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3 years ago

Which expression allows you to determine the mechanical advantage of an inclined plane?

Physics

1 answer:

Leya [2.2K]3 years ago

4 0

Answer:

Explanation:

The mechanical advantage of an inclined plane can be determined using different variables. In this case, the geometry of the setup is relevant. The advantage is proportional to the length of the plane, and inversely proportional to the height: it is the ratio (length) / (height) of the plane. For example, given a desired, fixed height, a long inclined plane gives you a bigger mechanical advantage than a short inclined plane. In this example, pushing an object up the long plane will require a smaller force, than it would on the short plane.

Strictly speaking, (D) would also "allow you to determine the mechanical advantage" because you could simply invert the ratio listed under (D). However, (C) is the best, direct, answer.

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2 years ago

How much heat is necessary to warm 500g of water from 20°c to 65°c

Vsevolod [243]

Explanation:

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2 years ago

Finally, you are ready to answer the main question. Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour in 2.22 s

muminat

Answer:

Acceleration = 10.06 m/s²

Explanation:

1 mile = 1.6093km

1609.3m = 1 mile

1 m = $\frac{1}{1609}$ mile

50.0 miles/hour = $\frac{50 * 1609.3}{60 * 60}$ m/s

= 22.35m/s

from equation

S = Ut + 1/2 at²

v = U + at

22.35 = 0 + a * 2.22

a = 22.35 ÷ 2.22

= 10.06 m/s²

4 0

3 years ago

The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th

Inga [223]

According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

$E_i = E_f$

$0 = \frac{1}{2} (m_1+m_2)v_f^2-m_2gh+m_1gh$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

PART B) Replacing the values given as,

$h= 1.7m\\m_1 = 3.5kg\\m_2 = 4.3kg \\g = 9.8m/s^2 \\$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

$v_f = \sqrt{2(9.8)(1.7)(\frac{4.3-3.5}{3.5+4.3})}$

$v_f = 1.8486m/s$

Therefore the speed of the masses would be 1.8486m/s

6 0

3 years ago

A transformer has two sets of coils, the primary with N1 = 160 turns and the secondary with N2 = 1400 turns. The input rms volta

vovikov84 [41]

To solve the problem it is necessary to apply the concepts related to the voltage in a coil, through the percentage relationship that exists between the voltage and the number of turns it has.

So things our data are given by

$N_1 = 160$