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d1i1m1o1n [39]
1 year ago
10

José and Laurel measured the length of a stick's shadow during the day. Without knowing the length of the stick, which of their

measurements would you ask José and Laurel to examine more closely? A. 50 cm at 6:00 a.m. B. 75 cm at 12:30 p.m. C. 42 cm at 5:45 p.m. D. 55 cm at 7:15 p.m. dont speak spanish answer with english plz???
Physics
1 answer:
skelet666 [1.2K]1 year ago
5 0

Answer:

Option B.

Explanation:

Assuming the stick is in vertical position, its shadow depends on two factors: its length and the angle between the sun rays and the stick. When the angle is bigger, the lenght of the shadow increases, and vice versa. So, when the sun rays are parallel to the stick, the shadow may be small. Since they are nearly perpendicular to the Earth's surface at 12 o'clock, the shadow of the stick at that time should be minimal. It means that the measured shadow of 75 cm at 12:30 p.m. is almost impossible (Option B).

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Two point charges of values +3.4 and +6.6 μc are separated by 0.10 m. what is the electrical potential at the point midway betwe
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 To solve this problem, we should remember that:

Energy = Force x Distance

Since we are talking about charges, therefore we make use of Coulumb’s law for the electrical force between the two charges:

F = k q1 q2 / d^2

Where,

k = Coulumb’s constant = 9 x 10^9 N m^2/ c^2

q = charge

d = distance between the charges

Plugging back into the energy equation:

E = (k q1 q2 / d^2) * d

E = k q1 q2 / d

Solving for E using the given values:

E = (9 x 10^9 N m^2/ c^2) (3.4 E -6 c) (6.6 E -6 c) / 0.10 m

<span>E = 2.02 N m = 2.02 J</span>

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2 years ago
Karyotypes are done by matching up _____________________________ so that they are paired up. Question 11 options:
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Homologous Chromosomes

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2 years ago
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A traditional set of cycling rollers has two identical, parallel cylinders in the rear of the device that the rear tire of the b
mars1129 [50]

Answer:

ω2  =  216.47 rad/s

Explanation:

given data

radius r1 =  460 mm

radius r2 = 46 mm

ω =  32k rad/s

solution

we know here that power generated by roller that  is

power = T. ω    ..............1

power = F × r × ω

and this force of roller on cylinder is equal and opposite force apply by roller

so power transfer equal in every cylinder so

( F × r1 × ω1)  ÷ 2 = (  F × r2 × ω2 )  ÷  2    ................2

so

ω2  =  \frac{460\times 32}{34\times 2}

ω2  =  216.47

8 0
2 years ago
A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of
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Explanation:

When Michelson-Morley apparatus is turned through 90^{o} then position of two mirrors will be changed. The resultant path difference will be as follows.

      \frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}

Formula for change in fringe shift is as follows.

          n = \frac{2lv^{2}}{\lambda c^{2}}

       v^{2} = \frac{n \lambda c^{2}}{2l}

             v = \sqrt{\frac{n \lambda c^{2}}{2l}}

According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

             l = 11 m

    \lambda = 5.9 \times 10^{-7} m

           c = 3.0 \times 10^{8} m/s

Hence, putting the given values into the above formula as follows.

            v = \sqrt{\frac{n \lambda c^{2}}{2l}}

               = \sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}

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Thus, we can conclude that velocity deduced is 2.41363 \times 10^{9} m/s.

3 0
2 years ago
Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci
Rudiy27

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W

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2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]

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specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

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quality of steam x2

h2 = hf + x2(hfg)

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x2 = 0.99

6 0
2 years ago
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