The Image distance and Magnification of The Image will be 30 cm and 3.
<h3>What is focal length?</h3>
The focal length of the lens, which is often expressed in millimeters, is the distance between the lens and the image sensor when the subject is in focus.
Given data;
Focal length,f=?
Image distance,v=?
Object distance,u= 10 cm
Magnification,m= 2.85
The focal length is half of the radius;
f=R/2
f=30 Cm/2
f= 15 Cm
The mirror equation is found as;
The magnification of the lens is found as;
Hence, the image distance and magnification of The image will be 30 cm and 3.
To learn more about the focal length refer;
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First one, holding a basketball in the air. Potential energy is the energy it has mostly from gravity. The further you go from the center of mass, the more energy.
Answer:
18.4 m
Explanation:
(a)
The known variables in this problem are:
u = 1.40 m/s is the initial vertical velocity (we take downward direction as positive direction)
t = 1.8 s is the duration of the fall
a = g = 9.8 m/s^2 is the acceleration due to gravity
(b)
The vertical distance covered by the life preserver is given by
If we substitute all the values listed in part (a), we find
F=MA
F=(8 kg)(9.8 m/s)
F= 78.4 N
W=FD
W=(78.4 N)(7 m)
W=548.8 J
How this helps
Answer:
1.11 V
Explanation:
Given that the Einstein photoelectric equation states that;
KE = E - Wo
E = energy of incident photon
Wo= work function of the metal
E = hf = 6.64 x 10-34 * 6 x 1014
E = 39.84 * 10^-20 J or 3.98 * 10^-19 J
KE = 3.98 * 10^-19 J - 2.2 x 10-19J
KE = 1.78 * 10^-19J
We convert this value of KE to electron volts
KE = 1.78 * 10^-19J/1.6 x 10-19C
KE = 1.11 eV
Hence; 1.11 V will be just sufficient to stop electrons emitted by the sodium photo-plate reaching the collector plate.
