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2 years ago

Can someone list 6 advantages of renewable energy.

Physics

1 answer:

igor_vitrenko [27]2 years ago

4 0

Answer:

Solar energy, Wind energy, Hydro energy, Tidal energy, Geothermal energy, Biomass Energy.

Explanation:

I hope that it helps you...

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A lever has an input arm 50 centimeters long and an output arm 40 centimeters long. What is the mechanical advantage of the leve

taurus [48]

Answer:

Explanation:

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2 years ago

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The difference between experimental technique and procedure

kaheart [24]

A procedure is all the steps used to do an experiment in order.
<span>the experiment is when you test your hypothesis and is designed to answer your question. </span>
<span>the procedure is all the steps of the experiment.</span>

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3 years ago

What is the attractive force between all matter in the universe?

kvv77 [185]

The attractive force between all matter in the universe is gravity.

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2 years ago

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A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(

Natasha2012 [34]

Answer:

a) $t=2s$

b) $h_{max}=1024ft$

c) $v_{y}=-256ft/s$

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

$v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2$

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$0=(64ft/s)^2-2(32ft/s^2)(y-960ft)$

$y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft$

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

$h=-16t^2+64t+960$

$1024=-16t^2+64t+960$

$0=-16t^2+64t-64$

Solving the quadratic equation

$t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

$a=-16\\b=64\\c=-64$

$t=2s$

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s$

Since the projectile is going down the velocity at the instant it reaches the ground is:

$v=-256ft/s$

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2 years ago

During which segments are two states of matter present? the answer is D

Tomtit [17]

Answer:

2,4

Explanation:

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2 years ago