When the object is at the focal point the angular magnification is 2.94.
Angular magnification:
The ratio of the angle subtended at the eye by the image formed by an optical instrument to that subtended at the eye by the object when not viewed through the instrument.
Here we have to find the angular magnification when the object is at the focal point.
Focal length = 6.00 cm
Formula to calculate angular magnification:
Angular magnification = 25/f
= 25/ 8.5
= 2.94
Therefore the angular magnification of this thin lens is 2.94
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Atomic mass = number of protons + number of neutrons = 4+5 = 9 amu
From conservation of energy, the height he will reach when he has gravitational potential energy 250J is 0.42 meters approximately
The given weight of Elliot is 600 N
From conservation of energy, the total mechanical energy of Elliot must have been converted to elastic potential energy. Then, the elastic potential energy from the spring was later converted to maximum potential energy P.E of Elliot.
P.E = mgh
where mg = Weight = 600
To find the height Elliot will reach, substitute all necessary parameters into the equation above.
250 = 600h
Make h the subject of the formula
h = 250/600
h = 0.4167 meters
Therefore, the height he will reach when he has gravitational potential energy 250J is 0.42 meters approximately
Learn more about energy here: brainly.com/question/24116470
Answer:
32.46m/s
Explanation:
Hello,
To solve this exercise we must be clear that the ball moves with constant acceleration with the value of gravity = 9.81m / S ^ 2
A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are the follow
Where
Vf = final speed
Vo = Initial speed
=7.3m/S
A = g=acceleration
=9.81m/s^2
X = displacement
=51m}
solving for Vf
the speed with the ball hits the ground is 32.46m/s
Answer:
Explanation:
The charge alters that space, causing any other charged object that enters the space to be affected by this field. The strength of the electric field is dependent upon how charged the object creating the field is and upon the distance of separation from the charged object.
