Question

Calculate the power of an engine required to lift 105 kg of coal per hour from a mine 360m deep. [ g=10m/s2 ]. (2)​

Answer 1

Answer:

mass=105kg,h=360m,time=3600sec

Explanation:

p=w/t

p=mgh/t

p=105*10*360/3600

p=105watt

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Answer 2

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★ Correct question :-

Calculate the power of an engine required to lift 10⁵ kg of coal per hour from a mine 360m deep. [ g=10m/s² ].

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★ Solution :-

Given ,

• Mass = 10⁵ kg
• Height = 360 m
• Acceleration due to gravity = 10 m/s²
• Time = 1 hr = 3600s

We need to find ,

• Power of engine = ?

To find, the power of engine firstly we need find the work done .

W = mgh

Where,

• W → work = ?
• m → mass = 10⁵ kg
• h → height = 360 m
• g → acceleration due to gravity = 10 m/s²

Substituting the values we have

➥ W = 10⁵ × 360 × 10

➥ W = 10⁵ × 3600 J

Now , finding power

P = W/t

Where ,

• P → power = ?
• W → work = 10⁵ × 3600
• t → time = 1 hr = 3600s

Substituting the values we have ,

⇒ P = 10⁵ × 3600 ÷ 3600

⇒ Power = 10⁵ J/s or watt

Hence , power of the engine = 10⁵ watts

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