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olga nikolaevna [1]
3 years ago
5

Which quantity and unit are INCORRECTLY paired?

Physics
1 answer:
liq [111]3 years ago
5 0
Force, the unit is Newton, newton is the force to accelerate a mass. so it should be kg m/s^2

joule (J) is equal to Nm not Ns

the unit of work is J and it is correct.
the unit of power is J/s which is equal to W
the unit of of energy is the same with work, which is J which equivalent to kgm2/s2
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A parallel-plate capacitor has plates of area 0.40 m2 and plate separation of 0.20 mm. The capacitor is connected to a 9.0 V bat
mafiozo [28]

Answer:

a) E = 4.5*10⁴ V/m

b) C= 17.7 nF

c) Q = 159. 3 nC  

Explanation:

a)

  • By definition, the electric field is the electrostatic force per unit charge, and since the potential difference between plates is just the work done by the field, divided by the charge, assuming a uniform electric field, if V is the potential difference between plates, and d is the separation between plates, the electric field can be expressed as follows:

       E = \frac{V}{d} = \frac{9.0V}{2*10-4m} =4.5 * 10e4 V/m (1)

b)

  • For a parallel-plate capacitor, applying the definition of capacitance as the quotient between the charge on one of the plates and the potential difference between them, and assuming a uniform surface charge density σ, we get:

       Q = \sigma* A (2)

        From (1), we know that V = E*d, but at the same time, applying Gauss'

        Law at a closed surface half within the plate, half outside it , it can be

        showed than E= σ/ε₀, so finally we get:

       C = \frac{Q}{V} =\frac{\sigma*A}{E*d}  = \frac{\sigma*A}{\frac{\sigma}{\epsilon_{o} } d} =\frac{\epsilon_{0}*A}{d} = \frac{8.85e-12F/m*0.4m2}{2e-4m} = 17.7 nF (3)

c)    

  • From (3) we can solve for Q as follows:

       Q = C* V = 17.7 nF * 9.0 V = 159.3 nC  (4)

6 0
3 years ago
What happens to the average kinetic energy of water molecules as water freezes
Mekhanik [1.2K]

Answer:

C. It decreases as the water releases energy to its surroundings

Explanation:

The energy that is possessed in the body because of the motion is said to be kinetic energy. When the water freezes, the kinetic energy of the water molecules decreases as the particles stop motion. The energy is released in the surrounding. Freezing takes place when the liquid molecules slow down and lose the kinetic energy to form the solid crystals.

6 0
3 years ago
if a free falling rock were equipped with a speedometer, by how much would it's speed readings increase with each second IF it w
Kaylis [27]
Accelleration(a) is changing of velocity in second.
free fall a = g
speed increase = a = g = 20 (m/s) / s
7 0
3 years ago
A parallel-plate capacitor has plates with an area of 451 cm2 and an air-filled gap between the plates that is 2.51 mm thick. Th
Nostrana [21]

To solve this problem we will apply the concepts related to Energy defined in the capacitors, as well as the capacitance and load. From these three definitions we will build the solution to the problem by defending the energy with the initial conditions, the energy under new conditions and finally the change in the work done to move from one point to the other.

Energy in a capacitor can be defined as

E = \frac{1}{2}CV^2 = \frac{1}{2}\frac{Q^2}{C}

Here,

V = Potential difference across the capacitor plates

Q = Charge stored on the capacitor plates

At the same time capacitance can be defined as,

C = \epsilon_0 (\frac{A}{d})

Here,

\epsilon_0 =  Vacuum permittivity constant

A = Area

d = Distance

Replacing with our values we have that,

C = (8.85*10^{-12})(\frac{0.0451}{2.51*10^{-3}})

C = 1.5901*10^{-10}F

PART A) Energy stored in the capacitor is

E = \frac{1}{2} CV^2

E = \frac{1}{2} (1.5901*10^{-10})(575)^2

E = 2.628*10^{-5}J

PART B) We know first that everything that the load can be defined as the product between voltage and capacitance, therefore

Q = CV

Q = (1.59*10^{-10})(575)

Q = 9.1425*10^{-8}C

Now if d = 10.04*10^{-3}m we have that the capacitance is

C = \epsilon_0 (\frac{A}{d})

C = (8.85*10^{-12})(\frac{0.0451}{10.04*10^{-3}})

C = 3.9754*10^{-11}F

Then the energy stored is

E = \frac{1}{2} \frac{Q^2}{C}

E = \frac{1}{2} (\frac{(9.1425*10^{-8})^2}{3.9754*10^{-11}})

E = 1.051*10^{-4} J

PART C) The amount of work or energy required to carry out this process is the difference between the energies obtained, therefore

W = 1.051*10^{-4} J -2.628*10^{-5}J

W = 7.882*10^{-5} J

6 0
3 years ago
A 2 kg body is dropped from a height of 3 m. Calculate:
NikAS [45]

Answer:

   a) 6.26 m/s

   b) 7.67 m/s

Explanation:

The potential energy at height h0 is initially ...

  PE0 = mgh0

At height h1, the potential energy is ...

  PE1 = mgh1

The difference in potential energy is converted to kinetic energy:

  PE0 -PE1 = KE1 = (1/2)m(v1)^2

Solving for v1, we have ...

  mg(h0 -h1) = (1/2)m(v1)^2

  2g(h0 -h1) = (v1)^2

  v1 = √(2g(h0 -h1))

__

a) When the body is 1 m high, its speed is ...

  v = √(2(9.8)(3 -1)) ≈ 6.26 m/s . . . at 1 m high

__

b) When the body is 0 m high, its speed is ...

  v = √(2(9.8)(3 -0)) ≈ 7.67 m/s . . . when it reaches the ground

3 0
3 years ago
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