Answer:
The car stops in 7.78s and does not spare the child.
Explanation:
In order to know if the car stops before the distance to the child, you take into account the following equation:
(1)
vo: initial speed of the car = 45km/h
a: deceleration of the car = 2 m/s^2
t: time
xo: initial distance to the child = 25m
x: final distance to the child = 0m
It is necessary that the solution of the equation (1) for time t are real.
You first convert the initial speed to m/s, then replace the values of the parameters and solve the quadratic polynomial for t:
You take the first value t1 because it has physical meaning.
The solution for t is real, then, the car stops in 7.78s and does not spare the child.
To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,
Through the aforementioned formula we will have to
The particulate part of the rest, so the final speed would be
Now from Newton's second law we know that
Here,
m = mass
a = acceleration, which can also be written as a function of velocity and time, then
Replacing we have that,
Therefore the force that the water exert on the man is 1386.62
Answer:
J
Explanation:
The daughter moves with greater acceleration backwards because of her weight.
The acceleration of the ball is 5 m/s^2. This can be calculated using a formula that relates the change in velocity, acceleration, and time. This formula is:
Vf = Vi + at
where:
Vf = final velocity
Vi = initial velocity
a = acceleration
t = time
Substituting the values gives:
30 = 20 + a(2)
<span>a = 5 m/s^2 --> Final Answer</span>
Answer:
Explanation:
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