Answer:
The transverse wave will travel with a speed of 25.5 m/s along the cable.
Explanation:
let T = 2.96×10^4 N be the tension in in the steel cable, ρ = 7860 kg/m^3 is the density of the steel and A = 4.49×10^-3 m^2 be the cross-sectional area of the cable.
then, if V is the volume of the cable:
ρ = m/V
m = ρ×V
but V = A×L , where L is the length of the cable.
m = ρ×(A×L)
m/L = ρ×A
then the speed of the wave in the cable is given by:
v = √(T×L/m)
= √(T/A×ρ)
= √[2.96×10^4/(4.49×10^-3×7860)]
= 25.5 m/s
Therefore, the transverse wave will travel with a speed of 25.5 m/s along the cable.
Answer:
a=4,32m/s^2
Explanation:
Fnet = F1 - F2
= 12-1.2
= 10.8N
m=2.5kg
Fnet =ma
10.8=2.5a then divide both sides by 2.5 to get acceleration
Answer:0.000002
Explanation: I Looked It Up lol
Answer:
108.7 V
Explanation:
Two forces are acting on the particle:
- The external force, whose work is 
- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: 
where
q is the charge
is the potential difference
The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:
and since the charge starts from rest, 
, so the formula becomes
In this problem, we have
is the work done by the external force
is the charge
is the final kinetic energy
Solving the formula for , we find
Answer:
a) S = 1.69 10⁹ W/m², b) P = 5.63 Pa
, c) F = 20.6 10⁻¹² N
Explanation:
a) The intensity defined as the energy per unit area
S = U / A
Area of a circle is
W = 6.2 mw = 6.2 10-3 W
R = 1080 nm = 1080 10⁻⁹ m = 1.080 10⁻⁶ m
A = π R2
A = π (1,080 10⁻⁶)²
A = 3.66 10 -12 m²
S = 6.2 10-3 / 3.66 10-12
S = 1.69 10⁹ W / m²
b) The radiation pressure
P = 1 / c (dU / dt) / A
S = (dU / dt) / A
P = S / c
P = 1.69 10 9 / 3. 108
P = 5.63 Pa
c) the definition of pressure is force over area
P = F / A
F = P A
F = 5.63 3.66 10⁻¹²
F = 20.6 10⁻¹² N
d) for this we use Newton's second law
F = ma
a = F / m
