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Vesna [10]
2 years ago
15

Please help me on this!

Physics
1 answer:
Luda [366]2 years ago
4 0

Answer:

c or b

Explanation:

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What type of circuit has little or no resistance?
lions [1.4K]

Answer: A. short circuit

Explanation:

7 0
3 years ago
Read 2 more answers
Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ
bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; H_0 = 51km/s/Mly

Age of the universe; t = \ ?

We know that, the reciprocal of the Hubble's constant ( H_0 ) gives an estimate of the age of the universe ( t ). It is expressed as:

Age\ of\ Universe; t = \frac{1}{H_0}

Now,

Hubble's constant; H_0 = 51km/s/Mly

We know that;

1\ light\ years = 9.46*10^{15}m

so

1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m

Therefore;

H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly}  \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 =  5.39 *10^{-18}s^{-1}\\

Now, we input this Hubble's constant value into our equation;

Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t =  \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0
2 years ago
The initial temperature of a bomb calorimeter is 28.50°
solniwko [45]
The answer is 5,170 J
8 0
3 years ago
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The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete
oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\

make shear stress (τ) the subject of the formula

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of  0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

\tau = \frac{\delta P *r}{2L}  = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2

Part (b) shear stress at a distance of  0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

\tau = \frac{\delta P *r}{2L}  = \frac{86.4 *0}{2*12} =0

3 0
3 years ago
When a cube is inscribed in a sphere of radius r, the length Lof a side of the cube is . If a positive point charge Qis placed a
Nana76 [90]

Answer:

  Ф_cube /Ф_sphere = 3 /π

Explanation:

The electrical flow is

      Ф = E A

where E is the electric field and A is the surface area

Let's shut down the electric field with Gauss's law

       Фi = ∫ E .dA = q_{int} / ε₀

the Gaussian surface is a sphere so its area is

        A = 4 π r²

the charge inside is

        q_{int} = Q

we substitute

       E 4π r² = Q /ε₀

       E = 1 / 4πε₀   Q / r²

To calculate the flow on the two surfaces

* Sphere

       Ф = E A

        Ф = 1 / 4πε₀  Q / r² (4π r²)

        Ф_sphere = Q /ε₀

* Cube

Let's find the side value of the cube inscribed inside the sphere.

In this case the radius of the sphere is half the diagonal of the cube

          r = d / 2

We look for the diagonal with the Pythagorean theorem

         d² = L² + L² = 2 L²

         d = √2 L

         

we substitute

          r = √2 / 2 L

          r = L / √2

          L = √2  r

now we can calculate the area of ​​the cube that has 6 faces

          A = 6 L²

          A = 6 (√2  r)²

          A = 12 r²

the flow is

          Ф = E A

          Ф = 1 / 4πε₀  Q/r²  (12r²)

          Ф_cubo = 3 /πε₀  Q

the relationship of these two flows is

         Ф_cube /Ф_sphere = 3 /π

8 0
3 years ago
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