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Rus_ich [418]
3 years ago
6

A 1.60 m tall person lifts a 2.10-kg book from the ground so it is 2.20 m above the ground. What is the potential energy of the

book relative to (a) the ground and (b) the top of the person’s head? How is the work done by the person related to the answers in parts (a) and (b)?
Physics
1 answer:
Viefleur [7K]3 years ago
6 0

Answer:

Explanation:

Given

mass of book(m)=2.1 kg

height up to which book is lifted is (h)2.2 m

height of person (h_0)1.6 m

Potential energy of book relative to ground=mgh

PE=2.1\times 9.8\times 2.2=45.276 J

(b)PE w.r.t to person head =mg(h-h0)

=2.1\times 9.8\times (2.2-1.6)=12.348 J

work done by person in lifting box 2.2 m w.r.t floor

Word done =Potential Energy of box relative to floor=45.2 J

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Match the events related to the formation of the universe with the stages during which they occurred.
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3 years ago
After being struck by a bowling ball, a 1.8 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.8 kg
kaheart [24]

Answer:

a) v₂ = 4.2 m/s

b) v₂ = 5 m/s

Explanation:

a)

We will use the law of conservation of momentum here:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0.8 m/s

v₂ = speed of second after before collsion = ?

Therefore,

(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0.8\ m/s)+(1.8\ kg)(v_2)\\v_2 = 5\ m/s - 0.8\ m/s

<u>v₂ = 4.2 m/s</u>

<u></u>

b)

We will use the law of conservation of momentum here:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0 m/s

v₂ = speed of second after before collsion = ?

Therefore,

(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0\ m/s)+(1.8\ kg)(v_2)

<u>v₂ = 5 m/s</u>

5 0
3 years ago
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