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Fynjy0 [20]
4 years ago
12

What must be the distance between point charge 91 = 20 MC and point charge 02 -45,0 AC for the electrostatic force between them

to have a magnitude of 7.0 N?
Physics
1 answer:
Fynjy0 [20]4 years ago
7 0

Answer:

Explanation:

Q₁ = 20 X 10⁻⁶ C

Q₂ = -45 X 10⁻⁶ C

d is required distance.

F = Force between the = 7 N

F = k x Q₁ X Q₂ / d²

d² = k x Q₁ X Q₂ / F = 20 X 45 X 10⁻¹² X 9 X 10⁹  /7

=1.157

d = 1.075 m

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BigorU [14]

It shows that acceleration of particle M is directly proportional to its displacement and its direction is opposite to that of displament. Thus particle M performs simple harmonic motion but M is projection of particle performing U.C.M. hence S.H.M. is projection of U.C.M. along a diameter, of circle.

4 0
3 years ago
A(n) 69.8 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 60.4 m away from the shuttle
alukav5142 [94]

Answer:

The time taken is 6.7  min

Explanation:

Using the linear momentum conservation theorem, we have:

m_1*v_{o1}+m_2*v_{o2}=m_1*v_{f1}+m_2*v_{f2}

when she was 60.4m from the shuttle, she has zero speed, so the initial velocity is zero.

m_1*0+m_2*0=m_1*v_{f1}+m_2*v_{f2}\\m_1*_{f1}=-m_2*v_{f2}\\v_{f1}=-\frac{m_2*v_{f2}}{m_1}\\\\v_{f1}=-\frac{0.886kg*12m/s}{69.8kg}\\\\V_{f1}=-0.15m/s

That is 0.15m/s in the opposite direction of the camera.

the time taken to get to the shuttle is given by:

t=\frac{d}{v_{f1}}\\\\t=\frac{60.4m}{0.15m/s}\\\\t=403s\\t_{min}=403s*\frac{1min}{60s}=6.7min

6 0
3 years ago
What two kinds of motion combine to produce projectile motion?
solong [7]
The two kinds of motion are horizontal and vertical
3 0
3 years ago
A 2190 kg car moving east at 10.5 m/s collides with a 3220 kg car moving east. The cars stick together and move east as a unit a
Bezzdna [24]

To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:

m_1v_1+m_2v_2 = (m_1+m_2)V_f

Where,

m_{1,2}= Mass of each object

v_{1,2}= Initial Velocity of Each object

V_f= Final Velocity

Our values are given as

m_1 = 2190Kg

v_1 =10.5m/s

m_2 = 3220kg

V_f = 4.74m/s

Replacing we have that

m_1v_1+m_2v_2 = (m_1+m_2)V_f

(2190)(10.5)+(3220)v_2 = (2190+3220)(4.74)

v_2 = 0.8224m/s

Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s

8 0
4 years ago
A solenoid that is 66.1 cm long has a cross-sectional area of 13.8 cm2. There are 1410 turns of wire carrying a current of 8.01
Ymorist [56]

Answer:

a) Energy density of the magnetic field, u = 183.46 J/m³

b) Total energy, E = 0.167 J

Explanation:

a) Number of turns in the solenoid, N = 1410

Area, A = 13.8 cm² = 0.00138 m²

Current, I = 8.01 A

Length of the solenoid, l = 66.1 cm = 0.661 m

Energy density, u is given by the formula u = \frac{B^2}{2 \mu_{0} }

Where B is the magnetic field

The magnetic field in a solenoid is given by the formula, B = \frac{N \mu_{0} I }{l}

B = \frac{1410 * 8.01* \mu_{0}  }{0.661}

B = 17086.38 \mu_0 T

u = \frac{(17086.38 \mu_0)^2}{2 \mu_{0} }\\u = \frac{291944527.29 \mu_0^2}{2 \mu_{0} }\\u = \frac{291944527.29 \mu_0}{2  }\\u = \frac{291944527.29 * 4\pi * 10^{-7} }{2  }\\u = 183.46 J/m^3

b) The total energy = Energy density * Volume

E = u V

Volume = Area * Length

V = Al = 0.00138 * 0.661

V = 0.00091218 m³

E = 183.46 * 0.00091218

E = 0.167 J

4 0
3 years ago
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