Answer:
The angle is 
Explanation:
From the question we are told that
The distance of the dartboard from the dart is 
The time taken is 
The horizontal component of the speed of the dart is mathematically represented as

where u is the the velocity at dart is lunched
so

substituting values

=> 
From projectile kinematics the time taken by the dart can be mathematically represented as

=> 


=> 
![\theta = tan^{-1} [0.277]](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20tan%5E%7B-1%7D%20%5B0.277%5D)

Answer: sheet of charge
Explanation:
a )
Since the charge is negative , potential will be negative near it . At a far point potential will be less negative. So potential will virtually increase on going away from the sheet . At infinity it will become almost zero. Electric field will be towards the plate , so potential will decrease towards the plate.
b ) The shape of equi -potential surface will be plane parallel to the sheet of charge because electric field will be perpendicular to the sheet of charge and almost uniform near the sheet of charge. The equi- potential surface is always perpendicular to electric field.
C ) Electric field which is almost uniform near the sheet of charge is equal t the following
E = σ / ε₀ where σ is charge density of surface and ε₀ is permittivity of medium whose value is 8.85 x 10⁻¹²
E = 3 x 10⁻⁹ / 8.85 x 10⁻¹²
= .3389 x 10³
= 338.9 V / m
spacing between 1 V
= 1 / 338.9 m
= 2.95 X 10⁻3 m
= 2.95 mm.
Heat supplied to the gold will raise the temperature of the gold from 20 degree Celsius to 90 degree Celsius.
Mass of the gold (m) = 0.072 kg
Temperature change (ΔT) = 90 - 20 = 70 degree Celsius
Specific heat capacity of the gold (c) = 136 J/kg C
Heat supplied = m × c × ΔT
Heat supplied = 0.072 × 136 × 70
Heat supplied = 685.44 Joules
Hence, the heat supplied to the gold to raise the temperature from 20 degree Celsius to 90 degree Celsius = 685.44 Joules
Explanation:
Given that,
Initial speed of the sports car, u = 80 km/h = 22.22 m/s
Final speed of the runner, v = 0
Distance covered by the sports car, d = 80 km = 80000 m
Let a is the acceleration of the sports car. It can be calculated using third equation of motion as :




Value of g, 


Hence, this is required solution.
Answer: Option (D) is the correct answer.
Explanation:
The given elements Li, C and F are all second period elements. So, when we move from left to right across a period then there occurs increase in number of valence electrons as there occurs increase in total number of electrons.
So, it means more electrons are added to the same energy level.
Thus, we can conclude that a property of valence electrons for each element is located in the same energy level is common in the given elements.