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oksian1 [2.3K]
3 years ago
10

Which expression shows the amount of work a 6-kilogram mass is capable of performing if it is dropped from a height of 2 meters?

The acceleration of gravity is .
Physics
2 answers:
galina1969 [7]3 years ago
6 0
To answer this question you need to know that work = force x distance. You know the distance is 2m. Weight is a force and you can find that with the information given.
Weight = mass x gravity
In this case we're taking about acceleration of gravity which equals 9.8m/s/s
Weight = 6kg x 9.8m/s/s
W= 58.8 newtons or kg-m/s/s
So now that you know force and distance you can find work
Work= force x distance
Work = 58.8 newtons x 2 meters
Work = 117.6 n-m or joules
Hope this helps!
Ipatiy [6.2K]3 years ago
3 0
The acceleration of gravity is 9.8 m/s^2
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1\ km/h=\frac{1}{3.6}\ m/s

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Time=\frac{55}{\frac{100}{9}}=4.95\ seconds

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The speed of the water in the wider part will be 1.194 m/sec. Speed is a time-based quantity. Its SI unit is m/sec.

<h3> What is speed?</h3>

Speed is defined as the rate of change of the distance or the height attained.

The given data in the problem is;

The initial diameter is,\rm d_1 = 2.2 \ cm

initial radius,

r_1 = \frac{d_1}{2} \\\\ r_1 = \frac{2.2}{2} \\\\ r_1 = 1.1\ cm

The initial crossection area;

\rm A_1 = \pi r_1^2 \\\\ \rm A_1 = 3.14 \times  (1.1\times 10^{-2})^2 \\\\ \rm A_1 =3.8 \times 10^{-4} \ m^2

The final crossection area;

\rm A_2 = \pi r_2^2 \\\\ \rm A_2 = 3.14 \times ( 2 \times 10^{-2})^2 \\\\ \rm A_2 = 12.56 \ m^2

The initial flow rate is;

R = density ×velocity ×area

\rm R = \rho A V \\\\ 1.5 = 1000 \times V_1 \times 3.8 \times 10^{-4} \\\\ V_1  = 3.947 \ m/sec

The speed of the water in the wider part will be;

From the continuity equation;

\rm A_1 V_1 = A_2V_2  \\\\\ 3.8 \times 10^{-4} \times 3.947 = 12.56 \times 10^{-4} \times V_2 \\\\ V_2= 1.194 \ m/sec

Hence, the speed of the water in the wider part will be 1.194 m/sec.

To learn more about the speed, refer to the link;

brainly.com/question/7359669

#SPJ1

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