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qaws [65]
3 years ago
15

Point charge q1 of 30 nC is separated by 50 cm from point charge q2 of -45 nC. As shown in the diagram, point a is located 30 cm

from q1 and 20 cm q2. What are the magnitude and direction of the electric field at point a? Let the electrostatic constant k= 8.99 x 10^9 N x m^2/C^2

Physics
1 answer:
Angelina_Jolie [31]3 years ago
6 0

Answer:

E1 =  2996.667N/C E2 = 11237.5N/C

Explanation:

E1 = kQ1/r^2

  =8.99 x 10^9 x 30 x 10^-9/(30x10^-2)^2

  = 2996.667N/C

E2 = kQ2/r^2

      = 8.99 x 10^9 x 50 x 10^-9/(20x10^-2)^2

      = 11237.5N/C

The direction are towards the point a

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A 20 kilogram ball rolls down a 10 meter ramp at the rate of 15 meters per second. The kinetic energy is joules
kobusy [5.1K]

Answer:2250J

Explanation:

mass(m)=20kg

velocity(v)=15m/s

Kinetic energy=(m x v^2)/2

Kinetic energy =(20 x 15^2)/2

Kinetic energy =(20x15x15)/2

Kinetic energy=4500/2

Kinetic energy=2250J

3 0
3 years ago
A particle of mass m moves under an attractive central force F(r) = -Kr4 with angular momentum L. For what energy will the motio
docker41 [41]

Answer:

Angular velocity is same as frequency of oscillation in this case.

ω = \sqrt{\frac{7K}{m} } x [\frac{L^{2}}{mK}]^{3/14}

Explanation:

- write the equation F(r) = -Kr^{4} with angular momentum <em>L</em>

- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.

- Write the energy of the orbit in relative to r = 0, and solve for "E".

- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.

- Solve for effective potential

- ω = \sqrt{\frac{7K}{m} } x [\frac{L^{2}}{mK}]^{3/14}

3 0
3 years ago
Is a truck braking to avoid an accident moving at constant
frutty [35]

Answer:

its not moving at a constant velocity because it is slowing down

Explanation:

3 0
2 years ago
Read 2 more answers
PLEASE HELP WILL GIVE BRAINLIEST!!!
Elan Coil [88]

The spring constant is 181.0 N/m

Explanation:

We can solve the problem by applying the law of conservation of energy. In fact, the elastic potential energy initially stored in the compressed spring is completely converted into gravitational potential energy of the dart when the dart is at its maximum height. Therefore, we can write:

\frac{1}{2}kx^2 = mgh

where the term on the left represents the elastic potential energy of the spring while the term on the right is the gravitational potential energy of the dart at maximum height, and where

k is the spring constant of the spring

x = 2.08 cm = 0.0208 m is the compression of the spring

m = 12.3 g = 0.00123 kg is the mass of the dart

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

h = 3.25 m is the maximum height of the dart

Solving for k, we find:

k=\frac{2mgh}{x^2}=\frac{2(0.00123)(9.8)(3.25)}{(0.0208)^2}=181.0 N/m

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

6 0
3 years ago
2. An alternating current is represented by the equation I=20sin 100mt.
Sladkaya [172]

Explanation:

The general equation of an AC current is given by :

I=I_o sin\omega t

Where

I₀ is the peak value of current

\omega is angular frequency

As\ \omega=2\pi f

So,

f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{100\pi}{2\pi}\\\\f=50\ Hz

We know that,

I_{rms}=\dfrac{I_0}{\sqrt{2}}\\\\=\dfrac{20}{\sqrt{2}}\\\\I_{rms}=14.14\ A

So, the frequency is 50 Hz and the maximum rms value of current is 14.14 A.

3 0
3 years ago
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