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2 years ago

What is the mechanical advantage of a hockey stick that is gripped at point D and hits the puck at point B?

Mathematics

2 answers:

ladessa [460]2 years ago

7 0

Answer: Mechanical advantage = Effort distance/Resistance distance

Effort distance = D - Fulcrum = 0.5 m

Resistance distance = A - Fulcrum = 2.0 m

Then,

Mechanical advantage = 0.5/2 = 0.25

Hockey stick is an example of third class where mechanical advantage is less than one as the speed of the end of point A is critical.

Step-by-step explanation:

RideAnS [48]2 years ago

5 0

Answer:

Mechanical advantage = Effort distance/Resistance distance

Effort distance = D - Fulcrum = 0.5 m

Resistance distance = A - Fulcrum = 2.0 m

Then,

Mechanical advantage = 0.5/2 = 0.25

Hockey stick is an example of third class where mechanical advantage is less than one as the speed of the end of point A is critical.

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aleksandr82 [10.1K]

Answer:

Step-by-step explanation:

Here's how this works:

Get everything together into one fraction by finding the LCD and doing the math. The LCD is sin(x) cos(x). Multiplying that in to each term looks like this:

$[sin(x)cos(x)]\frac{sin(x)}{cos(x)}+[sin(x)cos(x)]\frac{cos(x)}{sin(x)} =?$

In the first term, the cos(x)'s cancel out, and in the second term the sin(x)'s cancel out, leaving:

$\frac{sin^2(x)}{sin(x)cos(x)}+\frac{cos^2(x)}{sin(x)cos(x)}=?$

Put everything over the common denominator now:

$\frac{sin^2(x)+cos^2(x)}{sin(x)cos(x)}=?$

Since $sin^2(x)+cos^2(x)=1$ , we will make that substitution:

$\frac{1}{sin(x)cos(x)}$

We could separate that fraction into 2:

$\frac{1}{sin(x)}$ × $\frac{1}{cos(x)}$

$\frac{1}{sin(x)}=csc(x)$ and $\frac{1}{cos(x)}=sec(x)$

Therefore, the simplification is

sec(x)csc(x)

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4 years ago

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3 years ago

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3 years ago

The base of a triangle is 9/4 times of its height. If the area of triangle is 144 cm2. Find the height and the base.

Mazyrski [523]

Step-by-step explanation:

Given that the base of triangle is 9/4 times of its height . And area is 144 cm² . Let the height be h , then its base be 9h/4 .

• According to Question :-

$\tt\to Area_{(\triangle)}=\dfrac{base\times height}{2}\\\\\tt\to 144cm^2 = \dfrac{9h\times h}{4\times 2 } \\\\\tt\to h^2 =\dfrac{12\times 12 \times 2 \times 2 }{3\times 3 }\\\\\tt\to h =\sqrt{\dfrac{12\times 12 \times 2 \times 2 }{3\times 3 }}\\\\\tt\to h =\dfrac{(12)(2)}{3} \\\\\tt\to \boxed{\orange{\tt h = 8 cm }}$

Therefore ,

Height = h = 8cm .
Base ,= 9h/4 = 18cm .

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3 years ago