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Alexeev081 [22]
3 years ago
7

How to reduce friction of moving a heavy load .

Physics
1 answer:
Vilka [71]3 years ago
3 0
● Use lubricants such as oil.
● Place the load on a piece of cloth as it can ease the friction
● Place the load on a cart with wheels so it can be converted to rolling friction and can be pushed easily.
● Place the load on an inclined plane
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A 3 kg model airplane is traveling at a speed of 33 m/s. The operator then increases the speed up to 45 m/s in 2 seconds. How mu
nlexa [21]

Answer:

Force = 18N

Explanation:

Force = [ mass ( final velocity - initial velocity ) ] / time taken

using the formula, here mass is 3 kg, final velocity = 45 m/s , initial velocity = 45 m/s , time taken = 2 seconds

Force = [ 3 ( 45 - 33 ) ] / 2

Force = 18N

8 0
2 years ago
If a lever has an input arm of 80 cm and an output arm of 20 cm, what is its ideal mechanical advantage?
Alexeev081 [22]

Answer:

16

Explanation:

6 0
1 year ago
In a crash or collision, why is it advantageous for an occupant to extend the time during which the collision is taking place?
Veseljchak [2.6K]

Answer:

Explanation:

During a car collision momentum of vehicle ceases within a fraction of seconds so Force due to the impulse is huge.

Impulse is defined as the product of average force and time. If we can increase the period of collision for the same impulse then the average force imparted will be less.

If we can increase the time period then damage due to collision will be less.

                     

3 0
3 years ago
Animals need oxygen to complete the many chemical reactions taking place in the body. An animal inhales oxygen and exhales _____
mezya [45]
Carbon dioxide and nitrogen
5 0
3 years ago
Read 2 more answers
How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe
Rus_ich [418]

Explanation:

(a)  For an isothermal process, work done is represented as follows.

             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

              = 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c)   We know that for an isothermal process,

               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

                    = 11 atm

Hence, the required pressure is 11 atm.

(d)   For adiabatic process,  

          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0
3 years ago
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