¿Qué distancia recorre un móvil que lleva una aceleración de 5m/s durante 10seg?
1 answer:
Answer:
250 m
Explanation:
The car in this problem is moving of uniform accelerated motion, so we can use the following suvat equation:
where
s is the distance covered
u is the initial velocity
t is the time
a is the acceleration
Assuming the car starts from rest,
u = 0
Also we know that
a = 5 m/s^2 (acceleration of the car)
t = 10 s
Substituting, we find the distance covered:
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Answer:
V=15.3 m/s
Explanation:
To solve this problem, we have to use the energy conservation theorem:
the elastic potencial energy is given by:
The work is defined as:
this work is negative because is opposite to the movement.
The gravitational potencial energy at 2.5 m aboves is given by:
the gravitational potential energy at the ground and the kinetic energy at the begining are 0.
Answer:
x = 6.94 m
Explanation:
For this exercise we can find the speed at the bottom of the ramp using energy conservation
Starting point. Higher
Em₀ = K + U = ½ m v₀² + m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
½ m v₀² + m g h = ½ m v²
v² = v₀² + 2 g h
Let's calculate
v = √(1.23² + 2 9.8 1.69)
v = 5.89 m / s
In the horizontal part we can use the relationship between work and the variation of kinetic energy
W = ΔK
-fr x = 0- ½ m v²
Newton's second law
N- W = 0
The equation for the friction is
fr = μ N
fr = μ m g
We replace
μ m g x = ½ m v²
x = v² / 2μ g
Let's calculate
x = 5.89² / (2 0.255 9.8)
x = 6.94 m
The correct answer is:
D. Electromagnetic waves.
The arrows represent electromagnetic waves.