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Schach [20]
2 years ago
14

Lydia is often described as having a positive outlook on life. She assumes the best of people and situations. Lydia exemplifies

________.
Physics
2 answers:
HACTEHA [7]2 years ago
5 0
Lydia is an example of an optimist. Optimists always look at the positive side of a particular situation. They believe in a positive result. Optimist always see "the bright side" as opposed to the pessimist. The word optimism comes from latin word "optimum" which means "the best".

Artyom0805 [142]2 years ago
5 0

Answer:

<h2>Optimism.</h2>

Explanation:

<em>Optimism could be defined as a spiritual and mental disposal for positive outcomes</em>, refers to the best future coming, the opposite of pessimism.

A optimistic person is who have a positive view of situation, even if it isn't seem like that, even these kind of people see positiveness in others, like Lydia.

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The potential energy for a certain mass moving in one dimension is given by U(x)=(2.0J/m3)x3−(15J/m2)x2+(36J/m)x−23JU(x)=(2.0J/m
Angelina_Jolie [31]

Answer:x=2 and x=3

Explanation:

Given

Potential Energy for a certain mass is

U(x)=2x^3-15x^2+36x-23

and we know force is given by

F=-\frac{\mathrm{d} U}{\mathrm{d} x}

F=-(2\times 3x^2-15\times 2x+36)

For Force to be zero F=0

\Rightarrow 6x^2-30x+36=0

\Rightarrow x^2-5x+6=0

\Rightarrow x^2-2x-3x+6=0

\Rightarrow (x-2)(x-3)=0

Therefore at x=2 and x=3 Force on particle is zero.

8 0
2 years ago
2.27 A gas is compressed from V1= 0.3 m3, p1=1 bar to V2= 0.1 m3, p2 =3 bar. The pressure and
Georgia [21]

Answer:

-40 kJ

80 kJ

Explanation:

Work is equal to the area under the pressure vs volume graph.

W = ∫ᵥ₁ᵛ² P dV

2.27) Pressure and volume are linearly related.  When we graph P vs V, the area under the line is a trapezoid.  So the work is:

W = ½ (P₁ + P₂) (V₂ − V₁)

W = ½ (100 kPa + 300 kPa) (0.1 m³ − 0.3 m³)

W = -40 kJ

2.29) Pressure and volume are inversely proportional:

pV = k

The initial pressure and volume are 500 kPa and 0.1 m³.  So the constant is:

(500) (0.1) = k

k = 50

The final pressure is 100 kPa.  So the final volume is:

(100) V = 50

V = 0.5

The work is therefore:

W = ∫ᵥ₁ᵛ² P dV

W = ∫₀₁⁰⁵ (50/V) dV

W = 50 ln(V) |₀₁⁰⁵

W = 50 (ln 0.5 − ln 0.1)

W ≈ 80 kJ

5 0
2 years ago
A rectangular block weighs 240 N. the area of the block in contact with the floor is 20 cm2.calculate the pressure on the floor(
maks197457 [2]

Answer:

12 N/cm²

Explanation:

From the question given above, the following data were obtained:

Weight (W) of block = 240 N

Area (A) = 20 cm²

Pressure (P) =?

Next, we shall determine the force exerted by the block. This can be obtained as follow:

Weight (W) of block = 240 N

Force (F) =.?

Weight and force has the same unit of measurement. Thus, we force applied is equivalent to the weight of the block. Thus,

Force (F) = Weight (W) of block = 240 N

Force (F) = 240 N

Finally, we shall determine the pressure on the floor as follow:

Force (F) = 240 N

Area (A) = 20 cm²

Pressure (P) =?

P = F/A

P = 240 / 20

P = 12 N/cm²

Therefore, the pressure on the floor is 12 N/cm².

7 0
2 years ago
Two long straight wires enter a room through a window. One carries a current of 2.9A into the room, while the other carries a cu
Degger [83]

Answer and Explanation:

curents i = 2.9 A

           i ' = 4.4 A

the magnitude (in T.m) of the path integral of B.dl around the window frame = μo * current enclosed

          = μo* ( i '- i )

Since from Ampere's law

where μ o = permeability of free space = 4π * 10 ^-7 H / m

plug the values we get the magnitude (in T.m) of the path integral of B.dl = ( 4π*10^-7 ) (2.9+4.4)

                                 = 1.884 * 10^-6 Tm

4 0
2 years ago
If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g
EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

F_G = G\frac{M_1M_2}{R^2}

where G =6.67408 × 10^{-11} m^3/kgs^2 is the gravitational constant on Earth. M_1, M_2 are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}

\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}

\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}

\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2

Since M_2 = 2m_2 and r = R/3

\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5

So gravity would have been decreased by a factor of 4.5  

8 0
2 years ago
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