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Yakvenalex [24]
4 years ago
15

A test charge of +3 µC is at a point P where the electric field due to the other charges is directed to the right and has a magn

itude of 4 106 N/C. If the test charge is replaced with a charge of -3 µC, what happens to the electric field at P?
Physics
1 answer:
umka2103 [35]4 years ago
5 0

Answer:

Electric field will remain same.

Explanation:

Given that

At initial condition ,charge q = +3μ C

Electric field E = 4106 N/c

As we know that

Electric field due to charge q

E=\dfrac{kq}{r^2}  

Now when charge is replaced by new charge q'= -3μ C

From the expression of electric field we can say that electric field will remain same from same quantity of electric charge.

So we can say that electric field will remain same.

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The total volume of hydrogen gas needed to fill the hindenburg was 2.09 × 108 l at 1.00 atm and 25.1°
Inessa05 [86]
The solution for this problem:

PV = nRT 

But we are looking for n so, alter the original formula:
n = PV / RT

= (1.00 atm) x (2.09 x 10^8 L) / ((0.08205746 L atm/K mol) x  (25.1 + 273.15 K)) 

= 8539801.26 mol H2 
total volume = (8539801.26 mol H2) x (-286 kJ/mol H2) = -2.44 x 10^9 k J
3 0
3 years ago
Read 2 more answers
1) [25 pts] A 90-kg merry-go-round of radius 2.0 m is spinning at a constant speed of 20 revolutions per minute. A kid standing
Ray Of Light [21]

Answer:

(A) 180 kg·m²

(B) 0.111 rad/s²

(C) The number of revolutions the merry-go-round will complete until it finally stops is 3.142 or π rev

Explanation:

The equation of the moment of inertia of a solid cylinder is presented as follows;

I = \frac{1}{2}MR^2

Where:

I = Moment of inertia of the merry-go-round

M = Mass of the merry-go-round

R = Radius of the merry-go-round

Therefore, I = 1/2×90×2² = 180 kg·m²

(B) For the angular acceleration we have;

Therefore, since the force × radius = The torque, we have, angular acceleration is found as follows

F × R = τ

10.0 × 2.0 = 20 = I×α = 180×α

α = 20/180 = 0.111 rad/s².

angular acceleration = 0.111 rad/s².

(C) Here we have ω₀ = 20 rev/ min = 20×2×π rad/min = π·40/60 rad/s

2/3·π rad/s

ω = ω₀ - α×t

∴ t = ω₀/α = (2/3·π rad/s)/(0.111 rad/s²) = 18.85 s

Hence we have

θ = ω₀·t + 1/2·α·t², plugging in the values, we have;

θ = 2/3·π×18.85 - 1/2·0.111·18.85²

θ = 19.74 rad

Therefore, since 2·π radian = 1 revolution

The number of revolutions the merry-go-round will complete until it stops is 19.74/(2·π) = 3.142 or π revolutions.

5 0
3 years ago
A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0° below the horizontal. The negligent
anzhelika [568]

Answer:

The position is  P  =  47.4 \ m relative to the base of the ocean

Explanation:

From the question we are told that

    The angle made by the incline with the horizontal is \theta =  24.0 ^o

    The constant acceleration is a =  3.82 \ m/s^2

    The distance covered is d =  60.0 \ m

    The height of the cliff is h  =  50 .0 \ m

The velocity of the car is mathematically represented as

      v^2 = u^2 +  2ad

The initial velocity of the car is  u= 0

So

     v^2 =  2ad

substituting values

     v^2 = 2 *  3.82 * 60

    v =  21.4 \ m/s

The vertical component of this velocity is

    v_v  =  -v * sin(\theta )

substituting values

    v_v  = -21.4 * sin(24.0)

    v_v  = -8.7 \ m/s

The negative sign is because is moving in the negative direction of the y-axis

The horizontal  component of this velocity is

     v_h  =  v * cos (\theta)

    v_h  =  21.4 * cos (24.0)

    v_h  = 19.5 \ m/s

Now according to equation of motion we have

     h  =  v_v*t  - \frac{1}{2} *  g t^2

substituting  values

    50  =  -8.7 t  - \frac{1}{2} *  9.8 t^2

    4.9t^2  +8.7t -50 = 0

using quadratic equation we have that

  t_1 = 2.42\ s \ and\   t_2 =  -4.20\ s

given that time cannot be negative

      t = 2.42 \ s

The  car’s position relative to the base of the cliff when the car lands in the ocean is mathematically evaluate as

           P  =  v_h * t

substituting values

          P  =  19.5 *  2.43

         P  =  47.4 \ m

5 0
3 years ago
A skydiver jumps out of a hovering helicopter, so there is no forward velocity. Ignore wind resistance for this exercise. 1. Wha
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1. What is the skydiver's acceleration?

Initially, like any falling object, a skydiver's downward acceleration is 9.8 meters/seconds^2, or about 28-35 feet per second squared. This acceleration reduces over a few seconds and approaches zero as the skydiver reaches terminal velocity.
3 0
3 years ago
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Select. The star cycle that is accurate
melomori [17]

One stellar-mass star, red giants, white dwarfs, or planetary nebulae will make up the correct star cycle. Option C is correct.

<h3 /><h3>What is the solar system?</h3>

The satellites of the planet, countless comets, asteroids, and meteoroids, as well as the interplanetary medium, make up the solar system.

The complete question is;

"Choose the correct star life cycle.

A. Supernova, star as well as red giant, the nebula is incorrect.

B. Nebula, white dwarf, planetary nebula That is incorrect.

C. Planetary nebula, red giant, white dwarf, and star with a single stellar mass That is true!

D. Nebula, a star of one stellar mass, is not correct. It is correct if the star had four or more stellar masses."

Star of one stellar mass, red giant, white dwarf, and planetary nebula make up the proper life cycle.

Hence option C is correct.

To learn more about the solar system refer to the link;

brainly.com/question/1207587

#SPJ1

5 0
2 years ago
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