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ludmilkaskok [199]

3 years ago

11

If the torque required to loosen the nut that is holding a flat tire in a place on a car has a magnitude of 35 N•m what minimum

force must be exerted by the mechanic at the end of a 27 cm long wrench to loosen the nut

2 answers:

EleoNora [17]3 years ago

8 0

Answer:

129.63 N

Explanation:

Torque : This can be defined as a force that tend to rotate or twist a body. The S.I unit of torque is N.m. The expression of torque is given as,

T = F×d ....................... Equation 1

Where T = Torque on the car's nut. F = The minimum force exerted on the car, d= distance between the force and the nut.

make F the subject of the equation,

F = T/d............... Equation 2

Given: T = 35 N.m, d = 27 cm = 0.27 m.

Substitute into equation 2

F = 35/0.27

F = 129.63 N.

Hence the minimum force exerted by the mechanic to loosen the nut = 129.63 N

VashaNatasha [74]3 years ago

4 0

Answer: F = 130 N

Explanation: Solution:

Convert first 27 cm to m.

27 cm x 0.01 m / 1 cm = 0.27 m

Calculate the torque using T = Fd

Derive to find force F

F = T /d

= 35 N.m / 0.27 m

= 130 N

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B) The space time coordinate t of the collision in Earth's reference frame is

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Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

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The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

$x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y'=y$

$z'=z$

$t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y=y'$

$z=z'$

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

First we calculate the expression in the denominator

$\frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}$

$\sqrt{1-\frac{v^{2}}{c^{2}}} =0,714$

then we calculate t

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}$

$t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}$

$t=\frac{190s+79,388s}{0,714}$

finally we get that

$t=377,29s$

then we calculate x

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+39872396914m}{0,714}}$

$x=\frac{73872396914m}{0,714}}$

$x=103462740775,91m$

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$x \approx 103,46x10^{9} m$

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