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d1i1m1o1n [39]
3 years 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]3 years 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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In the Bohr model of the hydrogen atom, the speed of the electron is approximately 2.2 106 m/s.
Murrr4er [49]

The central force acting on the electron as it revolves in a circular orbit is 9.52 \times 10^{-8} \ N.

The given parameters;

  • <em>speed of electron, v = 2.2 x 10⁶ m/s</em>
  • <em>radius of the circle, r = 4.63 x 10⁻¹¹ m</em>

<em />

The central force acting on the electron as it revolves in a circular orbit is calculated as follows;

F = \frac{M_e v^2}{r} \\\\

where;

M_e is mass of electron = 9.11 x 10⁻³¹ kg

F = \frac{(9.11 \times 10^{-31}) \times(2.2\times 10^6)^2 }{4.63 \times 10^{-11}} \\\\F = 9.52 \times 10^{-8} \ N

Thus, the central force acting on the electron as it revolves in a circular orbit is 9.52 \times 10^{-8} \ N.

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3 years ago
( timed for this!! please help!!) Which best defines scientific question?
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3 years ago
An object is made of glass and has the shape of a cube 0.13 m on a side, according to an observer at rest relative to it. Howeve
Vlad1618 [11]

Answer:

v=0.9833\ c

Explanation:

The density changes means that the length in the direction of the motion is changed.

Therefore,

$\text{Density} = \frac{m}{lwh}$

Given :

Side,  b = h = 0.13 m

Mass, m = 3.3 kg

Density = 8100 kg/m^3

So,

$8100=\frac{3.3}{l \times 0.13 \times 0.13}$

$l=\frac{3.3}{8100 \times 0.13 \times 0.13}$

l = 0.024 m

Then for relativistic length contraction,

$l= l' \sqrt{1-\frac{v^2}{c^2}}$

$0.024= 0.13 \sqrt{1-\frac{v^2}{c^2}}$

$0.184=  \sqrt{1-\frac{v^2}{c^2}}$

$0.033=  1-\frac{v^2}{c^2}}$

$\frac{v^2}{c^2}= 0.967$

$\frac{v}{c}=0.9833$

v=0.9833\ c

Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).

3 0
3 years ago
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