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Alecsey [184]
2 years ago
15

Which of the following is a limited resource, which can take longer to be replenished than it takes to be used up?

Physics
2 answers:
Zarrin [17]2 years ago
8 0
What is the list of choices?
sasho [114]2 years ago
7 0

A. Energy from wind

B. Natural gas

C. Fertile soil

D. Oxygen in air


The answer is C. Fertile soil

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If you place a balloon in the freezer, it will __________.
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The answer is "Deflate because volume is directly proportional to temperature"
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3 years ago
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A 125kg man buy a 6kg watermelon, a 3kg cantaloupe, and 6kgs of potatoes, he walks home with his purchases in a large bag. His w
Stells [14]

Answer:

350N

Explanation:

Given parameters:

Mass of the man = 125kg

Mass of the watermelon = 6kg

Mass of cantaloupe  = 3kg

Mass of potatoes  = 6kg

Acceleration  = 2.5m/s²  

Unknown:

Force required to get home  = ?

Solution:

To find this force we use;

    Force  = mass x acceleration

mass  = 125 + 6 + 3 + 6  = 140kg

 So;

      Net force = 140 x 2.5  = 350N

5 0
3 years ago
According to the chart, one gram of copper and
ad-work [718]

Answer:

C) three

Explanation:

Let gram of gold required be m . Let temperature change in both be Δ t .

heat absorbed = mass x specific heat x change in temperature

for copper

heat absorbed = 1 x .385 x Δt

for gold

heat absorbed = m x .129 x Δt

So

m x .129 x Δt = 1 x .385 x Δt

m = 2.98

= 3 g approximately .

4 0
2 years ago
Material speed of light
Cloud [144]
The question is poor. Light doesn't refract on its way THROUGH anything. It refracts at the boundary BETWEEN two different media. The effect is greatest where the ratio of the speeds of light in the two media is greatest. On your list, that would be at the boundary between air or space and glass.
3 0
3 years ago
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A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas
Radda [10]

Answer:

change in momentum, \Delta p=7.65 \,kg.m.s^{-1}

  • \Delta p_x= 6.6251 \,kg.m.s^{-1}
  • \Delta p_y= 3.825 \,kg.m.s^{-1}

Average Force, F=144.3396\,N

  • F_x=125.0018\,N
  • F_y=72.1698\,N

Explanation:

Given:

angle of kicking from the horizon, \theta= 30^{\circ}

velocity of the ball after being kicked, v=18 m.s^{-1}

mass of the ball, m=0.425\, kg

time of application of force, t=5.3\times 10^{-2}\,s

We know, since body is starting from the rest

\Delta p=m.v.....................(1)

\Delta p=0.425\times 18

\Delta p=7.65 \,kg.m.s^{-1}

Now the components:

\Delta p_x= 7.65\times cos 30^{\circ}

\Delta p_x= 6.6251 \,kg.m.s^{-1}

similarly

\Delta p_y= 7.65\times sin 30^{\circ}

\Delta p_y= 3.825 \,kg.m.s^{-1}

also, impulse

I=F\times t.........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

F\times t=m.v

F\times 5.3\times 10^{-2}= 7.65

F=144.3396\,N

Now, the components

F_x=144.3396\times cos 30^{\circ}

F_x=125.0018\,N

&

F_y=144.3396\times sin 30^{\circ}

F_y=72.1698\,N

6 0
3 years ago
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