Answer:
At 81. 52 Deg C its resistance will be 0.31 Ω.
Explanation:
The resistance of wire =
Where
=Resistance of wire at Temperature T
= Resistivity at temperature T ![=\rho_0 \ [1 \ + \alpha\ (T-T_0\ )]](https://tex.z-dn.net/?f=%3D%5Crho_0%20%5C%20%5B1%20%5C%20%2B%20%5Calpha%5C%20%28T-T_0%5C%20%29%5D)
Where 
l=Length of the wire
& A = Area of cross section of wire
For long and thin wire the resistance & resistivity relation will be as follows

![\frac{0.25}{0.31}=\frac{1}{[1+\alpha(T-20)]}](https://tex.z-dn.net/?f=%5Cfrac%7B0.25%7D%7B0.31%7D%3D%5Cfrac%7B1%7D%7B%5B1%2B%5Calpha%28T-20%29%5D%7D)



T = 81.52 Deg C
Answer:
15193.62 m/s
Explanation:
t = Time taken = 6.5 hours
u = Initial velocity = 0 (Assumed)
m = Mass of rocket = 1380 kg
F = Thrust force = 896 N
v = Final velocity
a = Acceleration of the rocket
Force

Equation of motion

The velocity of the rocket after 6.5 hours of thrust is 15193.62 m/s
About 12 hours is the time between a morning high tide and the next high tide
Explanation:
The Earth’s rotation happens between two tidal bulges
The “periodic rise and fall” of the surface water levels of the ocean is called tides. The gravitational action and interaction on the earth by the sun and the moon causes these tides. Different regions of the World experiences different patterns of tides like the diurnal, semi-diurnal etc.
When there is one high and one low tide occurring on a lunar day, then it is diurnal pattern. Semi-diurnal pattern occurs when there are two equal high and low tides on a single lunar day.
Since the Earth’s rotation happens between two tidal “bulges” on each lunar day, the coastal areas can experience two high and two low tides in every 24 hours plus 50 minutes.
Accordingly the time between two high tides would be 12 hours plus 25 minutes. Similarly, the time gap between a high to low tide would be 6 hours plus 12.5 minutes.
Answer:

Explanation:
By Einstein's Equation of photoelectric effect we know that

here we know that
= energy of the photons incident on the metal
= minimum energy required to remove photons from metal
= kinetic energy of the electrons ejected out of the plate
now we know that it requires 351 nm wavelength of photons to just eject out the electrons
so we can say

here we know that

now we have

now by energy equation above when photon of 303 nm incident on the surface




