Name three methods that can be used to change the overall density of an object.

A spring whose spring constant is 270 lbf/in has an initial force of 100 lbf acting on it. Determine the work, in Btu, required

cricket20 [7]

Answer:

0.02585 BTU

Explanation:

Given: Spring constant, k = 270 lbf/in

Initial force, f =100 lbf

Compression, x = 1 in

Work done can be calculated as follows:

$W = \int {(f+kx)} \, dx \\W = fx + \frac{1}{2}kx^2\\W= (100 lbf)(1 in)+ \frac{1}{2}(270 lbf/in)(1 in)^2\\W= 100+135 = 235 lbf in\\W=235 \times 0.00011 BTU = 0.02585 BTU$

6 0

3 years ago

Vector ~A has a magnitude of 29 units and points in the positive y-direction. When vector ~B is added to ~A, the resultant vecto

timurjin [86]

Answer:

The magnitude of vector B is 43 units and it points in the negative y-direction.

Explanation:

Resultant of vectors = vector sum of all the vectors

Vector A = 29j

Vector B = ?

Resultant of vector A and B = R = -14j

R = A + B

-14j = 29j + B

B = -14j - 29j = - 43j

Hence, the magnitude of vector B is 43 units and it points in the negative y-direction.

4 0

3 years ago

What is ozone? How and where is it formed in the atmosphere?

mrs_skeptik [129]

Ozone gas is made up of oxygen molecules that have three atoms. it exists in polluted air and ozone layer. Ozone layer is formed in stratosphere(part of atmosphere).

4 0

3 years ago

If we wanted to describe and better understand the structure of the rings of Saturn, we might develop a

Svetradugi [14.3K]

The Greek philosopher Aristotle and the Roman Catholic Church also believed the sun revolved around the earth. In 1543, Nicolaus Copernicus<span> published a new theory stating the earth revolves around the sun. This is known as the Copernican theory.</span>

6 0

3 years ago

Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless

drek231 [11]

Answer:

$v_f = 15 \frac{m}{s}$

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum $\vec{L}$ is

$\vec{L} = \vec{r} \times \vec{p}$

where $\vec{r}$ is the position and $\vec{p}$ the linear momentum.

We also know that the torque is

$\vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt} ( \vec{r} \times \vec{p} )$

$\vec{\tau} = \frac{d}{dt} \vec{r} \times \vec{p} + \vec{r} \times \frac{d}{dt} \vec{p}$

$\vec{\tau} = \vec{v} \times \vec{p} + \vec{r} \times \vec{F}$

but, as the linear momentum is $\vec{p} = m \vec{v}$ this means that is parallel to the velocity, and the first term must equal zero

$\vec{v} \times \vec{p}=0$

so

$\vec{\tau} = \vec{r} \times \vec{F}$

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

$\vec{\tau}_{rod} = 0$