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Lesechka [4]
3 years ago
13

If the torque required to loosen a nut on a wheel has a magnitude of 40.0 N·m and the force exerted by a mechanic is 133 N, how

far from the nut must the mechanic apply the force?
Question 7 options:

1.) 0.15 m
2.) 0.6 m
3.) 0.3 m
4.) 1.2 m
Physics
1 answer:
MakcuM [25]3 years ago
3 0

Answer:

3.) 0.3 m

Explanation:

Given parameters:

Torque  = 40Nm

Force = 133N

Unknown:

Distance from the nut that the mechanic must apply the force  = ?

Solution:

Torque is the force that causes an object to rotate on its axis;

  Torque  = force x distance

   Distance  = \frac{40}{133}  = 0.3m

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Read 2 more answers
A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N
Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

r=\sqrt{(r_{1})^2+(r_{2})^2}

Put the value into the formula

r=\sqrt{(9.8)^2+(2.1)^2}

r=10.02\ cm

We need to calculate the force

Using formula of force

F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}

Force F₁₂,

F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}

F_{12}=13.33\ N

F_{21}=-13.33\ N

Force F₂₃,

F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}

We need to calculate the value of third charge

F_{net}=F_{21}+F_{23}

4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}

q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}

q_{3}=7.99\times10^{-7}\ C

q_{3}=0.8\times10^{-6}\ C

Hence, The value of third charge is 0.8μC.

4 0
3 years ago
What is the term for hardening or curing the sealant or composite material by exposing the material to a curing light?
natka813 [3]

Answer:

Photopolymerization is the term for hardening or curing the sealant or composite material by exposing the material to a curing light

Explanation:

It the process through which a photopolymer reacts to the radiations in presence of the Ultra-violet light. Photopolymers refers to those materials and liquid or the plastic resins that are hardened in presence of light source such as the lasers or the lamps. When these materials are irradiated, they undergo chemical reactions and they are hardened and become solid, and hence the process called photopolymerization. Photopolymerization is majorly used in the coating and printing industries and in also in dentistry to fill the top of teeth and prevent the cavities

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3 years ago
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