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3 years ago

Oil _____.

Physics

2 answers:

Ostrovityanka [42]3 years ago

6 0

Answer: The correct answer for the blank is -

must be mined from underground.

Oil is considered as a fossil fuel as it is formed by the dead remains of plants and animals that lived on earth millions of years back.

It is a non renewable source of energy because it can not be replenished readily by natural means and to renew, it would again require millions of years.

Mining of oil is a process of extraction of fossil fuels from underground, which is based on the digging of underground mining excavations.

Thus, oil must be mined from underground is the right answer.

atroni [7]3 years ago

4 0

Oil <span>must be mined from underground

Oil is NOT a renewable resource
Oil does NOT release more toxins than coal
Oil also is NOT easily replenished

</span>

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Convert 5.5 kilometers into millimeters.

dimaraw [331]

Answer:

5500000 millimeters

Explanation:

1 kilometre= 1000 meter

5.5 km=5.5 * 1000

=5500

Now,

1 metre = 1000 millimetres

5500 metre=1000*5500

=5500000 mm

4 0

1 year ago

Can someone pleaseEeeeeeee ASAP answer this❗️❗️❗️❗️ I really need help it is my second time posting this❗️❗️❗️❗️

belka [17]

Answer:

Data:-vi=om/s (b/c as in question penny is dropped from building means before coming to ground its initial state or velocity was considered as zero ) now distance or height h=380m and now we have to find the final velocity vf=? and the time t=?

Explanation:

So applying second eq of motion s=vit+1/2×gt² (here we have taken a gravity b/c when ever body is in vertical position then acceleration due to gravity is applied ) s=0×t+1/2×gt² , s=0+1/2×9.8×t² ,380=4.9t² we have to find t so 4.9t²=380 , t²=380÷4.9 , t²=77.55 now sq root on b/s

$\sqrt{t } = \sqrt{77.55}$

so t=8.806s and now apply 1st eq o²f motion to find out vf so vf=vi+gt , vf=0+9.8×8.806 ,vf=86.298 and if you want to verify that either this is answer is correct or not so put the value of t in second eq of motion and if you got distance same as give in the question so your value of t is considered as correct likewise s=vit+1/2gt² , s=0+1/2×9.8(8.806)²,s=4.9×77.55 ,s=380m (proved) I hope it would be helpfull

3 0

3 years ago

1.78 km = __m<br> Help plz

Sauron [17]

Answer:

1780

Explanation:

move decimal point to the right 3 times.

6 0

3 years ago

Read 2 more answers

Starting from rest, a particle moving in a straight line has an acceleration of a = (2t - 6) m/s^2, where t is in seconds. What

hjlf

To solve this problem we will use the given expression and derive it in order to find the algebraic expressions of velocity and position. These equations will be similar to those already known in the cinematic movement but will be subject to the previously given function. We start deriving the equation for velocity

$a = 2t-6$

$\frac{dv}{dt} = 2t-6$

Integrate acceleration equation

$\int dv = (2t-6)dt$

$v = 2(\frac{t^2}{2})-6t+C_1$

$v=t^2-6t+C_1$

At $t = 0, v = 0$

Replacing,

$0 = 0^2-6*0+C_1$

Therefore the value of the first Constant is

$C_1 = 0$

The expression can be escribed as,

$v = t^2-6t$

Calculate the velocity after 6s,

$v=t^2-6t$

$v = 6^2-6*6$

$v = 0m/s$

Now using the same expression we can derive the equation for distance

$v = t^2-6t$

$\frac{dx}{dt} =t^2-6t$

$\int dx = \int (t^2-6t)dt$

$x = \frac{t^3}{3}-6\frac{t^2}{2}+C_2$

At t=0, x=0

$0 = \frac{0^3}{3}-6(\frac{0^2}{2})+C_2$

Therefore the value of the second constant is

$C_2 = 0$

$x = \frac{t^3}{3}-\frac{6t^2}{2}$

Calculate the distance traveled after 11 s

At $t=11s$

$x = \frac{11^3}{3}-6(\frac{11^2}{2})$

$x = 80.667m$

6 0

3 years ago

Which of the following options correctly describe electromagnetic radiation? Select all that apply. Particles of electromagnetic

shutvik [7]

Answer:

Particles of electromagnetic radiation exhibit wave behavior

Electromagnetic radiation includes visible light

The specific amount of energy a particular photon possesses is called a quantum

Explanation:

Electromagnetic waves propagate through space perpendicularly to a oscillating magnetic and electric field producing it. Both of these fields oscillates perpendicular to one another, and to the direction of propagation of the wave.

Electromagnetic waves exhibit both wave and particle light behaviors, this phenomenon is known as the wave-particle duality.

Visible light is part of the broad spectrum of waves in the electromagnetic wave spectrum. And each particle of an electromagnetic wave is known as a photon, and they carry energy in discrete amounts called quantum.

7 0

2 years ago