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3 years ago

8

A system of pulleys is used to raise a load of bricks that weighs 1,700 newtons. The force applied to the pulley is 340 newtons.

What’s the mechanical advantage of this machine? 1 2 5 10

2 answers:

Alchen [17]3 years ago

6 0

Answer:

The answer is 5

Explanation:

I got this question on a test and this was the right answer

Please mark as brainliest :)

Law Incorporation [45]3 years ago

4 0

By definition, we have that the mechanical advantage is given by the following equation:
$MA = \frac{W}{T} 
$
Where,
W: is the load
T: is the tension
Substituting the values in the given equation we have:
$MA = \frac{1700}{340}$
$MA = 5
$
Therefore, the mechanical advantage is equal to 5.
Answer:
The mechanical advantage of this machine is:
MA = 5

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Solve the following quadratic equations:2x² - 4x - 9 = 0 x₊ = 1 + √22/2Answersx_ = 1 - √22/2

Pavel [41]

The root quadratic equation of $2x^{2} - 4x - 9 = 0$ is $x=\frac{2+\sqrt{22}}{2},\:x=\frac{2-\sqrt{22}}{2}$

What is the formula used to calculate the quadratic equation?

To solve the quadratic equation use the formula of $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Find the quadratic equation by using the formula ?

On comparing with $2x^{2} - 4x - 9 = 0$

We get , $a=2 , b=-4 , c=-9$

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By using the formula,

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$X==\frac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\c* \:2\left(-9\right)}}{2\c* \:2}\\\\\sqrt{\left(-4\right)^2-4\cdot \:2\left(-9\right)}\\\\$

Apply rule,

$-(-a)=a\\\\=\sqrt{\left(-4\right)^2+4\cdot \:2\cdot \:9}\\\\$

Apply exponent rule,

$(-a)^{n} =a^{n}$ if n is even

$(-4)^{2} =4^{2}$

$\sqrt{4^2+4\cdot \:2\cdot \:9}$

Multiply the numbers :

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Add the number $=\sqrt{88}$

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2 years ago

A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 30 m/s2 for 35 s , then ru

Gre4nikov [31]

Answer:

a) $h_m=74625\ m$

b) $t=265.55\ s$

Explanation:

Given:

mass of rocket, $m_r=200\ kg$

mass of fuel, $m_f=100\ kg$

acceleration of the rocket consuming fuel, $a=30\ m.s^{-2}$

time after which the fuel exhaust, $t_f=35\ s$

<u>During the phase of fuel exhaustion:</u>

<u>velocity attained by the rocket just as the fuel ends:</u>

$v_f=u+a.t_f$

where:

$u=$ initial velocity of the rocket = 0

$v_f=0+30\times 35$

$v_f=1050\ m.s^{-1}$ this will be the initial velocity for the phase of ascending of the rocket's height under the influence of gravity.

<u>height at which the fuel finishes:</u>

$v_f^2=u^2+2a.h_f$

$1050^2=0^2+2\times 30\times h_f$

$h_f=18375\ m$

<u>During the phase of ascend in height of rocket after the fuel is over:</u>

<u>Time taken to reach the top height after the fuel is over:</u>

$v=v_f+g.t'$

at top v = (final velocity during this course of motion )= 0 $m.s^{-1}$

$0=1050-9.8\times t'$

$t'=107.1429\ s$

<u>Height ascended by the rocket after the fuel is over:</u>

$v^2=v_f^2+2g.h'$

at the top height the velocity is zero

$0^2=1050^2-2\times 9.8\times h'$ (-ve sign denotes that the direction of motion is opposite to that of acceleration)

$h'=56250\ m$

<u>Therefore the maximum altitude attained by the rocket:</u>

$h_m=h_f+h'$

$h_m=18375+56250$

$h_m=74625\ m$

b)

time taken by the rocket to fall back to the earth:

$h_m=v.t_m+\frac{1}{2} g.t_m^2$

where:

$v=$ initial velocity of the rocket during the course of free fall from the top height.

$74625=0+4.9\times t_m^2$

$t_m=123.41\ s$

Now the total time for which the rocket is in the air:

$t=t_f+t'+t_m$

$t=35+107.1429+123.41$

$t=265.55\ s$

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