Suppose that 2 J of work are needed to stretch a spring from its natural length of 30 cm to a length of 42 cm. How far (in cm) b
1 answer:
A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched
<u>Explanation:</u>
Work, W = 2 J
Initial distance, = 30 cm
Final distance, = 42 cm
Force, F = 30 N
Stretched length, x = ?
We know,
W = 1/2 kΔx²
Δx = 42-30 cm = 12 cm = 0.12 m
2 J = 1/2 k X (0.12)²
k = 277.77 N/m
According to Hooke's law,
F = kx
30 N = 277.77 X x
x = 0.108 m
x = 10.8 cm
A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched.
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Answer:
Explanation:
The gravitational pull between two object is given by:
Where M and m are the masses of the object, r is the distance between the masses and G = 6.67× 10⁻¹¹ m³kg⁻¹ s⁻² is the gravitational constant.
We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.
Mass of Earth, Me = 5.98 × 10²⁴ kg
Mass of Venus, Mv = 0.815 Me
Mass of Jupiter, Mj = 318 Me
Mass of Saturn, Ms = 95.1 Me
closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU
closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU
closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU
where 1 AU = 1.5 × 10¹¹ m
Inserting the values:
Answer:
W / n = - 9133 J / mol, W / n = 3653 J / mol , e = 0.600
Explanation:
The Carnot cycle is described by
In this case they indicate that the final volume is
V = 3V₀
In the part of the heat absorption cycle from the source is an isothermal expansion
W = n RT ln (V₀ / V)
W / n = 8.314 1000 ln (1/3)
W / n = - 9133 J / mol
During the part of the isothermal compression in contact with the cold focus, as in a machine the relation of volumes is maintained in this part is compressed three times
W / n = 8.314 400 (3)
W / n = 3653 J / mol
The efficiency of the cycle is
e = 1- 400/1000
e = 0.600