I think the answer is D. permit the wave to travel through it
Horizontal forehead wrinkles are caused by contraction of the frontalis muscles. The frontalis muscles are two large fanlike muscles that extend from the eyebrow region to the top of the forehead. I think this is the answer
Answer:
The speed of the piñata immediately after being cracked by the stick is .
Explanation:
Using the conservation of linear momentum:
(1)
Here:
m(s) is the mass of the stick
m(p) is the mass of the piñata
v(is) is the initial velocity of the stick
v(fp) is the final velocity of the piñata
So, we just need to solve the equation (1) to v(fp).
I hope it helps you!
Answer:
Explanation:
Plotting the original location of the helicopter before it flies 25 km north, it would be at the origin, (0,0) then after it flies north, the y vertex gains 25 points, so it would be (0,25)
After it flies east, the x coordinate gains 5 points, so it would now be (5,25)
After it flies south, the y coordinate loses or is subtracted by 5 points. so it would now be (5,20)
After flying west, the x coordinate loses 15 points. So the final vertex would be at (-10,20)
East = Right
West = Left
South= Down
North = Up
I used mainly mathematical methods by adding and subtracting the x and y coordinate values, but this could be graphed easily since I gave the coordinates just incase!
Hope this helps!
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