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s344n2d4d5 [400]
3 years ago
7

To practice Problem-Solving Strategy 8.1 for circular-motion problems. A cyclist competes in a one-lap race around a flat, circu

lar course of radius 140 m . Starting from rest and speeding up at a constant rate throughout the race, the cyclist covers the entire course in 60 s . The mass of the bicycle (including the rider) is 76 kg . What is the magnitude of the net force Fnet acting on the bicycle as it crosses the finish line?
Physics
1 answer:
Anna007 [38]3 years ago
5 0

Answer:

Explanation:

distance travelled

s = 2πR

= 2 X 3.14 X 140

= 880 m

final velocity = v

initial velocity = u

distance travelled = s

time = 60 s

s = ut + 1/2 at²

880 = .5 x a x 60²

a = .244 m/s²

final velocity v = at

= .244 x 60

= 14.66

centripetal acceleration at final moment

v² /R

14.66 X 14.66 / 140

= 1.53 m/s⁻²

1.53 m/s²

this is centripetal acceleration which acts towards the centre.

tangential acceleration calculated a _t = .244

redial acceleration ( centripetal ) = 1.53

Resultant acceleration

R²= 1.53² + .244 ²

R = 1.55 m/s²

total force = 1.55 x 76

= 118 N  

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Answer:

1] 8500000 = <u>8.5 × 10⁶</u>

2] .000072 = <u>7.2 × 10⁻⁵</u>

3] 5.3 × 10⁴ = <u>53000</u>

4] 2.8 × 10⁻³ = <u>0.0028</u>

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 <u>V = 5 m/s</u>

6] Acceleration = \frac{V1-V2}{time}

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How much work in joules must be done to stop a 920-kg car traveling at 90 km/h?
tankabanditka [31]

Answer:

Work done, W = −287500 Joules

Explanation:

It is given that,

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We need to find the amount of work must be done to stop this car. The final velocity of the car, v = 0

Work done is also defined as the change in kinetic energy of an object i.e.

W=\Delta K

W=\dfrac{1}{2}m(v^2-u^2)

W=\dfrac{1}{2}\times 920\ kg(0-(25\ m/s)^2)

W = −287500 Joules

Negative sign shows the work done is done is opposite direction.

3 0
3 years ago
A girl is riding her bike and coasting down a hill. When she gets to the bottom of the hill, her bike continues to move without
ddd [48]

Answer:

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B

Explanation:

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3 years ago
In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r
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Answer:

the weight of the rider is 493.53 N

Explanation:

Given the data in the question and as illustrated in the image below,

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In the horizontal direction

F_{sailcos( 30° ) = Tcos ( 17° )

F_{sail = Tcos( 17° ) / cos( 30° )

F_{sail = 1900cos( 17° ) / cos( 30° )

F_{sail = 2098.07 N

Now, In the vertical direction,

F_{sail sin( 30° ) = W + T sin( 17° )

W = F_{sail sin( 30° ) - T sin( 17° )

W = 2098.07sin( 30° ) - 1900sin( 17° )

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W = 493.53 N

Therefore, the weight of the rider is 493.53 N

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