Answer:
Stress = F / A force per unit area
A = 3.00 cm^2 = 3 E-4 m^2
F = 2.4E8 N/m^2 * 3E-4 m^2 = 7.2E4 N max force applied
F/3 = 2.4E4 N if force not to exceed limit (= f)
f = M a
a = 2.4 E4 N / 1.2 E3 kg = 20 m / s^2 about 2 g
Answer:
a) a = 3.09 m/s²
b) aₓ = 2.60 m/s²
Explanation:
a) The magnitude of her acceleration can be calculated using the following equation:
<u>Where</u>:
: is the final speed = 8.89 m/s
: is the initial speed = 0 (since she starts from rest)
a: is the acceleration
d: is the distance = 12.8 m
Therefore, the magnitude of her acceleration is 3.09 m/s².
b) The component of her acceleration that is parallel to the ground is given by:
<u>Where</u>:
θ: is the angle respect to the ground = 32.6 °
Hence, the component of her acceleration that is parallel to the ground is 2.60 m/s².
I hope it helps you!
Answer:
Coefficient of static friction = 0.37
Explanation:
At the point the the quoll slides, quoll attains its maximum velocity.
So Ne = (mv^2)/r ....equa 1
And N =mg....equ 2
Where N vertical force of qoull acting on the surface, e = coefficient of friction, m=mass, g=9.8m/s^2, r =radius =1.6m, v= max velocity of quill = 2.4m/s
Sub equ 2 into equ 1
Mge= (mv^2)/r ...equa3
Simplfy equ3
e = v^2/(gr)...equ 4
Sub figures above
e = 5.76/(9.8*1.6)
e = 0.37
Explanation:
Her cells will not work well when they have low levels of oxygen.
