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Archy [21]
3 years ago
11

All the formulas for Friction​

Physics
1 answer:
notka56 [123]3 years ago
6 0

Answer:

f = \mu N

Explanation:

Newton’s 2nd Law

o Fnet = mass x acceleration = m * a

o If force is gravitational (Fg or weight↓),

 Weight = mass * gravitational acceleration

 Weight = m * g

Kinematic Equations (general)

o vf = v0 + at

o vf

2

= v0

2

+ 2ad

o d = v0t + ½ at2

Inclined Plane ( ) - F|| vs Ff

o F|| down plane (causing motion) = mg sin

o Ff up plane (friction force opposing motion) = mg cos

Friction

o Coefficient = Force of friction (Ff) (0 < <1)

Normal force (FN)

Ff = FN

Net Force (Fnet)

o Flat plane: Fnet = Fa – Ff = Fa – μmg

o Inclined plane: Fnet = F|| - Ff = mgsinθ = μmgcosθ

Pressure

o P = Applied force

area of application

o P = F/A (= mg/A when using a gravity or weight model)

Conversion

o cm2

x 10-4

= m2

o mm2

x 10-6

= m2

Units

o Force - Newtons

o Pressure - N/m2

or Pascals (Pa)

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Water flows through a garden hose which is attached to a nozzle. The water flows through hose with a speed of 1.81 m/s and throu
Serggg [28]

Answer:

a) 17.086m

b) 0.1671 m

Explanation:

Given data: speed of water through the hose  = 1.81 m/s

through the nozzle = 18.3 m/s

We know that maximum height of an object with upward velocity v is given by,

a) H = v^2/2g

where H is the maximum height water emerges  

= 18.3^2/(2×9.8) = 17.086 m answer

b) Again,  

H = v^2/2g

= 1.81^2/(2×9.8) = 0.1671 m

6 0
3 years ago
The work output of a machine divided by the work input is the ____ of the machine.
s344n2d4d5 [400]
<span>The work output of a machine divided by the work input is the "Efficiency" of the machine.

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7 0
3 years ago
At this radius, what is the magnitude of the net force that maintains circular motion exerted on the pilot by the seat belts, th
Ainat [17]

Answer:

Fc=5253 N

Explanation:

Answer:

Fc=5253 N

Explanation:

sequel to the question given, this question would have taken precedence:

"The 86.0 kg pilot does not want the centripetal acceleration to exceed 6.23 times free-fall acceleration. a) Find the minimum radius of the plane’s path. Answer in units of m."

so we derive centripetal acceleration first

ac (centripetal acceleration) = v^2/r

make r the subject of the equation

r= v^2/ac

 ac is 6.23*g which is 9.81

v is 101m/s

substituing the parameters into the equation, to get the radius

(101^2)/(6.23*9.81) = 167m

Now for part

( b) there are two forces namely, the centripetal and the weight of the pilot, but the seat is exerting the same force back due to newtons third law.

he net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth is the centripetal force.

Fc (Centripetal Force) = m*v^2/r  

So (86kg* 101^2)/(167) =

Fc=5253 N

4 0
3 years ago
2 Points
mezya [45]
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5 0
3 years ago
Which temperature is the hottest? 98 F or 39 C or 303K?<br> F= 1.8C + 32<br> C= (F-32)/1.8
sergejj [24]

Answer:

The hottest temperature is  T_2 = 39^o C

Explanation:

From the question we are given

    T_1 =  98 F

  T_2 =  39^oC

  T_3 =  303 \  K

Generally converting T_3 to  Fahrenheit

    T_3' =  (T_3 -273 ) * \frac{9}{5}  + 32

=> T_3' =  (303 -273 ) * \frac{9}{5}  + 32

=> T_3' = 86 F

Converting  T_2 to  Fahrenheit

      T_2' =  T_2 * \frac{9}{5}  + 32

=> T_2' =  39 * \frac{9}{5}  + 32

=> T_2' =102.2 F  

Now comparing  the temperature  in Fahrenheit we see that T_2  is the hottest

3 0
3 years ago
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