A man claims that he can hold onto a 15.0-kg child in a head-on collision as long as he has his seat belt on. Consider this man
1 answer:
Answer:
F= 5195.625 N
Explanation:
To obtain the force needed to hold the child, we need to know the aceleration in which the car is breaking.
Aceleration is equal to velocity divided by the time of breaking
In international system, velocity [m/s] is
v= (62 mi/h)*(1609 m/mi)*(1 h/3600 s)
v= 27.71 m/s
Now, we part the velocity by the time that is 0.08 seconds
a= v/t= (27.71 m/s)/(0.08 s)
a= 346.375 m/
The force in agreement with the Newton's second law is
F=m*a = 15 Kg*346.375 m/
F= 5195.625 N
(Note: 1 N = 1 Kg*m/)
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Answer:
A) 16.6 m/s i -17.2 m/s j B) 23.9 m/s c) 46º below horizontal.
Explanation:
A) Once released, the football is not under the influence of any external force in the horizontal direction, so it continues moving at a constant speed equal to the initial velocity, i.e., 16.6 m/s.
If we choose the horizontal direction to be coincident with the x-axis, and make positive the direction towards the right (assuming that this was the direction along which the football was thrown), we can write the horizontal component of the veelocity vector, as follows:
vₓ = 16.6 m/s i
In the vertical direction, the football, once released, is in free fall, starting from rest.
So, we can find the vertical component of the velocity vector, at a given point in time, applying the definition of acceleration, as follows:
vy = a*t = -g*t = -9.81 m/s²*1.75 s = -17.2 m/s
Assuming that the upward direction is the positive for the y-axis (perpendicular to the chosen x-axis), we can write the vertical component of the velocity vector, at t=1.75 s, as follows:
vy = -17.2 m/s j
So, the velocity vector, in terms of the unit vectors i and j, can be written in this way:
v = 16.6 m/s i -17.2 m/s j
b) The magnitude of this vector can be found applying trigonometry, as the magnitude is the hypotenuse of a triangle with sides equal to vx and vy, as follows:
v = 23.9 m/s
c) The direction of the vector (below the horizontal) can be found as the angle which tangent is given by the quotient between vy and vx, as follows:
tg θ =
⇒ θ = tg⁻¹ (-1.036) = 46º below horizontal.
Answer:
m = 2.01[kg]
Explanation:
This problem can be solved using Newton's second law which tells us that the force applied on a body is equal to the product of mass by acceleration.
where:
F = force = 12.5 [N]
m = mass [kg]
a = acceleration = 6.2 [m/s²]