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3 years ago

Please help correct answering get 20 points (physical science btw )

Physics

1 answer:

Zinaida [17]3 years ago

6 0

Answer:

Acceleration = 3.67 m/s²

Explanation:

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}$

Given the following data;

Initial velocity, u = 5 m/s

Final velocity, v = 27

Time, t = 6

To find the acceleration;

Acceleration = (v - u)/t

Acceleration = (27 - 5)/6

Acceleration = 22/6

Acceleration = 3.67 m/s²

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A 2 kg block slides on a rough horizontal surface with muk=0.6. It has an initial velocity of 5 m/s. Use g = 10 m/s2

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Two loudspeakers (A and B) are 3.20m apart and emitting a sound with a frequency of 400Hz. An observer is 2.10m directly in fron

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Answer:

The observer hears a loud sound

Explanation:

In order to know if the observer hears a loud or a quiet sound, you need to know if there is a constructive or destructive interference between the sound waves of the loudspeakers.

You first calculate the distance between the observer and the loudspeakers.

The distances are given by:

d1: distance to loudspeaker A = 2.10m

d2: distance to loudspeaker B

$d_2=\sqrt{(3.20m)^2+(2.10m)^2}=3.827m$

Next, you calculate the wavelength of the sound waves by using the following formula:

$\lambda=\frac{v_s}{f}$

vs: speed of sound = 343 m/s

f: frequency of the waves = 400Hz

λ: wavelength

$\lambda=\frac{343m/s}{400Hz}=0.8575m$

Next, you calculate the path difference between the distance from the observer to the loudspeakers:

$\Delta d=3.827m-2.10m=1.727m$

You obtain a constructive interference (loud sound) if the quotient between the wavelength of the sound and the difference path is an integer:

$\frac{\Delta d}{\lambda}=\frac{1.727m}{0.857}\approx2$

Then, there will be a constructive interference, and the sound who the observer hears is loud.

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3 years ago

A careful photographic survey of Jupiter's moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to

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The concept used to solve this problem is that given in the kinematic equations of motion. From theory we know that the change in velocities of a body is equivalent to twice the distance traveled by acceleration, in other words:

$v_f^2-v_i^2 = 2ax$

Where,

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x = Displacement

For the given case, the displacement is equivalent to the height (x = h) and the acceleration is the same gravitational acceleration (a = g). In turn we do not have initial speed therefore

$v_f^2 = 2hg$

$v_f = \sqrt{2hg}$

Our values are given as

$h = 70km = 70*10^3m$

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Replacing we have that,

$v_f = \sqrt{2hg}$

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Therefore the speed with which the liquid sulfur left the volcano is 529.15m/s

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3 years ago

The curvature of the helix r(t)equals(a cosine t )iplus(a sine t )jplusbt k (a,bgreater than or equals0) is kappaequalsStartF

4vir4ik [10]

Answer:

$\kappa = \frac{1}{2 b}$

Explanation:

The equation for kappa ( κ) is

$\kappa = \frac{a}{a^2 + b^2}$

we can find the maximum of kappa for a given value of b using derivation.

As b is fixed, we can use kappa as a function of a

$\kappa (a) = \frac{a}{a^2 + b^2}$

Now, the conditions to find a maximum at $a_0$ are:

$\frac{d \kappa(a)}{da} \left | _{a=a_0} = 0$

$\frac{d^2\kappa(a)}{da^2} \left | _{a=a_0} < 0$

Taking the first derivative:

$\frac{d}{da} \kappa = \frac{d}{da} (\frac{a}{a^2 + b^2})$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} \frac{d}{da}(a)+ a * \frac{d}{da} (\frac{1}{a^2 + b^2} )$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 + a * (-1) (\frac{1}{(a^2 + b^2)^2} ) \frac{d}{da} (a^2+b^2)$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 - a (\frac{1}{(a^2 + b^2)^2} ) (2* a)$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 - 2 a^2 (\frac{1}{(a^2 + b^2)^2} )$

$\frac{d}{da} \kappa = \frac{a^2+b^2}{(a^2 + b^2)^2} - 2 a^2 (\frac{1}{(a^2 + b^2)^2} )$

$\frac{d}{da} \kappa = \frac{1}{(a^2 + b^2)^2} (a^2+b^2 - 2 a^2)$

$\frac{d}{da} \kappa = \frac{b^2 - a^2}{(a^2 + b^2)^2}$

This clearly will be zero when

$a^2 = b^2$

as both are greater (or equal) than zero, this implies

$a=b$

The second derivative is

$\frac{d^2}{da^2} \kappa = \frac{d}{da} (\frac{b^2 - a^2}{(a^2 + b^2)^2} )$

$\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} \frac{d}{da} ( b^2 - a^2 ) + (b^2 - a^2) \frac{d}{da} ( \frac{1}{(a^2 + b^2)^2} )$

$\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} ( -2 a ) + (b^2 - a^2) (-2) ( \frac{1}{(a^2 + b^2)^3} ) (2a)$

$\frac{d^2}{da^2} \kappa = \frac{-2 a}{(a^2 + b^2)^2} + (b^2 - a^2) (-2) ( \frac{1}{(a^2 + b^2)^3} ) (2a)$

We dcan skip solving the equation noting that, if a=b, then

$b^2 - a^2 = 0$

at this point, this give us only the first term

$\frac{d^2}{da^2} \kappa = \frac{- 2 a}{(a^2 + a^2)^2}$

if a is greater than zero, this means that the second derivative is negative, and the point is a minimum

the value of kappa is

$\kappa = \frac{b}{b^2 + b^2}$

$\kappa = \frac{b}{2* b^2}$

$\kappa = \frac{1}{2 b}$

3 0

3 years ago