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1 year ago

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A bicycle rider accelerates from rest up to full speed on a flat, straight road. The frictional force between the road and the t

ires pushing her forward is f1. The air drag (and other frictional forces) pushing back is f2. What relationship would f1 and f2 have in the first few seconds of the ride?

2 answers:

Sergio [31]1 year ago

8 0

Answer:

$f_1 > f_2$

Explanation:

Frictional force due to road is pushing the cycle in forward direction

so it is given as f1

also the air drag is always in opposite direction of motion so it is given as f2

so here net force on the bicycle is given as

$F_{net} = f_1 - f_2$

now initially the speed of cycle is increasing with time

so here we can say that net force on the cycle must be positive in direction

so we have

$f_1 - f_2 > 0$

so we have

$f_1 > f_2$

Effectus [21]1 year ago

8 0

Answer:

$f_1>f_2$

Explanation:

The bicycle is able to move forward due to the frictional force between the Tyre of the bicycle and the road. The air drag and other frictional forces act in the opposite direction of motion. The net force is:

$F_{net}=f_1-f_2$ is non-zero due to which the bicycle is able to accelerate forward. Therefore, the frictional force between the Tyre and the road must be greater than the air drag.

Later on, as the bicylce attains full constant speed, the two forces would be equal.

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Answer:

option B

Explanation:

given,

Force exerted by the hydraulic jack piston = F₁ = 250 N

diameter of piston, d₁ = 0.02 m

r₁ = 0.01 m

diameter of second piston, d₂ = 0.15 m

r₂ = 0.075 m

mass of the jack to lift = ?

now,

$\dfrac{F_1}{A_1} =\dfrac{F_2}{A_2}$

$\dfrac{250}{\pi r_1^2} =\dfrac{F_2}{\pi r_2^2}$

$\dfrac{250}{0.01^2} =\dfrac{F_2}{0.075^2}$

$F_2= \dfrac{250}{0.01^2}\times {0.075^2}$

F₂ = 14062.5 N

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$m = \dfrac{F}{g}$

$m = \dfrac{14062.5}{9.8}$

m = 1435 Kg

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2 years ago

A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.

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Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

Tension in the vertical rope = ?

now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8 + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

T₁ = 710.5 N

from equation (1)

T₂ = 1176 - 710.5

T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

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2 years ago

A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit

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Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

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Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

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We need to find final velocity of the block( u )

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We have

2.87 = -1.442 + 1.8 u

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2 years ago

The position of a particle moving along the x axis may be determined from the expression x(t) = btu + ctv, where x will be in me

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As per given equation we have

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now as per left side of equation its dimension is same as length or meter

now we can say it should be meter on right side also

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$b*T^8 = M^0L^1T^0$

$b = M^0L^1T^{-8}$

similarly for other term we have

$ct^v = M^0L^1T^0$

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$c = M^0L^1T^{-7}$

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1 year ago

(Another tomato/skyscraper problem.) You are looking out your window in a skyscraper, and again your window is at a height of 45

Ivan

Answer:

1027.2 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.2 ft/s

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2 years ago