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vodomira [7]
3 years ago
12

Determjne the density of liquid whose relative density is 1.25 given that the density is 1000kgm-3​

Physics
1 answer:
nlexa [21]3 years ago
7 0

Explanation:

Relative density=density of liquid÷density of water

1.25=x ÷ 1000

x=1.25 × 1000

x=1250kgm^-3

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Hydrogen line spectrum lies entirely within visible range
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According to newtons third law if an eighth grader pushed against a large boulder with a force of 8 N how much force would the b
dexar [7]

Answer:

8n

Explanation

5 0
3 years ago
A beaker of water rests on an electronic balance that reads 875.0 g. A 2.75-cm-diameter solid copper ball attached to a string i
Firdavs [7]

Answer:

a  

The tension in the string is  T = 0.85 N

 b

 The new balance reading is  M_b  =  885.86 g

Explanation:

From the question we are told that

    The mass of the beaker of water is  m = 875 .0g

     The diameter of the copper ball is  d = 2.75 cm = \frac{2.75}{100} = 0.0275 m

There are two forces acting on the copper ball

   The first is the Buoyant force of the water pushing it up which is mathematically represented as

                     F = \rho V g

Where \rho is the density of water which has value of  \rho = 1000 kg/m^3

            g is the acceleration due to gravity g= 9.8 \  m/s^2

          V is the volume of water displaced by the copper ball  which is mathematically evaluated as

                             V = \frac{4}{3}  \pi r^3

The radius r is  r = \frac{d}{3}  = \frac{0.0275}{2} = 0.01375 m

Substituting value  

                        V = \frac{4}{3} * 3.142 * (0.01375)^3

                            V = 1.08 9 *10^{-5 } m^3  

   Substituting for  F

              F = 1000 * 1.089 *10^{-5} * 9.8

               F = 0.1067 N      

     The second force is the weight of the copper ball which is mathematically represented as

       W_c = mg

Now m is the mass which can be mathematically evaluated as

        m =  \rho_c * V

Where  is the density of copper with  value of  \rho_c = 8960 kg /m^3

So      m = 8960 * 1.089 *10^{-5}

         m = 0.0976

So the weight of copper is  

             W_c = 0.09756  *  9.8

            W_c = 0.956 N

Now the tension the string would be mathematically evaluated as

            T = W_c - F

So        T = 0.956 - 0.1067

           T = 0.85 N

From this value we that the string is holding only 0.85 N of the copper weight thus (0.956 - 0.85 = 0.1065 N ) is being held by the balance

Now the mass equivalent of this weight is mathematically evaluated as

             m_z = \frac{1.0645 }{9.8 }

             m_z = 0.01086 kg

Converting to grams  

                     m_z = 10.86 g

So the new balance reading is  

                  M_b  =  875.0 +10.86

                  M_b  =  885.86 g

5 0
3 years ago
An object in free fall is at heights y1, y2, and y3 at times t1, t2, and t3 respectively.
Dvinal [7]

Answer:

  a = (v₃₂ - v₂₁) / (t₃₂ -t₂₁)

Explanation:

This is an exercise of average speed, which is defined with the variation of the distance in the unit of time

         v = (y₃ - y₂) / (t₃-t₂)

the midpoint of a magnitude is the sum of the magnitude between 2

         t_mid = (t₂ + t₃) / 2

the same reasoning is used for the mean acceleration

         a = (v_f - v₀) / (t_f - t₀)

   

in our case

        a = (v₃₂ - v₂₁) / (t₃₂ -t₂₁)

5 0
3 years ago
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