Answer:
the resulting angular acceleration is 15.65 rad/s²
Explanation:
Given the data in the question;
force generated in the patellar tendon F = 400 N
patellar tendon attaches to the tibia at a 20° angle 3 cm( 0.03 m ) from the axis of rotation at the knee.
so Torque produced by the knee will be;
T = F × d⊥
T = 400 N × 0.03 m × sin( 20° )
T = 400 N × 0.03 m × 0.342
T = 4.104 N.m
Now, we determine the moment of inertia of the knee
I = mk²
given that; the lower leg and foot have a combined mass of 4.2kg and a given radius of gyration of 25 cm ( 0.25 m )
we substitute
I = 4.2 kg × ( 0.25 m )²
I = 4.2 kg × 0.0626 m²
I = 0.2625 kg.m²
So from the relation of Moment of inertia, Torque and angular acceleration;
T = I∝
we make angular acceleration ∝, subject of the formula
∝ = T / I
we substitute
∝ = 4.104 / 0.2625
∝ = 15.65 rad/s²
Therefore, the resulting angular acceleration is 15.65 rad/s²
Using the formula v=f times lambada
then v=the speed of light.
and f=what’s we’re looking for
and lambada=the wavelength.
so then you sub what you have (v and lambada) in the formula.
then multiply the frequency(f) by the given wavelength and then solve for f
They are positive and remain inside the nucleus.
Answer:
B) No.
Explanation:
Okay,so,
this is equation is y=mx +b
mx represents the slope
and b represents the y-intercept
in order to figure this out you need to plot the y-intercept first
that makes its (0,-6) because the 6 is negative in the equation
4x is also equal to 4/1 since we dont know what x is
we have to do rise over run for this
you go up 4 spots on the y intercept from -6 because 4 is positive
then you go to the right 1 time because 1 is positive.
this leaves you at (1,-2)
so, (2,2) is NOT a solution
Answer:
constant volicty of the pumper when they hit ground 7.03-/s
