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2 years ago

What do you need to calculate the mechanical advantage of a block and tackle

2 answers:

7 0

In a block and tackle, some friction in the pulleys will reduce the mechanical advantage of the machine. To include friction in a calculation of the mechanical advantage of a block and tackle, divide the weight of the object being lifted by the weight necessary to lift it.

Hope this helps

Viktor [21]2 years ago

3 0

The answer was "Number of rope segments that support the resistance weight".

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A skier starts from rest down a slope 500.0 M long, the skier accelerates at a constant rate of 2.00 m/s/s, what's the velocity

nevsk [136]

We can use the kinematic equation
$(v_f)^2 = (v_i)^2 + 2*a*d$
where Vf is what we are looking for
Vi is 0 since we start from rest
a is acceleration
and d is the distance

we get
(Vf)^2 = (0)^2 + 2*(2)*(500)
(Vf)^2 = 2000
Vf = about 44.721
or 44.7 m/s [if you are rounding this by significant figures]

8 0

2 years ago

Rosný bod závisí od ?

WITCHER [35]

Answer:

what did u say and what language are you speaking in

8 0

3 years ago

What is the period (in hours) of a satellite circling Mars 100 km above the planet's surface? The mass of Mars is 6.42 × 1023 kg

scZoUnD [109]

To solve this problem it is necessary to apply the concepts related to the Centrifugal Force and the Gravitational Force. Since there is balance on the body these two Forces will be equal, mathematically they can be expressed as

$F_c = F_g$

$\frac{mv^2}{r} = \frac{GmM}{r^2}$

Where,

m = Mass

G =Gravitational Universal Constant

M = Mass of the Planet

r = Distance/Radius

Re-arrange to find the velocity we have,

$v^2 = \frac{GM}{r}$

At the same time we know that the period is equivalent in terms of the linear velocity to,

$T = \frac{2\pi}{\frac{v}{r}}$

$v = \frac{2\pi r}{T}$

If our values are that the radius of mars is 3400 km and the distance above the planet is 100km more, i.e, 3500km we have,

$v^2 = \frac{GM}{r}$

$( \frac{2\pi r}{T})^2 = \frac{GM}{r}$

$T = \sqrt{\frac{4\pi^2 r^3}{GM}}$

Replacing we have,

$T = \sqrt{\frac{4\pi^2 (3500*10^3)^3}{(6.67430*10^{-11})(6.42*10^23)}}$

$T = 6285.09s (\frac{1min}{60s})(\frac{1hour}{60min})$

$T= 1.74hour$

Therefore the correct answer is C.

7 0

3 years ago

A 3.5-kg object placed on an inclined plane (angle 30? above the horizontal) is connected by a string going over a pulley to a 1

Blababa [14]

Answer:

a= 0.22 m/s²

Explanation:

Given that

M = 3.5 kg

θ = 30°

m = 1 kg

μ= 0.3

The force due to gravity

F₁= M g sinθ

F₁=3.5 x 10 x sin 30

F₁= 17.5 N

F₂ = m g

F₂ = 1 x 10 = 10 N

The maximum value of the friction force on the incline plane

Fr = μ M g cosθ

Fr = 0.3 x 2.5 x 10 cos30°

Fr= 6.49 N

Lets take acceleration of the system is a m/s²

F₁ - F₂ - Fr = (M+m) a

17.5 - 10 - 6.49 = (3.5+1)a

a= 0.22 m/s²

7 0

2 years ago

6A certain load and set of slings create a 20-degree angle between the load and each sling leg. Using a spreader for the same li

Otrada [13]

Solution:

The angle between the sling and the load is $20^{\circ}$

So the tension in each sling can be calculated as

$Sin \theta = Mg => T = \frac{Mg}{2Sin\theta}$

$Sin \theta=> \frac{Mg}{2Sin 20^{\circ}}$

Where

M is the mass of the load

The Horizontal reaction on the sling will be inward.

After using the spreader, the new angle between sling and load is $60^{\circ}$ , the tension in the sling will be

$T= \frac{Mg}{2 Sin 60^{\circ}}$ = $\frac{Mg}{2 Sin 20^{\circ}}$

The tension will be same as before in the sling move away through the spreader at an angle more than 90 degree the horizontal force will act opposite and will be outward

5 0

3 years ago