They can have a close similar appearance to the parents, have close relation of child reactions.
for example, everyone born in my father's side of the family had the tendency to bump their head on something as they fall asleep up to the point when you are a toddler.
Anything less dense than water will float, like oil. Anything more dense than water will sink, like rock.
Answer:
√(6ax)
Explanation:
Hi!
The question states that during a time t the motorcyle underwent a displacement x at constant acceleration a starting from rest, mathematically we can express it as:
x=(1/2)at^2
Then the we need to find the time t' for which the displacement is 3x
3x=(1/2)a(t')^2
Solving for t':
t'=√(6x/a)
Now, the velocity of the motorcycle as a function of time is:
v(t)=a*t
Evaluating at t=t'
v(t')=a*√(6x/a)=√(6*x*a)
Which is the final velocity
Have a nice day!
Answer:
the shooting angle ia 18.4º
Explanation:
For resolution of this exercise we use projectile launch expressions, let's see the scope
R = Vo² sin (2θ) / g
sin 2θ = g R / Vo²
sin 2θ = 9.8 75/35²
2θ = sin⁻¹ (0.6)
θ = 18.4º
To know how for the arrow the tree branch we calculate the height of the arrow at this point
X2 = 75/2 = 37.5 m
We calculate the time to reach this point since the speed is constant on the X axis
X = Vox t
t2 = X2 / Vox = X2 / (Vo cosθ)
t2 = 37.5 / (35 cos 18.4)
t2 = 1.13 s
With this time we calculate the height at this point
Y = Voy t - ½ g t²
Y = 35 sin 18.4 1.13 - ½ 9.8 1,13²
Y = 6.23 m
With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch
Given that the rope is not moving (acceleration is zero), by the second Law of Newton (F=m*a), the net force acting on the rope is zero.
Then, the force applied by the team B equals the force applied by the tema A: 103 N.
