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jasenka [17]
3 years ago
5

When in use, a lamp is designed to draw

Physics
1 answer:
Ganezh [65]3 years ago
8 0

Answer:

R=∆V/I=12/0.6=20ohm....

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A proton (mass=1.67x10^-27 kg, charge= 1.60x10^-19 C) moves from point A to point under the influence of an electrostatic force
Tom [10]

Answer:

VB -  VA  =  - 33.4

Explanation:

Generally the workdone in moving the proton is mathematically represented as

     W  =  KE_f  - KE_i

Where KE_i \ and \  KE_f \  are\  the\  initial  \  and  \  final \  kinetic \  energy

So

    KE_i  =  \frac{1}{2} m v_a^2

Here v_a is the velocity at A with value  50 m/s

So

    KE_i  =  \frac{1}{2} (1.67*10^{-27}) * 50^2

    KE_i  = 2.09 *10^{-24} \  J

Also  

     KE_f  =  \frac{1}{2} m v_b^2

Here v_a is the velocity at A with value 80 km/s = 80000 m/s

=>   KE_f  =  \frac{1}{2} (1.67*10^{-27}) * 80000^2

=>   KE_f  = 5.34 *10^{-18} \  J

 So

    W  =   5.34 *10^{-18}  - 2.09 *10^{-24}

     W  =   5.34 *10^{-18}  m/s

Now this workdone is also mathematically represented as

     W =  q *  V

So  

    q *  V =   5.34 *10^{-18}

Here  q =  1.60*10^{-19} C

So

        V =   \frac{5.34 *10^{-18} }{1.60*10^{-19}}

         V =   33.4 \  V

Generally proton movement is in the direction of the electric field it means that  VA>VB

So

    VB -  VA  =  - 33.4

8 0
3 years ago
A varying force is given by F=Ae ^-kx, where x is the position;A and I are constants that have units of N and m^-1 , respectivel
Burka [1]
W = ∫ (x from 0.1 to +oo) F dx

= ∫ (x from 0.1 to +oo) A e^(-kx) dx

= A/k x [ - e^(-kx) ](between 0.1 and +oo)

= A/k x [ 0 + e^(-k * 0.1) ]

<span> = A/k x e^(-k/10) </span>
4 0
3 years ago
What do you think will happen to Charlie now that he is smart? Explain.
postnew [5]
.... I don’t know but, he will be able to make smarter choices, he will be able to think before he does something, honestly don’t know
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2 years ago
this refers to the process of manufacturing that introduced powered machinery to the production of goods​
Sidana [21]
This had to do with gain power and trade inequality business
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3 years ago
How much time will it take to do 33j of work with 11 w of power
Yuri [45]

Time required : 3 s

<h3>Further explanation </h3>

Power is the work done/second.

\tt P=\dfrac{W}{t}\\\\P=power,j/s,watt\\\\W=work, J\\\\t=times=s

 

To do 33 J of work with 11 W of power

P = 11 W

W = 33 J

\tt t=\dfrac{W}{P}\\\\t=\dfrac{33}{11}\\\\t=\boxed{\bold{3~s}}

8 0
3 years ago
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