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3 years ago

Why would adding oil to the moving parts of a machine increase its efficiency?

Physics

1 answer:

luda_lava [24]3 years ago

5 0

Adding oil reduces friction between the moving parts
Less energy is used

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The nearest star to the Earth is the red dwarf star Proxima Centauri, at a distance of 4.218 light-years. Convert this distance

alukav5142 [94]

Explanation:

The nearest star to the Earth is the red dwarf star Proxima Centauri, at a distance of 4.218 light-years.

Light year is the unit of distance covered by the heavenly bodies. 1 light year is equal to :

$1\ light\ year=3.72\times 10^{17}\ inches$

So, $4.218\ light\ year=4.218\times 3.72\times 10^{17}\ inches$

$4.218\ light\ year=1.56\times 10^{18}\ inches$

We need to convert 4.218 light-years barley corns.

Since, 1 barleycorn = 1/3 inch

$1\ inch=3\ barleycorn$

$1.56\times 10^{18}\ inches=3\times 1.56\times 10^{18}=4.68\times 10^{18}\ barleycorn$

So, the nearest star to the Earth is at a distance of $4.68\times 10^{18}\ barleycorn$ . Hence, this is the required solution.

3 0

3 years ago

A 30 kg box is placed on a skateboard and pushed across the room with a velocity of 4 m/s. The pusher applies more force, increa

Reptile [31]

Answer:

Dawg

Explanation:

Oh yea actually u suck

3 0

3 years ago

1. Applying _____ indicates that layers were repositioned from a flat-lying orientation.

saveliy_v [14]

Answer:

1. the principle of original horizontality

2. the principle of crosscutting relationships

3. the law of superposition

4. older

5. younger

5 0

3 years ago

The moment of inertia for a 5500 kg solid disc is 12100 kg-m^2. Find the radius of the disc? (a) 2.111 m (b) 2.579 m (c) 1.679

Soloha48 [4]

Answer:

The radius of the disc is 2.098 m.

(e) is correct option.

Explanation:

Given that,

Moment of inertia I = 12100 kg-m²

Mass of disc m = 5500 kg

Moment of inertia :

The moment of inertia is equal to the product of the mass and square of the radius.

The moment of inertia of the disc is given by

$I=\dfrac{mr^2}{2}$

Where, m = mass of disc

r = radius of the disc

Put the value into the formula

$12100=\dfrac{5500\times r^2}{2}$

$r=\sqrt{\dfrac{12100\times2}{5500}}$

$r= 2.098\ m$

Hence, The radius of the disc is 2.098 m.

8 0

3 years ago

By what factor will the Electrostatic Force between two charged objects change when the amount of charge on both objects doubles

Mademuasel [1]

Answer:

F' = (4/9)F

Explanation:

The electrostatic force between two charged objects is given by Coulomb's Law:

F = kq₁q₂/r² -------------------- equation (1)

where,

F = Electrostatic Force

k = Coulomb's Constant

q₁ = magnitude of first charge

q₂ = magnitude of second charge

r = distance between charges

Now, when the charges and distance altered as follows:

q₁' = 2q₁

q₂' = 2q₂

r' = 3r

Then,

F' = kq₁'q₂'/r'²

F' = k(2q₁)(2q₂)/(3r)²

F' = (4/9)kq₁q₂/r²

using equation (1):

7 0

3 years ago