The answer is option C. SWOT analysis. The compilation of decisions made by managers to aid the organization in reaching its objectives is referred to as SWOT Analysis.
Here are the options. A. Strategy B. Scenario planning C. SWOT analysis D. Diversification E. Related diversification
Explanation:
Data provided:
Area A = 10 cm×2 cm = 20×10⁻⁴ m²
Separation distance d between the plates = 1 mm = 1×10⁻³ m
Battery voltage, or emf = 100 V
Resistance = 1025 ohm
Solution:
In an RC circuit, the voltage across the plates varies with time t. At the outset, the voltage matches that of the battery, V₀ = emf = 100V. However, after a certain time t, both the resistance and capacitance alter this, leading to a final voltage V expressed as

Applying the natural logarithm to both sides,

(1)
Next, we can determine the capacitance using the plates' area.
C = ε₀A/d
= 
= 18×10⁻¹²F
We can now find the time it takes for the voltage to drop from 100 to 55 V by substituting C, V₀, V, and R values into equation (1)

= -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)
= 15×10⁻⁹s
= 15 ns
Answer:
1/7 kg
Explanation:
Refer to the attached diagram for enhanced clarity regarding the question.
One of the blocks weighs 1.0 kg and accelerates downward at 3/4g.
g denotes the acceleration due to gravity.
Let M represent the block with known mass, while 'm' signifies the mass of the other block and 'a' refers to the acceleration of body M.
Given M = 1.0 kg and a = 3/4g.
By applying Newton's second law; 
For the body with mass m;
T - mg = ma... (1)
For the body with mass M;
Mg - T = Ma... (2)
Combining equations 1 and 2 gives;
+Mg -mg = ma + Ma
Ma-Mg = -mg-ma
M(a-g) = -m(a+g)
Substituting M = 1.0 kg and a = 3/4g into this equation leads to;
3/4 g-g = -m(3/4 g+g)
3/4 g-g = -m(7/4 g)
-g/4 = -m(7/4 g)
1/4 = 7m/4
Multiplying gives: 28m = 4
m = 1/7 kg
Hence, the mass of the other box is 1/7 kg
Answer:
Explanation:
a) La fuerza neta que actúa sobre la caja en la dirección vertical es:
Fnet=Fg−f−Fp *sin45 °
aquí Fg representa la fuerza gravitacional, f es la fuerza de fricción, y Fp es la fuerza de empuje.
Fnet=ma
ma=Fg−f−Fp *sin45 °
a=
=0.24 m/s²
Vf =Vi +at
=0.48+0.24*2
Vf=2.98 m/s
b)
Fnet=Fg−f−Fp *sin45 °
=Fg−0.516Fp−Fp *sin45 °
=30-1.273Fp
Fnet=0 (Ya que la velocidad es constante)
Fp=30/1.273
=23.56 N