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saw5 [17]
2 years ago
9

HELP PLEASE

Engineering
1 answer:
Sveta_85 [38]2 years ago
8 0

..........23÷357=0.0644257703........

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An object whose mass is 251 kg is located at an elevation of 24 m above the surface of the earth. For g-9.78 ms, determine the g
Harlamova29_29 [7]

Answer:

Gravitational Potential =58.914 KJ

Explanation:

We know that

Gravitational Potential Energy = mass\times g\times Height

Given mass = 251 kg

Height= 24 m

g is acceleration due to gravity = 9.78m/s^{2}

Applying values in the equation we get

Gravitational Potential Energy=251X9.78X24 Joules

Gravitational Potential Energy=58914.72 Joules

Gravitational Potential Energy =\frac{58914.72}{1000}KJ= 58.914KJ

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A mass of 5 kg of saturated water vapor at 100 kPa is heated at constant pressure until the temperature reaches 200°C.
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g Let the charges start infinitely far away and infinitely far apart. They are placed at (6 cm, 0) and (0, 3 cm), respectively,
irina1246 [14]

Answer:

a) V =10¹¹*(1.5q₁ + 3q₂)

b) U = 1.34*10¹¹q₁q₂

Explanation:

Given

x₁ = 6 cm

y₁ = 0 cm

x₂ = 0 cm

y₂ = 3 cm

q₁ = unknown value in Coulomb

q₂ = unknown value in Coulomb

A) V₁ = Kq₁/r₁

where   r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m

V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁

V₂ = Kq₂/r₂

where   r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m

V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂

The electric potential due to the two charges at the origin is

V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)

B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows

U = Kq₁q₂/r₁₂

where

r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m

then

U = 9*10⁹q₁q₂/(3√5/100)

⇒ U = 1.34*10¹¹q₁q₂

5 0
2 years ago
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