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svet-max [94.6K]
3 years ago
6

A daring 510N swimmersdives off a cliff with a running horizontal lead.what must be her mimimum speed just as she leaves the top

of cliff so that she will miss the ledge at the bottom which is 1.75 m wide and 9m below the top of cliff

Physics
2 answers:
goldfiish [28.3K]3 years ago
8 0

Answer:

The running horizontal speed should be larger than 1.29 m/s.

Explanation:

In order for the swimmer to just miss the bone-breaking ledge, her horizontal speed must be

v > \frac{1.75m}{t_{fall}}

in which we need to know how long we she be diving through the air. To determine that, recall the formula for the distance made by an object with acceleration (in this case it is the gravitational acceleration) with no initial (vertical) velocity:

s = \frac{1}{2}gt^2

from which it follows that (for non-negative t)

t = \sqrt{\frac{2\cdot9m}{9.8\fram{m}{s^2}}}\approx1.36s

This result can be used in the initial inequality:

v > \frac{1.75m}{t_{fall}}=\frac{1.75m}{1.36s}=1.29\frac{m}{s}

The diving lady better gets a speed larger than 1.29 m/s to avoid landing on the ledge.

Andre45 [30]3 years ago
4 0

looks like its projectiles = swimmer. see encls may help

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Yuki888 [10]

1) The potential energy is the most at the highest position and the least at the equilibrium position

2) The kinetic energy is the most at the equilibrium position and  the least at the highest position

Explanation:

1)

The potential energy of an object is the energy possessed by the object due to its position in a gravitational field; mathematically, it is given by

PE=mgh

where

m is the mass of the object

g is the strength of the gravitational field

h is the height of the object relative to the ground

For the pendulum in this problem, m is the mass of the bob, and h is the height of the above relative to the ground. We see from the formula that the potential energy is directly proportional to the height:

PE\propto h

This means that:

  • The potential energy is the most when the bob is at the highest position
  • The potential energy is the least when the bob is at the equilibrium position,  which is the lowest position

2)

We can solve this part by applying the law of conservation of energy: in fact, the total mechanical energy of the pendulum (sum of potential and kinetic energy) is constant at any time during the motion,

E=KE+PE=const.

where KE is the kinetic energy.

From the equation above, we observe that:

  • When PE is maximum, KE must be at minimum
  • When PE is minimum, KE must be maximum

Therefore, this implies that:

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  • The kinetic energy is the least when the potential energy is the most, i.e. at the highest position

Learn more about kinetic and potential energy:

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6 0
3 years ago
A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an
3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

F_s = am = 1.31*16 = 20.92 N

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N

So the maximum acceleration on the block is

a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2

4)As the box slides, it is now subjected to kinetic friction, which is

F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

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3 years ago
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