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Levart [38]
2 years ago
6

When a point charge of q is placed on one corner of a square, an electric field strength of 2 N/C is observed at the center of t

he square. Suppose three identical charges of q are placed on the remaining three corners of the square. What is the magnitude of the net electric field at the center of the square
Physics
1 answer:
bearhunter [10]2 years ago
3 0

Answer:

Explanation:

Net electric field at the centre will be zero .

Since all the charges are equal and they all are symmetrically situated around the centre . So the electric field produced by each will cancel  out each other and hence the resultant electric field will be zero . It happens because electric field is a vector quantity and therefore it adds up vectorially . All the four electric field will form two pairs , in each pair electric fields are acting in opposite direction . So they all cancel out to zero .

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Answer:

c=10\ J/kg^{\circ} C

Explanation:

Given that,

Heat required, Q = 1200 J

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We need to find the specific heat of the object. The heat required to raise the temperature is given by :

Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{1200}{20\times 6}\\\\c=10\ J/kg^{\circ} C

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vodomira [7]

Answer:

We kindly invite you to read carefully the explanation and check the image attached below.

Explanation:

According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:

v_{f} = v_{o}+a\cdot (t-t_{o}) (1)

Where:

v_{o} - Initial velocity, measured in meters per second.

v_{f} - Final velocity, measured in meters per second.

a - Acceleration, measured in meters per square second.

t_{o} - Initial time, measured in seconds.

t - Final time, measured in seconds.

Now we obtain the kinematic equations for thrust and free fall stages:

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v = 900-9.81\cdot (t-30) (3)

Now we created the graph speed-time, which can be seen below.

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Power=F.V

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