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4 years ago

10.

Physics

1 answer:

nlexa [21]4 years ago

3 0

Answer:

Cell Membrane

Explanation:

The cell membrane controls what goes in and out of a cell, and keeps it shape, much like a city limit.

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Express 1010 miners in examiners. Answer in units of Eminer.

Lady_Fox [76]

Since 1 examiner = 10^18miner
then 10^10miners =10^(10-18)=10^-8examiner

7 0

3 years ago

The distance between two objects is increased by three times the oringinal distance. How will this change the force of attractio

tigry1 [53]

<span>The distance between two objects is increased by three times the oringinal distance. Since they were already separated by one time the original distance,
the additional three times the oringinal distance now puts them four times the original distance apart.

Whether we're talking about the gravitational forces of attraction or
the electrical forces of attraction, either one is inversely proportional
to the square of the distance between the objects.

So changing the distance to four times the original distance causes
the forces to become 1/4</span>² as strong as they were originally.

The forces become 1/16 of their original magnitude.<span>
</span>

8 0

3 years ago

(a) If the coefficient of kinetic friction between tires and dry pavement is 0.80, what is the shortest distance in which you ca

Colt1911 [192]

a) 52.5 m

b) 16.0 m/s

Explanation:

The motion of a car slowed down by friction is a uniformly accelerated motion, so we can use the following suvat equation:

$v^2-u^2=2as$

where

v = 0 is the final velocity (the car comes to a stop)

u = 28.7 m/s is the initial velocity of the car

a is the acceleration

s is the stopping distance

For a car acted upon the force of friction, the acceleration is given by the ratio between the force of friction and the mass of the car, so:

$a=\frac{-\mu mg}{m}=-\mu g$

where:

$\mu=0.80$ is the coefficient of friction

$g=9.8 m/s^2$ is the acceleration due to gravity

Substituting and solving for s, we find:

$s=\frac{v^2-u^2}{-2\mu g}=\frac{0-(28.7)^2}{-2(0.80)(9.8)}=52.5 m$

In this case, the car is moving on a wet road. Therefore, the coefficient of kinetic friction is

$\mu=0.25$

Here we want the stopping distance of the car to remain the same as part a), so

$s=52.5 m$

We can use again the same suvat equation:

$v^2-u^2=2as$

And since the final velocity is zero

u = 0

We can find the initial velocity of the car:

$v=\sqrt{2as}=\sqrt{2\mu gs}=\sqrt{2(0.25)(9.8)(52.5)}=16.0 m/s$

7 0

3 years ago

What is the maximum acceleration of a platform that oscillates at amplitude 4.70 cm and frequency 6.50 Hz?

mylen [45]

Answer:

78.315 m/s²

Explanation:

Amplitude, A = 4.70 cm = 0.047 m

Frequency, f = 6.50 Hz

Angular frequency, ω = 2 π f = 2 x 3.14 x 6.50 = 40.82 rad/s

Maximum acceleration, a = ω² A

a = 40.82 x 40.82 x 0.047

a = 78.315 m/s²

8 0

4 years ago

What mass of silver can be plated onto an object in 33.5 minutes at 8.70 a of current? ag (aq e- ? ag(s what mass of silver can

Schach [20]

To determine the mass plated, we use Faraday's Law of Electrolysis. We calculate as follows:

q = It
q = 8.70 (33.5) (60)
q = 17487 C

mass = 17487 C ( 1 mol e- / 96500 C) ( 1 mol / 2 mol e-) (107.9 g /mol)
mass = 9.78 g

Hope this helps.

4 0

4 years ago

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