Explanation:
u=166m/s, v=0(at it's highest point final velocity is zero), a=9.8m/s², t=8.6s
by the formula, S=ut+½at².
S=[166×8.6+½.×9.8×(8.6)²]. ...by calculation
S = 1427.6+362.404
S=1790.004m
hope this helps you.
Answer:
the gauge pressure at the upper face of the block is 116 Pa
Explanation:
Given the data in the question;
A cubical block of wood, 10.0 cm on a side.
height h = 1.50 cm = ( 1.50 × ( 1 / 100 ) ) m = 0.0150 m
density ρ = 790 kg/m³
Using expression for the gauged pressure;
p-p₀ = ρgh
where, p₀ is atmospheric pressure, ρ is the density of the substance, g is acceleration due to gravity and h is the depth of the fluid.
we know that, acceleration due to gravity g = 9.8 m/s²
so we substitute
p-p₀ = 790 kg/m³gh × 9.8 m/s² × 0.0150 m
= 116.13 ≈ 116 Pa
Therefore, the gauge pressure at the upper face of the block is 116 Pa
Answer:
The starting velocity for ball 1 is 1.00 meter/second. Its ending velocity is 0.25 meter/second.
The change in velocity for ball 1 is 0.25 – 1.00 = -0.75 meter/seconds
Answer:
The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet
Explanation:
The given parameters are;
The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M
The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R
The weight of the cadet on Earth = 800 N
Therefore, for the weight of the cadet on the exoplanet, W₁, we have;
The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.
D is the point where the planet moves the fastest. This is because it is in the perihelion, where the planet is moving at it’s fastest pace
