Based on the situation above the the work done was 400 Joules. <span>Q = FS cos(theta) is the so-called work function. It's important to learn the work physics; you'll see it over and over in science/physics class. Theta is the angle between the force vector F and the distance vector S. In your problem we assume theta = 0, the two vectors were assumed aligned.</span>
It takes 2 years of graduate study in a subfeild, practicum experience, and a thesis.
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Answer:
Trial 1 is the largest, trial 3 is the smallest
Explanation:
Given:
<em>Trial 1</em>
M₁ = 6·10²² kg
d₁ = 3 500 km = 3.5·10⁶ м
<em>Trial 2</em>
M₂ = 6·10²² kg
d₂ = 7 000 km = 7·10⁶ м
<em>Trial 3</em>
M₃ = 3·10²² kg
d₃ = 7 000 km = 7·10⁶ м
___________
F - ?
Gravitational force:
F₁ = G·m·M₁ / d₁² = m·6.67·10⁻¹¹·6·10²² / (3.5·10⁶)² = 0.37·m (N)
F₂ = G·m·M₂ / d₂² = m·6.67·10⁻¹¹·6·10²² / (7·10⁶)² = 0.08·m (N)
F₃ = G·m·M₃ / d₃² = m·6.67·10⁻¹¹·3·10²² / (7·10⁶)² = 0.04·m (N)
Trial 1 is the largest, trial 3 is the smallest
<span>The average weather of a particular place is "Climate"
In short, Your Answer would be Option B
Hope this helps!</span>
Answer:
Given a tube of diameter d, = 3cm = 0.03m
Pressure Balance
Mercury pressure at the tube bottom Pₓ = Pa + ρgh
where
Pa = Atmospheric pressure = 101kpa
ρ = Density of mercury = 13,546kg/m3
g = acceleration due to gravity
h = height of the tube?
Given
Bottom pressure in excess of the atmospheric pressure = 48kPa = Pₓ - Pa
Therefore, 48kPa = ρgh
h = 48(kN/m2)/ρg
h = 48,000kgms⁻²m⁻²/(13546kgm⁻³ x 9.81ms⁻²)
h = 0.36m
the tube is 36cm tall