Physics 5

(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup

KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

$W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx$

Where

$x_{o}$ , $x_{f}$ - Initial and final position, respectively, measured in meters.

$F(x)$ - Force as a function of position, measured in newtons.

Given that $F = k\cdot x$ and the fact that $F = 25\,N$ when $x = 0.3\,m - 0.2\,m$ , the spring constant ( $k$ ), measured in newtons per meter, is:

$k = \frac{F}{x}$

$k = \frac{25\,N}{0.3\,m-0.2\,m}$

$k = 250\,\frac{N}{m}$

Now, the work function is obtained:

$W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx$

$W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]$

$W = 0.313\,J$

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be $r(\theta) = 2\cdot \sin 5\theta$ . The area of the region enclosed by one loop of the curve is given by the following integral:

$A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta$

$A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta$

By using trigonometrical identities, the integral is further simplified:

$A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta$

$A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta$

$A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta$

$A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)$

$A = 4\pi$

The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

5 0

2 years ago

What are generated at transform fault plate boundaries?

tankabanditka [31]

Answer: The Earth's layer, which has the covering and layer, is made of a progression of things, or structural plates, that creep after some time. Along these lines, at intersecting limits, mainland outside is made and maritime covering is devastated. 2 plates slippy past each other structures a redesign plate limit.

5 0

2 years ago

Match each branch of science with a related career.

Lapatulllka [165]

A: geologist
b: physical science
c: space science
d: ecologist

8 0

2 years ago

Read 2 more answers

a student balances a 1.5kg of broom by placing her finger 1.4m from the end of the broom handle. How far from the broom handle w

Delicious77 [7]

The broom handle that she have to balance if she hung a 400g mass from the end of the broom handle is 5.24m

This problem is centered on moment. Moment is the turning effect of a force about a point. It is expressed as:

Moment = Force× Distance

According to principle of moment, the sum of clockwise moment is equal to sum of anticlockwise moment at shown

M1d1 = M2d2

Given the following

M1 = 1.5kg

d1 = 1.4m

M2 = 400g = 0.4kg

d2 is required

Substitute

1.5(1.4) = 0.4d2

2.1 = 0.4d2

d2 = 2.1/0.4

d2 = 5.24m

Hence the broom handle that she have to if she hung a 400g mass from the end of the broom handle is 5.24m

Learn more here: brainly.com/question/21945515

4 0

2 years ago

An airplane accelerates from a speed of 88m/s to a speed of 132 m/s during a 15 second time interval. How far did the airplane t

Gelneren [198K]

Answer:

$1650\:\mathrm{m}$

Explanation:

We can use the following kinematics equations to solve this problem:

$v_f=v_i+at,\\{v_f}^2={v_i}^2+2a\Delta x$ .

Using the first one to solve for acceleration:

$132=88+a(15),\\15a=44,\\a=\frac{44}{15}=2.9\bar{3}\:\mathrm{m/s^2}$ .

Now we can use the second equation to solve for the distance travelled by the airplane:

$132^2=88^2+2\cdot2.9\bar{3}\cdot \Delta x,\\\Delta x= \frac{9680}{2\cdot2.9\bar{3}},\\\Delta x =\fbox{$ 1650\:\mathrm{m}$}$ (three significant figures).

6 0

2 years ago