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Alekssandra [29.7K]

1 year 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]1 year 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 metal has a strength of 414 MPa at its elastic limit and the strain at that point is 0.002. Assume the test specimen is 12.8-m

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To solve this problem, we will start by defining each of the variables given and proceed to find the modulus of elasticity of the object. We will calculate the deformation per unit of elastic volume and finally we will calculate the net energy of the system. Let's start defining the variables

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Yield Strain of the Specimen

$\epsilon_{el} = 0.002$

Diameter of the test-specimen

$d_0 = 12.8mm$

Gage length of the Specimen

$L_0 = 50mm$

Modulus of elasticity

$E = \frac{S_{el}}{\epsilon_{el}}$

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Considering that the net strain energy of the sample is

$U_{el} = U_{el}' \cdot (\text{Volume of sample})$

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1 year ago

You are called as an expert witness to analyze the following auto accident: Car B, of mass 2100 kg, was stopped at a red light w

Artemon [7]

Answer:

Explanation:

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work done by friction = 13377 x 7.30 = 97652.1 J

If v be the common velocity of both the cars after collision

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so , applying work - energy theory ,

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v = 7.365 m /s

This is the common velocity of both the cars .

To know the speed of car A , we shall apply law of conservation of momentum .Let the speed of car A before collision be v₁ .

So , momentum before collision = momentum after collision of both the cars

1500 x v₁ = ( 1500 + 2100 ) x 7.365

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( b )

yes Car A was crossing speed limit by a difference of

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7 0

1 year ago

In coordinates with the origin at the barn door, the cow walks from x 0 to x 6.9 m as you apply a force with x component Fx 320.

Stella [2.4K]

Answer:

-209.42J

Explanation:

Here is the complete question.

A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component Fx=−[20.0N+(3.0N/m)x]. How much work does the force you apply do on the cow during this displacement?

Solution

The work done by a force W = ∫Fdx since our force is variable.

Since the cow moves from x₁ = 0 m to x₂ = 6.9 m and F = Fx =−[20.0N+(3.0N/m)x] the force applied on the cow.

So, the workdone by the force on the cow is

W = ∫₀⁶°⁹Fx dx = ∫₀⁶°⁹−[20.0N+(3.0N/m)x] dx

= ∫₀⁶°⁹−[20.0Ndx - ∫₀⁶°⁹(3.0N/m)x] dx

= −[20.0x]₀⁶°⁹ - [3.0x²/2]₀⁶°⁹

= -[20 × 6.9 - 20 × 0] - [3.0 × 6.9²/2 - 3.0 × 0²/2]

= -[138 - 0] - [71.415 - 0] J = (-138 - 71.415) J

= -209.415 J ≅ -209.42J

5 0

1 year ago