<u>Answer:</u> The voltage needed is 35.7 V
<u>Explanation:</u>
Assuming that the resistors are arranged in parallel combination.
For the resistors arranged in parallel combination:

We are given:

Using above equation, we get:

Calculating the voltage by using Ohm's law:
.....(1)
where,
V = voltage applied
I = Current = 3.75 A
R = Resistance = 
Putting values in equation 1, we get:

Hence, the voltage needed is 35.7 V
Answer:
gas is dioatomic
T_f = 330.0 K

Explanation:
Part 1
below equation is used to determine the type Gas by determining
value

where V_i and V_f is initial and final volume respectively
and P_i and P_f are initial and final pressure


\gamma = 1.38
therefore gas is dioatomic
Part 2
final temperature in adiabatic process is given as
](https://tex.z-dn.net/?f=T_f%20%3D%20T_i%2A%5B%5Cfrac%7Bv_i%7D%7BV_f%7D%5D%28%5E%5Cgamma-1%29)
substituing value to get final temperature
![T_f = 260*[\frac{151}{80.6}]^ {(1.38-1)}](https://tex.z-dn.net/?f=T_f%20%3D%20260%2A%5B%5Cfrac%7B151%7D%7B80.6%7D%5D%5E%20%7B%281.38-1%29%7D)
T_f = 330.0 K
Part 3
determine number of moles by using following formula



Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so
= A × ω
= 2.2×10⁻³ × 2722.69 m/s
= 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
= A' × ω
= 1.555×10⁻³ × 2722.69
= 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Answer:
Lithium and Sodium
Explanation:
Lithium and Sodium is in the same group 1A
m = mass held by mr. Z above his head = 200 kg
g = acceleration due to gravity = 9.8 m/s²
F = force applied by mr. Z to hold the mass
Using equilibrium of force , force equation is given as
F = mg
F = (200) (9.8)
F = 1960 N
Since the mass is not moved,
d = displacement of the mass = 0 m
we know that , work done is given as
W = F d
inserting the values
W = (1960) (0)
W = 0 J