Answer:
In metric prefixes, if a meter were divided into 100 equal-sized subunits, a single unit would be called a centimeter.
Explanation:
Here we asked to divide 1 meter in to 100 equal parts.
Let us find out what is the length of piece when 1 m is divided in to 100 equal parts.
Length
That is length of 1 m divided into 100 equal parts is 0.01m.
We know that 0.01 m is 1 centimeter.
So, in metric prefixes, if a meter were divided into 100 equal-sized subunits, a single unit would be called a centimeter.
Explanation:
to develop the technologies required for designing, planning, management and operations of an interplanetary mission. The secondary objective is to explore Mars' surface features, morphology, mineralogy and Martian atmosphere using indigenous scientific instruments
Answer:
L2 = 1.1994 m
the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m
Explanation:
Given;
Initial length L1 = 1.2m
Initial temperature T1 = 27°C
Final temperature T2 = 0.0°C
Linear expansion coefficient of brass x = 1.9 × 10^-5 /°C
The change i length ∆L;
∆L = L2 - L1
L2 = L1 + ∆L ...........1
∆L = xL1(∆T)
∆L = xL1(T2 - T1) ......2
Substituting the given values into equation 2;
∆L = 1.9 × 10^-5 /°C × 1.2m × (0 - 27)
∆L = 1.9 × 10^-5 /°C × 1.2m × (- 27)
∆L = -6.156 × 10^-4 m
From equation 1;
L2 = L1 + ∆L
Substituting the values;
L2 = 1.2 m + (- 6.156 × 10^-4 m)
L2 = 1.2 m - 6.156 × 10^-4 m
L2 = 1.1993844 m
L2 = 1.1994 m
the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m
Answer:
Explanation:
Inductance L = 1.4 x 10⁻³ H
Capacitance C = 1 x 10⁻⁶ F
a )
current I = 14 .0 t
dI / dt = 14
voltage across inductor
= L dI / dt
= 1.4 x 10⁻³ x 14
= 19.6 x 10⁻³ V
= 19.6 mV
It does not depend upon time because it is constant at 19.6 mV.
b )
Voltage across capacitor
V = ∫ dq / C
= 1 / C ∫ I dt
= 1 / C ∫ 14 t dt
1 / C x 14 t² / 2
= 7 t² / C
= 7 t² / 1 x 10⁻⁶
c ) Let after time t energy stored in capacitor becomes equal the energy stored in capacitance
energy stored in inductor
= 1/2 L I²
energy stored in capacitor
= 1/2 CV²
After time t
1/2 L I² = 1/2 CV²
L I² = CV²
L x ( 14 t )² = C x ( 7 t² / C )²
L x 196 t² = 49 t⁴ / C
t² = CL x 196 / 49
t = 74.8 μ s
After 74.8 μ s energy stored in capacitor exceeds that of inductor.
