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harkovskaia [24]
3 years ago
12

Interconic attraction theory states what​

Chemistry
1 answer:
iogann1982 [59]3 years ago
7 0

Explanation:

Interionic Attractions are when an ion is surrounded by an ionic atmosphere which has a net charge opposite for its own. ... The ionic atmosphere cannot created nor destroyed. In solutions with weak electrolytes the number of ions is not large, therefore the effect of the interionic attraction is small.

<h3>Hope this is fine for you</h3>
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1) A car is traveling down the interstate at 37.1 m/s. The driver sees a cop and quickly slows down. If the driver slows to 29.8
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1)

The acceleration of the car is the rate of change of velocity of the car; it can be calculated as:

a=\frac{v-u}{t}

where

u is the initial velocity

v is the final velocity

t is the time taken for the velocity of the car to change from u to v

In this problem, for this car we have:

u = 37.1 m/s

v = 29.8 m/s

t = 3 s

So, the acceleration is:

a=\frac{29.8-37.1}{3}=-2.43 m/s^2

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The work done in lifting the box is equal to the potential energy transferred to the box during the process; it is given by:

W=Fd

where

F is the force applied

d is the displacement of the box

Here we have:

F = 87.3 N is the force applied

d = 2.04 m is the displacement of the box

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W=(87.3)(2.04)=178.1 J

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The power is the rate of work done per unit time. It is calculated as:

P=\frac{W}{t}

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W = 1250 J is the work done by the child running up the stairs

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The kinetic energy of an object is the energy possessed by the object due to its motion. Mathematically, it is given by

KE=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

For the rabbit in this problem, we have:

m = 8.642 kg is the mass of the rabbit

KE = 125.6 is its kinetic energy

Solving the formula for v, we find the speed of the rabbit:

v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(125.6)}{8.642}}=5.4 m/s

5)

The efficiency of a machine is the ratio between the energy produced in output by the machine and the work done in input. Mathematically, it is given by

\eta = \frac{E_{out}}{W_{in}}\cdot 100

where

E_{out} is the energy in output

W_{in} is the work in input

For the machine in this problem,

W_{in}=120 J is the work in input

E_{out}=93 J is the energy in output

Therefore, the efficiency of this machine is:

\eta=\frac{93}{120}\cdot 100=77.5\%

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c=f\lambda

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c=3.0\cdot 10^8 m/s

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