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Elanso [62]
3 years ago
10

Why are marble and granite used so often as a building materials?

Physics
1 answer:
Katyanochek1 [597]3 years ago
3 0

Answer:

Granite is an excellent building material that provides some outstanding interior design opportunities whilst being incredibly sturdy and reliable. It has become one of the most popular building materials in modern construction, from kitchen surfaces through to paving slabs.You Can Expect Durability and Longevity. Firstly, marble is so popular around the world because of its durability in a wide variety of weather conditions. Structures that are over hundreds of years old made from marble are still standing to this day, and look as pristine as they day they were crafted.

Explanation:

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An example of fluid friction is:
ASHA 777 [7]

Answer:

I would say C.

8 0
2 years ago
Read 2 more answers
a straight wire that is 0.60m long and carrying a current of 2.0a is placed at an angle with respect to a magnetic field of stre
Rashid [163]

Answer:

The  angle is  \theta =  30^o

Explanation:

From the question we are told that

    The length of the wire is  l = 0.60 \ m

    The current is  I  =  2.0 \ A

    The magnetic field strength is  B  =  0.30 \ T

     The magnitude of the magnetic force is  F _b  =  0.18  \ N

Generally the magnetic force exerted on the wire is mathematically represented as

      F_b  =  I  *  l  *  B * sin \theta

Making \theta the subject

      \theta =sin^{-1} [ \frac{ F_b }{I  *  l  *  B } ]

substituting values    

      \theta =sin^{-1} [ \frac{ 0.18 }{ 2.0   *   0.6  *   0.3 } ]

     \theta =  30^o

8 0
2 years ago
An electrostatic paint sprayer has a 0.200-m-diameter metal sphere at a potential of 25.0 kV that repels paint droplets onto a g
NeX [460]

The solution is in the attachment

3 0
2 years ago
What is the specific heat of silver if the temp ig a 15.4g sample of silver is increased frin 20 degrees celsius to 31.2 degrees
Genrish500 [490]

heat = mass x spec heat x temp rise

40.5=15.4x10^-3xspec heatx11.2

4 0
2 years ago
There is given an ideal capacitor with two plates at a distance of 3 mm. The capacitor is connected to a voltage source with 12
Licemer1 [7]

The kinematic energy of the positive charge is 2 10⁻⁸ J

This electrostatics exercise must be done in parts, the first part: let's start by finding the charge of the capacitor, the capacitance is defined by

        C = \frac{Q}{\Delta V}

        C = ε₀ \frac{A}{d}

we solve for the charge (Q)

        \frac{Q}{\Delta V} = \epsilon_o \frac{A}{d}

indicates that for the initial point d₁ = 3 mm = 0.003 m and the voltage is DV₁ = 12

         Q = \epsilon_o \  \frac{A \ \Delta V_1 }{d_1}

Now the voltage source is disconnected so the charge remains constant across the ideal capacitor.

For the second part, the condenser is separated at d₂ = 5mm = 0.005 m

         Q = \epsilon_o \  \frac{A \ \Delta V_2 }{d_2}

we match the expressions of the charge and look for the voltage

          \frac{\Delta V_1}{d_1} = \frac{\Delta V_2}{d_2}

          ΔV₂ = \frac{d_2}{d_1 } \ \Delta V_1

The third part we use the concepts of conservation of energy

starting point. With the test load (q = 1 nC = 1 10⁻⁹ C) next to the left plate

          Em₀ = U = q DV₂

          Em₀ = q  \frac{d_2}{d_1 } \ \Delta V_1

           

final point. Proof load on the right plate

         Em_f = K

energy is conserved

         Em₀ = em_f

         q  \frac{d_2}{d_1 } \ \Delta V_1 = K

   

we calculate

         K = 1 10⁻⁹  12  \frac{0.005}{0.003}  

         K = 20 10⁻⁹ J

In this exercise, as the conditions at two different points of separation give, the area of ​​the condenser is not necessary and with conservation of energy we find the final kinetic energy of 2 10⁻⁸ J

3 0
2 years ago
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