If it is the same vehicle, then the 60mph vehicle has more kinetic energy since it is moving faster. Therefore, it requires more energy to stop, and if it is the same car with the same beak system, the braking distance of the 30mph car will be significantly shorter than the 60mph car.
Answer:
a) F = 3.2 10⁻¹⁰ N
, b) v = 9.9 10⁷ m / s
Explanation:
a) The electric force is
F = q E
The electric field is related to the potential reference
V = E d
E = V / d
Let's replace
F = e V / d
Let's calculate
F = 1.6 10⁻¹⁹ 28 10³ / 1.4 10⁻²
F = 3.2 10⁻¹⁰ N
b) For this part we can use kinematics
v² = v₀ + 2 a d
v = √ 2 ad
Acceleration can be found with Newton's second law
e V / d = m a
a = e / m V / d
a = 1.6 10⁻¹⁹ / 9.1 10⁻³¹ 28 10³ / 1.4 10⁻²
a = 3,516 10⁻¹⁷ m / s²
Let's calculate the speed
v = √ (2 3,516 10¹⁷ 1.4 10⁻²)
v = √ (98,448 10¹⁴)
v = 9.9 10⁷ m / s
Answer:
a) 4 289.8 J
b) 4 289.8 J
c) 6 620.1 N
d) 411 186.3 m/s^2
e) 6 620.1 N
Explanation:
Hi:
a)
The kinetic energy of the bullet is given by the following formula:
K = (1/2) m * v^2
With
m = 16.1 g = 1.61 x 10^-2 kg
v = 730 m/s
K = 4 289.8 J
b)
the work-kinetic energy theorem states that the work done on a system is the same as the differnce in kinetic energy of the same. Since the initial state of the bullet was at zero velocity (it was at rest) Ki = 0, therefore:
W = ΔK = Kf - Ki = 4 289.8 J
c)
The work done by a force is given by the line intergarl of the force along the trayectory of the system (in this case the bullet).
If we consider a constant force (and average net force) directed along the trayectory of the bullet, the work and the force will be realted by:
W = F * L
Where F is the net force and L is the length of the barrel, that is:
F = (4 289.8 J) / (64.8 cm) = (4 289.8 Nm) / (0.648 m) = 6620.1 N
d)
The acceleration can be found dividing the force by the mass:
a = F/m = (6620.1 N) /(16.1 g) = 411 186.3 m/s^2
e)
The force will have a magnitude equal to c) and direction along the barrel towards the exit
El beneficio que obtuvo es de 2 L
