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3 years ago

9

The figure (Figure 1) shows the velocity of a solar-powered motorhome (RV) as a function of time. The driver accelerates from a

stop sign, cruises for 20 s at a constant speed of 60 km/h, and then brakes to come to a stop 40 s after leaving the stop sign.
*****What is the instantaneous acceleration at t=35s??????*********

1 answer:

SpyIntel [72]3 years ago

6 0

Explanation :

It is given that, the driver accelerates from a stop sign, cruises for 20 s at a constant speed of 60 km/h, and then brakes to come to a stop 40 s after leaving the stop sign.

We know that acceleration is defined as the rate of change of velocity.

$a=\dfrac{dv}{dt}$

Where

dv is the change in velocity, dv = 0 - 60 m/s = -60 m/s

dt is the change in time, dt = 40 s - 30 s = 10 s

So, $a=\dfrac{-60\ m/s}{10\ s}$

$a = -6\ m/s^2$

From the graph it is clear that, from 30 s to 40 s the car is decelerating. So, at every second within this time the value of acceleration will be same i.e. $-6\ m/s^2$ .

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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 $4\pi$ .

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3 years ago

En un momento dado , la nadadora de una prueba de natación de 100 m espalda está debajo de la cuerda falsa de salida. Indica a)

Virty [35]

Answer:

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Explanation:

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3 years ago

What characteristics determine how easily two substances change temperature

ValentinkaMS [17]

Answer;

Amount of time the two substances are in contact

Area in contact between the two substances

Specific heat of the material that makes up the substances

Explanation;

The change in temperature of a substance is caused by heat energy. The change in temperature will depend on factors such as mass of the substance, the type of material it is made from, the time taken , specific heat of the material that makes the substance, and also the area of contact.

The amount of time the two substances are in contact affect the change in temperature such that if the two bodies are in contact for a longer time then a bigger change in temperature will be observed.

Specific heat capacity also determines the change in temperature that will be observed, such that a substance with a bigger specific heat capacity will record a small change in temperature.

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3 years ago

What is the intensity of a sound with a sound intensity level (SIL) 67 dB, in units of W/m^2?

vfiekz [6]

Answer:

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Explanation:

We have expression for sound intensity level (SIL),

$L=10log_{10}\left ( \frac{I}{I_0}\right )$

Here we need to find the intensity of sound (I).

$L=10log_{10}\left ( \frac{I}{I_0}\right )\\\\log_{10}\left ( \frac{I}{I_0}\right )=0.1L\\\\\frac{I}{I_0}=10^{0.1L}\\\\I=I_010^{0.1L}$

Substituting

L = 67 dB and I₀ = 10⁻¹² W/m² in the equation

$I=I_010^{0.1L}=10^{-12}\times 10^{0.1\times 65}\\\\I_0=10^{-12}\times 10^{6.5}=10^{-5.5}=3.16\times 10^{-6}W/m^2$

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3 years ago

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Alex787 [66]

Let us take east and north as the positive x and y-axes should the motion be plotted in a cartesian plane. Thus, the x value is 45 miles and the y value is 20. The tangent of an angle is equal to the ratio of y to x.

tanθ = y / x

Substituting,

tanθ = 20/45 = 0.44

The value of θ is 23.96°.

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3 years ago