Answer:
The magnitude of the force that the 6.3 kg block exerts on the 4.3 kg block is approximately 41.9 N
Explanation:
Forces on block 4.3 kg are:
63N to the right and R21 (contact force from the 6.3 kg block) to the left
Net force on 4.3 kg block is: 63 N - R21
Forces on the 6.3 kg block are:
R12 to the right (contact force from the 4.3 kg block) and 11 N to the left.
So net force on the 6.3 kg block is: R12 - 11 N
According to the action-reaction principle the contact forces R21 and R12 must be equal in magnitude (let's call them simply "R").
Then, since the blocks are moving with the SAME acceleration, we equal their accelerations:
a1 = (63 N - R)/4.3 = (R - 11 N)/6.3 = a2
solve for R by cross multiplication
6.3 (63 - R) = 4.3 (R - 11)
396.9 - 6.3 R = 4.3 R - 47.3
369.9 + 47.3 = 10.6 R
444.2 = 10.6 R
R = 444.2 / 10.6
R = 41.90 N
KE= (1/2) mv²
So, the variables we need to include in our question would be a varable for a mass(m) of an object at some velocity(v).
My Answer:
(This is just an example question, yours can be different)
What is the Kinetic Energy experienced by an bouncy ball rolling at 7m/s (that's your velocity) across a frictionless surface that has a mass(m) of 10 grams?
Answer:
t = 0.37 seconds
Explanation:
t = (1/4)T
Maximum acceleration is;
a_max = Aω²
In simple harmonic motion, we know that v_max = Aω
Thus, a_max = v_max•ω
ω = a_max/v_max
We know that Period is given by;
T = 2π/ω
From initially, t = (1/4)T so, T = 4t
Thus, 4t = 2π/(a_max/v_max)
t = (2π/4)(v_max/a_max)
We are given;
Maximum velocity;v_max = 1.47 m/s
Max acceleration;a_max =6.24 m/s²
Thus,
t = (2π/4)(1.47/6.24)
t = 0.37 seconds
Answer:
the knee extensors must exert 15.87 N
Explanation:
Given the data in the question;
mass m = 4.5 kg
radius of gyration k = 23 cm = 0.23 m
angle ∅ = 30°
∝ = 1 rad/s²
distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m
using the expression;
ζ = I∝
ζ = mk²∝
we substitute
ζ = 4.5 × (0.23)² × 1
ζ = 0.23805 N-m
so
from; ζ = rFsin∅
F = ζ / rsin∅
we substitute
F = 0.23805 / (0.03 × sin( 30 ° )
F = 0.23805 / (0.03 × 0.5)
F F = 0.23805 / 0.015
F = 15.87 N
Therefore, the knee extensors must exert 15.87 N
If I am reading this correct, the energy content would decrease for the liquid to solidify