Answer:
Option C
Crimp terminals
Explanation:
It's possible to crimp terminals using a multipurpose wiring tool. Since the tool selected for use during crimping also depends on the volume of work, the multipurpose wiring tool is recommended for use when the volume is small to medium. Basically, crimping tools are sized according to the wire gauge that they can fit. Since multipurpose has different sizes, that's why it's used for crimping tools.
Input heat, Qin = 4 x 10⁵ J
Output heat, Qout = 3.5 x 10⁵ J
From the first Law of thermodynamics, obtain useful work performed as
W = Qin - Qout
= 0.5 x 10⁵ J
By definition, the efficiency is
η = W/Qin
= 100*(0.5 x 10⁵/4 x 10⁵)
= 12.5%
Answer: The efficiency is 12.5%
Answer: The correct answer is graph A.
Explanation:
See Khan Academy.
Answer:
a) a = 3.72 m / s², b) a = -18.75 m / s²
Explanation:
a) Let's use kinematics to find the acceleration before the collision
v = v₀ + at
as part of rest the v₀ = 0
a = v / t
Let's reduce the magnitudes to the SI system
v = 115 km / h (1000 m / 1km) (1h / 3600s)
v = 31.94 m / s
v₂ = 60 km / h = 16.66 m / s
l
et's calculate
a = 31.94 / 8.58
a = 3.72 m / s²
b) For the operational average during the collision let's use the relationship between momentum and momentum
I = Δp
F Δt = m v_f - m v₀
F =
F = m [16.66 - 31.94] / 0.815
F = m (-18.75)
Having the force let's use Newton's second law
F = m a
-18.75 m = m a
a = -18.75 m / s²
Answer:

Explanation:
given data
Radius of sphere 3.0 cm
charge Q = 2.0 m C
We know that maximum electric field is given as

electric field inside the sphere can be determine by using below relation


