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MA_775_DIABLO [31]
3 years ago
10

What is effort arm don't say the answer of gogle ​

Physics
2 answers:
luda_lava [24]3 years ago
4 0

Answer:

effort arm is a class of lever

Explanation:

Classes of lever are pivot effort and load

nikklg [1K]3 years ago
3 0

Answer:

effort arm mean the use of any work by using your hand force motion or by hand power

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6X-6=9<br><br> Solve for X<br><br> Round to TWO decimal places
Brums [2.3K]

Answer:

X=2.50

Explanation:

6x-6=9

6x= 9+6

X=15/6

X= 2.50

8 0
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Velocity is a vectors quantity that has both magnitudes and directions. Using complete sentences, described the object's velocit
Alenkasestr [34]

Answer:

Explanation:

Velocity is defined as the rate of change of displacement.

velocity is a vector quantity, that means it requires both magnitude and direction to completely explain the velocity.

For example, the velocity is 5 ms due east, it means an object is moving with speed 5 ms in the direction of east. We can say that the object covers the displacement of 5 m in one second due east.

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4 years ago
Suppose an object’s initial velocity is 10 m/s and its final velocity is 4 m/s. Mass is constant.
Ostrovityanka [42]

The correct answer is:

Work is negative, the environment did work on the object, and the energy of the system decreases.

In fact, the work-energy theorem states that the work done by the system is equal to its variation of kinetic energy:

W=\Delta K=K_f -K_i

In this problem, the variation of kinetic energy \Delta K is negative (because the final velocity is less than the initial velocity), so the work is negative, and this means that the environment did work on the object, and its energy decreased.

3 0
3 years ago
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Strontium 3890Sr has a half-life of 28.5 yr. It is chemically similar to calcium, enters the body through the food chain, and co
patriot [66]

Answer:

Thus the time taken is calculated as 387.69 years

Solution:

As per the question:

Half life of ^{3890}Sr\, t_{\frac{1}{2}} = 28.5 yrs

Now,

To calculate the time, t in which the 99.99% of the release in the reactor:

By using the formula:

\frac{N}{N_{o}} = (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}

where

N = No. of nuclei left after time t

N_{o} = No. of nuclei initially started with

\frac{N}{N_{o}} = 1\times 10^{- 4}

(Since, 100% - 99.99% = 0.01%)

Thus

1\times 10^{- 4} = (\frac{1}{2})^{\frac{t}{28.5}}}

Taking log on both the sides:

- 4 = \frac{t}{28.5}log\frac{1}{2}

t = \frac{-4\times 28.5}{log\frac{1}{2}}

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3 years ago
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