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masya89 [10]
2 years ago
14

A person pushes on the handle of a lawnmower with a force of 280. n. if the handle makes an angle of 40.0 degrees with the groun

d, calculate the coefficient of friction if the lawnmower weighs 350. n and is moving at a constant velocity.
Physics
2 answers:
Serggg [28]2 years ago
8 0
<span>Force F = 280 N Angle with the ground = 40 degrees Weight of the Lawnmower = 350 N Velocity is constant so Acceleration is 0 So Forward force Ff = F cos theta = 280 cos40 Frictional force with resists to back Fb = (u x Force from pressure) + vertical component of Force, where u is the coefficient of friction. Fb = (u x m x g) + (u x 280sin40) AS Ff = Fb => 280 cos40 = u x ((m x g) + 280sin40) u = 280 cos40 / ((350 x 9.81) + 280sin40) = 214.49 / () = 0.405 So the coefficient of friction u = 0.405</span>
Lorico [155]2 years ago
8 0

Answer:

µ= 0.405

Explanation:

Acceleration is 0 (constant V), so all the forces in the horizontal direction must add up to 0. Specifically those are these two:

Horizontal component of the force of (F1). It acts FORWARD.  

F1 = F*cosθ = 280cos40 N

Force of friction

F(fr) = µ*(total pressure force).

Two forces are creating the pressure force (Fp).

The other is the vertical component of F It's equal to

F*sinθ = 280sin40.

Then the friction force

F(fr) = µ*(mg + 280sin40)

Since

F1 = F(fr),

280cos40 = µ*(mg + 280sin40)

µ = 280cos40 / (mg + 280sin40)

µ= 0.405

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Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

                               j_{H_2O}=\frac{D}{l} \bigtriangleup c

where

  • j_{H_2O} is the mass flux of the water which is to be calculated.
  • D is diffusion coefficient which is given as 0.25 cm^2/s
  • l is the thickness of the film which is 0.15 cm thick.
  • \bigtriangleup c is given as

                                \bigtriangleup c= \frac{P_{sat}-P_a}{RT}

In this

  • P_{sat} is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa
  • P_a is the air pressure which is given as 0.5 times of P_{sat}
  • R is the universal gas constant as 8.314 kJ/kmol-K
  • T is the temperature in Kelvin scale which is 20+273= 293K

By substituting values in the equation

                                    \bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3

Converting \bigtriangleup c into cm^3/cm^3

As 1 mole of water 18 cm^3 so

                               \bigtriangleup c= 0.48 mol/m^3 \\ \bigtriangleup c= 0.48 \times 18 \times 10^{-6}  cm^3/cm^3 \\ \bigtriangleup c= 8.64 \times 10^{-6}  cm^3/cm^3

Putting this in the equation of mass flux equation gives

                            j_{H_2O}=\frac{D}{l} \bigtriangleup c \\ j_{H_2O}=\frac{0.25}{0.15} \times 8.64 \times 10^{-6} \\ j_{H_2O}=14.4 \times 10^{-6}  cm/s

For calculation of water level drop in a day, converting mass flux as

                     j_{H_2O}=14.4 \times 10^{-6}  \times 24 \times 3600  cm/day\\ j_{H_2O}=1.24  cm/day

So the water level will drop by about 1.24 cm in 1 day.

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An egg falls from a nest at a height of 4m. What speed will it have when it is 1m from the ground? Neglect air resistance and ta
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Answer: 7.66 m/s

Explanation:

This situation is related to free fall (vertical motion). Hence, this can be considered a one-dimension problem and we can use the following equation:

V^{2}=V_{o}^{2}+2gd

Where:

V is the final velocity of the egg at the asked height

V_{o}=0 m/s is the initial velocity of the egg

g=9.8 m/s^{2} is the acceleration due gravity

d=4 m- 1m= 3m is the distance at which the egg is from the nest, when it is 1 m from the ground

Isolating V:

V=\sqrt{V_{o}^{2}+2gd}

Substituting the known values:

V=\sqrt{2(9.8 m/s^{2})(3 m)}

V=7.66 m/s This is the final velocity of the egg

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Explanation:

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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
nydimaria [60]

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

B=2T

Lenght of rod is 1mm

L=1/1000=0.001m

Diameter of rod=1.5mm

d=1.5/1000=0.0015m

Radius is given as

r=d/2=0.0015/2

r=0.00075m

Area of the circle is πr²

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

We can calculate the resistance of the cylinder by using

R= ρl/A

R=0.6×0.001/1.77×10^-6

R=339.4Ω

Then the current can be calculated, using ohms law

V=iR

i=V/R

i=0.1/339.4

i=2.95×10^-4 A

i=29.5 mA

The force in a magnetic field of a wire is given as

B=μoI/2πR

Where

μo is a constant and its value is

μo=4π×10^-7 Tm/A

Then,

B=4π×10^-7×2.95×10^-4/(2π×0.00075)

B=8.43×10^-8 T

Then, the force is given as

F=iLB

Since B=2T

F=iL(2B)

F=2.95×10^-4×2×8.34×10^-8

F=4.97×10^-11N

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2 years ago
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