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andriy [413]
3 years ago
14

Two point charges each have a value of 3.0 c and are separated by a distance of 4.0 m. what is the electric field at a point mid

way between the two charges?
Physics
1 answer:
swat323 years ago
4 0
 <span>Place a test charge in the middle. It is 2cm away from each charge. 
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point. 
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out. 
THIS IS A TRICK QUESTION. 
THe electric field exactly midway between them = 0/Q = 0. 
But if the point moves even slightly you need the following formula 
F= (1/4Piε)(Q1Q2/D^2) 
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>
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Avg. acceleration = -5 m/s^2.
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4 years ago
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A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 30 m/s2 for 35 s , then ru
Gre4nikov [31]

Answer:

a) h_m=74625\ m

b) t=265.55\ s

Explanation:

Given:

  • mass of rocket, m_r=200\ kg
  • mass of fuel, m_f=100\ kg
  • acceleration of the rocket consuming fuel, a=30\ m.s^{-2}
  • time after which the fuel exhaust, t_f=35\ s

<u>During the phase of fuel exhaustion:</u>

<u>velocity attained by the rocket just as the fuel ends:</u>

v_f=u+a.t_f

where:

u= initial velocity of the rocket = 0

v_f=0+30\times 35

v_f=1050\ m.s^{-1} this will be the initial velocity for the phase of ascending of the rocket's height under the influence of gravity.

<u>height at which the fuel finishes:</u>

v_f^2=u^2+2a.h_f

1050^2=0^2+2\times 30\times h_f

h_f=18375\ m

<u>During the phase of ascend in height of rocket after the fuel is over:</u>

<u>Time taken to reach the top height after the fuel is over:</u>

v=v_f+g.t'

at top v = (final velocity during this course of motion )= 0 m.s^{-1}

0=1050-9.8\times t'

t'=107.1429\ s

<u>Height ascended by the rocket after the fuel is over:</u>

v^2=v_f^2+2g.h'

at the top height the velocity is zero

0^2=1050^2-2\times 9.8\times h' (-ve sign denotes that the direction of motion is opposite to that of acceleration)

h'=56250\ m

<u>Therefore the maximum altitude attained by the rocket:</u>

h_m=h_f+h'

h_m=18375+56250

h_m=74625\ m

b)

time taken by the rocket to fall back to the earth:

h_m=v.t_m+\frac{1}{2} g.t_m^2

where:

v= initial velocity of the rocket during the course of free fall from the top height.

74625=0+4.9\times t_m^2

t_m=123.41\ s

Now the total time for which the rocket is in the air:

t=t_f+t'+t_m

t=35+107.1429+123.41

t=265.55\ s

4 0
3 years ago
The coil of an ac generator has 50 loops and a cross-sectional area of LaTeX: 0.2~m^20.2 m 2 . What is the maximum emf that can
lisov135 [29]

Answer:

Maximum emf in the coil is 100 volt    

Explanation:

We have given number of loops in the coil N = 50

Cross sectional area A=0.2m^2

Angular speed is given \omega =5rad/sec

Magnetic field is given B=0.2T

We have to find the maximum emf in the coil

Maximum emf in the coil is given by e=NBA\omega

e=50\times 2\times 0.2\times 5=100volt

So maximum emf induced in the coil is 100 volt

7 0
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Elena L [17]

Answer:

nerve pathways

Explanation:

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3 years ago
A particle moves along a straight line and its position at time t is given by
Alisiya [41]

(A) Use interval notation to indicate the time interval or union of time intervals when the particle is moving forward and backward

the particle is speeding up

<span>if  a(t) >0, it means 12t-48>0, and t >4,  I = ]4, infinity[</span>

the particle is speeding down

<span>if  a(t) < 0, it means 12t-48>0, and t < 4,  I = ] -infinity, 4[</span>

 

when the particle is moving forward and backward

that is depending of the sign of v(t)

moving forward if v(t)> 0

<span>moving backward if v(t)<  0</span>

but v(t) = 6t² - 48 t +90, this quadratic equation has exactly two solutions, t =1, and t =3

<span>so  after taking care the sign of </span>6t² - 48 t +90,

 

<span>v(t)> o when  t is in I = ] –inf, 1[ U ]4, inf[,the particle is moving forward</span>

<span>v(t) < o when  t is in I = ]1, 4[,the particle is moving backward</span>

<span> </span>

3 0
3 years ago
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