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Nookie1986 [14]
3 years ago
8

Which of the following is not accurate when describing solids? A. Solids are better conductors of heat than liquids B. The amoun

t of pressure exerted by a solid is solely dependent on its mass. C. Solids with many air pockets are good insulators. D. Molecules in solids are closer together than molecules in liquids.
Physics
2 answers:
Vinvika [58]3 years ago
4 0
<h3><u>Answer</u>;</h3>

B. The amount of pressure exerted by a solid is solely dependent on its mass.

<h3><u>Explanation</u>;</h3>
  • Pressure is the force per unit area exerted by a substance on a surface.
  • <em><u>Solids exerts pressure on surfaces, which does not depend on the mass of the solid, but depends on the area of contact on the solid and the force applied.</u></em>
  • <em><u>Since pressure refers to force exerted per unit area. Pressure can be increased by either increasing the size of the force or reducing the area over which the force acts.</u></em>
  • <em><u>A solid with a larger surface area where the force acts will exert less pressure compared to a solid surface with a small area where the force acts. </u></em>
Elza [17]3 years ago
3 0
B. The amount of pressure exerted by a solid is solely dependent on its mass
i just took the test
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8 0
3 years ago
Acceleration of a free-falling object in a frictionless environment increases as a function of time.
Digiron [165]

A free-falling object is an object moving under the effect of gravitational forces alone

The correct option to select for the True or False question is False

The reason the above selected option is correct is as follows:

According to Newton's second law of motion, we have;

Force = Mass × Acceleration

The force of gravity is F_{g} =G \cdot \dfrac{M \cdot m}{r^{2}}

Where;

G \cdot \dfrac{M }{r^{2}} = Acceleration \ due \ to \ gravity , \ g \approx 9.81 m/s^2

m  = The mass of the object

∴ The force acting on an object in free fall, F_g = m × g

Therefore the acceleration of an object in free fall is the constant acceleration due to gravity, and it therefore, does not change with time

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2 years ago
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3 years ago
At one point in a pipeline, the water's speed is 3.57 m/s and the gauge pressure is 68.7 kPa. Find the gauge pressure at a secon
ArbitrLikvidat [17]

Answer:

The  pressure at point 2 is P_2  = 254.01 kPa

Explanation:

From the question we are told that

   The speed at point 1  is  v_1  =  3.57 \ m/s

   The  gauge pressure at point 1  is  P_1  =  68.7kPa =  68.7*10^{3}\  Pa

    The density of water is  \rho  = 1000 \ kg/m^3

Let the  height at point 1 be  h_1 then the height at point two will be

      h_2  =  h_1  -  18.5

Let the  diameter at point 1 be  d_1 then the diameter at point two will be

      d_2  =  2 * d_1

Now the continuity equation is mathematically represented as  

         A_1 v_1  =  A_2 v_2

Here A_1 , A_2  are the area at point 1 and 2

    Now given that the are is directly proportional to the square of the diameter [i.e A=  \frac{\pi d^2}{4}]

   which can represent as

             A \ \  \alpha \ \  d^2

=>         A = c   d^2

where c is a constant

  so      \frac{A_1}{d_1^2}  =  \frac{A_2}{d_2^2}

=>          \frac{A_1}{d_1^2}  =  \frac{A_2}{4d_1^2}

=>        A_2  =  4 A_1

Now from the continuity equation

        A_1  v_1  =  4 A_1 v_2

=>     v_2  =  \frac{v_1}{4}

=>     v_2  =  \frac{3.57}{4}

       v_2  =  0.893 \  m/s

Generally the Bernoulli equation is mathematically represented as

       P_1 + \frac{1}{2}  \rho v_1^2 +  \rho *  g * h_1  =  P_2 + \frac{1}{2}  \rho v_2^2 +  \rho *  g * h_2

So  

         P_2  =  \rho  * g  (h_1 -h_2 )+P_1  +  \frac{1}{2}  *  \rho (v_1^2 -v_2 ^2 )  

=>    P_2  =  \rho  * g  (h_1 -(h_1 -18.3)  + P_1  +  \frac{1}{2}  *  \rho (v_1^2 -v_2 ^2 )

substituting values

        P_2  =  1000  * 9.8  (18.3) )+ 68.7*10^{3}  +  \frac{1}{2}  *  1000 ((3.57)^2 -0.893 ^2 )

       P_2  = 254.01 kPa

 

8 0
2 years ago
How much heat is required to convert 2.55g of water at 28 degrees c to steam?
pantera1 [17]
There are two different processes here:
1) we must add heat in order to bring the temperature of the water from 28^{\circ}C to 100^{\circ}C (the temperature at which the water evaporates)
2) other heat must be added to make the water evaporates

1) The heat needed for process 1) is
Q_1=m C_s \Delta T
where 
m=2.55 g is the water mass
C_s = 4.18 g/J^{\circ}C is the water specific heat
\Delta T=100^{\circ}C-28^{\circ}C=72^{\circ}C is the variation of temperature of the water
If we plug the numbers into the equation, we find
Q_1 = (2.55 g)(4.18 J/g^{\circ}C)(72^{\circ}C)=767.4 J

2) The heat needed for process 2) is
Q=m L_e
where 
m=2.55 g is the water mass
L_e = 2264.7 J/g is the latent heat of evaporation of water
If we plug the numbers into the equation, we find
Q_2=(2.55 g)(2264.7 J/g)=5775.0 J

So, the total heat needed for the whole process is
Q=Q_1+Q_2=767.4 J+5775.0 J=6542.4 J
4 0
2 years ago
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