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Alexxandr [17]
3 years ago
5

A mechanic uses a screw driver to install a 1⁄4-20 UNC bolt into a mechanical brace. What is the mechanical advantage of the sys

tem? What is the resistance force if the effort force is 5 lb
Physics
1 answer:
Serggg [28]3 years ago
8 0

Answer:

15.7 ; 78.5

Explanation:

Given that

The mechanic use a screw driver to install 1⁄4-20 UNC bolt

And, the effort force is 5lb

We need to find out the mechanical advantage

And, the resistance force

As we know that

The Mechanical advantage of a screw = Circumference ÷ pitch

where,  

Circumference = pi × d

 pi = 3.142, D = diameter

So,  

Circumference = 3.142 × (1 ÷ 4)

= 0.785 in

Now

Pitch = 1 ÷ TPI

Here TPI (thread per inch) = 20

Pitch = 1 ÷  20 = 0.05

So,

Mechanical advantage = 0.785 ÷ 0.05

= 15.7

Now Resistance force if effort force is 5lb

We know that

Mechanical advantage = Fr ÷ Fe

Here

Fe = effort force, Fr = resistance force

15.7 = Fr ÷ 5

Fr = 15.7 × 5

= 78.5 lbs

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Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

5 0
3 years ago
A soldier fires a gun and another boy at a distance of 1020 m hears the sound of firing the gun 33 seconds after seeing its smok
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Answer:

Speed of sound is 30.91m/s

Explanation:

Speed is given by:

Speed = distance travelled / time taken

Given:

Distance travelled = 1020m

Time taken = 33seconds

Speed = 1020m/33s

Speed = 30.91m/s

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If a power utility were able to replace an existing 500 kV transmission line with one operating at 1 MV, it would change the amo
mestny [16]

Answer:

It would change the amount of heat produced in the transmission line to four times the previous value.

Explanation:

Given;

initial voltage in the transmission line, V₁ = 500 kV = 500,000 V

Final voltage in the transmission line, V₂ = 1 MV = 1,000,000

The power lost in the transmission line due to heat is given by;

P = \frac{V^2}{R}

Power lost in the first wire;

P_1 = \frac{V_1^2}{R}

R = \frac{V_1^2}{P_1}

Power lost in the second wire

P_2 = \frac{V_2^2}{R}\\\\ R = \frac{V_2^2}{P_2}

Keeping the resistance constant, we will have the following equation;

\frac{V_2^2}{P_2} = \frac{V_1^2}{P_1} \\\\P_2 = \frac{V_2^2P_1}{V_1^2}\\\\

P_2 = \frac{(1,000,000)^2P_1}{(500,000)^2}\\\\P_2 =4P_1

Therefore, it would change the amount of heat produced in the transmission line to four times the previous value.

8 0
3 years ago
A 5.7 kg particle starts from rest at x = 0 and moves under the influence of a single force Fx = 4.5 + 13.7 x − 1.5 x 2 , where
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Answer:

The work done by the force on the particle is 29.85 J.

Explanation:

The work is given by:  

W = ^{x_{2}}_{x_{1}}\int F_{x} dx

Where:

x₁: is the lower limit = 0 m    

x₂: is the upper limit = 1.9 m

Fₓ: is the force in the horizontal direction =  (4.5 + 13.7x - 1.5x²)N

W = ^{1.9}_{0}\int (4.5 + 13.7x - 1.5x^{2}) dx  

W = 4.5x|^{1.9}_{0} + \frac{13.7}{2}x^{2}|^{1.9}_{0} - \frac{1.5}{3}x^{3}|^{1.9}_{0}  

W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3}    

W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3}      

W = 29.85 J

Therefore, the work done by the force on the particle is 29.85 J.

I hope it helps you!                                

6 0
3 years ago
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