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wolverine [178]
3 years ago
8

In which situation can you be at rest and moving at then same time

Physics
1 answer:
anyanavicka [17]3 years ago
4 0

If you are stationary, but in/on a moving vehicle/object you can be at rest and moving at then same time.

<u>Explanation</u>:

  • A particle, when viewed from a given frame of reference, cannot be both at rest and in motion. However, in one frame of reference, a particle can be in motion whereas in another frame of reference the particle is in motion.
  • For example, if you are seated in a plane, the plane is stationary in that reference frame and the Earth moves under it, but in the reference frame of the Earth, the plane is moving concerning the Earth. When you are standing still on Earth, in your frame of reference, the Earth is stationary, and the Sun and stars move around the Earth.
  • However, in the frame of reference of the center of our solar system, the Earth orbits the Sun and the Sun are perturb slightly by the rest of the planets, but the rest of the galaxy orbits our solar system. Of course, in rest from our Galaxy, our solar system orbits a giant black hole at its center.
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2.5 miles an hour but yea hope that helps
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2 years ago
Alice and Marge are studying the properties of matter. The girls placed some an iron nail in a beaker containing water. Iron is
tatuchka [14]

Answer:

When substances made of iron are exposed to oxygen and moisture (water), rusting takes place. Rusting removes a layer of material from the surface and makes the substance weak. Rusting is a chemical change.

7 0
3 years ago
A rod extending between x = 0 and x = 13.0 cm has uniform cross-sectional area A = 8.00 cm2. Its density increases steadily betw
mezya [45]

Answer:

The mass is  m  = 3.45408 kg

Explanation:

From the question we are told that

    The extension  of the rod is from , \   x_1 = 0 \to x_2 = 13.0

     The area is  A =  8.0 cm^2

      The density increase as follows from  \  \rho_1 =2.5 g/cm^2 \to  \rho_2= 19.0 g/cm^3

    The equation  \rho =  B + Cx

at  x_1= 0  \rho_1 =2.5 g/cm^2

So

      2.5  =  B  + 0

=>  B =2.5

    So at x_2 = 13.0 ,  \rho_2= 19.0 g/cm^3

So

            19.0 = 2.5 + C(13)

       =>   C = 1.27

Now  

       m  =  8   \int\limits^{13}_{0} {2.5 + 1.27x} \, dx

      m  =  8   [{2.5 +\frac{ 1.27x^2}{2} } ]\left  | 13} \atop {0}} \right.

      m  =  8   [{2.5 +\frac{ 1.27(13)^2}{2} } ]

      m  = 3454.08 g

        m  = 3.45408 kg

         

8 0
3 years ago
6) 100 ml of water is initially at 20°C. 30,000 J of heat is added to the water. What is temperature change for the water?
vovangra [49]
<h2>Δt = 71.67 °C</h2>

The temperature change of water is equal to 71.67 °C

<h3>Explanation:</h3>

Given:

Amount of transferred energy = 30,000 K J

Mass of water = 100 ml

Initial temperature = 20°C

To find the change in temperature of water.

Formula for Heat capacity is given by

Q = m×c×Δt ........................................(1)

where:

Q = Heat capacity of the substance (in J)

m=mass of the substance being heated in grams(g)

c = the specific heat of the substance in J/(g.°C)

Δt = Change in temperature (in °C)

Δt = (Final temperature - Initial temperature) = T(f) - T(i)

Q = 30,000 J

Mass of water = m = 100 ml

1 ml = 1 g ................................................(2)

Therefore m = 100 ml = 100 g

Specific heat of water is c = 4.186 J /g.

Δt = ?

Substituting these in equation (1), we get

Q = m×c×Δt

Rearranging the terms for Δt,

Δt = \frac{Q}{m\times c}

Δt = \frac{30,000}{100\times 4.186}  = \frac{30,000}{418.6}= 71.67\°C

Δt = 71.67 °C

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Answer:

A book containing information on various types of science related topics.

4 0
2 years ago
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