If you travel 30 miles somewhere and then come home again your distance is 60 miles. Your displacement is 0 because it is the ending position minus the beginning position, which are the same place (home). In other words, displacement is a vector and distance is a scalar.
Answer: 20.4 miles
Explanation:
Here we need to use the equation:
Velocity = Distance/Time.
Initially we have that he can travel 30 miles in 2 hours, so the velocity is:
V = 30mi/2h = 15mph
Now, we reduce the velocity by 3 mph, so the new velocity is 15mph - 3 mph = 12mph.
Now we want to know the distance traveled in 1.7 hours with this velocity, this is.
Velocity*Time = Distance
12mi/h*1.7h = 20.4 miles
We have that for the Question "Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be said its equation is
From the question we are told
Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e
<h3>An E
xpression for the <em>magnitude </em>of charge moved</h3>
Generally the equation for the <em>magnitude </em>of charge moved, Q is mathematically given as
Therefore
An expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be
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<span>The answer is letter D.
The most important step in formulation a scientific inquiry is to first formulate a question. All answers sprung from a question that scratched the mind. Experimentation is not possible without the need to question, with this, the drive to find the answers is what motivates scientists or people involved in an experiment to find the answers that would suffice their curiosity. Questions are the basis of all the other choices above and is the most crucial step in the scientific inquiry.<span>
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Answer:
distance between object and image = 18.9 cm
Explanation:
given data
radius of curvature = 18 cm
focal length = 1/2 radius of curvature
magnification = 40%
to find out
distance between object and image
solution
we know lens formula that is
1/f = 1/v + 1/u ....................1
here f = 18 /2 and v and u is object and image distance
and we know m = 40% = 0.40
so 0.40 = -v / u
so here v = - 0.40 u
so from equation 1
1/f = 1/v + 1/u
2/18 = - 1/0.40u + 1/u
u = -13.5 cm ..................2
and
v = -0.40 (- 13.5)
v = 5.4 cm ......................3
so from equation 2 and 3
distance between object and image = 5.4 + 13.5
distance between object and image = 18.9 cm
