Answer:
K = kg/m
Explanation:
as velocity has units of m/s V³ has units of m³/s³
power has units of J/s = N•m/s = kg•m/s²•m/s = kg•m²/s³
kg•m²/s³ = K(m³/s³)
K = kg•m²/s³•s³/m³
K = kg/m
Use: dsin(θ) = mλ where d is slit separation, m is fringe order (1 here), and
θ = 0.183
Now λ = dsin(θ) /m = (0.215e-3)(sin(0.183))/1 = 6.867e-7 or λ = 687.7nm
(red laser)
Answer:
The height is the same
Explanation:
Because they were at the same height but they fell at different velocities
Answer: option D) 42.4 N
The weight of the frame is balanced by the vertical component of tension.
W = T sin θ + T sin θ = 2 T sin θ
The tension in each cable is T = 30 N
Angle made by the cables with the horizontal, θ = 45°
⇒ W = 2×30 N × sin 45° = 2 × 30 N × 0.707 = 42.4 N
Hence, the weight of the frame is 42.4 N. Correct option is D.
Answer:
F = - 3.53 10⁵ N
Explanation:
This problem must be solved using the relationship between momentum and the amount of movement.
I = F t = Δp
To find the time we use that the average speed in the contact is constant (v = 600m / s), let's use the uniform movement ratio
v = d / t
t = d / v
Reduce SI system
m = 26 g ( 1 kg/1000g) = 26 10⁻³ kg
d = 50 mm ( 1m/ 1000 mm) = 50 10⁻³ m
Let's calculate
t = 50 10⁻³ / 600
t = 8.33 10⁻⁵ s
With this value we use the momentum and momentum relationship
F t = m v - m v₀
As the bullet bounces the speed sign after the crash is negative
F = m (v-vo) / t
F = 26 10⁻³ (-500 - 630) / 8.33 10⁻⁵
F = - 3.53 10⁵ N
The negative sign indicates that the force is exerted against the bullet
