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3 years ago

When the fulcrum is shifted to the right, which side has a mechanical advantage and which side has a mechanical disadvantage?

1 answer:

8 0

The lever is a movable bar that pivots on a fulcrum attached to or positioned on or across a fixed point. The lever operates by applying forces at different distances from the fulcrum, or pivot.

A lever amplifies an input force to provide a greater output force, which is said to provide leverage. The ratio of the output force to the input force is the mechanical advantage of the lever.

The mechanical advantage of the lever is increased by moving the fulcrum closer to the load i.e., by shifting the fulcrum towards the right. Moving the fulcrum away from the load has mechanical disadvantage as the distance between the load and the fulcrum increases, the input force becomes greater than the output force.

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Consider an opaque horizontal plate that is well insulated on its back side. The irradiation on the plate is 2500 W/m2, of which

GuDViN [60]

Answer:

i. 0.34

ii. 0.4

iii. 1700 w/m²

iv. 2211.36 w/m²

Explanation:

Given that

Irradiation of the plate, G = 2500 w/m²

Reflected rays, p = 500 w/m²

Emissive power, E = 1200 w/m²

See attachment for calculations

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3 years ago

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lesya692 [45]

A velocidade mínima é 0. A velocidade máxima é inferior ou igual a 620 km/h.
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3 years ago

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oksian1 [2.3K]

Answer:

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3 years ago

Read 2 more answers

A rocket travels at a speed of 14,000 m/s. What is the distance travelled by the rocket in 150s?

kati45 [8]

1) The distance travelled by the rocket can be found by using the basic relationship between speed (v), time (t) and distance (S):

$v=\frac{S}{t}$

Rearranging the equation, we can write

$S=vt$

In this problem, v=14000 m/s and t=150 s, so the distance travelled by the rocket is

$S=vt=(14000 m/s)(150 s)=2.1 \cdot 10^6 m$

2) We can solve the second part of the problem by using the same formula we used previously. This time, t=300 s, so we have:

$S=vt=(14000 m/s)(300 s)=4.2 \cdot 10^6 m/s$

5 0

3 years ago

Two mass m1 and m2 lie on a frictionless surface. Between the two masses is a compressed spring, with spring constant k. The sys

max2010maxim [7]

Answer:

The spring was compressed the following amount:

$\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }$

Explanation:

Use conservation of energy between initial and final state, considering that the surface id frictionless, and there is no loss in thermal energy due to friction. the total initial energy is the potential energy of the compressed spring (by an amount $\Delta x$ ), and the total final energy is the addition of the kinetic energies of both masses:

$E_i=\frac{1}{2} k\,(\Delta x)^2\\\\E_f=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2$

$E_i=E_f\\$

$\frac{1}{2} k\,(\Delta x)^2=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2\\k\,(\Delta x)^2=m_1\,v_1^2+ m_2\,v_2^2\\(\Delta x)^2=\frac{m_1\,v_1^2+ m_2\,v_2^2}{k} \\\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }$

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3 years ago