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evablogger [386]
3 years ago
11

Two thermometers are calibrated, one in degrees Celsius and the other in degrees Fahrenheit.

Physics
1 answer:
pochemuha3 years ago
5 0

Answer:

temperature for which two scales are same is given as

x = - 40^oC

Explanation:

As we know that temperature scales are linear function

so we can compare two temperature scales by

\frac{T_c - T_f}{T_b - T_f} = constant

now we have

\frac{T_c - 0}{100 - 0} = \frac{T_f - 32}{212 - 32}

here we know that

T_c = T_f = x

so we have

\frac{x - 0}{100} = \frac{x - 32}{180}

1.8 x = x - 32

x = - 40^oC

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Topic Gravitational force amd firld strength.. help me please
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The gravitational force between <em>m₁</em> and <em>m₂</em> has magnitude

F_{1,2} = \dfrac{Gm_1m_2}{x^2}

while the gravitational force between <em>m₁</em> and <em>m₃</em> has magnitude

F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}

where <em>x</em> is measured in m.

The mass <em>m₁</em> is attracted to <em>m₂</em> in one direction, and attracted to <em>m₃</em> in the opposite direction such that <em>m₁</em> in equilibrium. So by Newton's second law, we have

F_{1,2} - F_{1,3} = 0

Solve for <em>x</em> :

\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}

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Read 2 more answers
For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103
Alona [7]

Answer:

a) P = 44850 N

b) \delta l =0.254\ mm

Explanation:

Given:

Cross-section area of the specimen, A = 130 mm² = 0.00013 m²

stress, σ = 345 MPa = 345 × 10⁶ Pa

Modulus of elasticity, E = 103 GPa = 103 × 10⁹ Pa

Initial length, L = 76 mm = 0.076 m

a) The stress is given as:

\sigma=\frac{\textup{Load}}{\textup{Area}}

on substituting the values, we get

345\times10^6=\frac{\textup{Load}}{0.00013}

or

Load, P = 44850 N

Hence<u> the maximum load that can be applied is 44850 N = 44.85 KN</u>

b)The deformation (\delta l) due to an axial load is given as:

\delta l =\frac{PL}{AE}

on substituting the values, we get

\delta l =\frac{44850\times0.076}{0.00013\times103\times 10^9}

or

\delta l =0.254\ mm

3 0
2 years ago
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