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Posts Tagged with ‘physics homework’

Find the critical numbers of f (x) = x^2-6x. Find also the open intervals on which the function is increasing or decreasing and locate all relative extrema.

Saturday, July 24th, 2010

Solution:

A number a in the domain of a given function f is called a critical number  ‘(a) = 0 or f ‘ is undefined at x = a.

Relative extrema are the minimums or maximum points on a part of a curve, while absolute extrema are the minimums and maximum points along the entire curve.

Now we are looking at the following curve:

f (x) = 

To find the extrema, we take the first derivative and figure out at what value of x, the first derivate of the function equals to zero:

f’(x) = 2x-6

f’(3) = 0

As you can see the function is decreasing from negative infinity to 3, and increasing from 3 to infinity. The critical number is 3. There is only one relative (or local) minimum at (3, -9).

A 925 N crate is being pushed across a level floor by a force of 325 N at an angle of 25 degrees above the horizontal. The coefficient of kinetic friction is 0.25. Find the magnitude of the acceleration of the crate.

Friday, July 23rd, 2010

Solution:

First, let’s define what a Newton (N) is: “The Newton (N) is defined as the amount of force that, when acting on a 1 kg mass, produces an acceleration of 1 m/s/s (one meter per second per second). Therefore, 1 N = 1 kg =D7 1 m/s/s.”

The force of friction opposing the 325 N force is .25 * 925 N (which is equal to 231.25 N). The force pulling the crate is 325 N * cos(25), i.e. 294.55 N. (This is the component of force acting parallel to the floor.)

Take the difference between these two and you will have the net force pulling the crate (which is 63.3 N). Using Newton’s equation F = ma, you will find the acceleration by dividing the net force by the mass of the crate (the mass is 925 =/(one standard gravity [9.80665 m/s/s], or 94.32 kg). The acceleration is therefore 63.3 N / 94.32 kg, which equals 0.671 m/s/s.

How much energy would it take to raise 4 grams of water by 6 degrees Celcius?

Friday, July 16th, 2010

Solution:

Specific heat capacity for water = 4.18 Joules/(gram*degree Celcius)
(how much energy is required per gram per change in degrees C)

mass = 4g
Change in temperature = 6 deg C

energy required = mass * change in temperature * specific heat capacity
= 4g * 6 deg C * 4.18 J/(g*deg C)
= 100.32 Joules

What is the difference between a scalar and vector Quantity?

Friday, July 16th, 2010

Vector quantities have two characteristics: magnitude and a direction. Scalar quantities only have magnitude.