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3 years ago

Which automobile is faster a chevrolet or a ford? Also explain pls

Physics

2 answers:

antiseptic1488 [7]3 years ago

6 0

Answer:

depends on what type of car it is

Explanation:

saveliy_v [14]3 years ago

3 0

Answer:

It depends which model.

Explanation:

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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

3 years ago

If you apply an equal force to a larger mass, it will take longer to accelerate. <br> Tru or false

Makovka662 [10]

Answer:

true

Explication:

The acceleration of an object depends on the mass of the object, and the amount of force applied

5 0

3 years ago

A 40-W lightbulb is 1.7 m from a screen. What is the intensity of light incident on the screen? Assume that a lightbulb emits ra

Sonja [21]

Answer:

Intensity, $I=1.101\ W/m^2$

Explanation:

Power of the light bulb, P = 40 W

Distance from screen, r = 1.7 m

Let I is the intensity of light incident on the screen. The power acting per unit area is called the intensity of the light. Its formula is given by :

$I=\dfrac{P}{A}$

$I=\dfrac{P}{4\pi r^2}$

$I=\dfrac{40\ W}{4\pi (1.7\ m)^2}$

$I=1.101\ W/m^2$

So, the intensity of light is $1.101\ W/m^2$ .

6 0

3 years ago

Why must the electric field be normal to the surface at every point of a charged conductor?

solong [7]

-- If the field were inclined to the surface, then it would have
some component parallel to the surface.

-- Then, since we're talking about a conductor, the charges
on the object would move in response to that component
of the field, until there was no longer any component of the
field trying to move them.

3 0

3 years ago

xeze [42]

D) Submarine is your answer. Have a great rest of your day!

7 0

3 years ago

Read 2 more answers