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2 years ago

A person is walking down the street maintain the same speed and direction. The person is accelerating.

Physics

1 answer:

balandron [24]2 years ago

6 0

Answer:

no change in speed, therefore the body cannot be accelerated. a=0

Explanation:

When a person is accelerating his speed must change, if the speed is in the same direction as the acceleration the speed increases and if the acceleration is in the opposite direction to the speed it decreases.

In this case there is no change in speed, therefore the body cannot be accelerated.

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(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

An engineer is designing a runway for a witch. Several brooms will use the runway and the engineer must design it so that it is

jonny [76]

Answer:

1170 m

Explanation:

Given:

a = 3.30 m/s²

v₀ = 0 m/s

v = 88.0 m/s

x₀ = 0 m

Find:

v² = v₀² + 2a(x - x₀)

(88.0 m/s)² = (0 m/s)² + 2 (3.30 m/s²) (x - 0 m)

x = 1173.33 m

Rounded to 3 sig-figs, the runway must be at least 1170 meters long.

6 0

3 years ago

Name the property illustrated be each statement. If t - 13 = 52, then 52 = t - 13

Natali5045456 [20]

The property illustrated to each statement is commutative because it shows that for some value of t both statements are equal.

3 0

2 years ago

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What is the momentum of a 950 kg car moving at 10.0 m/s?

pentagon [3]

Answer:

<h3>The answer is 9500 kgm/s</h3>

Explanation:

The momentum of an object can be found by using the formula

<h3>momentum = mass × velocity</h3>

From the question

mass = 950 kg

velocity = 10.0 m/s

We have

momentum = 950 × 10

We have the final answer as

Hope this helps you

8 0

3 years ago

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What can be found in every skeletal muscle?

anastassius [24]

You can find muscle fibers, nerves, connective tissue, and blood vessels in every skeletal system. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

7 0

2 years ago

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