the water specific heat will remain at 4.184.
Answer:
Explanation:
Given that,
Mass m = 6.64×10^-27kg
Charge q = 3.2×10^-19C
Potential difference V =2.45×10^6V
Magnetic field B =1.6T
The force in a magnetic field is given as Force = q•(V×B)
Since V and B are perpendicular i.e 90°
Force =q•V•BSin90
F=q•V•B
So we need to find the velocity
Then, K•E is equal to work done by charge I.e K•E=U
K•E =½mV²
K•E =½ ×6.64×10^-27 V²
K•E = 3.32×10^-27 V²
U = q•V
U = 3.2×10^-19 × 2.45×10^6
U =7.84×10^-13
Then, K•E = U
3.32×10^-27V² = 7.84×10^-13
V² = 7.84×10^-13 / 3.32×10^-27
V² = 2.36×10^14
V=√2.36×10^14
V = 1.537×10^7 m/s
So, applying this to force in magnetic field
F=q•V•B
F= 3.2×10^-19 × 1.537×10^7 ×1.6
F = 7.87×10^-12 N
Answer:
a
b
Explanation:
Generally the force of attraction between this two irons is mathematically represented as
![F = \frac{k * [Q_{Li} ] * [Q_{O} ] }{ r^2}](https://tex.z-dn.net/?f=F%20%3D%20%20%5Cfrac%7Bk%20%2A%20%20%5BQ_%7BLi%7D%20%20%5D%20%2A%20%5BQ_%7BO%7D%20%20%5D%20%20%7D%7B%20r%5E2%7D)
Here k is known as the proportionality constant with value 
substituting -2 for 
i.e the charge on oxygen , +1 for 
i.e the charge on Lithium and ![[0.140 + 0.068 ] nm= 0.208 nm = 0.208*10^{-9}](https://tex.z-dn.net/?f=%20%5B0.140%20%2B%200.068%20%5D%20nm%3D%200.208%20nm%20%3D%20%200.208%2A10%5E%7B-9%7D%20%20%20)
for r
So
Generally the force of repulsion will be the magnitude but different direction to the force o attraction
So Force of repulsionn is
The vertical component is = vsinx m/s
If you know the angle, substitute the value of x.
If you know the velocity at which it is moving, substitute it for v
Hope it helps :)
