The graph represents the force applied on an 3.00kg crate while it moved 5.0m. A. How much total work is done on the crate? B. I
f the ball was moving at 6.00m/s when the force was first applied, what is its final velocity?
1 answer:
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Answer:
1.Theimage will be located at -0.13m or -13 cm
2.The height of the image will be 0.052m or 5.2cm
Explanation:
Given that;
Height of object, h=20 cm = 0.2m
Object distance in front of convex mirror, o,= 50 cm =0.5m
Radius of curvature, r, =34 cm =0.34m
Let;
Image distance, i,=?
Image height, h'=?
You know that focal length,f, is half the radius of curvature,hence
f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)
f= -0.17m
Apply the relationship that involves the focal length;
Re-arrange to get i
This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror
Apply the magnification formula
magnification, m=height of image÷height of object
substitute the values to get the height of image h'
The electric field strength is
Explanation:
The strength of the electric field produced by a single point charge is given by:
where
is the Coulomb's constant
q is the magnitude of the charge
r is the distance from the charge at which the field strength is calculated
For the charge in the problem, we have:
is the charge
Therefore, the electric field strength is
Learn more about electric fields:
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