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mote1985 [20]
3 years ago
10

A point charge with a charge q1 = 3.60 μC is held stationary at the origin. A second point charge with a charge q2 = -4.60 μC mo

ves from the point x= 0.100 m , y= 0 to the point x= 0.250 m , y= 0.250 m .How much work W is done by the electric force on the moving point charge?
Physics
1 answer:
nikdorinn [45]3 years ago
4 0

Answer:

W = -1.069 J

Explanation:

To find the work (W) of the electric force, we can use the formula:

W = q.k₀.Q(1/r₁ - 1/r₂)

q = moving point charge

k₀ = 9.10⁹ Nm²/C²

Q = stationary charge

r₁ = distance from origin to starting point

r₂ = distance from origin to ending point

r₁ = 0.100 - 0 = 0.1

r₂ = √((0.25-0)² + (0.25-0)²) = √(0.25² + 0.25²) = √0.125 = 0.3536

W = -4.6*10⁻⁶*9*10⁹*3.6*10⁻⁶(1/0.1 - 1/0.3536)

W = -149.04*10⁻³(10 - 2.8284)

W = -149.04*10⁻³*7.1716 = -1068.85*10⁻³ = -1.069 J

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I need help with this too. (im not good at science or math)
mr_godi [17]

Answer:

i belive 1 m/s

Explanation:

dividing displacement from time it should be 1 cuz 5/5 is 1

please tell me if right!

6 0
3 years ago
Read 2 more answers
Two particles, one with charge − 3.77 μC −3.77 μC and one with charge 4.39 μC, 4.39 μC, are 4.34 cm 4.34 cm apart. What is the m
Fudgin [204]

Answer:

the magnitude of the force that one particle exerts on the other is 79.08 N

Explanation:

given information:

q₁ = 3.77 μC = -3.77 x 10⁻⁶ C

q₂ = 4.39 μC = 4.39 x 10⁻⁶ C

r = 4.34 cm = 4.34 x 10⁻² m

What is the magnitude of the force that one particle exerts on the other?

lFl = kq₁q₂/r²

   = (9 x 10⁹) (3.77 x 10⁻⁶) (4.39 x 10⁻⁶)/(4.34 x 10⁻²)²

   = 79.08 N

8 0
3 years ago
Please show steps as to how to solve this problem <br> Thank you!
liubo4ka [24]

Answer:

Torques must balance

F1 * X1 = F2 * Y2

or M1 g X1 = M2 g X2

X2 = M1 / M2 * X1 = 130.4 / 62.3 * 10.7

X2 = 22.4 cm

Torque = F1  * X2 =

62.3 gm* 980 cm/sec^2  * 22.4 cm = 137,000 gm cm^2 / sec^2

Normally x cross y   will be out of the page

r X F  for F1 will be into the page so the torque must be negative

6 0
3 years ago
An alligator swims to the left with a constant velocity of 5 \,\dfrac{\text{m}}{\text s}5
dexar [7]

Answer:

The alligator will take t = 10 s to reach the final speed of 35 m/s

Explanation:

As we know that the initial speed of the alligator is 5 m/s

then it accelerate by given acceleration to reach the final speed of 35 m/s

so we will have

v_i = 5 m/s

v_f = 35 m/s

a = 3m/s^2

now we have

v_f = v_i + at

35 = 5 + 3 t

t = 10 s

5 0
3 years ago
An artificial satellite of mass 900kg Is launched at a speed of 11,000 m/s from its launching station. How much is the kinetic e
Marina CMI [18]

Answer:

KE = 54450 J

Explanation:

Data:

  • m = 900 kg
  • v = 11.000 m/s
  • KE = ?

Use the formula:

  • \boxed{\bold{K_{E}=\frac{m*(v)^{2}}{2}}}

Replace and solve:

  • \boxed{\bold{K_{E}=\frac{900\ kg*(11.000\frac{m}{s})^{2}}{2}}}
  • \boxed{\bold{K_{E}=\frac{900\ kg*121\frac{m^{2}}{s^{2}}}{2}}}
  • \boxed{\bold{K_{E}=\frac{108900\ J}{2}}}
  • \boxed{\boxed{\bold{K_{E}=54450\ J}}}

A kinetic energy of <u>54450 Newtons</u> is imparted to it.

Greetings.

6 0
3 years ago
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