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miv72 [106K]
3 years ago
5

Describe a situation that includes no less than four charges of any magnitude, but they combine so that another location, p, has

no net electric field at that point. Describe where those charges could be or what magnitudes they could be

Physics
1 answer:
Degger [83]3 years ago
6 0

Answer:

Four charges of equal magnitude sitting at the vertices of a square

Explanation:

We can arrive at such a situation by thinking of a simple example first, a configuration of two charges. The force acting on the middle point of a straight line joining the two points(charges) will be zero. That is, the net Electric field will be zero as they cancel out being equal in magnitude and opposite in direction.

Now, we can extend this idea to a square having charge q at each vertex. If we put 'p' at the geometric center, we can see that the Electric fields along the diagonals cancel out due to the charges at the diagonally opposite vertices(refer to the figure attached). Actually, the only requirement is that the diagonally opposite charges are equal.

We can further take this to 3 dimensions. Consider a cube having charges of equal magnitude at each vertex. In this case, the point 'p' will yet again be the geometric center as the Electric field due to the diagonally opposite charges will cancel out.

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Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block (W), measured in newtons, is:

W = m\cdot g (1)

Where:

m - Mass of the block of ice, measured in kilograms.

g  - Gravitational acceleration, measured in meters per square second.

If we know that m = 14\,kg and g = 9.807\,\frac{m}{s^{2}}, the magnitudes of the weight and normal force of the block of ice are, respectively:

N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

N = W = 137.298\,N

And the pull force is:

F_{pull} = 98\,N

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f} (2)

Where:

v_{o}, v_{f} - Initial and final speeds of the block, measured in meters per second.

\Sigma F - Horizontal net force, measured in newtons.

\Delta t - Impact time, measured in seconds.

Now we clear the final speed in (2):

v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}

If we know that v_{o} = 0\,\frac{m}{s}, m = 14\,kg, \Sigma F = 98\,N and \Delta t = 1.40\,s, then final speed of the ice block is:

v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}

v_{f} = 9.8\,\frac{m}{s}

The final speed of the block of ice is 9.8 meters per second.

6 0
2 years ago
If a plane is moving at a constant velocity what is happening to the acceleration?
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The plane is not accelerating.

Hope this helps!
4 0
3 years ago
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A goose with a mass of 2.0 kg strikes a commercial airliner with a mass of 160,000 kg head-on. Before the collision, the goose w
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Answer:

The change in momentum of the goose during this interaction is 33.334 m/s

Explanation:

Given;

mass of goose, m₁ = 2.0 kg

mass of commercial airliner, m₂ = 160,000 kg

initial velocity of the bird, u₁ = 60 km/hr  = 16.667 m/s

initial velocity of the airliner, u₂ = 870 km/hr = 241.667 m/s

Change in momentum is given as;

ΔP = mv - mu

where;

u is the initial velocity of the bird

v is the final velocity of the bird

Apply the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the final velocity of bird and airliner after collision;

(2 x 16.667) + (160,000 x 241.667) = v (2 + 160,000)

38,666,753.334 = 160,002v

v = 38,666,753.334 / 160,002

v = 241.664 m/s

Thus, the final velocity of the bird is negligible compared to final  velocity of the airliner.

ΔP = mv - mu

ΔP = m(v - u)

ΔP = 2(0 - 16.667)

ΔP = -33.334 m/s

The negative sign implies a deceleration of the bird after the impact.

Therefore, the change in momentum of the goose during this interaction is 33.334 m/s

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Two students are watching a person riding a skateboard up and down a ramp. Each student shares what they think about the energy
Setler79 [48]

Answer:

The correct option is;

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Explanation:

The total energy, also known as the total mechanical energy, is the sum of the kinetic and potential energies of the skateboarder

Given that the potential energy is the energy gained due to elevation, the maximum potential energy is obtained at the top of the ramp, while the maximum kinetic energy, which is the energy due to motion, is at the bottom of the ramp where the skateboarder moves fastest.

However, by the energy conservation principle, the kinetic energy of he skateboarder comes from the conversion of the potential energy, such that the total energy is the same at any particular point on the ramp.

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