What is the basic building block of all matter

(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

3 years ago

Which person will most likely hear the loudest sound?

Nataliya [291]

Person standing on A will hear the loudest sound

Explanation:

The intensity of a sound wave (which is proportional to the loudness of the sound) follows an inverse square law, which is:

$I\propto \frac{1}{r^2}$

where

I is the intensity of the wave

r is the distance from the source of the sound

This equation means that the intensity of the sound wave (and therefore, its loudness) is inversely proportional to the square of the distance from the source: therefore,

As we get closer to the source of sound, the loudness increases
As we move away from the source of sound, the loudness decreases

Therefore, the person that will hear the loudest sound is the one standing closer to the source, and therefore person A.

Learn more about sound waves:

brainly.com/question/4899681

#LearnwithBrainly

6 0

3 years ago

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After driving a portion of the route, the taptap is fully loaded with a total of 27 people including the driver, with an average

lara [203]

Answer:

compression of spring is x = 0.12 m

Assumed k = 160,000 N/m ........ Truck's suspension system

Explanation:

Given:

- The mass of average person m_p = 69 kg

- Total number of persons n_p = 27

- The mass of each goat m_g = 15 kg

- The total number of goats n_g = 3

- The mass of each chicken m_c = 3 kg

- The total number of goats n_c = 5

- The total mass of bananas m_b = 25 kg

Find:

How much are the springs compressed?

Solution:

- Using equilibrium equation on the taptap in vertical direction:

F_net = F_spring - F_weight = 0

- Compute the force due to all the weights on the taptap:

F_weight = (n_p*m_p + n_g*m_g + n_c*m_c + m_b)*9.81

F_weight = (69*27 + 3*15 + 5*3 + 25)*9.81

F_weight = 19109.88 N

- The restoring force of a spring is given by:

F_spring = k*x

Where, k is the spring stiffness and x is the displacement:

F_weight = F_spring

19109.88 = k*x

x = 19109.88 / k

We need to assume the spring stiffness we will take k = 160,0000 N/m (trucks suspension systems). The value of the stiffness must be high enough to sustain a load of 1.911 tonnes.

x = 19109.88 / 160,000

x = 0.1194 m ≈ 0.12 m = 12 cm

- A compression of 12 cm seems reasonable for a taptap to carry 1.911 tonnes of load. Hence, the assumption of spring stiffness was reasonable. Hence, the compression of spring is x = 0.12 m.

8 0

3 years ago

2. A rock is shot straight up into the air with a slingshot that had been stretched 0.30 m. Assume

Luba_88 [7]

Answer:

e

Explanation:

4 0

3 years ago

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How is energy transformed to chemical energy?

morpeh [17]