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Korvikt [17]
3 years ago
11

SHOW ADEQUATE WORKINGS IN THIS SECTION

Physics
1 answer:
cluponka [151]3 years ago
3 0

Answer:

12 i. The work done by Wale = 107.910 kJ

The work done by Lekan = 117.720 kJ

Total work done = 225.36 kJ

ii. Wale's power =  4.3164 kW

Lekan's power = 3.924 kW

Wale has more power and is more powerful than Lekan

13. 313.92 N

Explanation:

i. The work done, W = Force, F × Distance moved by the force, D

The given parameters are

The mass of Wale = 55 kg

The mass of Lekan = 60 kg

The acceleration due to gravity, g =9.81 m/s²

The motion force of Wale and Lekan are;

Motion force of Wale = 9.81 × 55 = 539.55 N

Motion force of Lekan = 9.81 × 60 = 588.6 N

The work done by Wale = 539.55 × 200 = 107910 J = 107.910 kJ

The work done by Lekan= 588.6 × 200 = 117720 J = 117.720 kJ

107910 + 117720 =225630 J = 225.36 kJ

ii. Power = Work done/time

Wale finished the race in 25 s, therefore, his power = 107910/25 = 4316.4 W

Lekan finished the race in 30 s, therefore, his power = 117720/30 = 3924 W

Wale has more power and is more powerful than Lekan

13. The velocity ratio = 5

V. R. = Distance moved by effort/(Distance moved by load)

Efficiency = 80%

Work done by effort = x

Work done by machine = Efficiency × Work done by effort  = 0.8 × x

Distance moved by effort, E = V. R. × Distance moved by load, D = 5 × D

Work done by effort = Force × Distance moved = 200×9.81× E

Work done by effort = 1962×E = 1962×E = 1962×5×D

Work done by machine = 1962 × D, when D = 1, we have;

0.8 × 1962×1 = 1569.6 J

Work done by effort = Force × Distance moved

Work done by effort = Force × 5×D = Force × 5 (D = 1)

From the principle of conservation of energy, we have;

Energy is neither created nor destroyed

Therefore

Work done by effort = Force × 5 = 1569.6 J

Force = 1569.6 /5 = 313.92 N.

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Answer:

The water is stored in ice sheets and as snow

Explanation:

Temperature reduces with an increase in altitudes. The standard laps rate is 6.5°C per 1,000 m gained in elevation

At very high elevations, therefore,  the air is usually very cold such that when an elevation of 4,500 meters is reached at the equator, it is possible to observe snowfall and the water remain temporarily stored on the surface of the mountain as ice and snow

7 0
3 years ago
a 2000 kg truck is moving eastward at 25 m/s. it collides inelastically with a 1500 kg truck traveling southward at 30 m/s. they
Otrada [13]

According to the given statement Final velocity when they stick together is 8.735i^ + 11.25j^​

<h3>What is collision and momentum?</h3>

The unit of momentum is kg ms -1. Momentum is a vector parameter that is influenced by the object's direction. During collisions involving objects, momentum is a relevant concept. The final velocity before a collision between two objects equals the total motion after the impact (in the absence of external forces).

<h3>Briefing:</h3>

From conservation of momentum

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m u +M U =(m+M) V

2000×25 i^ +1500×30 j^​ =(2000+1500) V

V = 8.735i^ + 11.25j^​

Final velocity when they stick together is 8.735i^ + 11.25j^​

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The complete question is -

A 2000 kg truck is moving eastward at 25 m/s. it collides inelastically with a 1500 kg truck traveling southward at 30 m/s. they collide at the intersection. Find the direction and magnitude of velocity of the wreckage after the collision, assuming the vehicles stick together after the collision.

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1 year ago
You leave the doctor's office after your annual checkup and recall that you weighed 688 N in her office. You then get into an el
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Answer:

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a=1.283\ m.s^{-2} downwards

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Given:

weight of the person, w=688\ N

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m=\frac{688}{9.81}

m=70.132\ kg

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<u>Then the difference between the two weights is :</u>

\Delta w=w_a-w

\Delta w=726-688

\Delta w=38\ N is the force that acts on the body which generates the acceleration.

Now the corresponding acceleration:

a=\frac{\Delta w}{m}

a=\frac{38}{70.132}

a=0.5418\ m.s^{-2} upwards, because the normal reaction that due to the weight of the body is increased here.

  • Now if the apparent weight in the elevator, w_a= 598\ N

<u>Then the difference between the two weights is :</u>

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a=\frac{\Delta w}{m}

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Answer:

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Explanation:

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