Answer:
The change in temperature is
Explanation:
From the question we are told that
The temperature coefficient is ![\alpha = 4 * 10^{-3 }\ k^{-1 }](https://tex.z-dn.net/?f=%5Calpha%20%20%3D%20%204%20%2A%2010%5E%7B-3%20%7D%5C%20%20k%5E%7B-1%20%7D)
The resistance of the filament is mathematically represented as
![R = R_o [1 + \alpha \Delta T]](https://tex.z-dn.net/?f=R%20%20%3D%20%20R_o%20%5B1%20%2B%20%5Calpha%20%20%5CDelta%20T%5D)
Where
is the initial resistance
Making the change in temperature the subject of the formula
![\Delta T = \frac{1}{\alpha } [\frac{R}{R_o} - 1 ]](https://tex.z-dn.net/?f=%5CDelta%20T%20%3D%20%5Cfrac%7B1%7D%7B%5Calpha%20%7D%20%5B%5Cfrac%7BR%7D%7BR_o%7D%20-%201%20%5D)
Now from ohm law
![I = \frac{V}{R}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BV%7D%7BR%7D)
This implies that current varies inversely with current so
![\frac{R}{R_o} = \frac{I_o}{I}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7BR_o%7D%20%3D%20%5Cfrac%7BI_o%7D%7BI%7D)
Substituting this we have
![\Delta T = \frac{1}{\alpha } [\frac{I_o}{I} - 1 ]](https://tex.z-dn.net/?f=%5CDelta%20T%20%20%3D%20%5Cfrac%7B1%7D%7B%5Calpha%20%7D%20%5B%5Cfrac%7BI_o%7D%7BI%7D%20-%201%20%5D)
From the question we are told that
![I = \frac{I_o}{8}](https://tex.z-dn.net/?f=I%20%20%3D%20%5Cfrac%7BI_o%7D%7B8%7D)
Substituting this we have
![\Delta T = \frac{1}{\alpha } [\frac{I_o}{\frac{I_o}{8} } - 1 ]](https://tex.z-dn.net/?f=%5CDelta%20T%20%20%3D%20%5Cfrac%7B1%7D%7B%5Calpha%20%7D%20%5B%5Cfrac%7BI_o%7D%7B%5Cfrac%7BI_o%7D%7B8%7D%20%7D%20-%201%20%5D)
=> ![\Delta T = \frac{1}{3.9 * 10^{-3}} (8 -1 )](https://tex.z-dn.net/?f=%5CDelta%20T%20%20%3D%20%5Cfrac%7B1%7D%7B3.9%20%2A%2010%5E%7B-3%7D%7D%20%288%20-1%20%29)
Answer: An atom that gains or loses an electron becomes an ion. If it gains a negative electron, it becomes a negative ion. If it loses an electron it becomes a positive ion
Answer:
39.6 m/s
Explanation:
Taking down to be positive:
Given:
v₀ = 0 m/s
a = 9.8 m/s²
Δy = 80 m
Find: v
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (9.8 m/s²) (80 m)
v = 39.6 m/s
Answer/solution:
Given :
Mass =5kg
T 1 =20 C,T 2 =100 ∘C
ΔT=100−20=80 ∘C
Q=m×C×ΔT
where C= specific heat capacity of water
=4200J/(kgK)
Q=5×4200×80
=1680000 Joule.
=1680KJ
What’s the weight and how high is the clif