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andreyandreev [35.5K]
2 years ago
6

What voltage is required to move 6A through 20?​

Physics
1 answer:
Marina CMI [18]2 years ago
7 0

Answer:

120V

Explanation:

Given parameters:

Current  = 6A

Resistance  = 20Ω

Unknown:

Voltage = ?

Solution:

According to ohms law;

           V = IR

Where V is the voltage

            I is the current

            R is the resistance

Now, insert the parameters and solve;

    V  = 6 x 20  = 120V

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A transverse wave is traveling on a string stretched along the horizontal x-axis. The equation for the
maxonik [38]

Answer:

A) 0.33 m/s

Explanation:

The standard form of a transverse wave is given by  

y = a  cos ( ω t − kx ) ,   k =  2 π  / λ

Amplitude,   a =  0.002  m

Wavenumber (k)=47.12 and wavelength  ( λ )  =  0.133 m

Time period(T)=0.0385 s and angular frequency  ( ω )  =  52 π  rad/s

Maximum speed of the string is given by  aw

Therefore ; max. speed = 0.002 x 52 π = 0.327 m/s

 

6 0
2 years ago
In the Bohr model of the hydrogen atom, an electron (massm=9.1×10−31kg) orbits a proton at a distance of 5.3×10−11m. The proton
Marysya12 [62]

Answer:

\omega = 6.557 \times 10^{16}\ rev/s

Explanation:

GIVEN,

mass of electron =  9.1 x 10 kg

Radius = 5.3 x 10 m

pulling force = 8.2 x 10 N

Required centripetal for (Fe) for circular motion will be provided with electrical force (F)

      F = m_e\omega^2 r

      \omega = \sqrt{\dfrac{F}{m_e\ r}}

      \omega = \sqrt{\dfrac{8.2 \times 10^{-8}}{9.1 \times 10^{-31}\times 5.3 \times 10^{-11}}}

      \omega = \sqrt{0.17 \times 10^{34}}

       ω = 4.12 x 10¹⁶ rad/s

\omega = \dfrac{4.12 \times 10^{16}}{2\pi}

\omega = 6.557 \times 10^{16}\ rev/s

6 0
2 years ago
I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a
attashe74 [19]
<span>The angular momentum of a particle in orbit is 

l = m v r 

Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2" 

m_1 v_1 r_1 = m_2 v_2 r_2 

Assuming that the mass did not change, conservation of angular momentum demands that 

v_1 r_1 = v_2 r_2 

or 

v1 = v_2 (r_2/r_1) 

Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have 

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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3 0
2 years ago
A 0.0500-kg golf ball heads off from the tee with an initial speed of 78.2 m/s and reaches to its maximum height of 37.8 m. If a
Marianna [84]

(a) 134.4 J

The total mechanical energy of the ball at the moment of launch is just equal to its initial kinetic energy:

E=K_i = \frac{1}{2}mu^2

where

m = 0.05 kg is the mass of the ball

u = 78.2 m/s is the initial speed

Substituting,

E=\frac{1}{2}(0.05)(78.2)^2=152.9 J

At the pinnacle of its trajectory, the total mechanical energy is sum of kinetic energy and potential energy:

E=K_f + U_f = K_f + mgh

where

g=9.8 m/s^2 is the acceleration of gravity

h = 37.8 m is the maximum height

Since the total energy must be conserved,

E = 152.9 J

Therefore, we can solve to find the kinetic energy of the ball at the pinnacle:

K_f = E-mgh=152.9-(0.05)(9.8)(37.8)=134.4 J

b) 74.2 m/s

When the ball is 5.60 m below the pinnacle point, the heigth of the ball is

h=37.8-5.6=32.2 m

So its potential energy is

U=mgh=(0.05)(9.8)(32.2)=15.8 J

The total energy is again the sum of potential and kinetic energy:

E = K + U

So the kinetic energy at that point is

K=E-U=152.9-15.8=137.1 J

And since the kinetic energy is

K=\frac{1}{2}mv^2

We can find the speed:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(137.1)}{0.05}}=74.2 m/s

3 0
2 years ago
If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a 0.100
SCORPION-xisa [38]

Answer:

Charge enter a 0.100 mm length of the axon is 8.98\times 10^{-12} C

Explanation:

Electric field E at a point due to a point charge is given by

E=k \frac{q}{r^2}

where k is the constant =9.0 \times 10^9  Nm^2 / C^2

q is the magnitude of point charge and r is the distance from the point charge

Charges entering one meter of axon is 5.\times 10^{11} \times (+e)

Charges entering 0.100 mm of axon is 5.\times 10^{11} \times (+e) \times (0.1 \times 10^{-3}

substituting the value of +e=1.6\times 10^{-19} C in above equation, we get charge enter a 0.100 mm length of the axon is

q=5.\times 10^{11} \times1.6\times 10^{-19}  \times (0.1 \times 10^{-3}\\q=8.98\times 10^{-12} C

3 0
2 years ago
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