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klemol [59]
2 years ago
10

A rock is suspended by a light string. When the rock is in air, the tension in the string is 51.9 N . When the rock is totally i

mmersed in water, the tension is 31.6 N . When the rock is totally immersed in an unknown liquid, the tension is 11.4 N. What is the Density of the unknown liquid. -When I looked at this problem, I though we needed to know the volume of the rock. Can someone show me how to do it without the volume of this rock?

Physics
1 answer:
Luden [163]2 years ago
5 0

Answer:

\rho _{liquid}=1995.07kg/m^{3}

Explanation:

When the rock is immersed in unknown liquid the forces that act on it are shown as under

1) Tension T by the string

2) Weight W of the rock

3) Force of buoyancy due to displaced liquid B

For equilibrium we have T_{3}+B = W_{rock}

T_{3}+\rho _{Liquid}V_{rock}g=W_{rock}.....(\alpha)

When the rock is suspended in air for equilibrium we have

T_{1}=W_{rock}....(\beta)

When the rock is suspended in water for equilibrium we have

T_{2} + \rho _{water}V_{rock}g=W_{rock}.....(\gamma)

Using the given values of tension and solving α,β,γ simultaneously for \rho _{Liquid} we get

W_{rock}=51.9N\\31.6+1000\times V_{rock}\times g=51.9N\\\\11.4+\rho _{liquid}V_{rock}g=51.9N\\\\

Solving for density of liquid we get

\rho _{liquid}=\frac{51.9-11.4}{51.9-31.6}\times 1000

\rho _{liquid}=1995.07kg/m^{3}

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Wewaii [24]

A nitrogen laser generates a pulse containing 10.0 mj of energy at a wavelength of 340.0 nm and has 1785 x 10¹⁹ photons in the pulse.

<h3>How many photons are in the pulse?</h3>

Energy of a single photon is

E=hcλ

E=6.626×10⁻³⁴ J s×3×108 m/s /340×10⁻⁹ m

E=6.31×10⁻¹⁹  J

Number of photons in the laser is

n=Total Energy/Energy per photon

n=10⁷×10⁻³J /5.90×10⁻¹⁹J/photon

n= 1785 x 10¹⁹ photons

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6 0
1 year ago
A 30-mm-diameter copper rod is 1 m long with a yield strength of 70 MPa. Determine the axial force necessary to cause the diamet
ivolga24 [154]

Explanation:

Given data:

d = 30 mm = 0.03 m

L = 1m

S_{y} = 70 Mpa

Δd = -0.0001d

Axial force = ?

validity of elastic deformation assumption.

Solution:

O'₂ = Δd/d = (-0.0001d)/d = -0.0001

For copper,

v = 0.326      E = 119×10³ Mpa

O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶

∵δ = F.L/E.A    and σ = F/A so,

σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa

F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN

S_{y} = 70 MPa > σ = 36.5 MPa

∵ elastic deformation assumption is valid.

so the answer is

F = 25800 K N            and     S_{y} > σ

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3 years ago
A scientific theory _______.
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B. is not a validated bu experimentation

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2 years ago
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A 1-kg mass at the earth's surface weighs about? a. 1 n. b. 5 n. c. 10 n. d. 12 n. e. none of these
ValentinkaMS [17]

A 1-kg mass at the earth's surface weighs about C. 10N

The third planet from the Sun is the Earth. It is the seventh largest in terms of size and weighs roughly 5.98 1024 kg. The inherent quality of mass is unaffected by the environment of the object or the technique employed to quantify it.

Newton's law of gravitation can be used to estimate the mass of the Earth. This is set to the fundamental formula, which reads: force (F) = mass (m) times acceleration. Gravitational acceleration (G) is equal to 9.8 m/s2, the Earth's radius is 6.37 106 m, and the gravitational constant (G) is 6.673 1011 Nm2/kg2. The Earth has a mass of 5.96 1024 kg after rearranging the equation and entering all the numbers.

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1 year ago
The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou
Inessa05 [86]

Answer:

a) 4.40 s

b) 2.20 s

Explanation:

Given parameters are:

At constant power  ,

initial speed of the car, v_0=0

final speed of the car, v=32 mph

At full power,

initial speed of the car, v_0=0

final speed of the car, v=64 mph

a)

At constant power, KE = \frac{1}{2} mv^2

At full power, KE = \frac{1}{2} m(2v)^2

So KE_f = 4KE_i

So, time to reach 64 mph speed is 4 times more than the initial time

t = 4*1.10 =4.40 s

b)

v=v_0+at\\a=\frac{v-v_0}{t}=\frac{32-0}{1.1/3600}=104727.27 miles/hours^2

For final 64 mph speed,

v=v_0+at\\t=\frac{v-v_0}{a}=\frac{64-0}{104727.27} = 6.111*10^{-4} hours = 6.111*10^{-4}*3600=2.20 s

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