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2 years ago

Someone help me please.

Physics

1 answer:

Soloha48 [4]2 years ago

8 0

1/2*2.8*x^{2}=2.8*9.81*1.50

X=5.425

Because K.E=1/2*M*V² and it is equal to P.E in the conservation of energy which means its K.E=P.E

P.E=m*g*h

So K.E equals 1/2*2.8*(5.425)²=41.20

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Two loudspeakers are placed next to each other and driven by the same source at 500 Hz. A listener is positioned in front of the

stellarik [79]

To solve this problem we will apply the concepts related to wavelength as the rate of change of the speed of the wave over the frequency. Mathematically this is

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

Here,

v = Wave velocity

f = Frequency,

Replacing with our values we have that,

$\lambda = \frac{340}{500}$

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Therefore the minimum distance will be 0.34m

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2 years ago

What type of configuration is used in the HEVs available in the US today?

Citrus2011 [14]

Answer:

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1 year ago

Chinook salmon are able to move upstream faster by jumping out of the water periodically; this behavior is called porpoising. Su

deff fn [24]

Answer:he formula for average speed is (total distance/total time)

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the total time is ((time in water)+(time out of water)) since you dont have time you must eliminate it. to do this you need (distance)/(time)=velocity

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Explanation:

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3 years ago

Can anyone solve these for my by using unit vectors? Can you also please show your work

Oxana [17]

4. The Coyote has an initial position vector of $\vec r_0=(15.5\,\mathrm m)\,\vec\jmath$ .

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$\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2$

where $\vec a$ is the Coyote's acceleration vector at time $t$ . He experiences acceleration only in the downward direction because of gravity, and in particular $\vec a=-g\,\vec\jmath$ where $g=9.80\,\frac{\mathrm m}{\mathrm s^2}$ . Splitting up the position vector into components, we have $\vec r=r_x\,\vec\imath+r_y\,\vec\jmath$ with

$r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t$

$r_y=15.5\,\mathrm m-\dfrac g2t^2$

The Coyote hits the ground when $r_y=0$ :

$15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s$

4b. Here we evaluate $r_x$ at the time found in (4a).

$r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m$

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