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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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What is the dimensional formulae of "Mass per unit Area"?
Snowcat [4.5K]

Answer&Explanation:

Since dimensional formula is the way of expressing physical quantity in terms of it's dimension;

Mass per unit area will be,,

Mass=M

Area=L×L=L²

Therefore dimensional formula is=ML^-2 and not M/L²

5 0
3 years ago
What type of climate would be near cold ocean currents? PLEASE HELP QUICK!!
olya-2409 [2.1K]

Answer:

hot humid with lots of rain.

Explanation:

ocean currents act as conveyer belts of warm and cold water sending heat to the polar regions and helping the tropical areas cool off, thus influencing both weather and climate. the tropics are particularly rainy because heat absorption , and thus ocean evaporation, is highest.

5 0
4 years ago
(a) Calculate the number of free electrons per cubic meter for gold assuming that there are 1.5 free electrons per gold atom. Th
KonstantinChe [14]

Answer:

Part A:

n=8.85*10^{28}m^{-3}

Part B:

Electron Mobility=3.03*10^{-3} m^2/V

Explanation:

Part A:

To calculate the number of free electrons n we use the following formula::

n=1.5N-Au

Where N-Au is number of gold atoms per cubic meter

N-Au=\frac{Density*Avogadro Number}{atomic weight}

Density = 19.32g/cm^3

Avogadro Number=6.02*10^{23} atoms/mol

Atomic weight=196.97g/mol

So:

n=1.5*\frac{Density*Avogadro Number}{atomic weight}

n=1.5*\frac{19.32*6.02*10^{23}}{196.97}

n=8.85*10^{28}m^{-3}

Part B:

Electron Mobility=\frac{Elec-conductivity}{n * charge on electron}

n is calculated above which is 8.85*10^{28}m^{-3}

Charge on electron=1.602*10^{-19}

Elec- Conductivity= 4.3*10^{7}

Electron Mobility=\frac{4.3*10^{7}}{ 8.85*10^{28} * 1.602*10^{-19}}

Electron Mobility=3.03*10^{-3} m^2/V

4 0
4 years ago
You are a member of an alpine rescue team and must get a box of supplies, with mass 3 kg , up an incline of constant slope angle
denis23 [38]

Answer:

v = 9.04 m / s

Explanation:

For this exercise we can use the relation that the work of the non-conservative force (friction) is equal to the variation of the mechanical energy of the system.

          W = Em_f - Em₀         (1)

Starting point. Lower slope

        Em₀ = K = ½ m v²

highest point. Where is the skier at a height h

        Em_f = U = m g h

The work of rubbing

        W = -fr x

the negative sign is because the friction force opposes the movement.

Let's set a reference system where the x axis is parallel to the slope and the y axis is perpendicular

let's use trigonometry to break down the weight

        cos θ = W_y / W

        sin θ = Wₓ / W

        W_y = W cos θ

        Wₓ = W sin θ

Y axis

        N - Wₓ = 0

        N = mg sin  θ

X axis

         fr = m a

the friction force has the expression

         fr = μ N

         fr = μ mg sin θ

we look for the job

         W = - μ mg sin θ  x

where x is the distance along the slope

       

we substitute in 1

         -μ mg sin θ x = mg h - ½ m v²

let's use trigonometry to find the distance x

        tan 30 = h / x

        x = h / tan 30

we substitute

          -   \mu \ mg \ sin \theta \  \frac{h}{tan 30} \ x = m gh - ½ m v²

we use  

          tan 30 = sin30 / cos30

         

          v² = 2g h + 2 μ g h cos 30

          v = \sqrt{ 2gh \ (1+ cos 30}

let's calculate

          v = \sqrt{ 2 \ 9.8 \ 4 \ (1 + 0.05 \ cos \ 30)}

          v = 9.04 m / s

4 0
3 years ago
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