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Oxana [17]
2 years ago
8

A student mixed two clear liquids together in a beaker. A solid and a new liquid formed. The student forgot to write down the ma

ss of one of the reactants. The
rest of the data are shown in the table below.

What is the mass of liquid reactant A?
A 1.0 g
B 8.0 g
C 9.0 g
D 11.0 g
E 20.0 g

Physics
1 answer:
ziro4ka [17]2 years ago
3 0
The answer to the question is probably A
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Arrange the distances between Earth and various celestial objects in order from least to greatest. Use the conversion table to h
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distance to the star Betelgeuse: 640 ly

As we know that

1 ly = 63000 AU

also we know that

1AU = 1.5 \times 10^8 km

1 ly = 63000 (1.5 \times 10^8) = 9.45 \times 10^{12} km

So the distance of Betelgeuse = 640 ly

d_1 = 640 \times 9.45 \times 10^{12} = 6.05 \times 10^{15} m

distance to the star VY Canis Majoris: 3.09 × 10^8 AU

d_2 = 3.09\times 10^8 \times 1.5 \times 10^8 km

d_2 = 4.64 \times 10^{16} km

distance to the galaxy Large Magellanic Cloud: 49976 pc

1 pc = 3.262 ly = 3.262 \times 9.45 \times 10^{12} km

1pc = 3.08 \times 10^{13} km

now we have

d_3 = 49976 \times 3.08 \times 10^{13}

d_3 = 1.54 \times 10^{18} km

distance to Neptune at the farthest: 4.7 billion km

d_4 = 4.7 \times 10^9 km

now the order of distance from least to greatest is as following

1. distance to Neptune at the farthest

2. distance of Betelgeuse

3. distance to the star VY Canis Majoris

4. distance to the galaxy Large Magellanic Cloud

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An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b
lys-0071 [83]

Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

             S = ut + 0.5at²

     Here u = 0, and a = g

              S = 0.5gt²

    Distance traveled during the first second ( t =1 )

              S = 0.5 x 9.81 x 1² = 4.905 m

   Distance traveled during the first second = 4.905 m.

b)  We have equation of motion

            v² = u² + 2as

      Here u = 0, s= 75 m and a = g

           v² = 0² + 2 x g x 75 = 150 x 9.81

           v = 38.36 m/s

      Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

                   75 = 0.5 x 9.81 x t²

                    t = 3.91 s

   We need to find distance traveled last second

   That is

          S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

   Distance traveled during the last second of motion before hitting the ground = 33.45 m

       

3 0
3 years ago
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