The total energy of the system is 88.2 J
Explanation:
The total (mechanical) energy of an object is given by:
where
PE is the potential energy
KE is the kinetic energy
At the starting point, the object is at rest: this means that its kinetic energy is zero, so all its energy is potential energy, which is given by
where
m is the mass of the object
g is the acceleration of gravity
h is the height of the object
In this problem, we have (at point A):
m = 3.0 kg
h = 3.0 m
Substituting, we find:
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Metamorphic rocks form from the alteration of other rocks through pressure and temperature induced changes in the minerals.
Here is the highly detailed, arcane, complex, technical form of Ohm's Law that is needed in order to answer this question ===> V = I · R .
Voltage = (current) x (resistance)
Voltage = (2 Ampere) x (8 Ω)
<em>Voltage = 16 volts</em>
Answer:
Explanation:
Given that
The window height is 2m
And the window is 7.5m from the ground
Then the total height of the window from the ground is 7.5+2=9.5m
It takes the ball 0.32sec travelled pass the window.
When the ball get to the window, it has an initial velocity (u') and when it gets to the top of the window it has a final velocity ( v')
Now using the equation of free fall during this window travels
S=ut-½gt² against motion.
S=2, g=9.81, t=0.32sec
Then,
S=u't-½gt²
2=u'×0.32-½×9.81×0.32²
2=0.32u'-0.5023
2+0.5032=0.32u'
Then, 0.32u'=2.5032
u'=2.5032/0.32
u'=7.82m/s
This is the initial velocity as the ball got the the window
Now, let analyse from the window bottom to the ground which is a distance of 7.5m
Using the equation of free fall again
v²=u²-2gH
In this case the final velocity (v) is the velocity when the ball reach the bottom of the window i.e u'=7.82m/s,
While u is the original initial velocity from the throw of the ball
Then,
u'²=u²-2gH
7.82²=u²-2×9.81×7.5
61.146=u²-147.15
61.146+147.15=u²
Then, u²=208.296
So, u=√208.296
u=14.43m/s
The initial velocity of the ball form the throw is 14.43m/s
Answer:
The focal length fe of the eyepiece is <em>2.86 cm</em>
Explanation:
Since we are given the telescope's magnification and the length of the tube, we can use the expressions
<em>M = f_o/fe (1)</em> and
<em>l = f_o + fe (2)</em>
where
- M is the telescope's magnification
- l is the length of the tube
- fe is the focal length of the eye-piece
Rearranging equation (2) to make f_o the subject of the formula, we get
<em>f_o = l - fe</em>
Substituting the above equation into equation (1) we get
<em>M = (l - fe)/fe ⇒ fe = l/(M +1)</em>
<em> ⇒ fe = 60/(20 + 1)</em>
⇒ <em>fe = 2.86 cm</em>
