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3 years ago

Which illustration represent the substance at a higher temperature? Explain.

2 answers:

7 0

The addition of heat energy to a system always causes the temperature of that system to increase. This is always true because you are adding heat of a substance to increase its temperature. For example, you are going to drink a cup of coffee. And you wanted it hot to boost your attention. So you have to use hot water. In order for your water to become hot or warm, you need boil it in a kettle. Note that you are going to use an electric stove. The electric stove gets it energy from the source giving it a hotter temperature to the water in the kettle. You are applying heat energy to warm the water. So, the statement is true.

vladimir2022 [97]3 years ago

7 0

Answer:

Explanation:

the big paragrah

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If the coefficient of kinetic friction between a 22 kg kg crate and the floor is 0.27, what horizontal force is required to move

alina1380 [7]

Answer:

58.27 N

Explanation:

the data we have is:

mass: $m=22kg$

coefficient of friction: $\mu =0.27$

and we also know the acceleration of gravity is $g=9.81m/s^2$

We need to do an analysis of horizontal and vertical forces acting on the object:

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Vertically the forces acting on the object:

Normal force $N$ (acting up from the object)

weight: $w=mg$ (acting down from)

so the sum of forces in the vertical axis "y" are:

$F_{y}=N-w\\F_{y}=N-mg$

from Newton's second Law we know that $F=ma$ , so:

$ma_{y}=N-mg$

and since the object is not accelerating in the vertical direction (the movement is only horizontal) $a_{y}=0$ , and:

$0=N-mg\\N=mg$

-----------

now let's analyze the horizontal forces

frictional force: $f= \mu N$ and since $N=mg$ --> $f=\mu mg$

force to move the object: $F$

and the two forces just mentioned must be opposite, thus the sum of forces in the "x" axis is:

$F=ma_{x}=F-f\\ma_{x}=F-\mu mg$

and we are told that the crate moves at a steady speed, thus there is no acceleration: $a_{x}=0$

and we get:

$0=F-\mu mg\\F=\mu mg$

substituting known values:

$F=(0.27)(22kg)(9.81m/s^2)\\F=58.27N$

3 0

3 years ago

Consider a portion of a cell membrane that has a thickness of 7.50 nm and 1.4 µm ✕ 1.4 µm in area. A measurement of the potentia

Alla [95]

Answer:

Part a: The current is 1.49x10⁻¹² A

Part b: The current is decreased by a factor of 4.

Explanation:

Part a

The area is given as

A =(1.4*10⁻⁶)² =1.96*10⁻¹² m²

The resistance is given as where resistivity of the membrane material is 1.30 x 10⁷ ohms*m

R=pL/A =(1.3*10⁷)(7.5*10⁻⁹)/(1.96*10⁻¹²)

R=4.97x10¹⁰ ohms

So the resistance is 4.97x10¹⁰ ohms.

I=V/R =86.0x10⁻³/5.77x10¹⁰

I=1.49x10^⁻¹² A

So the current is 1.49x10⁻¹² A

S=So/2=1.4/2 =0.7μm

A=(0.7*10^-6)^2=4.9*10⁻¹³ m²

R=pL/A =(1.3*10⁷)(7.5*10⁻⁹)/(4.9*10⁻¹³)

R=1.98x10¹¹ ohms

So the resistance is 1.98x10¹¹ ohms.

I=V/R =86.0x10⁻³/1.98x10¹¹

I2=4.52x10^⁻¹³ A

So the ratio of the new current vs the old current is as

I2/I=4.52x10^⁻¹³/1.49x10⁻¹²=0.25

So the current is decreased by a factor of 4.

7 0

4 years ago

Since acids are so corrosive they are not found anywhere in the human body true/false

pantera1 [17]

the answer is in fact True

5 0

4 years ago

Read 2 more answers

The scale on the horizontal axis is 8 s per division and on the vertical axis 5 m per division. What is the time represented by

Nadya [2.5K]

Answer:

24 s

Explanation:

8 s / tic * 3 tic = 24 sec from origin

8 0

2 years ago

A force of 5.0 N acts on a 15 kg body initially at rest. Compute the work done by the force in (a) the first, (b) the second, an

Leokris [45]

Answer:

(a) 0.833 j

(b) 2.497 j

(d) 4.995 watt

Explanation:

We have given force F = 5 N

Mass of the body m = 15 kg

So acceleration $a=\frac{F}{m}=\frac{5}{15}=0.333m/sec^2$

As the body starts from rest so initial velocity u = 0 m/sec

(a) From second equation of motion $s=ut+\frac{1}{2}at^2$

For t = 1 sec

$s=0\times 1+\frac{1}{2}\times 0.333\times 1^2=0.1666m$

We know that work done W =force × distance = 5×0.1666 =0.833 j

(b) For t = 2 sec

$s=0\times 2+\frac{1}{2}\times 0.333\times 2^2=0.666m$

We know that work done W =force × distance = 5×0.666 =3.33 j

So work done in second second = 3.33-0.833 = 2.497 j

$s=0\times 3+\frac{1}{2}\times 0.333\times 3^2=1.4985m$

We know that work done W =force × distance = 5×1.4985 =7.4925 j

So work done in third second = 7.4925 - 2.497 -0.833 = 4.1625 j

(d) Velocity at the end of third second v = u+at

So v = 0+0.333×3 = 0.999 m /sec

We know that power P = force × velocity

So power = 5× 0.999 = 4.995 watt

4 0

4 years ago