I’m assuming you’re supposed to calculate the resultant force?
425N (right) -300N (left)
=125 N to the right
Answer:
f1 = -3.50 m
Explanation:
For a nearsighted person an object at infinity must be made to appear to be at his far point which is 3.50 m away. The image of an object at infinity must be formed on the same side of the lens as the object.
∴ v = -3.5 m
Using mirror formula,
i/f1 = 1/v + 1/u
Where f1 = focal length of the contact lens, v = image distance = -3.5 m, u = object distance = at infinity(∞) = 1/0
∴ 1/f1 = (1/-3.5) + 1/infinity
Note that, 1/infinity = 1/(1/0) = 0/1 =0.
∴ 1/f1 = 1/(-3.5) + 0
1/f1 = 1/(-3.5)
Solving the equation by finding the inverse of both side of the equation.
∴ f1 = -3.50 m
Therefore a converging lens of focal length f1 = -3.50 m
would be needed by the person to see an object at infinity clearly
Answer:
The answer is ( D) i. e both b and c.
Answer:
lubricant (decrease friction).
control production speed.
use only when required.
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!!!