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baherus [9]
3 years ago
8

The largest single publication in the world is the 1112-volume set of British Parliamentary Papers for 1968 through 1972. The co

mplete set has a mass of 3.3 × 10^3 kg. Suppose the entire publication is placed on a cart that can move without friction. The cart is at rest, and a librarian is sitting on top of it, just having loaded the last volume. The librarian jumps off the cart with a horizontal velocity relative to the floor of 2.5 m/s to the right. The cart begins to roll to the left at a speed of 0.05 m/s. Assuming the cart’s mass is negligible,what is the librarian’s mass?
Physics
2 answers:
Marat540 [252]3 years ago
6 0

Answer:

m_l=550\ kg is the mass of librarian.

Explanation:

Given:

  • mass of the system, m_s=3.3\times 10^{3}\ kg
  • velocity of librarian relative to the ground, v_l=2.5\ m.s^{-1}
  • velocity of the cart relative to the ground, v_c=0.5\ m.s^{-1}

N<u>ow using the principle of elastic collision:</u>

Net momentum of the system is zero.

m_l\times v_l=(3300-m_l)\times v_c

m_l\times 2.5=(3300-m_l)\times 0.5

m_l=550\ kg is the mass of librarian.

Hunter-Best [27]3 years ago
3 0

Answer:

660 kg

Explanation:

Using conservation of momentum:

P_{f} =P_{i}

The initial momentum of cart + librarian is zero because at rest!

P_{i} = 0\\P_{f} = 0

The final momentum can be calculated as follows:

P_{f} = m_{publication}*v_{cart} +  m_{librarian}*v_{librarian} \\\\m_{librarian} = \frac{- m_{publication}*v_{cart}}{v_{librarian}} \\\\m_{librarian} = \frac{- 3.3*10^3*0.05}{-2.5} \\\\m_{librarian} = 660kg

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Answer:

Heating 100 g of water from 10◦C to 50◦C

Explanation:

m₁ ΔT₁ = 2000

m₂ ΔT₂ = 300

m₃ ΔT₃ = 4000

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3 0
3 years ago
What is the critical angle for the interface between water and light flint? nflint= 1.58, nwater=1.33?
Angelina_Jolie [31]
When light crosses the interface between a medium with higher refractive index and a medium with lower refractive index, there is a maximum value of the angle of incidence after which there is no refraction, but all the light is reflected, and this maximum value is called critical angle.

The critical angle is given by
\theta_C = \arcsin ( \frac{n_2}{n_1} )
where n1 is the refractive index of the first medium while n2 is the refractive index of the second medium. In our problem, n1=1.33 and n2=1.58, so the critical angle is
\theta_C = \arcsin ( \frac{1.33}{1.58} )=57.3 ^{\circ}

3 0
3 years ago
Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are
slavikrds [6]

Answer:

The correct answer is

p = p₁ + p₂

Explanation:

Newton's second law states that force = the change of momentum produced therefore since the collision is inelstic then the change of momentum of each car is p₁ and  p₂ and the force of the collition is proportional to p₁ + p₂ that is

F ∝ p₁ + p₂ and since force is directly proportional to p we have

p = p₁ + p₂

7 0
4 years ago
A sailboat took 25 hours to cover 1/4 of a journey. Then, it
ozzi

Answer:

32

Explanation:

4 0
3 years ago
Consider a air-filled parallel-plate capacitor with plates of length 8 cm, width 5.52 cm, spaced a distance 1.99 cm apart. Now i
prohojiy [21]

Answer:

The ratio of the new potential energy to the potential energy before the insertion of the dielectric is 0.58

Explanation:

Given that,

Length of plates = 8 cm

Width = 5.52 cm

Distance = 1.99 cm

Dielectric constant = 2.6

Length = 4.4 cm

Potential = 0.8 V

We need to calculate the initial capacitance

Using formula of capacitance

C=\dfrac{\epsilon_{0}A}{d}

Put the value into the formula

C=\dfrac{8.85\times10^{-12}\times8\times5.52\times10^{-4}}{1.99\times10^{-2}}

C=1.96\times10^{-12}

We need to calculate the final capacitance

Using formula of capacitance

C'=\dfrac{\epsilon_{0}A_{1}}{d}+\dfrac{k\epsilon_{0}A_{2}}{d}

Put the value into the formula

C'=(\dfrac{8.85\times10^{-12}}{1.99\times10^{-2}})((4.4\times5.52)+(3.6\times5.52)2.6)\times10^{-4}

C'=3.37\times10^{-12}

We need to calculate the  ratio of the new potential energy to the potential energy before the insertion of the dielectric

Using formula of energy

\dfrac{E}{E'}=\dfrac{\dfrac{1}{2}CV^2}{\dfrac{1}{2}C'V^2}

Put the value into the formula

\dfrac{E}{E'}=\dfrac{1.96\times10^{-12}}{3.37\times10^{-12}}

\dfrac{E}{E'}=0.58

Hence, The ratio of the new potential energy to the potential energy before the insertion of the dielectric is 0.58

4 0
3 years ago
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