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podryga [215]
3 years ago
6

If two bicycles of same masses move at different velocities,it will be easier to stop the bicycle that is moving at lower veloci

ty.Explain why?​
Physics
1 answer:
s344n2d4d5 [400]3 years ago
8 0

Answer:

It is easier to stop the bicycle moving at a lower velocity because it will require a <em>smaller force</em> to stop it when compared to a bicycle with a higher velocity that needs a<em> bigger force.</em>

Explanation:

The question above is related to "Newton's Law of Motion." According to the <em>Third Law of Motion</em>, whenever an object exerts a force on another object <em>(action force)</em>, an equal force is exerted against it. This force is of the same magnitude but opposite direction.

When it comes to moving bicycles, the force that stops their movement is called "friction." Applying the law of motion, the higher the speed, the higher the force<em> </em>that is needed to stop it while the lower the speed, the lower the force<em> </em>that is needed to stop it.

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The gravitational force on the woman is A) 500 N

Explanation:

There are two forces acting on the woman during her fall:

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According to Newton's second law, the net force acting on the woman is equal to the product between the woman's mass and her acceleration:

\sum F=ma

where m is the mass of the woman and a her acceleration.

The net force can be written as

\sum F = F_G - F_D

Also, we know that the woman falls at a constant velocity (5 m/s), this means that her acceleration is zero:

a=0

Combining the equations together, we get:

F_G-F_D = 0

which means that the magnitude of the gravitational force is equal to the magnitude of the air resistance:

F_G=F_D=500 N

Learn more about forces and Newton's second law:

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a. 0.6 m b. 0.385 m c. 3.6 m/s at 287.78° to the horizontal

Explanation:

a. Using s = ut - 1/2gt² for motion under gravity where s = vertical distance = height of table, u = initial vertical velocity of book = 0 m/s, t = time of flight = 0.350 s and g = acceleration due to gravity = 9.8 m/s².

Substituting these these values into s and taking the top of the table as position 0 m, we have.

0 - s = 0t - 1/2gt²

-s = -1/2gt²

s = 1/2gt²

s = 1/2 × 9.8 m/s² × (0.350 s)²

s = 0.6 m

b. Using d = v't where d = horizontal distance from table, v' = horizontal velocity of book = 1.10 m/s and t = time of flight = 0.350 s

d = v't = 1.10 m/s × 0.350 s = 0.385 m

c. Using v² = u² - 2gs where u = initial vertical velocity of book = 0 m/s and g = 9.8 m/s², s = -0.6 m (negative since we are at the bottom and 0 m is at the top)and v = final vertical velocity of book

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So, the magnitude of the resultant velocity is V = √(v² + v'²)

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= √12.97 m²/s²

= 3.6 m/s

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= tan⁻¹(-3.43 m/s/1.10 m/s)

= tan⁻¹(-3.1182)

= -72.22°

Ф = -72.22°+ 360 = 287.78° since it is in the third quadrant

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