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Travka [436]
3 years ago
12

If the torque required to loosen a nut that holds a wheel on a car has a magnitude of 55 n·m, what force must be exerted at the

end of a 0.47 m lug wrench to loosen the nut when the angle is 48°?
Physics
1 answer:
erastova [34]3 years ago
6 0

Either 175 N or 157 N depending upon how the value of 48° was measured from.    
You didn't mention if the angle of 48° is from the lug wrench itself, or if it's from the normal to the lug wrench. So I'll solve for both cases and you'll need to select the desired answer.    
Since we need a torque of 55 N·m to loosen the nut and our lug wrench is 0.47 m long, that means that we need 55 N·m / 0.47 m = 117 N of usefully applied force in order to loosen the nut. This figure will be used for both possible angles.    
Ideally, the force will have a 0° degree difference from the normal and 100% of the force will be usefully applied. Any value greater than 0° will have the exerted force reduced by the cosine of the angle from the normal. Hence the term "cosine loss".     
If the angle of 48° is from the normal to the lug wrench, the usefully applied power will be:  
U = F*cos(48)  
where  
U = Useful force  
F = Force applied    
So solving for F and calculating gives:  
U = F*cos(48)  
U/cos(48) = F  
117 N/0.669130606 = F  
174.8537563 N = F    
So 175 Newtons of force is required in this situation.    
If the 48° is from the lug wrench itself, that means that the force is 90° - 48° = 42° from the normal. So doing the calculation again (this time from where we started plugging in values) we get  
U/cos(42) = F  
117/0.743144825 = F  
157.4390294 = F    
Or 157 Newtons is required for this case.
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OverLord2011 [107]

Answer:

Earth attract the Moon with a force that is greater.

Explanation:

According to the law of gravitation, the gravitational force between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Mathematically, F1 = Gm1m2/r²... 1

Let m1 be the mass of the earth and m2 be that of the moon

If the Earth is much more massive than is the Moon, the new force of attraction between them will become;

F2= G(2m1)m2/r²

F2 = 2Gm1m2/r² ... (2)

Dividing eqn 1 by 2 we have;

F1/F2 = (Gm1m2/r²)÷(2Gm1m2/r²)

F1/F2 = Gm1m2/r²×r²/2Gm1m2

F1/F2 = 1/2

F2=2F1

This shows that that the earth will attract the moon by a force 2times the initial force of the masses(i.e a much greater force)

6 0
3 years ago
During a clinic visit a4 year old girl suddenly yells don't sit on erin the parents wispers that erin is an imaginary friend wha
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3 years ago
A rectangular block of copper has sides of length 15 cm, 26 cm, and 43 cm. If the block is connected to a 5.0 V source across tw
nexus9112 [7]

Answer:

the case is the one  with the greatest current, L=15 cm ,   i = 2.19 10⁸  A

Explanation:

Ohm's law is

          V = i R

Resistance is

         R = ρ L / A

Where L is the length of the electrons pass and A the area perpendicular to the current

      i = V / R

      i = V (A / ρ L)

      i = V / ρ  (A / L)

We can calculate the relationship between the area and the length to know in which direction the maximum currents

Case 1

      L = 0.15 m

      A = 0.26 0.43 = 0.1118 m2

      A / L = 0.1118 / 0.15

      A / L = 0.7453 m

Case 2

        L = 0.26 m

        A = 0.15 0.43 = 0.0645 m2

        A / L = 0.248 m

Case 3

       L = 0.43 m

       A = 0.15 0.26 = 0.039 m2

        A / L = 0.0907 m

We can see that the case is the one  with the greatest current, L=15 cm

Let's calculate the current

     i = 5 / 1.7 10⁻⁸ (0.7453)

      i = 2.19 10⁸  A

3 0
3 years ago
You are given a parallel plate capacitor that has plates of area 27 cm^2 which are separated by 0.0100 mm of nylon (dielectric c
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Answer:

Applied voltage should be 13.5396 kV

Explanation:

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Q=\frac{K\epsilon _{o} A}{d}V

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V=\frac{Qd}{K\epsilon _{o}A}

thus to store a charge of 0.11mC we solve for V as follows

Applying values we get

V=\frac{0.11\times 10^{-3}\times 0.01\times 10^{-3}}{27\times 10^{-4}\times 3.4\times 8.85\times 10^{-12}}

\therefore V=13.5396kV

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