Answer:
0.34148 m
Explanation:
= Resistivity of tungsten = 
d = Diameter = 0.0018 inch
r = Radius = 

= Temperature coefficient of tungsten = 
Power is given by

We have the equation
![R_2=R_1[1+\alpha(T_2-T_1)]\\\Rightarrow R_1=\dfrac{R_2}{1+\alpha(T_2-T_1)}\\\Rightarrow R_1=\dfrac{144}{1+0.0045(2550-25)}\\\Rightarrow R_1=11.64812\ \Omega](https://tex.z-dn.net/?f=R_2%3DR_1%5B1%2B%5Calpha%28T_2-T_1%29%5D%5C%5C%5CRightarrow%20R_1%3D%5Cdfrac%7BR_2%7D%7B1%2B%5Calpha%28T_2-T_1%29%7D%5C%5C%5CRightarrow%20R_1%3D%5Cdfrac%7B144%7D%7B1%2B0.0045%282550-25%29%7D%5C%5C%5CRightarrow%20R_1%3D11.64812%5C%20%5COmega)
Resistance is given by

The length of the filament is 0.34148 m
D, all of these are voltage sources. A battery can power small things, generators can power larger things, and outlets can charge, or power things.
Answer:
charges of the beads is 1.173 ×
C
Explanation:
given data
mass = 3.8589 g = 0.003859 kg
spring length = 5 cm = 0.05 m
extend spring x = 1.5747 cm = 0.15747 m
spring's extension = 0.0116 m
to find out
charges of the beads
solution
we know that force is
force = mass × g
force = 0.003859 × 9.8
force = 0.03782 N
so we know force for mass
force = -kx
so k = force / x
put here force and x value
k = -0.03782 / 0.1575
k = -0.24 N/m
and
force for spring's extension
force = -kx
force = -0.24 ( 0.0116) = 0.002784 N
so here
total length L = 0.05 + 0.0116 = 0.0616
so charges of the beads = force × L² / ke
charges of the beads = 0.002784 × (0.0616)² / (9 ×
)
so charges of the beads = 1.173 ×
C
Answer:
(A) Q = 321.1C (B) I = 42.8A
Explanation:
(a)Given I = 55A−(0.65A/s2)t²
I = dQ/dt
dQ = I×dt
To get an expression for Q we integrate with respect to t.
So Q = ∫I×dt =∫[55−(0.65)t²]dt
Q = [55t – 0.65/3×t³]
Q between t=0 and t= 7.5s
Q = [55×(7.5 – 0) – 0.65/3(7.5³– 0³)]
Q = 321.1C
(b) For a constant current I in the same time interval
I = Q/t = 321.1/7.5 = 42.8A.
If the temperature increases, then pressure increases too. (T<span>he molecules in the gas move faster, exerting a greater force. This </span>increases t<span>he </span>pressure<span>.)</span>