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2 years ago

Why the bicycle stops when the cyclist stops pedaling?

2 answers:

5 0

There is no kintentic energy being giving to the bike through the action of pedaling therefore the bike will stop moving

mr_godi [17]2 years ago

3 0

Answer:

The bicycle slows down because of the frictional force acting on the tyres of the bicycle. When this frictional force overcomes the force applied by paddling, the bicycle stops.

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Physics

mixas84 [53]

Answer:

3 and 2

May be

ok may be subscribeIsenberg Isenberg Ignatius

3 0

2 years ago

In __________ waves, the motion of the particles in a medium is across the direction of the wave (perpendicular).

Alex_Xolod [135]

In longitudinal waves, the motion of the particles in a medium is across the direction of the wave.

4 0

2 years ago

Traumatic brain injury such as a concussion results when the head undergoes a very large acceleration. Generally an acceleration

eimsori [14]

The complete text of the problem is:

"Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.43 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.1 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. "

Solution:

1) Acceleration: $-2336 m/s^2$ on the hardwood floor, $-382 m/s^2$ on the carpeted floor

First of all, we need to calculate the speed of the child just before he hits the floor. This can be done by using the equation

$v^2 - u^2 = 2ad$

where

v is the final speed

u = 0 is the initial speed (the child starts from rest)

a = g = 9.8 m/s^2 is the acceleration of gravity

d = 0.43 m is the distance covered by the child as he falls from the bed

Solving for v,

$v=\sqrt{2ad}=\sqrt{2(9.8)(0.43)}=2.9 m/s$

Now we can analyze the moment of the collision. The child hits the floor with an initial speed of v = 2.9 m/s, and he comes to a stop, so the final speed is v' = 0. If the floor is hardwood, the stopping distance is

$d = 1.8 mm = 0.0018 m$

So we can find the acceleration by using again the equation

$v'^2 - v^2 = 2ad$

Solving for a,

$a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.0018)}=-2336 m/s^2$

For the carpeted floor instead,

$d=1.1 cm = 0.011 m$

therefore the acceleration is

$a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.011)}=-382 m/s^2$

2) Duration: 1.24 ms for the hardwood floor, 7.59 ms for the carpeted floor

We can find the duration of the collision in both cases by using the equation of the acceleration

$a=\frac{v'-v}{t}$

where

v' = 0

v = 2.9 m/s

For the hardwood floor,

$a=-2336 m/s^2$

So the duration of the collision is

$t = \frac{v'-v}{a}=\frac{0-2.9}{-2336}=0.00124 s = 1.24 ms$

For the carpeted floor,

$a=-382 m/s^2$

So the duration of the collision is

$t = \frac{v'-v}{a}=\frac{0-2.9}{-382}=0.00759 s = 7.59 ms$

We can now comment the results using the initial statement of the problem:

"Generally an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s2 lasting for at least 1ms will cause injury"

Therefore, the fall on the hardwood floor can result in injury (since the acceleration is greater than 1,000 m/s2 for more than 1 ms), while the fall on the carpeted floor is not dangerous (much less than 1000 m/s^2).

8 0

2 years ago

A stone is dropped from a tower 100 meters above the ground. The stone falls past ground level and into a well. It hits the wate

lilavasa [31]

Take the stone's position at ground level to be the origin, and the downward direction to be negative. Then its position in the air $y$ at time $t$ is given by

$y=100\,\mathrm m-\dfrac g2t^2$

Let $d$ be the depth of the well. The stone hits the bottom of the well after 5.00 s, so that

$-d=100\,\mathrm m-\dfrac g2(5.00\,\mathrm s)^2\implies d=\boxed{22.6\,\mathrm m}$

7 0

3 years ago

Imagine that you are working as a roller coaster designer. You want to build a record breaking coaster that goes 70.0 m/s at the

Rzqust [24]

Wow ! This is not simple. At first, it looks like there's not enough information, because we don't know the mass of the cars. But I"m pretty sure it turns out that we don't need to know it.

At the top of the first hill, the car's potential energy is

PE = (mass) x (gravity) x (height) .

At the bottom, the car's kinetic energy is

   KE = (1/2) (mass) (speed²) .

You said that the car's speed is 70 m/s at the bottom of the hill,
and you also said that 10% of the energy will be lost on the way
down. So now, here comes the big jump. Put a comment under
my answer if you don't see where I got this equation:

   KE = 0.9 PE

(1/2) (mass) (70 m/s)² = (0.9) (mass) (gravity) (height)

Divide each side by (mass):

   (0.5) (4900 m²/s²) = (0.9) (9.8 m/s²) (height)

(There goes the mass. As long as the whole thing is 90% efficient,
the solution will be the same for any number of cars, loaded with
any number of passengers.)

Divide each side by (0.9):

   (0.5/0.9) (4900 m²/s²) = (9.8 m/s²) (height)

Divide each side by (9.8 m/s²):

   Height = (5/9)(4900 m²/s²) / (9.8 m/s²)

      = (5 x 4900 m²/s²) / (9 x 9.8 m/s²)

      = (24,500 / 88.2) (m²/s²) / (m/s²)

      = 277-7/9 meters
      (about 911 feet)

3 0

2 years ago