Answer:
Danny hits the water with kinetic energy of 5000 J.
Explanation:
Given that,
The Weight of Danny Diver,
F = 500 N
m*g= 500 N
He steps off a diving board 10 m above the water.
h=10 m
when Danny diver hits water he generates the kinetic energy.
We need to find the kinetic energy of the water.
Let kinetic energy is K.
K = m*g*h
Where g is acceleration due to gravity.
that g= 9.8 m/s^2
now substituting the values in above equation
K= (500) * 10
K= 5000 J
Hence,
he hits the water with kinetic energy of 5000 J.
Learn more about Kinetic energy here:
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Winds transfer energy in the form heat from the air to the ground!
Explanation:
Initial velocity is 25m/s i.e. u. Final velocity would be zero since the bus stopped. It is noted that the bus deccelartion of 4m/s, so the acceleration should be considered negative of 4m/s I.e. 4m/s. So using third law of motion i.e.
v^2 - u^2= 2as
(0)^2 - (25)^2= -2×4×s {Note here I used minus sign in acceleration}
-625 = -8×s
s=625 ÷ 8
s= 125 m
So bus would cover a distance of 125m before coming to rest.
Answer:
89.6 cm
Explanation:
From the question,
Volume of the rectangular object = Mass/Density.
V = m/D.................. Equation 1
Given: m = 1.278 kg, D = 4.98 g/cm³ = 4980 kg/m³
Substitute into equation 1
V = 1.278/4980
V = 2.57×10⁻⁴ m³.
But,
V = lwh............... Equation 2
Where l = length of the rectangular object, w = width of the rectangular object, h = height of the rectangular object.
make h the subject of the equation
h = V/lw........... Equation 3
Given: V = 2.57×10⁻⁴ m³, l = 0.047 m, w = 0.061 m.
Substitute into equation 3
h = 2.57×10⁻⁴/(0.047×0.061)
h = 0.896 m
h = 89.6 cm
The trickiest part of this problem was making sure where the Yakima Valley is.
OK so it's generally around the city of the same name in Washington State.
Just for a place to work with, I picked the Yakima Valley Junior College, at the
corner of W Nob Hill Blvd and S16th Ave in Yakima. The latitude in the middle
of that intersection is 46.585° North. <u>That's</u> the number we need.
Here's how I would do it:
-- The altitude of the due-south point on the celestial equator is always
(90° - latitude), no matter what the date or time of day.
-- The highest above the celestial equator that the ecliptic ever gets
is about 23.5°.
-- The mean inclination of the moon's orbit to the ecliptic is 5.14°, so
that's the highest above the ecliptic that the moon can ever appear
in the sky.
This sets the limit of the highest in the sky that the moon can ever appear.
90° - 46.585° + 23.5° + 5.14° = 72.1° above the horizon .
That doesn't happen regularly. It would depend on everything coming
together at the same time ... the moon happens to be at the point in its
orbit that's 5.14° above ==> (the point on the ecliptic that's 23.5° above
the celestial equator).
Depending on the time of year, that can be any time of the day or night.
The most striking combination is at midnight, within a day or two of the
Winter solstice, when the moon happens to be full.
In general, the Full Moon closest to the Winter solstice is going to be
the moon highest in the sky. Then it's going to be somewhere near
67° above the horizon at midnight.
