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docker41 [41]
2 years ago
3

A truck traveling at 75 mi/h has a braking efficiency of 70%. The coefficient of road adhesion is 0.80. Ignoring aerodynamic res

istance, determine the theoretical stopping distance on a level grade.
Physics
1 answer:
Greeley [361]2 years ago
7 0

Answer:100.36 m

Explanation:

Given

speed of truck u=75 mi/hr\approx 33.528\ m/s

Braking efficiency of 70 %

i.e. braking declaration will be 70 % of frictional force

\mu =0.8

a_{max}=\mu g=0.8\times 10=8 m/s^2

effective acceleration a_{eff}=0.7\times 8=5.6\ m/s^2

using equation of motion

v^2-u^2=2 a s

where v=final velocity

u=initial velocity

a=acceleration or declaration

s=displacement

here v will be zero

-(33.528)^2=2(-5.6)(s)

s=\frac{33.528^2}{2\times 5.6}

s=100.36 m              

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For a magnetic field strength of 2T, estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical
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Answer:

Incomplete question

This is the complete question

For a magnetic field strength of 2 T, estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical nerve that has a diameter of 1.5 mm. Assume that the entire nerve carries a current due to an applied voltage of 100 mV (that of a typical action potential). The resistivity of the nerve is 0.6ohms meter

Explanation:

Given the magnetic field

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Lenght of rod is 1mm

L=1/1000=0.001m

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d=1.5/1000=0.0015m

Radius is given as

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A=π×0.00075²

A=1.77×10^-6m²

Given that the voltage applied is 100mV

V=0.1V

Given that resistive is 0.6 Ωm

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V=iR

i=V/R

i=0.1/339.4

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B=μoI/2πR

Where

μo is a constant and its value is

μo=4π×10^-7 Tm/A

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B=4π×10^-7×2.95×10^-4/(2π×0.00075)

B=8.43×10^-8 T

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F=iL(2B)

F=2.95×10^-4×2×8.34×10^-8

F=4.97×10^-11N

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Which equation is most likely used to determine the acceleration from a velocity vs:time graph?
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An archer draws her bow and stores 34.8 J of elastic potential energy in the bow. She releases the 63 g arrow, giving it an init
elena-14-01-66 [18.8K]

Answer:

Approximately 71\%.

Explanation:

The formula for the kinetic energy \rm KE of an object is:

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m = 63\; \rm g = 0.063\; \rm kg.

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That's the same as the energy output of this bow. Hence, the efficiency of energy transfer will be:

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