Answer:
Explanation:
t = Time taken =
i = Current = 3 A
q(0) = Initial charge =
Charge is given by
The magnitude of the net electric charge of the capacitor is
The focal length of given concave lens will be -26.85 cm
The height of an image to the height of an object is the ratio that is used to determine a lens' magnification. Additionally, it is provided in terms of object and image distance. It is equivalent to the object distance to image distance ratio.
Given concave lens creates a virtual image at -47.0 cm and a magnification of +1.75.
We have to find focal length
The focal length can be found out by following way:
Magnification = m = +1.75
m = hi/h
hi = -47 cm
1.75 = -47/h
h = -26.85 cm
So the focal length of given concave lens will be -26.85 cm
Learn more about magnification factor here:
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Answer:
Because the Moon casts a smaller shadow than Earth does, eclipses of the Sun tightly constrain where you can see them. If the Moon completely hides the Sun, even for a moment, the eclipse is considered total.
Explanation:
Change minutes to hrs, divide by 60:
30 min = .50 hrs
45 min = .75 hrs
12 min = .20 hrs
----------------
total + 1.45 hrs, total travel time
:
let a = average speed for the trip
:
Write a dist equation, dist = speed * time
:
80(.5) + 100(.20) + 40(.75) = 1.45a
40 + 20 + 30 = 1.45a
90 = 1.45a
a =
a = 62.069 km/h, for the entire trip
and
90 km is the total distance
Answer:
distance traveled is 15 mi
displacement is 5 mi
Explanation:
Distance takes time into account and adds up all the tiny displacements during the entire period of the trip.
Displacement ignores time and looks only at the change in position from the starting point to the ending point.