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.
<span>An object roating at one revokution per second has an angular velocity of 360 degrees per second or 2pi radians per second. This is found by taking the number of revolutions over a period of time and than dividing by the chosen period of time to get the velocity. There are 360 degrees or 2pi radians in one revolution.</span>
The final temperature of the seawater-deck system is 990°C.
<h3>What is heat?</h3>
The increment in temperature adds up the thermal energy into the object. This energy is Heat energy.
The deck of a small ship reaches a temperature Ti= 48.17°C seawater on the deck to cool it down. During the cooling, heat Q =3,710,000 J are transferred to the seawater from the deck. Specific heat of seawater= 3,930 J/kg°C.
Suppose for 1 kg of sea water, the heat transferred from the system is given by
3,710,000 = 1 x 3,930 x (T - 48.17)
T = 990°C to the nearest tenth.
The final temperature of the seawater-deck system is 990°C.
Learn more about heat.
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At point C because it is at the lowest position.
Answer:
period of oscillations is 0.695 second
Explanation:
given data
mass m = 0.350 kg
spring stretches x = 12 cm = 0.12 m
to find out
period of oscillations
solution
we know here that force
force = k × x .........1
so force = mg = 0.35 (9.8) = 3.43 N
3.43 = k × 0.12
k = 28.58 N/m
so period of oscillations is
period of oscillations = 2π × 
................2
put here value
period of oscillations = 2π × 
period of oscillations = 0.6953
so period of oscillations is 0.695 second
