How do i do this question? Please help...

Mathematics

1 answer:

Kitty [74]3 years ago

5 0

y = -2x + 2

slope = -2

perpendicular slope = 1/2 (-ve inverse of the slope)

coordinates (0, 1)

y - y1 = m(x -x1)

y - 1 = 1/2(x - 0)

y = 1/2x + 1

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A uniform 41.0 kg scaffold of length 6.6 m is supported by two light cables, as shown below. A 74.0 kg painter stands 1.0 m from

Nimfa-mama [501]

Step-by-step explanation:

Let

$m_p$ = mass of the painter

$m_s$ = mass of the scaffold

$m_e$ = mass of the equipment

$T$ = tension in the cables

In order for this scaffold to remain in equilibrium, the net force and torque on it must be zero. The net force acting on the scaffold can be written as

$3T = (m_p + m_s + m_e)g\:\:\:\:\:\:\:(1)$

Set this aside and let's look at the net torque on the scaffold. Assume the counterclockwise direction to be the positive direction for the rotation. The pivot point is chosen so that one of the unknown quantities is eliminated. Let's choose our pivot point to be the location of $m_e$ . The net torque on the scaffold is then

$T(1.4\:\text{m}) + m_sg(1.9\:\text{m}) + m_pg(4.2\:\text{m}) - 2T(5.2\:\text{m}) = 0$

Solving for T,

$9T = m_sg(1.9\:\text{m}) + m_pg(4.2\:\text{m})$

$T = \frac{1}{9}[m_sg(1.9\:\text{m}) + m_pg(4.2\:\text{m})]$

$\:\:\:\:= 423.3\:\text{N}$

To solve for the the mass of the equipment $m_e$ , use the value for T into Eqn(1):

$m_e = \dfrac{3T - (m_p + m_s)g}{g} = 14.6\:\text{kg}$

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3 years ago

Phrase cost per minute

gogolik [260]

"cost per minute" refers to a unit rate of something that costs per minute to do. for example, a net cafe making you pay $0.30 per minute when you use a computer

4 0

3 years ago

Need help asap please

saveliy_v [14]

Answer:

$\displaystyle no$