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Allisa [31]
3 years ago
7

What is the energy equivalent of 5.0kg of mass?

Physics
2 answers:
DiKsa [7]3 years ago
7 0

Answer : The energy equivalent if 5 kg of mass is 4.5\times 10^{17}\ J

Explanation :

Given that,

Mass, m = 5 kg

From Einstein's mass - energy equivalence relation :

E=mc^2

E is energy

m is the mass of the object

c is the speed of light

E=5\ kg\times (3\times 10^8\ m/s)^2

E=4.5\times 10^{17}\ J

Hence, this is the required solution.

Ira Lisetskai [31]3 years ago
3 0
Ah hah !  Sounds to me like an  " E = m c² "  question.

c = 3 x 10⁶  m/s

c² = 9 x 10¹² m²/s²

                       E  =  (5 kg)  x  (9 x 10¹² m²/s²)

                           =       45 x 10¹²  kg-m²/s²

                           =       4.5 x 10¹³  newton-m

                           =        4.5 x 10¹³ Joules  . 

That's the energy needed to

-- run a 1000-watt blow-dryer for roughly  1,427 years .
or
-- run a 400-horsepower truck, wide open, pedal to the metal,
   for   4.78 years . 
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REDDIT You look at a circular loop of wire such that the plane of the loop is perpendicular to your line of vision. The loop has
sergiy2304 [10]

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The circular loop experiences a constant force which is always directed towards the center of the loop and tends to compress it.

Explanation:

Since the magnetic field, B points in my direction and the current, I is moving in a clockwise direction, the current is always perpendicular to the magnetic field and will thus experience a constant force, F = BILsinФ where Ф is the angle between B and L.

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<u>So, the circular loop experiences a constant force which is always directed towards the center of the loop and tends to compress it.</u>

3 0
3 years ago
The parallel plates in a capacitor, with a plate area of 8.00 cm2 and an air-filled separation of 2.70 mm, are charged by a 8.70
MrRa [10]

Answer:

a)  ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J,  d)  W = 150 10⁻¹² J

Explanation:

Let's find the capacitance of the capacitor

         C = \epsilon_o \frac{A}{d}

         C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³

         C = 2.62 10⁻¹² F

for the initial data let's look for the accumulated charge on the plates

          C = \frac{Q}{\Delta V}

          Q₀ = C ΔV

           Q₀ = 2.62 10⁻¹² 8.70

           Q₀ = 22.8 10⁻¹² C

a) we look for the capacity for the new distance

          C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³

          C₁ = 1.04 10⁻¹² F

       

          C₁ = Q₀ / ΔV₁

          ΔV₁ = Q₀ / C₁

          ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²

          ΔV₁ = 21.9 V

b) initial stored energy

          U₀ = \frac{Q_o}{ 2C}

          U₀ = (22.8 10⁻¹²)²/(2  2.62 10⁻¹²)

          U₀ = 99.2 10⁻¹² J

c) final stored energy

          U_f = (22.8 10⁻¹²) ² /(2  1.04 10⁻⁻¹²)

          U_f = 249.9 10⁻¹² J

d) the work of separating the plates

as energy is conserved work must be equal to energy change

          W = U_f - U₀

          W = (249.2 - 99.2) 10⁻¹²

          W = 150 10⁻¹² J

note that as the energy increases the work must be supplied to the system

6 0
3 years ago
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