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chubhunter [2.5K]
2 years ago
12

An electron with a charge value of 1.6 x 10-19 C is moving in the presence of an electric field of 400 N/C. What force does the

electron experience?
Physics
1 answer:
Vsevolod [243]2 years ago
4 0

Answer:

4.0×10⁻¹⁷ N

Explanation:

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Yes. a physical change is only modifying the structure of it.

a chemical change is permanently changing something.

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A small airplane has to reach a speed if 27.8 m/s to takeoff. It can accelerate at 2.00 m/s^2. What is the minimal length of run
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Using the constant acceleration formula v^2 = u^2 + 2as, we can figure out that it would take a distance of 193.21m to reach 27.8m/s

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3 years ago
Bats have poor vision and use echo-location to find their way.The frequency produced by bats is 100kHz. If the speed of the soun
irga5000 [103]

The distance covered is 25.9 m.

<h3>How deep is the cave?</h3>

We know that the speed of sound refers to the speed with which an sound moves in an object.

Given that;

speed of sound = 345m/s

Time taken =  0.15s

We know that;

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v = speed of sound

d = distance

t = time taken

vt = 2d

d = vt/2

d = 345m/s *  0.15s/2

d = 25.9 m

Learn more about speed of sound:brainly.com/question/15137350

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5 0
2 years ago
A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca
svetoff [14.1K]

Answer:

q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Explanation:

Given that L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V, we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000  \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}

#To find the particular solution:

Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Hence the charge at any time, t is q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

6 0
3 years ago
Which device uses a rotating magnetic field to produce an electric current?
Kay [80]
<span>The correct answer is C) a motor.
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</span>\phi=BAcos\theta<span>
where B is the intensity of the magnetic field, A is the area enclosed by the coil and </span>\theta<span> is the angle between the direction of B and the perpendicular to the plane of the coil). For Faraday-Newmann-Lenz law, this change in flux induces an electromotive force (emf) into the coil, according to:
</span>emf=- \frac{d \Phi}{dt}<span>
 where the numerator is the variation of magnetic flux and dt is the time interval. This emf in the coil produced an electrical current in the circuit.</span>
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3 years ago
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