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antiseptic1488 [7]
3 years ago
6

The effect of itching power what’s the Independent variable and dependent and constants?

Physics
1 answer:
kenny6666 [7]3 years ago
7 0
The independent variable is what type of powder is used, the dependent variable is how long itching lasts and the constant is the amount of powder used (50 grams)
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A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant inten
Lyrx [107]

Answer:

a) the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴  

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

(a) Determine the maximum shear stress tmax in the bar

Maximum Applied torque  T_max = tL

we know that;

shear stress τ = 16T/πd³

where d is the diameter

so

τ_{max} = 16T_{max} /πd³

Therefore, the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

(b) Determine the angle of twist between the ends of the bar.

let theta (\theta) be the angle of twist

polar moment of inertia I_p} = πd⁴/32

now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d\theta = T(x)dx / GI_{p}

so

d\theta = tx.dx / G(πd⁴/32)

d\theta = 32tx.dx / πGd⁴

so total angle of twist \theta will be;

\theta =  \int\limits^L_0  \, d\theta

\theta =  \int\limits^L_0  \, 32tx.dx / πGd⁴

\theta = 32t / πGd⁴  \int\limits^L_0  \, xdx

\theta = 32t / πGd⁴ [ L²/2]

\theta = 16tL² / πGd⁴  

Therefore,  the angle of twist between the ends of the bar is 16tL² / πGd⁴  

7 0
3 years ago
A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk
Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration\frac{m}{s^{2} }

at: truck acceleration\frac{m}{s^{2} })

ac = at= \frac{vf-vi}{t-ti}  equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 \frac{m}{s}, t = final time =5 s

We replaced the known information in the equation(1):

ac = at = \frac{13-0}{15-0}

ac=ac=\frac{13}{15}  \frac{m}{s}

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

Wc=mc*g

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 \frac{m}{s^{2} }

Wc= 2230*9.8

Wc=21854 N

Normal force calculation:

Newton's first law

sum Fy= 0

N-W=0

N=W

N=21854 N

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N  Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

Ff=0.373*21854

Ff=8151.54 N

Calculation of the tension force in the rope (T):

Newton's Second law

sum Fx= mc*ac

T-Ff=mc*ac

T=2230(\frac{13}{15}) + 8151.54

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

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It is 3.2 because you add 2.3 and 4.1 then divide your answer by 2.
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