Use newton's second law, F = ma.
Force is 34 N and a is 4 m/s/s
34 = m(4), solve for m.
34/4 = 8.5 g
F=ma
For the first (10kg) cart,
12=10a
a=6/5 m/s^2 to the left
For the second (5kg) cart,
8=5a
a=8/5 m/s^2 to the left
Therefore, the lighter (5kg) cart experiences a greater acceleration.
Answer:
y = - (½ g / v₀²) x²
Explanation:
This is a projectile launch exercise where there is no acceleration on the x-axis so
x = v₀ₓ t
v₀ₓ = v₀ cos tea
y =
t - ½ g t2
v_{oy} = v₀ sin θ
as the sphere is thrown horizontally, the angle is tea = 0º, so the initial velocity remains
v₀ₓ = v₀
v_{oy} = 0
we substitute in our equations
x = v₀ t
y = - ½ g t²
we eliminate the time from these equations, we substitute the first in the second
y = - ½ g (x / v₀)²
y = - (½ g / v₀²) x²
this is the equation of a parabola
The hawk’s centripetal acceleration is 2.23 m/s²
The magnitude of the acceleration under new conditions is 2.316 m/s²
radius of the horizontal arc = 10.3 m
the initial constant speed = 4.8 m/s
we know that the centripetal acceleration is given by
= 
= 23.04/10.3
= 2.23 m/s²
It continues to fly but now with some tangential acceleration
= 0.63 m/s²
therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration
so
= 
= 
= 2.316 m/s²
So the magnitude of net acceleration will become 2.316 m/s².
learn more about acceleration here :
brainly.com/question/11560829
#SPJ4
Answer:
I guess the answer is any device that makes work easier.