To solve this problem, we have to use the formula:
E = h f
where E is total energy, h is Plancks constant
6.626x10^-34 J s, f is frequency
f = E / h
f = 3.686 × 10−24 J / (6.626x10^-34 J s)
<span>f = 5.56 x 10^9 Hz</span>
The magnitude of the electric force on the charge is 5 N.
<h3>Magnitude of force on the charge</h3>
The magnitude of force on the charge is calculated as follows;
F = Eq
where;
- E is electric field
- q is magnitude of the charge
F = 100 N/C x 0.05 C
F = 5 N
Thus, the magnitude of the electric force on the charge is 5 N.
Learn more about electric force here: brainly.com/question/20880591
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V^2=u^2+2as
V=0
a =-u^2/2s
a=[4]^2/2[4]
a=-2m/s^2
The magnification of the ornament is 0.25
To calculate the magnification of the ornament, first, we need to find the image distance.
Formula:
- 1/f = u⁻¹+v⁻¹.................... Equation 1
Where:
- f = Focal length of the ornament
- u = image distance
- v = object distance.
make u the subject of the equation
- u = fv/(f+v)................ Equation 2
From the question,
Given:
Substitute these values into equation 2
- u = (12×4)/(12+4)
- u = 48/16
- u = 3 cm.
Finally, to get the magnification of the ornament, we use the formula below.
- M = u/v.................. Equation 3
Where
- M = magnification of the ornament.
Substitute these values above into equation 3
Hence, The magnification of the ornament is 0.25
Answer:
true
Explanation:
The law of conservation of charge states that whenever electrons are transferred between objects, the total charge remains the same.
