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shusha [124]
3 years ago
13

If the battery of your phone can provide 2 mA of current to your phone and holds a charge of 130 C, how long will it take a full

y charged phone to die?
Physics
1 answer:
Naya [18.7K]3 years ago
5 0

Answer: 65000 seconds

Explanation:

Given that,

Current (I) = 2 mA

(Since 1 mA = 1 x 10^-3A

2 mA = 2 x 10^-3A)

Charge (Q) = 130 C

Time taken for a fully charged phone to die (T) = ?

Recall that the charge is the product of current and time taken.

i.e Q = I x T

130C = 2 x 10^-3A x T

T = 130C / (2 x 10^-3A)

T = 65000 seconds (time will be in seconds because seconds is the unit of time)

Thus, it will take a fully charged phone 65000 seconds to die

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Wolfgang pauli hypothesized an exclusion principle. This principle says two electrons in an atom cannot have the same what?.
tatiyna

No two electrons in an atom or molecule may have the same four electronic quantum numbers, according to the Pauli Exclusion Principle. Only two electrons can fit into an orbital at a time, hence they must have opposing spins.

<h3>What is Pauli's exclusion principle ?</h3>

According to Pauli's Exclusion Principle, no two electrons in the same atom can have values for all four of their quantum numbers that are exactly the same. In other words, two electrons in the same orbital must have opposing spins and no more than two electrons can occupy the same orbital.

  • The reason it is known as the exclusion principle is because it states that all other electrons in an atom are excluded if one electron in the atom has the same specific values for all four quantum numbers.

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3 0
2 years ago
A sharp edged orifice with a 60 mm diameter opening in the vertical side of a large tank discharges under a head of 6 m. If the
Ierofanga [76]

Answer:

The discharge rate is Q = 0.0192 \  m^3 /s

Explanation:

From the question we are told that

   The  diameter is  d =  60 \ mm   =  0.06 \ m

    The  head is  h  =  6 \ m

     The  coefficient of contraction is  Cc  =  0.68

     The  coefficient of  velocity is  Cv  =  0.92

The radius is mathematically evaluated as

         r =  \frac{d}{2}

substituting values

        r =  \frac{ 0.06 }{2}

        r =  0.03 \ m

The  area is mathematically represented as

      A =  \pi r^2

substituting values

      A =  3.142 *  (0.03)^2

      A = 0.00283 \ m^2

 The  discharge rate is mathematically represented as

        Q =  Cv *Cc  *  A  *  \sqrt{ 2 * g *  h}

substituting values

       Q = 0.68 *  0.92*   0.00283  *  \sqrt{ 2 * 9.8 *  6}

       Q = 0.0192 \  m^3 /s

6 0
3 years ago
Explain what parallel medium boundaries do to the path of the wave after it travels through both boundaries. (air and water)
Brilliant_brown [7]

Answer:

The boundaries cause the waves to change direction an effect called <u>refraction.</u>

Explanation:

When a wave crosses a boundary between different materials, the speed of the wave and its wavelength changes.When passing from air to water the two properties (speed and wavelength) decreases, and the wave is observed to change direction as it crosses the boundary between the two material.The bending of the wave is called refraction.

7 0
3 years ago
an auditorium measures 30.0 m ✕ 15.0 m ✕ 5.0 m. the density of air is 1.20 kg/m3. (a) what is the volume of the room in cubic fe
Fynjy0 [20]

dimension = 30.0 m ✕ 15.0 m ✕ 5.0 m.

density = 1.20 kg/m3

(a)volume = lenght * breadth * height

      = 30 * 15 * 5

      = 2250 metre cube = 2.25 cubic meter

(b)   mass of air = density * volume

        mass of air = 1.2 * 2250

mass of air = 2700kg

weight  = mass * 9.8

             = 2700 * 9.8

             = 26,460 N

  • The definition of Density is the amount of matter in a given space, or volume
  • Density = mass/volume
  • units for density kg/m^3
  • Density of water 1g/ml
  • Salt water is denser that is why  don't sink as easily.

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5 0
1 year ago
7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N
artcher [175]

Answer:

20.4 m/s^{2}

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 m/s^{2}

7 0
3 years ago
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