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dusya [7]
2 years ago
10

Hey how has ur morning gone?

Physics
1 answer:
victus00 [196]2 years ago
6 0

Answer:

good

Explanation:

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The patellar tendon attaches to the tibia at a 20 deg angle 3 cm from the axis of rotation at the knee. If the force generated i
gregori [183]

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

8 0
3 years ago
Can y’all help me with 5 plsssss
Alona [7]

Answer:

Bar graph

Explanation:

each day collects data so a bar graph would work.

4 0
3 years ago
Read 2 more answers
If a system has 225 kcal of work done to it, and releases 5.00 × 102 kj of heat into its surroundings, what is the change in int
vovikov84 [41]

We can solve the problem by using the first law of thermodynamics:

\Delta U = Q-W

where

\Delta U is the change in internal energy of the system

Q is the heat absorbed by the system

W is the work done by the system on the surrounding


In this problem, the work done by the system is

W=-225 kcal=-941.4 kJ

with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is

Q=-5 \cdot 10^2 kJ=-500 kJ

with a negative sign as well because it is released by the system.


Therefore, by using the initial equation, we find

\Delta U=Q-W=-500 kJ+941.4 kJ=441.4 kJ

8 0
3 years ago
When a 25-kg crate is pushed across a frictionless horizontal floor with a force of 200 N, directed 20 below the horizontal, th
Fofino [41]

Answer:

Option E is correct 310N

Explanation:

Given that the force used to push the crate is F = 200N

The force directed 20° below the horizontal

Mass of crate is m = 25kg

Weight of the crate can be determine using

W = mg

g is gravitational constant =9.8m/s²

W = 25×9.8

W = 245 N

Check attachment. For free body diagram and better understanding

Using newton second law along the vertical axis since we want to find the normal force

ΣFy = m•ay

ay = 0, since the body is not moving in the vertical or y direction

N—W—F•Sin20 = 0

N = W+F•Sin20

N = 245+ 200Sin20

N = 245 + 68.4

N = 313.4 N

The normal force is approximately 310 N to the nearest ten

3 0
3 years ago
Consider a venturi with a small hole drilled in the side of the throat. This hole is connected via a tube to a closed reservoir.
Marina CMI [18]

Answer:

(P_1-P_2)=1913.31 N/m^2

Explanation:

given:

\frac{A_t}{A_1}=0.85

V_1=90 m/s

γ∞=1.23 kg/m^3

solution:

since outside pressure is atm pressure vaccum can be defined by (P_1-P_2)

V_1=√2(P_1-P_2)/γ∞[\frac{A_t}{A_1}^2-1]

(P_1-P_2)=1913.31 N/m^2

6 0
3 years ago
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