Answer:
4.14 eV
Explanation:
f = 1.0 ×10^15 Hz
h= 6.63×10^-34 J s ( this is called PLANCK 'S CONSTANT)
ENEGY = E = ?
E = hf ( THIS IS FORMULA FOR ENERGY OF ONE QUANTA OR ONE PHOTON )
E= 6.63×10^-34×1.0 ×10^15
E = 6.63×10^-19 J
As 1eV = 1.6×10^-19 J so changing energy in eV from joules we will divide energy by 1.6×10^-19
hence E in eV = 6.63×10^-19/(1.6×10^-19)
E = 4.14 eV
Explanation:
Below is an attachment containing the solution.
The statement about "<span>efficiency compared the output work to the output force" is false. Efficiency can be compared from the input work to the output work.</span>
The net force = sum of all forces acting on the body
If we take left side as -ve and right side as +ve,
then,
The net force here would be equal to,
10N + (- 3N)
= 7N.
Therefore, a net force of +7N ( + indicates it's moving towards right) is acting on the book of mass 2kg.
Answer:
vf = 11.2 m/s
Explanation:
m = 10 Kg
F = 2*10² N
x = 4.00 m
μ = 0.44
vi = 0 m/s
vf = ?
We can apply Newton's 2nd Law
∑ Fx = m*a (→)
F - Ffriction = m*a ⇒ F - (μ*N) = F - (μ*m*g) = m*a ⇒ a = (F - μ*m*g)/m
⇒ a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²
then , we use the equation
vf² = vi² + 2*a*x ⇒ vf = √(vi² + 2*a*x)
⇒ vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s
