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VMariaS [17]
3 years ago
14

Name two ways that friction is harmful and two ways that friction is helpful to you when riding a bicycle

Physics
1 answer:
lawyer [7]3 years ago
6 0
<u>Harmful</u><u />

-causes heat; if it's not used correctly, it can cause injury
-when falling, the friction between your knee/leg and the ground can cause scrapes
<u>
</u>
<u>Helpful</u><u />

-helps stop the bike, the brakes rub against the tire which slows it down to a stop
-it can also help move the bike; the chains rub against tires which accelerate the bike
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An Olympic discus thrower (~100 kg) launches the 2.0 kg discus by spinning rapidly (~4 times per second) with arm outstretched (
vladimir1956 [14]

Answer:

F = 1263.03 N

Explanation:s

given,                      

mass of the disk thrower = 100 Kg

mass of the disk = 2 Kg                

angular speed of the disk  = 4 rev/s

arm outstretched = 1 m                  

centripetal force of the disk in the circular path

F = m ω² r                        

ω = 4 x 2 x π        

ω = 25.13 rad/s

F = m ω² r                      

F = 2 x 25.13² x 1

F = 1263.03 N                                              

hence, centripetal force equal to the F = 1263.03 N

6 0
3 years ago
Continuous and aligned fiber-reinforced composite with cross-sectional area of 340 mm2 (0.53 in.2) is subjected to a longitudina
Alecsey [184]

(a) 23.4

The fiber-to-matrix load ratio is given by

\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}

where

E_f = 131 GPa is the fiber elasticity module

E_m = 2.4 GPa is the matrix elasticity module

V_f=0.3 is the fraction of volume of the fiber

V_m=0.7 is the fraction of volume of the matrix

Substituting,

\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4 (1)

(b) 44,594 N

The longitudinal load is

F = 46500 N

And it is sum of the loads carried by the fiber phase and the matrix phase:

F=F_f + F_m (2)

We can rewrite (1) as

F_m = \frac{F_f}{23.4}

And inserting this into (2):

F=F_f + \frac{F_f}{23.4}

Solving the equation, we find the actual load carried by the fiber phase:

F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N

(c) 1,906 N

Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:

F=F_f + F_m (2)

Using

F = 46500 N

F_f = 44594 N

We can immediately find the actual load carried by the matrix phase:

F_m = F-F_f = 46,500 N - 44,594 N=1,906 N

(d) 437 MPa

The cross-sectional area of the fiber phase is

A_f = A V_f

where

A=340 mm^2=340\cdot 10^{-6}m^2 is the total cross-sectional area

Substituting V_f=0.3, we have

A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2

And the magnitude of the stress on the fiber phase is

\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa

(e) 8.0 MPa

The cross-sectional area of the matrix phase is

A_m = A V_m

where

A=340 mm^2=340\cdot 10^{-6}m^2 is the total cross-sectional area

Substituting V_m=0.7, we have

A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2

And the magnitude of the stress on the matrix phase is

\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa

(f) 3.34\cdot 10^{-3}

The longitudinal modulus of elasticity is

E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa

While the total stress experienced by the composite is

\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa

So, the strain experienced by the composite is

\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}

3 0
3 years ago
Three resistances 2 ohm ,3ohm and 5 ohm are connected in parallel and a
hammer [34]

The potential difference across 3 Ohm resistor is 20V.

The resistors are connected in parallel which means all the three resistances have a fully potential difference of 20V.

7 0
3 years ago
A machine whose efficiency is 75% is used to lift a load of 100m.calculate the effort put into the machine if it has a velocity
aliina [53]

Explanation:

M.A = load / Effort

efficiency = M.A/V.R X 100

75 = M.A / 4 X 100

75 = 25 X M.A

M.A = 75/25 = 3

M.A = load / effort

3 = 100/E

E = 100/3 = 33.333

8 0
3 years ago
The diagram below shows a boat moving north in a river at 3m/s while the current in the river moves south at 1 m/s.
salantis [7]

Answer it will move slower:

Explanation: cause the force that’s opposite of it

4 0
2 years ago
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