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Maurinko [17]
3 years ago
14

you ride your bike for a distance of 30km. you travel ata a speed of 0.75km/minute.how many minutes does it take

Physics
2 answers:
PilotLPTM [1.2K]3 years ago
8 0

                 (30 km) x (1 minute / 0.75 km)

             =  (30 x 1 / 0.75)  (km/minute)

             =       40 minutes .

The math is correct, but I wonder about the real world.
I don't know much about competitive biking, but you've got
this dude pumping along at almost  25 mph for 40 minutes
solid.   Is this realistic ?

mihalych1998 [28]3 years ago
7 0
Distance for which the bike is ridden = 30 km
Speed at which the bike is driven = 0.75 km/minute
Let us assume the number of minutes taken to travel the distance of 30 km = x
Now we already know the formula of speed can be written as
Speed = Distance traveled/ Time taken
0.75 = 30/x
0.75x = 30
x = 30/0.75
  = 40 minutes
So the time taken for riding a distance of 30 km will be 40 minutes. I hope this procedure is simple enough for you to understand.
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A 115 g hockey puck sent sliding over ice is stopped in 15.1 m by the frictional force on it from the ice.
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(a) Ff = 0.128 N

(b μk = 0.1135

Explanation:

kinematic analysis

Because the hockey puck  moves with uniformly accelerated movement we apply the following formulas:

vf=v₀+a*t Formula (1)

d= v₀t+ (1/2)*a*t² Formula (2)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s

Calculation of the acceleration of the  hockey puck

We apply the Formula (1)

vf=v₀+a*t      v₀=5.8 m/s ,  vf=0

0=5.8+a*t

-5.8 = a*t

a= -5.8/t   Equation (1)

We replace a= -5.8/t in the Formula (2)

d= v₀*t+ (1/2)*a*t²   ,  d=15.1 m ,  v₀=5.8 m/s

15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²  

15.1= 5.8*t-2.9*t

15.1= 2.9*t

t = 15.1 / 2.9

t= 5.2 s

We replace t= 5.2 s in the equation (1)

a= -5.8/5.2

a= -1.115 m/s²

(a) Calculation of the  frictional force (Ff)

We apply Newton's second law

∑F = m*a    Formula (3)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Look at the free body diagram of the  hockey puck in the attached graphic

∑Fx = m*a     m= 115g * 10⁻³ Kg/g = 0.115g    ,  a= -1.12 m/s²

-Ff = 0.115*(-1.115)  We multiply by (-1 ) on both sides of the equation

Ff = 0.128 N

(b) Calculation of the coefficient of friction (μk)

N: Normal Force (N)

W=m*g= 0.115*9.8= 1.127 N : hockey puck  Weight

g: acceleration due to gravity =9.8 m/s²

∑Fy = 0

N-W=0

N = W

N =  1.127 N

μk = Ff/N

μk = 0.128/1.127

μk = 0.1135

8 0
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