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dolphi86 [110]
3 years ago
13

How long would i t take to walk around the eath nonstop?

Physics
1 answer:
zaharov [31]3 years ago
4 0

Divide (25,000) by (the number of miles you can walk in 1 hour).
The answer you get is the number of hours it would take you to walk around the Earth once, IF you were able to walk on water too.

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Which one of the following formulas is used to find the voltage of a circuit? 
Sedbober [7]
D , since Voltage is one joule per coulomb
8 0
3 years ago
A nonconducting sphere is made of two layers. The innermost section has a radius of 6.0 cm and a uniform charge density of −5.0C
Leno4ka [110]

Answer:

a) E =0, b)   E = 1,129 10¹⁰ N / C , c)    E = 3.33 10¹⁰ N / C

Explanation:

To solve this exercise we can use Gauss's law

        Ф = ∫ E. dA = q_{int} / ε₀

Where we must define a Gaussian surface that is this case is a sphere; the electric field lines are radial and parallel to the radii of the spheres, so the scalar product is reduced to the algebraic product.

           E A = q_{int} /ε₀

The area of ​​a sphere is

          A = 4π r²

         E = q_{int} / 4πε₀ r²

         k = 1 / 4πε₀

         E = k q_{int} / r²

To find the charge inside the surface we can use the concept of density

        ρ = q_{int} / V ’

         q_{int} = ρ V ’

         V ’= 4/3 π r’³

Where V ’is the volume of the sphere inside the Gaussian surface

 Let's apply this expression to our problem

a) The electric field in center r = 0

     Since there is no charge inside, the field must be zero

          E = 0

b) for the radius of r = 6.0 cm

In this case the charge inside corresponds to the inner sphere

        q_{int} = 5.0  4/3 π 0.06³

         q_{int} = 4.52 10⁻³ C

        E = 8.99 10⁹  4.52 10⁻³ / 0.06²

         E = 1,129 10¹⁰ N / C

c) The electric field for r = 12 cm = 0.12 m

In this case the two spheres have the charge inside the Gaussian surface, for which we must calculate the net charge.

     The charge of the inner sphere is q₁ = - 4.52 10⁻³ C

The charge for the outermost sphere is

       q₂ =  ρ 4/3 π r₂³

       q₂ = 8.0 4/3 π 0.12³

       q₂ = 5.79 10⁻² C

The net charge is

     q_{int} = q₁ + q₂

     q_{int} = -4.52 10⁻³ + 5.79 10⁻²

     q_{int} = 0.05338 C

The electric field is

        E = 8.99 10⁹ 0.05338 / 0.12²

        E = 3.33 10¹⁰ N / C

8 0
3 years ago
A small object with mass 1.30 kg is mounted on one end of arod
Julli [10]

Answer:

(a) I_{system} = 1.014\ kg.m^{2}

(b) \tau = 0.0179\ N-m

Solution:

As per the question:

Mass of the object, m = 1.30 kg

Length of the rod, L = 0.780 m

Angular speed, \omega = 5010\ rev/min

Now,

(a) To calculate the rotational inertia of the system about the axis of rotation:

Since, the rod is mass less, the moment of inertia of the rotating system and that of the object about the rotation axis will be equal:

I_{system} = ML^{2} = 1.30\times (0.780)^{2} = 0.791\ kg.m^{2}

(b) To calculate the applied torque required for the system to rotate at constant speed:

Drag Force, F = 2.30\times 10^{- 2}\ N

\tau = FLsin\theta 90 = 2.30\times 10^{- 2}\times 0.780\times 1 = 0.0179\ N-m

3 0
3 years ago
What is the frequency of UV light that has an energy of 2.39 × 10 -18 J?
Setler [38]

Answer:

3.60 × 10^{15} Hz

Explanation:

\frac{2.39 * 10^{-18} J}{6.63 * 10^{-34} J.s}

4 0
3 years ago
A friend asks you how much pressure is in your car tires. You know that the tire manufacturer recommends 30 psi, but it's been a
Bezzdna [24]

Answer:

25 psi

Explanation:

The weight of the car is:

W = mg

W = 1550 kg * 9.8 m/s²

W = 15,190 N

Divided by 4 tires, each tire supports:

F = W/4

F = 15,190 N / 4

F = 3797.5 N

Pressure is force divided by area, so:

P = F / A

P = (3797.5 N) / (0.16 m × 0.14 m)

P ≈ 170,000 Pa

101,325 Pa is the same as 14.7 psi, so:

P ≈ 170,000 Pa × (14.7 psi / 101,325 Pa)

P ≈ 25 psi

8 0
3 years ago
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