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Jlenok [28]
3 years ago
11

In order to decrease the friction an object experiences, you can either change

Physics
1 answer:
Kipish [7]3 years ago
8 0

Answer:

Reduce its weight

Explanation:

For an object on a flat surface, the maximum force of friction experienced by the object is given by:

F = \mu W

where

\mu is the coefficient of friction

W is the weight of the object

As we can see from the formula, there are two ways of reducing the  friction:

- Decreasing \mu --> this coefficient depends on the material and on the roughness of the surface, so this can be changed by changing the properties of the surface

- Decreasing W, the weight --> this can be done by reducing the mass of the object

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3 years ago
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A coaxial cable consists of alternating coaxial cylinders of conducting and insulating material.
olga2289 [7]

Answer:

True

Explanation:

A coaxial cable is a type of cable that has an inner conductor surrounded by an insulating layer, surrounded by a conductive shielding.

4 0
3 years ago
A body with mass of 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time tequals0 the bod
11Alexandr11 [23.1K]

Answer:

X(t) = 13/13 cos(12t+α)

C =13/13

π/6 s

Explanation:

(A) A body with mass 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time t = 0 the body is pulled 1 m in to the right, stretching the spring, 3 set in motion with an initial velocity of 5 m/s to the left.  

(a) Find X(t) in the form c • cos(w_o*t— α)  

(b) Find the amplitude 3 Period of motion of the body 1  

mass: m = 200g =  0.200 kg  

displacement: ΔX = 20 cm =  0.20 m

Spring Constant: K =  9/0.20 = 45 N/m

IV:   X(0) = 1m V(0) = -5 m/s

Simple Harmonic Motion: c•cos(cosw_t— α) = X(t)  

Circular Frequency: w_o = √k/m= √36/(0.20) = 13 rad/s

X(0) = 1m =c_1

X'(0) = V(0) = c_2*w_o/w_o

        = -5/12 =   c_2

"radians Technically Unitless"  

Amplitude: c = √ci^2 + c^2 ==> √1^2 + (-5/12)^2 = 1 m =13/13 = c

X(t) = 13/13 cos(12t+α)

since, C>0 : damped forced vibration c_1>0, c_2>0

phase angle 2π+tan^-1(c_2/c_1)

                        =2π+tan^-1(-5/12/1)= 5.884

period: T =2π/w_o

                =π/6 s

6 0
4 years ago
A bicyclist of mass 60kg supplies 340W of power while riding into a 5 m/s head wind. The frontal area of the cyclist and bicycle
NikAS [45]

Answer:

87.1 mph

Explanation:

We are given that

Mass,m=60 kg

Power,P=340 W

Speed,v=5 m/s

Area,A=0.344 m^2

Drag coefficient,C_d=0.88

Coefficient of rolling resistance,\mu_r=0.007

Friction force,f=\mu_rmg=0.007\times 60\times 9.8=4.1 N

Where g=9.8 m/s^2

Let speed of cyclist=v'

Drag force,F_d=\frac{1}{2}\rho_{air}AC_dv^2

Density of air,\rho_{air}=1.225 kg/m^3

F_d=\frac{1}{2}\times 1.225\times 0.344\times 0.88(5)^2=4.635N

Power,P=(F_d+f)\times v'

340=(4.1+4.635) v'=8.735v'

v'=\frac{340}{8.735}=38.9m/s

v'=87.1 mph

1 m=0.00062137 miles

1 hour=3600 s

7 0
4 years ago
A 5 kilogram cat is resting on top of a bookshelf that is 3 meters high. What is the cat’s gravitational potential energy relati
Free_Kalibri [48]
U=mgh
U=5kg*9.8N/kg*3m
U=147J
5 0
3 years ago
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