Answer:
The magnitude of the electric field equal to 2.40 N/C at 1.1537m from the wire.
Explanation:
using Guass law,
(guessing that a cylinder of radius r and length L around wire such that wire is at centre )
E. A = qin / e0
E ( 2πr L ) = (1.56 x 
x L) / (8.85 x 
)
E = (1.56 x ) / (2πr x 8.85 x )
so 2.40 = (1.54 x ) / (2πr x 8.85 x )
2.40 (6.284r) = 0.174 x 10²
15.0816r = 17.4
r = 1.1537m
Answer:
c
Explanation:
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Answer:Falling objects form an interesting class of motion problems. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. By applying the kinematics developed so far to falling objects, we can examine some interesting situations and learn much about gravity in the process.
Explanation:
Acceleration = force/mass
f=5
m=25
5/25 = 0.2
Answer:
Description 1 matches C.
Description 2 matches A.
Description 3 matches D.
Description 4 matches B.
Explanation:
Description 1 matches C. From graph C, we can see that the <em>distance remains constant at a particular value</em>. Hence, the car is <em>not traveling</em> and hence it is <em>stopped.</em>
Description 2 matches A. <em>Speed is determined by the gradient of a distance-time graph</em>. A depicts a <em>linear graph</em> which tells us that the <em>gradient is constant </em>and hence the car is at <em>constant speed</em>. The fact that the<em> distance is increasing</em> shows that the<em> car is moving forward</em>.
Description 3 matches D. Graph of D becomes <em>less steeper as time progresses</em>. This tells us that the <em>gradient of the graph is decreasing </em>and hence the <em>speed of the car is decreasing.</em>
<em />
Description 4 matches B. Graph B shows <em>distance being decreased.</em> This tells us that the <em>car is coming back</em>.
