We have energy E = hc/λ, where h is Planck's constant c is speed of light and λ is the wavelength.
So Energy ,
Energy of one mol =
Energy of one mol of photons generated from this device = 225 kJ
Newton's 2nd law: F = m a
Divide each side by m :. a = F/m
Plug in the 2 given numbers:
a = 744N / 205kg
Acceleration = 3.63 m/s^2 north
Answer:
the average force exerted by seatbelts on the passenger is 5625 N.
Explanation:
Given;
initial velocity of the car, u = 50 m/s
distance traveled by the car, s = 20 m
final velocity of the after coming to rest, v = 0
mass of the passenger, m = 90 kg
Determine the acceleration of the car as it hit the pile of dirt;
v² = u² + 2as
0 = 50² + (2 x 20)a
0 = 2500 + 40a
40a = -2500
a = -2500/40
a = -62.5 m/s²
The deceleration of the car is 62.5 m/s²
The force exerted on the passenger by the backward action of the car is calculated as follows;
F = ma
F = 90 x 62.5
F = 5625 N
Therefore, the average force exerted by seatbelts on the passenger is 5625 N.
Answer:
the tension in the part of the cord attached to the textbook is 7.4989 N
Explanation:
Given the data in the question;
As illustrated in the image below;
first we determine the value of the acceleration,
along vertical direction; we use the second equation of motion;
y = ut + at²
we substitute;
0 m/s for u, 1.29 m for y, 0.850 s for t,
1.29 = 0×0.850 + ×a×(0.850)²
1.29 = 0.36125a
a = 1.29 / 0.36125
a = 3.5709 m/s²
Now when the text book is moving with acceleration , the dynamic equation will be;
T₁ = m₁a
where m₁ is the mass of the text book ( 2.10 kg )
a is the vertical acceleration ( 3.5709 m/s² )
so we substitute
T₁ = 2.10 × 3.5709
T₁ = 7.4989 N
Therefore, the tension in the part of the cord attached to the textbook is 7.4989 N