Answer:
they carry a heavier load
Explanation:
Answer:
a)W= 1 J
b)W= 1 J
c)W= 0 J
Explanation:
Given that
m = 5 kg
Coefficient of kinetic friction,K=0.5
d= 10 cm
We know that if velocity is constant it means that the acceleration of the system is zero.Or we can say that all forces are balance in the system.
We know that work is the dot product of force and displacement
W= F.d
The friction force on the box
Fr= K m g
Fr= 0.5 x 5 x 10 = 25 N ( take g =10 m/s²)
Fr=25 N
a)
Acceleration a= 0
So horizontal force F = Fr
W = Fr.d
W=10 x 0.1 J
W= 1 J
b)
The work done by friction force
W= 1 J
c)
The net force on the system is zero because acceleration is zero.
F= 0
So
W= 0 J
Incomplete question as the charge density is missing so I assume charge density of 3.90×10^−12 C/m².The complete one is here.
An electron is released from rest at a distance of 0 m from a large insulating sheet of charge that has uniform surface charge density 3.90×10^−12 C/m² . How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point 3.00×10−2 m from the sheet?
Answer:
Work=1.06×10⁻²¹J
Explanation:
Given Data
Permittivity of free space ε₀=8.85×10⁻¹²c²/N.m²
Charge density σ=3.90×10⁻¹² C/m²
The electron moves a distance d=3.00×10⁻²m
Electron charge e=-1.6×10⁻¹⁹C
To find
Work done
Solution
The electric field due is sheet is given as
E=σ/2ε₀
Now we need to find force on electron
Now for Work done on the electron
Answer: 15.6 metres
Explanation:
Given that:
length of wave (λ)= ?
Frequency of wave F = 28 Hertz
Speed of wave (V) = 437 m/s
The wavelength is the distance covered by the wave in one complete cycle. It is measured in metres, and represented by the symbol λ.
So, apply V = F λ
Make λ the subject formula
λ = V / F
λ = 437 m/s / 28 Hertz
λ = 15.6 m
Thus, the length of the wave is 15.6 metres
Smoother because it will increase energy and when the energy increases it’ll create heat also . Example: A car racing on a smooth road it’ll go faster than a Car speeding on a bumpy and rough road , Hope that helps .
