Data:
F (force) = ? (Newton)
k (<span>Constant spring force) = 50 N/m
x (</span>Spring deformation) = 15 cm → 0.15 m
Formula:

Solving:



Data:
E (energy) = ? (joule)
k (Constant spring force) = 50 N/m
x (Spring deformation) = 15 cm → 0.15 m
Formula:

Solving:(Energy associated with this stretching)




Answer:
74.86°C
Explanation:
P₂ = Vapour pressure of water at sea level = 760 mmHg
P₁ = Pressure at base camp = 296 mmHg
T₂ = Temperature of water = 373 K
ΔH°vap for H2O = 40.7 kJ/mol = 40700 J/mol
R = Gas constant = 8.314 J/mol K
From Claussius Clapeyron equation

T₁ = 347.996 K = 74.86°C
∴Water will boil at 74.86°C
Answer: 6.67 x 10^-11 m^3•kg^-1•s^-2
Answer:
19.99 kg m²/s
Explanation:
Angular Momentum (L) is defined as the product of the moment of Inertia (I) and angular velocity (w)
L = m r × v.
r and v are perpendicular to each other,
where r = lsinθ.
l = 2.4 m
θ= 34°
g = 9.8 m/s² and m = 5 kg
resolving using newtons second law in the vertical and horizontal components.
T cos θ − m g = 0
T sin θ − mw² lsin θ = 0
where T is the force with which the wire acts on the bob
w = √g / lcosθ
= √ 9.8 / 2.4 ×cos 34
= 2.2193 rad/s
the angular momentum L = mr× v
= mw (lsin θ)²
= 5 × 2.2193 (2.4 ×sin 34°)²
=19.99 kg m²/s