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Alekssandra [29.7K]

3 years ago

8

A solid sphere of mass 8.6 kg, made of metal whose density is 3,400 kg/, hangs by a cord. When the sphere is immersed in a liqui

d of unknown density, the tension in the cord is 38 N. The density of the liquid is closest to which one of the following values?A) 1900 kg/m³B) 1600 kg/m³C) 1700 kg/m³D) 1400 kg/m³E) 1500 kg/m³

1 answer:

creativ13 [48]3 years ago

4 0

Answer:

A. $1900 kg/m^3$

Explanation:

We are given that

$m=8.6 kg$

Density, $\rho_s=3400 kg/m^3$

Tension,T=38 N

We have to find the density of liquid.

$T=mg-\rho_l Vg$

$g=9.8 m/s^2$

Volume,V= $\frac{m}{\rho_s}$

$38=8.6\times 9.8-\rho_l\times \frac{8.6}{3400}\times 9.8$

$\rho_l\times \frac{8.6}{3400}\times 9.8=8.6\times 9.8-38$

$\rho_l=\frac{(8.6\times 9.8-38)\times 3400}{8.6\times 9.8}$

$\rho_l=1867kg/m^3\approx 1900 kg/m^3$

Option A is true.

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A 60.0 kg girl stands up on a stationary floating raft and decides to go into shore. She dives off the 180 kg floating raft with

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Momentum, p = m.v
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A) Momentum of the girl as she is diving:
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B) momentum of the raft = - momentum of the girl = -24.0 N/s

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3 years ago

One mole of iron (6 x 10^23 atoms) has a mass of 56 grams, and its density is 7.87 grams per cubic centimeter, so the center-to-

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Answer:

Explanation:

Given that:

length l = 2.3 m

a = 0.12 cm = $0.12 \times 10^{-2} \ m$

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$\text{Finally; the stiffness of a single interatomic spring is:}$

$k_{si} =\dfrac{N_{bond}}{N_{chain}}\times k_s$

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3 years ago

You have been hired as a technical consultant for an early-morning cartoon series for children to make sure that the science is

katen-ka-za [31]

The initial potential energy of the wagon containing gold boxes will enable

it roll down the hill when cut loose.

The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

$K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}$

Where;

v = The velocity of the wagon a the bottom of the cliff

Therefore;

$\dfrac{1}{2} \times 166 \times v^2 = 17457.0912$

$v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5$

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

$\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2) }}$

Which gives;

$\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66) } \approx 8.1$

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$Time = \dfrac{Distance}{Velocity}$

$\mathrm{The \ time \ the\ Lone \ Ranger \ and \ Tonto \ have, \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s$

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

Learn more here:

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