Answer:
Distance between two point charges, r = 0.336 meters
Explanation:
Given that,
Charge 1, 
Charge 2, 
Electric potential energy, U = -0.5 J
The electric potential energy at a point r is given by :
r = 0.336 meters
So, the distance between two point charges is 0.336 meters. Hence, this is the required solution.
Answer:
Actual Mechanical Advantage
Explanation:
AMA=mg/F(applied)
Answer:
7.9060 m²
8.57 Volts
5.142×10⁻⁶ Joule
1.2×10⁻⁶ Coulomb
Explanation:
C = Capacitance between plates = 0.14 μF = 0.14×10⁻⁶ F
d = Distance between plates = 0.5 mm = 0.5×10⁻³ m
Q = Charge = 1.2 μC = 1.2×10⁻⁶ C
ε₀ = Permittivity = 8.854×10⁻¹² F/m
Capacitance
∴ Area of each plate is 7.9060 m²
Voltage
∴ Potential difference between the plates if the capacitor is charged to 1.2 μC is 8.57 Volts.
Energy stored
E=0.5CV²
⇒E = 0.5×0.14×10⁻⁶×8.57²
⇒E = 5.142×10⁻⁶ Joule
∴ Stored energy is 5.142×10⁻⁶ Joule
Charge
Q = CV
⇒Q = 0.14×10⁻⁶×8.57
⇒Q = 1.2×10⁻⁶ C
∴ Charge the capacitor carries before a spark occurs between the two plates is 1.2×10⁻⁶ Coulomb
Answer:
0.00034 m
Explanation:
Since the length of the aluminium bar, L is given by , L = 1.0000 + 2.4 × 10⁻⁵T and T = 14.1°C, we substitute the value of T into L. So, we have L = 1.0000 + 2.4 × 10⁻⁵ × 14.1°C = 1.0000 + 0.0003384 = 1.0003384 m. The change in length is thus 1.0003384 - 1.0000 = 0.0003384 m ≅ 0.00034 m
Answer:
Option C is correct.
Modulus of elasticity of the composite perpendicular to the fibers = (12 × 10⁶) psi
Explanation:
For combination of materials, the properties (especially physical properties) of the resulting composite is a sum of the fractional contribution of each material thay makes up the composite.
In this composite,
The fibres = 20 vol%
Aluminium = 80 vol%
Modulus of elasticity of the composite
= [0.2 × E(fibres)] + [0.8 × E(Al)]
Modulus of elasticity of the fibers = E(fibres) = (55 × 10⁶) psi. =
Modulus of elasticity of aluminum = E(Al) = (10 × 10⁶) psi.
But modulus of elasticity of the composite perpendicular to the fibers is given in the expression.
[1 ÷ E(perpendicular)]
= [0.2 ÷ E(fibres)] + [0.8 ÷ E(Al)]
[1 ÷ E(perpendicular)]
= [0.2 ÷ (55 × 10⁶)] + [0.8 ÷ (10 × 10⁶)]
= (3.636 × 10⁻⁹) + (8.00 × 10⁻⁸)
= (8.3636 × 10⁻⁸)
E(perpendicular) = 1 ÷ (8.3636 × 10⁻⁸)
= 11,961,722.5 psi = (11.96 × 10⁶) psi
= (12 × 10⁶) psi
Hope this Helps!!!
