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
Answer:
Density of the object = 1.9399g/mL
Explanation:
Mass of object = 10.01g
Volume of water = 3.90mL
Volume of Object + Water = 9.06mL
Therefore, volume of Object = Volume of Object + Water - Volume of Water
= 9.06mL - 3.90mL
= 5.16mL
Density by definition is the mass per unit volume of a substance.
Density of the object = mass/volume
= 10.01/5.16
= 1.9399g/mL or 1.94kg/m3
Answer:
a)Q=71.4 μ C
b)ΔV' = 10.2 V
Explanation:
Given that
C ₁= 8.7 μF
C₂ = 8.2 μF
C₃ = 4.1 μF
The potential difference of the battery, ΔV= 34 V
When connected in series
1/C = 1/C ₁ + 1/C₂ + 1/C₃
1/ C= 1/8.4 +1 / 8.4 + 1/4.2
C=2.1 μF
As we know that when capacitor are connected in series then they have same charge,Q
Q= C ΔV
Q= 2.1 x 34 μ C
Q=71.4 μ C
b)
As we know that when capacitor are connected in parallel then they have same voltage difference.
Q'= C' ΔV'
C'= C ₁+C₂+C₃ (For parallel connection)
C'= 8.4 + 8.4 + 4.2 μF
C'=21 μF
Q'= C' ΔV'
Q'=3 Q
3 x 71.4= 21 ΔV'
ΔV' = 10.2 V
Answer:

Explanation:
As we know that electric field due to infinite line charge distribution at some distance from it is given as

now we need to find the electric field at mid point of two wires
So here we need to add the field due to two wires as they are oppositely charged
Now we will have

now plug in all data



now we have



B. they both involve wave interaction.