Answer:
work which is required to move an electron from the positive terminal to the negative terminal is
-4μJ
W = -4μJ
Explanation:
Work required to move the charge
W = qΔV
initial point V₁ = 150V
Final point V₂ = -50V
W = q(V₂ -V₁)
= 20 × 10⁹(-50 - 150)
W = -4μJ
work which is required to move an electron from the positive terminal to the negative terminal is
-4μJ
Wing D (high camber and surface area, least weight) generated the most lift force and Wing C (most streamlined) generated the least drag force. It is concluded that the shape of a wing does affect the lift and drag of an airplane.
Answer:
"Test Phase
" is the correct choice.
Explanation:
- DevSecOps seems to be a community as well as experience of corporate data science which encompasses software design, regulation, including operational activities. This same main feature of DevSecOps has always been to strengthen customer achievement as well as expedition importance by computerizing, supervising as well as implementing data protection at all stages of the development including its development tools.
- The testing method throughout the test phase would then help make sure that the controller is designed mostly under the responsibilities forecasted. The test focuses on either the reaction times, dependability, use of resources but instead interoperability of applications.
Answer:
Magnitude of force P = 25715.1517 N
Explanation:
Given - The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel.
To find - Determine the magnitude of force P so that the rigid beam tilts 0.015∘.
Proof -
Given that,
Diameter = 12 mm = 0.012 m
Length = 0.6 m
= 0.015°
Youngs modulus of elasticity of 34 stainless steel is 193 GPa
Now,
By applying the conditions of equilibrium, we have
∑fₓ = 0, ∑
= 0, ∑M = 0
If ∑
= 0
⇒
×0.9 - P × 0.6 = 0
⇒×3 - P × 2 = 0
⇒ = 
If ∑
= 0
⇒
×0.9 = P × 0.3
⇒ ×3 = P
⇒ = 
Now,
Area, A = 
= 1.3097 × 10⁻⁴ m²
We know that,
Change in Length , 
= 
Now,
= 9.1626 × 10⁻⁹ P
= 1.83253 × 10⁻⁸ P
Given that,
= 0.015°
⇒ = 2.618 × 10⁻⁴ rad
So,
⇒2.618 × 10⁻⁴ = ( 1.83253 × 10⁻⁸ P - 9.1626 × 10⁻⁹ P) / 0.9
⇒P = 25715.1517 N
∴ we get
Magnitude of force P = 25715.1517 N
