It is easier to climb up a slanted slope than a vertical slope

2 answers:

8 0

Yes it is easier to clime a gentle slope than a vertical or more steep slope. This is because gravity is always pulling objects down so when you climb upward gravity weighs you down, you have to use effort to move yourself in the opposite direction that gravity is pulling you that is why it is harder to climb up a slope. When you climb a gentle slope you still are using energy but it takes longer to reach the same height on a gentle slope versus a vertical slope. On a vertical slope you are climbing higher instead of farther so on each step gravity weighs you down much more than on a gentle slope. When climbing a gentle slope you don't rise as rapidly so gravity doesn't take as much of a toll on you and you have a chance to recover.

Bas_tet [7]2 years ago

7 0

Answer:

Yes, it is easier to climb a slanted slope than a vertical or more steep slope.

Explanation:

On a vertical slope, you are climbing higher instead of farther so on each step gravity weighs you down much more than on a gentle slope

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A sample contains 16 g of a radioactive isotope. how much radioactive

mafiozo [28]

After one half-life, 8 g of radioactive isotope will remain in the sample.

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The act of producing radiation spontaneously is known as radioactivity. This is accomplished by an unstable atomic nucleus that want to give up some energy in order to move to a more stable form.

The following formula is used to compute the number of half lives elapsed:

$\rm N=\frac{N_0}{2^n} \\\\ N=\frac{16}{2} \\\\ N= 8 \ gram$

Hence,8 gram of radioactive isotope remains in the sample after 1 half-life.

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2 years ago

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the acceleration of a rocket traveling upward is given by a=(6+.02s)m/s^s, where s is in meters. Determine the rocket's velocity

NikAS [45]

Answer:

a) velocity v = 322.5m/s

b) time t = 19.27s

Explanation:

Note that;

ads = vdv

where

a is acceleration

s is distance

v is velocity

Given;

a = 6 + 0.02s

so,

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Remember that

$v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t= (5\sqrt{2} ) ln \frac{| [s + 300 + \sqrt{(s^{2} + 600s)} ] |}{300} .......2$