Rewrite the square root as what is inside the parentheses to the 1/2 power.

Then you do the usual process of multiplying by the exponent, bringing the 1/2 power exponent down by one, to become -1/2 and then doing the chain rule.

Intro Physics & Physics Concepts

Rewrite the square root as what is inside the parentheses to the 1/2 power.

Then you do the usual process of multiplying by the exponent, bringing the 1/2 power exponent down by one, to become -1/2 and then doing the chain rule.

Filed Under: Calculus

This is an interesting problem.

What I would do first is try plugging in a somewhat large number, like 1000 for x.

It’s not infinity, but it’s large.

I removed the interior sets of parentheses since parentheses are implied in a fraction like that.

The expression then becomes a number a bit larger than 1, to an exponent of 2000.

If you go ahead and use a calculator, it becomes

about 21917

That seems distinctly non zero to me!

So I tried a bigger number, x=1,000,000

You get

about 22026

**It’s not immediately apparent what that value is. (you might try taking the natural log of this number though)**

An approach which uses an early definition of e can help you solve this problem.

There is another approach you could use. One of the first steps is to put a base of e with the natural log of the expression. And we will need to use a Taylor Expansion later.

You may have seen the approximation that sinx ~ x for small values, there is a similar approximation for ln(1+x), 5/infinity qualifies as a small value. The first term of the Taylor Expansion is the most dominant for small values.

Filed Under: Calculus

My process would be to first rewrite the square root by using a fractional exponent (1/2).

From there, you multiply by the exponent in front and then subtract one from the exponent. Then you do the chain rule.

You can rewrite the answer using square roots if you would like.

Filed Under: Calculus

0.6 kg of ice at zero degrees Celsius.

0.2 kg of steam at 100 degrees Celsius.

What will the final temperature be?

Using the latent heat of fusion, the latent heat of vaporization, and the specific heat. All of water.

Filed Under: Physics

Here’s a diagram with the information given, plus the necessary statement that the sum of the angles of a four sided figure is 360 degrees. For a triangle it’s 180.

(n-2)180 for the sum of the interior angles where n is the number of sides of the figure.

Filed Under: Geometry

Thanks for posting the photo of your work George, that makes it much easier to follow. Especially because the syntax in your typed version of the equations is not quote right.

If you followed exactly what you typed, it would be like this:

Because of missing parentheses. I thought that was probably not what you meant.

However, with the photo, it’s clear what the original problem is and what to fix:

And it’s also good to see your work.

Essentially, in the problem you have the same base for all the terms (2/5) and similar exponents.

For the first variation, you have an additional negative sign in the exponent of the left term.

The blue is the original, orange is simplified.

For the second variation, you have an additional negative sign in the exponent of the right term.

**Corrections**

**Left Side:**

Your work is mostly right, but here’s the first mistake I see on the left side:

You state that the exponent on the left term should be greater than the exponent on the right term. That would be true if the base was greater than 1, but it’s (2/5). If for example, you square (2/5) you get 4/25 which is a smaller number than 2/5.

Therefore you want the base on the left term to be less than the base on the right term. So switch the direction of the inequality sign.

-4x – 2 < 3x – 18

7x > 16

**x > 16/7**

**Right Side:**

I would recommend keeping the base the same for all the terms, 2/5 for this problem. At least for me, it’s easier to work through that way.

It seems like on the right side of the board, you turned the 2/5 into 5/2 which you would do if you had the exponent of -1 like you did on the left side of the board, but you don’t have that exponent of -1 in this variation. So it should be 2/5 as the base, not 5/2 for the left term.

The other difference for the right side of the board is the negative sign in exponent of the right term (in front of the 3).

You basically turned the base of the right term into 5/2, which is okay.

But you added a negative sign to the exponent on the left term, which isn’t there, to turn that base into 5/2 as well.

Let’s say we do go with 2/5 as the base for both terms.

Again, if the base is less than 1, a smaller exponent will result in a larger number.

4x + 2 < -3x + 18

Result:

7x < 16

**x < 16/17**

Obediah Jones was correct in one of his comments to you, although

“The fact that they stemmed from the same inequality is why the value of x is either less than or greater than the exact same number, 16/7 16/7.”

Filed Under: Precalculus

Newton’s Second Law states that

**F **= m **a**

(Vectors in bold)

Weight = gravitational force (units of Newtons in SI units)

**F **= m **g**

g being the acceleration due to gravity.

On Earth, g ~ -9.8 m/s^2

**Changes in weight**

But that acceleration will change depending on position on the Earth and altitude and will also change on the moon, in space, and in other places.

You can find the acceleration due to gravity by dividing the gravitational force by the mass.

There is an acceleration due to gravity whenever there are two masses, they can be a person sized mass and a planet sized mass, two smaller masses, or any combination of two masses.

Therefore the weight is less or more depending on where it is.

There is less gravity on the moon, so the weight is less there (for the same mass) compared to on Earth.

Let’s say on a bigger planet ‘X’, the mass of the planet is more than Earth. Therefore the acceleration due to gravity is greater and weight would be greater (for the same mass) compared to Earth.

If there are no other masses nearby, in space for instance, things are weightless.

You can use the Law of Gravitation for various situations.

Here’s a diagram

Filed Under: Physics

If you know the length of a car’s tires skidmarks and the coefficient of friction, you can figure out the starting velocity.

Filed Under: Physics

We started by looking at some problems involving conservation of energy and conversion of energy between different types.

The problem we went through a few times involved a car stopping using the brakes and how kinetic energy was converted to heat by the mechanism of friction.

We looked at the units of joules and how equations with work, gravitational potential energy, elastic potential energy, kinetic energy, etc all have the same units.

We also used the equation to derive the units of a spring constant.

Filed Under: Physics

The SI unit of charge is the Coulomb.

Both a proton and an electron have a charge of

1.60217662 × 10^{-19} coulombs

With a proton having a positive charge with this magnitude and an electron having a negative charge of this magnitude.

An Amphere-Hour is 3600 multiplied by one Coulomb.

Let me know if you have questions about the units of electric charge.

Filed Under: Physics

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- Next Page »

- How to Take a Derivative with a Square Root
- Solving a Somewhat Tricky Limit Problem
- Steps for Finding a Derivative of a Function with a Square Root
- A Song of Ice and Steam
- Angles of a Parallelogram
- Correcting a Problem with Bases and Exponents
- How to calculate the weight, given a mass
- Physics of Speeding Tickets & Skidmarks
- Tutoring Physics – Conversion of Energy, Kinetic Energy into Work Done by Friction
- Units of Electric Charge

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