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azamat
2 years ago
7

Bricks and insulation are used to construct the walls of a house. The

Physics
1 answer:
ira [324]2 years ago
7 0

Answer:

\dot Q=350.438\ W

Explanation:

Given:

<u>the thermal resistance in the form of </u>

R_1=\frac{x_1}{k_1} =0.095\ m^2.^{\circ}C.W^{-1}

R_2=\frac{x_2}{k_2} =0.704\ m^2.^{\circ}C.W^{-1}

where:

x_1\  \&\ x_2 are the thickness of the respective bricks

k_1\ \&\ k_2 are the respective coefficient of conductivity

temperature inside the house, T_h=24\ ^{\circ}C

temperature outside the house, \ T_c=10^{\circ}C

area of the wall, A=20\ m^2

Since the bricks and insulation are used to construct a wall then they must be used in series for better shielding.

<u>Using Fourier's law:</u>

\dot Q=k.A.\frac{dT}{x}

\dot Q={dT}\div {\frac{x}{k.A} }

in series the resistances get add up

\dot Q=dT\div (\frac{x_1}{k_1.A}+\frac{x_2}{k_2.A} )

\dot Q=(24-10)\div (\frac{0.095 }{20}+ \frac{0.704 }{20} )

\dot Q=350.438\ W

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A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.7 s later with an init
Darina [25.2K]

Answer:4.05 s

Explanation:

Given

First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s

Both hit the ground at the same time

Let h be the height of cliff and it reaches after time t

h=\frac{gt^2}{2}

For second stone

h=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}---2

Equating 1 &2 we get

\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}

\frac{g}{2}\left ( t-t+2.7\right )\left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

26.46t-35.721-52.92t+142.884=0

t=4.05 s

4 0
2 years ago
First find ∮RB⃗ ⋅dl⃗ , the line integral of B⃗ around a loop of radius R located just outside the left capacitor plate. This can
Allisa [31]

Answer:

the expression of current in the loop enclosed to the left of the capacitor plate is

I(t) = \frac{1}{\mu_0}\int B. dL

Explanation:

As we know by Ampere's law that line integral of magnetic field around a closed loop is proportional to the current enclosed in the path

So we will have

\int B. dL = \mu_0 I(t)

so we have

I(t) = \frac{1}{\mu_0}\int B. dL

so above is the expression of current in the loop enclosed to the left of the capacitor plate

5 0
2 years ago
You place a 500 g block of an unknown substance in an insulated container filled 2 kg of water. The block has an initial tempera
Nina [5.8K]

Answer:

3349J/kgC

Explanation:

Questions like these are properly handled having this fact in mind;

  • Heat loss = Heat gained

Quantity of heat = mcΔ∅

m = mass of subatance

c = specific heat capacity

Δ∅ = change in temperature

m₁c₁(∅₂-∅₁) = m₂c₂(∅₁-∅₃)

m₁ = mass of block = 500g = 0.5kg

c₁  = specific heat capacity of unknown substance

∅₂ = block initial temperature = 50oC

∅₁ = equilibrium temperature of block and water after mix= 25oC

m₂= mass of water = 2kg

c₂ = specific heat capacity of water = 4186J/kg C

∅₃ = intial temperature of water = 20oC

0.5c₁(50-25) = 2 x 4186(25-20)

And we can find c₁ which is the unknown specific heat capacity

c₁ = \frac{2*4186*5}{0.5*25}= 3348.8J/kg C≅ 3349J/kg C

4 0
2 years ago
Calculate the force between charges of 5.0 x 10^-8 c and 1.0 x 10^-7 if they are 5.0 feet apart
weqwewe [10]

Answer:

F = 19.375 x 10^-6  N

Explanation:

This problem can be solved by applying Coulomb's Law, which lets us determine the force between two electrically charged particles.

It is defined as

F = (ke * q1 * q2)/ r^2

Where,

ke =  is Coulomb's constant ≈ 9×10^9 N⋅m^2⋅C^−2

q1 = 5.0 x 10^-8 C

q2 = 1.0 x 10^-7 C

r = 5 ft = 1,524 m

F  = (9×10^9 N⋅m^2⋅C^−2)*(5.0 x 10^-8 C)*(1.0 x 10^-7 C)/ ((1,524 m)^2)

F  = (9×10^9 N⋅m^2⋅C^−2)*(5.0 x 10^-8 C)*(1.0 x 10^-7 C)/ ((1,524 m)^2)

F = 19.375 x 10^-6  N

6 0
2 years ago
8) A soccer goal is 2.44 m high. A player kicks the ball at a distance 10.0 m from the goal at an angle of 25.0°. The ball hits
White raven [17]

Answer:

The initial speed of the soccer ball is 16.38 m/s

Explanation:

given;

vertical distance y = 2.44 m

horizontal distance x = 10.0 m

angle of projection θ = 25.0°

Initial velocity has two components, Vₓ and V_y

Vₓ  = V_i cosθ

V_y = V_i sinθ

The horizontal distance = x = Vₓt + ¹/₂ ˣ g ˣ t², but g =0

x = Vₓt =  V_i cosθ *t

10 = V_i cos25 *t

10 = 0.906V_i*t

V_i*t = 10/0.906 = 11.038 m

The vertical distance (g = - g, because it upward motion against gravity)

y = V_y*t -¹/₂ ˣ g ˣ t²

2.44 = (V_i sinθ)t - ¹/₂ ˣ 9.8 ˣ t²

2.44 = (V_i*t)sinθ - ¹/₂ ˣ 9.8 ˣ t²

2.44 = (11.038)sin25° - 4.9t²

2.44 = (11.038)*0.4226 - 4.9t²

2.44 = 4.6647 - 4.9t²

4.9t² = 4.6647 - 2.44 = 2.2247

t² = 2.2247/4.9

t² = 0.454

t = √0.454

t = 0.674 s

Recall that V_i*t = 11.038 m

V_i*0.674 = 11.038 m, solve for V_i

V_i = 11.038/0.674

V_i = 16.38 m/s

Therefore, the initial speed of the soccer ball is 16.38 m/s

6 0
2 years ago
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