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3 years ago

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

= v0

+ 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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Afina-wow [57]

Answer:

$W = 1.22 \times 10^9 J$

Explanation:

Initial potential energy of the given spacecraft is given as

$U = -\frac{GM_e m}{r} - \frac{GM_m m}{r}$

so we have

$U = - \frac{Gm}{r}(M_e + M_m)$

so we have

$M_e = 5.98 \times 10^{24} kg$

$M_m = 7.35 \times 10^{22} kg$

$m = 1160 kg$

$r = 3.84 \times 10^8 m$

$U = - \frac{(6.67 \times 10^{-11})(1160)}{3.84 \times 10^8}(5.98 \times 10^{24} + 7.35 \times 10^{22})$

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2 years ago

Miles is camping in Glacier National Park. In the midst of a glacier canyon,

valentina_108 [34]

Answer:

t=1.623 sec

Explanation:

The distance traveled before the echo is had is:

$distance=2d, d=280\ m\\\\=280\times 2\\\\=560 \ m$

Given the speed of sound as v=345m/s, we use the speed equation to solve for t:

$v=\frac{d}{t}\\\\345\ m/s=\frac{560m}{t}\\\\t=\frac{560}{360}\\\\=1.623 \ s$

Hence, it takes 1.623 seconds to hear the echo.

8 0

2 years ago

Express the number in scientific notation: 4/100

ankoles [38]

Answer:

4*10^-2

Explanation:

for the scientific notation the first number must be between 1 and 10, so in this case it is 4. 4/100 is also equal to 0.04, and if we could the number of places before 4, there are two, therefore 4 times 10 to the power of -2

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2 years ago

The hydraulic oil in a car lift has a density of 8.30 102 kg/m³. The weight of the input piston is negligible. The radii of the

denpristay [2]

Answer:

(a) $F_i=68.58\ N$

(b) $F_i=69.903\ N$

Explanation:

Given:

density of hydraulic oil, $\rho=830\ kg.m^{-3}$

radius of input piston, $r_i=6.3\times 10^{-3}\ m$

radius of output plunger, $r_o=0.125\ m$

force to be supported, $F_o=27000\ N$

(a)

Condition:The bottom surfaces of piston and plunger at the same level.

According to Pascal's law the pressure of a fluid is exerted equally in all directions against the walls of its container.

Mathematically:

$\frac{F_i}{A_i} =\frac{F_o}{A_o}$

putting respective values

$\frac{F_i}{\pi\times r_i^2} =\frac{27000}{\pi\times r_o^2}$

$\frac{F_i}{\pi\times (6.3\times 10^{-3})^2} =\frac{27000}{\pi\times 0.125^2}$

$F_i=68.58\ N$

(b)

Condition:The bottom surface of the output plunger is 1.30 m above that of the input piston.

Given:

$h=1.3\ m$

Now,

$P_i=P_o+\rho.g.h$

$\frac{F_i}{\pi\times (6.3\times 10^{-3})^2} =\frac{27000}{\pi\times 0.125^2} +830\times 9.8\times 1.3$

$F_i=69.903\ N$

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2 years ago

The sound intensity a distance d1 = 15.0 m from a chain saw is 0.260 W/m2. What is the sound intensity a distance d2 = 27.0 m fr

bagirrra123 [75]

Answer:

0.08024 W/m²

Explanation:

P = Power

$d_1$ = Distance = 15 m

$I_1$ = Intensity at 15 m = 0.26 W/m²

$d_2$ = Distance = 27 m

$I_1$ = Intensity at 27 m

Sound intensity is given by

$I=\frac{P}{4\pi d^2}$

$\\\Rightarrow I\propto \frac{1}{d^2}$

So, we have the relation

$\frac{I_1}{I_2}=\frac{d_2^2}{d_1^2}\\\Rightarrow I_2=\frac{d_1^2\times I_1}{d_2^2}\\\Rightarrow I_2=\frac{15^2\times 0.26}{27^2}\\\Rightarrow I_2=0.08024\ W/m^2$

The sound intensity at the given distance is 0.08024 W/m²

8 0

3 years ago

All the formulas for Friction​

All the formulas for Friction