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2 years ago

Determine the cutting force f exerted on the rod s in terms of the forces p applied to the handles of the heavy-duty cutter.

2 answers:

6 0

CASE 1: Direction vertically downward on the blade

Assume pin force at E acts downward on the blade and CW is a positive moment 
Ey[3.4] - F[1.7] = 0 
Ey = F/2 

Case 2: Direction vertically upward on the blade . Assume CW is positive moment 

Ey[1.5sin19] – P[21 – 1.5sin19] = 0 
(F/2)[1.5sin19] = P[21 – 1.5sin19] 
F = 2P[21 – 1.5sin19] / [1.5sin19] 
F = 84P

Alexus [3.1K]2 years ago

5 0

If you use the next formula with the data given in the exercise you are asking:
Ey[3.4] - F[1.7] = 0
Ey = F/2
and after that what you need to do is sum the moments of the handle about D to zero asumming it is a positive moment and we proceed like this
Ey[1.5sin19] – P[21 – 1.5sin19] = 0
(F/2)[1.5sin19] = P[21 – 1.5sin19] 
F = 2P[21 – 1.5sin19] / [1.5sin19] 
F = 84P

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A student practicing for a cross country meet runs 250 m in 30 s. What is the average speed

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Explanation:

A student practicing for a track meet, ran 250 m in 30 sec. a. What was her average speed? 250 m = 8.3 m/sec 30 sec.

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1 year ago

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Water runs into a fountain, filling all the pipes, at a steady rate of 0.750 m3>s. (a) How fast will it shoot out of a hole 4

kati45 [8]

Answer:

velocity = 472 m/s

velocity = 52.4 m/s

Explanation:

given data

steady rate = 0.750 m³/s

diameter = 4.50 cm

solution

we use here flow rate formula that is

flow rate = Area × velocity .............1

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solve it we get

velocity = 472 m/s

and

when it 3 time diameter

put valuer in equation 1

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1 year ago

Some of the fastest dragsters (called "top fuel) do not race for more than 300-400m for safety reasons. Consider such a dragster

Masja [62]

Answer:

1.10261 times g

416.17506 mph

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 400=0\times 8.6+\frac{1}{2}\times a\times 8.6^2\\\Rightarrow a=\frac{400\times 2}{8.6^2}\\\Rightarrow a=10.81665\ m/s^2$

Dividing by g

$\dfrac{a}{g}=\dfrac{10.81665}{9.81}\\\Rightarrow \dfrac{a}{g}=1.10261\\\Rightarrow a=1.10261g$

The acceleration is 1.10261 times g

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 10.81665\times 1.6\times 10^3+0^2}\\\Rightarrow v=186.04644\ m/s$

In mph

$186.04644\times \dfrac{3600}{1609.34}=416.17506\ mph$

The speed of the dragster is 416.17506 mph

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2 years ago

The velocity component with which a projectile covers certain horizontal distance is maximum at the moment of? a) Hitting the gr

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A) hitting the ground

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Ksenya-84 [330]

Answer:

Show attached picture

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