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suter [353]
3 years ago
11

You have a neutral balloon. What is its charge after 12000 electrons have been removed from it? The elemental charge is 1.6 × 10

−19 C. Answer in units of µC.
Physics
1 answer:
AleksandrR [38]3 years ago
6 0

Answer:

1.92×10^-9 microCoulomb

Explanation:

Elemental charge = 1.6×10^-19 Coulomb

Charge of balloon after 12,000 electrons have been removed from it = 12,000 × 1.6×10^-19 Coulomb = 1.92×10^-15 Coulomb = 1.92×10^-15/10^-6 = 1.92×10^-9 microCoulomb

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Sound exits a diffraction horn loudspeaker through a rectangular opening like a small doorway. Such a loudspeaker is mounted out
blondinia [14]

Answer:

\theta = 20.98 degree

Explanation:

As we know that the speed of the sound is given as

v = 332 + 0.6 t

now at t = 273 k = 0 degree

v = 332 m/s

so we have

a sin\theta = N\lambda

a sin\theta = N(\frac{v_1}{f})

now when temperature is changed to 313 K we have

t = 313 - 273 = 40 degree

now we have

v = 332 + (0.6)(40)

v_2 = 356 m/s

a sin\theta' = N(\frac{v_2}{f})

now from two equations we have

\frac{sin19.5}{sin\theta} = \frac{332}{356}

so we have

sin\theta = 0.358

\theta = 20.98 degree

7 0
3 years ago
A ______________ is very general in nature, while a ________________ specifies what we want to study more specifically, suggesti
vova2212 [387]

Answer:

research topic and research question (hypothesis)

Explanation:

6 0
3 years ago
An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference
kondor19780726 [428]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

K=qV          (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J

The kinetic energy of the particle is also:

K=\frac{1}{2}mv^2       (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

|F_B|=qvBsin(\theta)

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm

the radius of the trajectory of the electron is 10.1 cm

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3 years ago
When scientists observed the light from stars and galaxies, they noticed that their color shifted toward the end of the visible
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Color shifted towards the end is C red
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In and electric circuit, where do the electrons come from that flow in the circuit
Firdavs [7]
The electrons are already there. They are freely moving through the conductor.
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3 years ago
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