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White raven [17]
1 year ago
5

A vertical spring stretches 3.9 cm when a 10.-g object is hung from it. The object is replaced with a block of mass 25 g that os

cillates up and down in simple harmonic motion. Calculate the period of motion.
Physics
1 answer:
Dmitry_Shevchenko [17]1 year ago
4 0
Period of motion is approximately 0.5447
seconds
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Suppose a single electron orbits about a nucleus containing two protons (+2e), as would be the case for a helium atom from which
notka56 [123]

Answer:

a=5.66*10^{23} \frac{m}{s^2}

Explanation:

In this case we will use the Bohr Atomic model.

We have that: F=m*a

We can calculate the centripetal force using the coulomb formula that states:

F=k*\frac{q*q'}{r^2}

Where K=9*10^9 \frac{Nm^2}{C}

and r is the distance.

Now we can say:

m*a=k*\frac{q*q'}{r^2}

The mass of the electron is = 9.1*10^{-31} Kg

The charge magnitud of an electron and proton are= 1.6*10^{-19}C

Substituting what we have:

[tex]a=\frac{9*10^{9}*(1.6*10^{-19} )*(2(1.6*10^{-19} ))}{9.1*10^{-31}*(2.99*10^{-11})^2 }[/tex]

so:

a=5.66*10^{23} \frac{m}{s^2}

3 0
3 years ago
An object originally at rest, is accelerated uniformly along a straight line to a speed of 8m/s in 2s. What is the acceleration
Anna11 [10]

Answer:

4m/s²

Explanation:

Initial velocity (u) = 0 m/s

Final velocity (v) = 8 m/s

Time taken (t) = 2 sec

Acceleration (a) = ?

We know

a  =  \frac{v - u}{t}  \\  =  \frac{8 - 0}{2} \\  =  \frac{8}{2}  \\  = 4 \: m |s ^{2}

Hope it will help :)

3 0
3 years ago
What must happen to the electrons in a material to create an electric current?
Nesterboy [21]
<span>The electrons need to migrate from a region of lower electric potential to a region of higher electric potential. This flow of electrons is known as electron current (which is contrary to the widely-accepted perspective of conventional current).</span>
4 0
4 years ago
Read 2 more answers
What is the area under the curve for the histogram below
igor_vitrenko [27]

Answer:  B srry if i'm wrong

3 0
2 years ago
*photo attached* The diameters of the main rotor and tail rotor of a single-engine helicopter are 7.67 m and 1.01 m, respectivel
stiks02 [169]

(a) The speeds of the tips of both rotors; main rotor <u>178.3 m/s</u> and  tail rotor <u>218.4 m/s</u>.

(b) The speed of the main rotor is <u>0.52</u> speed of sound, and the speed of the tail rotor is <u>0.64</u> speed of sound.

<h3>Linear speed of main motor and tail rotor</h3>

v = ωr

where;

  • ω is the angular speed (rad/s)
  • r is radius (m)

v(main rotor) = (444 rev/min x 2π rad x 1 min/60s) x (0.5 x 7.67 m)

v(main rotor) = 178.3 m/s

v(tail rotor) = (4,130 rev/min x 2π rad x 1 min/60s) x (0.5 x 1.01 m)

v(tail rotor) = 218.4 m/s

<h3>Speed of the rotors with respect to speed of sound</h3>

% speed (main motor) = 178.3/343 = 0.52 = 52 %

% speed (tail motor) = 218.4/343 = 0.64 = 64 %

Thus, the speed of the main rotor is 0.52 speed of sound, and the speed of the tail rotor is 0.64 speed of sound.

Learn more about linear speed here: brainly.com/question/15154527

#SPJ1

5 0
2 years ago
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