The wheel and axle increases your force. You exert your input force over a long distance and the output force is increased over a shorter distance. (Because the wheel is larger than the axle, the axle rotates and exerts a large output force.) A simple machine with a grooved wheel with a rope or cable wrapped around it.
Answer:
if you look it up i think u can find it it could be a mealltiod Co 27
Explanation:
Answer:
a) the magnitude of the force is
F= Q() and where k = 1/4πε₀
F = Qqs/4πε₀r³
b) the magnitude of the torque on the dipole
τ = Qqs/4πε₀r²
Explanation:
from coulomb's law
E =
where k = 1/4πε₀
the expression of the electric field due to dipole at a distance r is
E(r) = , where p = q × s
E(r) = where r>>s
a) find the magnitude of force due to the dipole
F=QE
F= Q()
where k = 1/4πε₀
F = Qqs/4πε₀r³
b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces
τ = F sinθ × s
θ = 90°
note: sin90° = 1
τ = F × r
recall F = Qqs/4πε₀r³
∴ τ = (Qqs/4πε₀r³) × r
τ = Qqs/4πε₀r²
Answer:
The correct answer is B
Explanation:
To calculate the acceleration we must use Newton's second law
F = m a
a = F / m
To calculate the force we use the defined pressure and the radiation pressure for an absorbent surface
P = I / c absorbent surface
P = F / A
F / A = I / c
F = I A / c
The area of area of a circle is
A = π r²
We replace
F = I π r² / c
Let's calculate
F = 8.0 10⁻³ π (1.0 10⁻⁶)²/3 10⁸
F = 8.375 10⁻²³ N
Density is
ρ = m / V
m = ρ V
m = ρ (4/3 π r³)
m = 4500 (4/3 π (1 10⁻⁶)³)
m = 1,885 10⁻¹⁴ kg
Let's calculate the acceleration
a = 8.375 10⁻²³ / 1.885 10⁻¹⁴
a = 4.44 10⁻⁹ m/s² absorbent surface
The correct answer is B