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user100 [1]
3 years ago
8

Identifying the guilty party was mainly based on eyewitness accounts during what time period?

Physics
2 answers:
kondaur [170]3 years ago
8 0

The time period for guilty party was between 1900-1988.

laiz [17]3 years ago
5 0

Answer:

1900-1988

Explanation:

This was the time period during which identifying the guilty party was mainly based on eyewitness accounts. The use of eyewitness accounts became common as witnesses often provided information that no one else could provide. Moreover, witnesses were believed to be unbiased. However, this is not the main method used to identify the guilty party anymore.

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Suppose that the Sun started shrinking in size, without losing any mass. What would be the effect of the Sun's change on the orb
adell [148]

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F = G M m / R^2 gravitational force on planet of mass m.

None of these quantities change in the given hypothesis so

there will be no change in the orbit of mass m

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2 years ago
Give an example of a situation in which you would describe an
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3 years ago
Un alambre de teléfono de 120m de largo y de 2.2mm de diámetro se estira debido a una fuerza de 380 N cual es el esfuerzo longit
sergij07 [2.7K]

Respuesta: verifique amablemente la explicación

Explicación:

Dado lo siguiente:

Longitud (L) del cable = 120 m

Diámetro (d) = 2,2 mm (2,2 / 1000) = 2,2 * 10 ^ -3 m

Fuerza (F) = 380 N

Esfuerzo longitudinal = Fuerza / Área

Área = πd² / 4 = (π * (2.2 * 10 ^ -3) ^ 2) / 4

Área = (3.142 * 4.84 * 10 ^ -6)

Área = 0.00000380132 m²

Estrés = Fuerza / Área

Estrés = 380 / 0.00000380132

Esfuerzo longitudinal = 99952128.12 = 9.9952128 * 10^7 Nm^-2

Deformación longitudinal: extensión / longitud

Extensión = 0.10 m

Longitud = 120 m

Deformación longitudinal = 0,1 m / 120 m

Deformación longitudinal = 0.0008333 = 8.33 × 10 ^ -4

6 0
2 years ago
A 43.9-g piece of copper (CCu= 0.385 J/g°C) at 135.0°C is plunged into 254 g of water at 39.0°C. Assuming that no heat is lost t
Semmy [17]

Answer:

T = 40.501\,^{\textdegree}C

Explanation:

The interaction of the piece of copper and water means that the first one need to transfer heat in order to reach a thermal equilibrium with water. Then:

-Q_{out,Cu} = Q_{in,H_{2}O}

After a quick substitution, the expanded expression is:

-(43.9\,g)\cdot (0.385\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-135^{\textdegree}C) = (254\,g)\cdot (4.187\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-39\,^{\textdegree}C)

-16.902\,\frac{J}{^{\textdegree}C}\cdot (T-135^{\textdegree}C) = 1063.498\,\frac{J}{^{\textdegree}C} \cdot  (T-39^{\textdegree}C)

43758,192\,J = 1080.4\,\frac{J}{^{\textdegree}C}\cdot T

The final temperature of the system is:

T = 40.501\,^{\textdegree}C

8 0
2 years ago
Read 2 more answers
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