L   =  constant   * r^2 e

Now the ration between the first and the second wire is

         \frac{F_1}{F_2}  * \frac{L_1}{L_2} =  \frac{r*2_1}{r^2}  *  \frac{e_1}{e_2}

Since tension , radius are constant

   We have

           \frac{L_1}{L_2} =   \frac{e_1}{e_2}

substituting values

          \frac{3}{6} =   \frac{0.0012}{e_2}

          0.5 e_2 =  0.0012

         e_2 = \frac{ 0.0012  }{0.5}

          e_2 = 0.0024 \  m =  2.4 mm

3 0
4 years ago
Convert the number from scientific into standard notation: 5.9 x 10-2
guapka [62]
Move the decimal point to:
Left : (if the exponent of ten is a negative number -) ... OUR CASE HERE (-2)
or to
Right : (if the exponent is positive +).

You should move the point as many times as the exponent indicates.
Do not write the power of ten anymore.

So, standard form is:
Two points to the left {Exponent of Ten is Negative (-2)}
0.059 ... (without the 10)
6 0
3 years ago
"A short-wave radio antenna is supported by two guy wires, 150 ft and 170 ft long. Each wire is attached to the top of the anten
Nonamiya [84]

Answer:

The anchor points are 78.37 ft and 111.99 ft

Explanation:

If you look at the attached (Fig 1) you will find that the union of antenna and its guy wires forms two right triangles. To solve problems that involve this kind of triangles, you can apply trigonometric functions (sine, cosine, etc) and Pythagoras Theorem. Trigonometric functions states the relation between angles, sides and hypotenuse of a right triangle. If you look Fig2, considering α angle, "b" is the opposite side, "a" the adjacent side and "c" the hypotenuse. Then

a) Sine (α) = b/c it means opposite side/hypotenuse

b) Cosine (α)= a/c, it means adjacent side/hypotenuse

c) Tangent (α) = b/a opposite side/adjacent side.

Pythagoras theorem states that if you called "a" and "b", the sides of the right triangle, and "c" the hypotenuse, then:

                                      a² + b² = c²    

As the problem states the lengths 150 ft and 170 ft represents the value of the hypotenuse of each triangle and 65° is one of the angles of the triangle with 150 ft hypotenuse. So you can solve this using sin (65°) to find the height of the antenna (h) and then the two distances (x and y,).

Sine (65°) = h/ 150 ft ⇒ Sine (65°) x 150ft = h ⇒ h = 127.9 ft.

To find x : Cosine (65°) = x/ 150 ft ⇒ Cosine (65°) x 150 ft = x

⇒ x = 78.37 ft.

And finally, to find y we can apply Pythagoras theorem

(170 ft)² = (127.9 ft)² + y² ⇒ y² = (170 ft)² - (127.9 ft)² ⇒ y = 111.99 ft

Summarizing, the anchor points are 78.37 ft and 111.99 ft

4 0
3 years ago
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