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Gre4nikov [31]
2 years ago
11

An object which has a mass of 70 kg is sitting on a cliff 10 m high. Calculate the object's Potential energy. Given g = 10m/s2

Physics
1 answer:
Rina8888 [55]2 years ago
6 0
Answer is a.)

Given: Mass m = 70 Kg   Height  h = 10 m;        g = 10 m/s²

Required. Gravitational Potential Energy or P.E

Formula: P.E = mgh

                     = (70 Kg)(10 m/s²)( 10 m)

              P.E = 7,000 Kg.m²/s² or

              P.E = 7,000 J

     
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v= 2413.5 m/s

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[since the 80% of mass which is fuel is exhausted]

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timofeeve [1]

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ωf = ω₀ + α*t

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The angular velocity, and the linear speed, are related by the following expression:

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Applying the definition of linear acceleration (tangential acceleration in this case) and angular acceleration, we can find a similar relationship between the tangential and angular acceleration, as follows:

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In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co
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Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m

a)

The final velocity of cart 1 after collision is given as:

v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\  Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s

The final velocity of cart 2 after collision is given as:

v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\  Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s

b) Using the law of conservation of energy:

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