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andreyandreev [35.5K]
4 years ago
5

If we relied solely on the nonrenewable resources found in the U.S., which one would we run out of first at current usage levels

?
A. natural gas

B. oil

C. uranium

D. coal
Physics
2 answers:
lisabon 2012 [21]4 years ago
4 0

If we are depending on nonrenewable resources only then we will run out first at that resource which we are using at large level

that resource which is used at all levels at very fast level but we can't get it at that rate from natural sources

So out of all given choices we know that at most fastest rate we are using the OIL which we extract from the earth crust.

The rate with which we are extracting oil is very fast and if we use it at same rate then we will definitely run out of it

so correct answer would be

B. oil

Fantom [35]4 years ago
3 0

Answer:

B: Oil

Explanation:

If consumption of natural resources continues at the 1994 rate, then without importing resources from other countries, the U.S. has approximately enough oil to last for 23 years, enough natural gas to last for 68 years, enough uranium to last for 364 years, and enough coal to last for 7,007 years.

Reference: Energy Information Administration's State Energy Data Report 1994

( study island )

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Alika [10]

Answer:

t = 30 s

d = 1125 m

Explanation:

t = v/a = 75/2.5 = 30 s

d = v²/2a = 75²/(2(2.5)) = 1125 m

or

d = ½at² = ½(2.5)30² = 1125 m

4 0
3 years ago
A 650-kg satellite is in a circular orbit about Earth at a height above Earth equal to
MakcuM [25]

For a 650-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth’s mean radius, the resultant values are is mathematically given as

  • V=5.59*10^3m/s
  • T=3.98hours
  • F=1.47*10^3N

<h3>What are the satellite’s orbital speed, the period of its revolution, and the gravitational force acting on it.?</h3>

Generally, the equation for satellite orbital speed is mathematically given as

a)

V^2=\frac{Gm}{R+h}\\\\Therefore\\\\V^2=\frac{6.667*10^-11*5.98*10^{24}}{2*6.38*10^6}

V=5.59*10^3m/s

b)

V=2\pi(R+h)/T\\\\Hence\\\\T=2\pi(R+h)/V\\\\T=2\pi(R+h)/2*3.14*6.38*10^6/5.59*10^3

T=3.98hours

c)

F=\frac{GMM}{R+h^2}

Therefore

F=\frac{6.67*10^{-11}*5.98*10^{24}*650}{2*6.38*10^6}

F=1.47*10^3N

Read more about Force

brainly.com/question/26115859

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5 0
2 years ago
What is friction and the types withe examples.​
Blizzard [7]

Explanation:

The answer is In the picture. Thanks.

7 0
3 years ago
In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu
Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

4 0
4 years ago
What is the net force acting on a golf car traveling at a constant speed of 5 mph?
MAVERICK [17]
Technically, we can't answer the question with the given information.
Even though the cart's speed is constant, it may be turning an corner,
or driving along a curved path.  If it's not driving in a straight, then there
must be some force acting on it.

If the cart IS driving in a straight line, AND its speed is constant, then
there can't be any net force acting on it.  In that case, the correct choice
is  ' B '.  (I'm sure this is right IF the cart is driving in a straight line.)

3 0
3 years ago
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