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Misha Larkins [42]
3 years ago
15

Which of the following represents a concave lens? A. -di B. +di C. -f D. +f

Physics
2 answers:
seraphim [82]3 years ago
6 0

for a concave lens it should be -f

Nina [5.8K]3 years ago
4 0

Answer:

The answer is option D. +f

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whats the question?

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Explanation:

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3 years ago
Read this excerpt from Through the Looking-Glass by Lewis Carroll.
balandron [24]

Answer:

A

Explanation:

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7 0
3 years ago
Read 2 more answers
Betty weighs 400 N and she is sitting on a playground swing seat that hangs 0.21 m above the ground. Tom pulls the swing back an
disa [49]

Answer:

4.15 m/s

Explanation:

As the total energy must be conserved (neglecting air resistance) the change in gravitational potential energy, must be equal to the change in kinetic energy:

ΔE = ΔK + ΔU =0

If we take as a zero reference level for the gravitational potential energy, the height of the swing seat above the ground, (which is equal to 0.21 m), we can find the initial gravitational energy, considering the height of the point where the seat is released, regarding this point:

h₀ = 1.09 m -0.21 m = 0.88 m

⇒ U₀ = m*g*h₀ = 400 N*0.88 m = 352 J

As Uf = 0, ΔU = Uf -U₀ = -352 J

As the swing starts from rest, K₀=0, so we can say:

ΔK = Kf = \frac{1}{2} *m*vf^{2}  (1)

As ΔK = -ΔU ⇒ ΔK = 352 J (2)

From (1) and (2) we can solve for vf, as follows:

vf = \sqrt{\frac{2*352J}{40.8kg}} = 4.15 m/s

So, when the swing passes through its lowest position, Betty moves at 4.15 m/s.

5 0
4 years ago
an athlete runs 300 m up a hill at a steady speed of 3.0 m/s. She then immediately runs the same distance at 6.0 m/s . What is h
mina [271]

Answer:

4.0 m/s

Explanation:

In the first part of the run, the athlete runs a distance of

d_1 = 300 m

at a speed of

v_1 = 3.0 m/s

So, the time he/she takes is

t_1 = \frac{d_1}{v_1}=\frac{300}{3.0}=100 s

In the second part of the run, the athlete covers an additional distance of

d_2 = 300 m

with a speed

v_2 = 6.0 m/s

So, the time taken in this second part is

t_2 = \frac{d_2}{v_2}=\frac{300}{6.0}=50 s

So, the total distance covered is

d = 300 m + 300 m = 600 m

And the total time taken

t = 100 s + 50 s = 150 s

Therefore, the average speed for the entire trip is

v=\frac{d}{t}=\frac{600}{150}=4.0 m/s

4 0
4 years ago
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