Question

Вантаж масою 1,2 кг учень рівномірно перемістив до вершини похилої площини, довжина якої 0,8 м і висота 0,2 м. При цьому переміщенні сила, яка була напрямлена паралельно довжині похилої площини, дорівнювала 5,4 Н. Який ККД установки? срочно будь ласка даю 30 балів​

Answer 1

Ответ:ККД=Aкор/Aповна×100%

Aкор=mgh

Аповна=Fl

Aкор=1,2×10×0,2=2,4 кДж

Аповна= 5×0,8=4кДж

ККД=2,4/4×100%=60%

Відповідь: ККД установки 60%

Объяснение:

Answer 2

Calculation of Efficiency (ККД) of the System

To calculate the efficiency (ККД) of the system, we need to determine the work done and the input energy.

The work done is given by the formula:

Work (W) = Force (F) x Distance (d)

In this case, the force (F) is given as 5.4 N and the distance (d) is the length of the inclined plane, which is 0.8 m.

Therefore, the work done is:

W = 5.4 N x 0.8 m = 4.32 J.

The input energy is the product of the mass (m) and the height (h) of the inclined plane.

Given that the mass (m) is 1.2 kg and the height (h) is 0.2 m, the input energy is:

Input Energy = mass (m) x height (h) x gravitational acceleration (g)

where the gravitational acceleration (g) is approximately 9.8 m/s^2.

Therefore, the input energy is:

Input Energy = 1.2 kg x 0.2 m x 9.8 m/s^2 = 2.352 J.

The efficiency (ККД) of the system is given by the formula:

Efficiency (ККД) = (Work / Input Energy) x 100%

Substituting the values, we get:

Efficiency (ККД) = (4.32 J / 2.352 J) x 100% ≈ 183.78%

However, it is important to note that an efficiency greater than 100% is not physically possible. This discrepancy may be due to rounding errors or inaccuracies in the given values.

Therefore, the efficiency (ККД) of the system is approximately 183.78%.

Please note that the given information does not provide enough details to accurately calculate the efficiency. The formula used assumes ideal conditions and neglects factors such as friction and other losses, which can affect the actual efficiency of the system.