B. Electrons are negative so one single 1 would be negative 1
Time taken by the package to reach the sea level= 13.7 s
height=h=925 m
initial velocity along vertical= vi=0
acceleration due to gravity=g=9.8 m/s^2
using the kinematic equation h= Vi*t + 1/2 gt^2
925=0(t)+1/2 (9.8)t^2
4.9 t^2=925
t= 13.7 s
If the lightbulb A in the circuit shown in the image burned out, the path for the current to flow is disrupted because one of its terminals is connected direct to the source. So, there will be no current through the lightbulbs B, C, and D, and they will turn off. Similarly it will happen, if the lightbulb D burned out.
If the lightbulb B burned out the current will continue circulating through the lightbulbs A, C, and D, because lightbulb B is connected in parallel. Similarly it will happen, if the lightbulb C burned out.
Answer:
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Explanation:
In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:
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In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:
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The initial potential energy of the spring is given by the equation:
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the Kinetic energy of the block is then given by the equation:
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so we can now set them both equal to each other, so we get:
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This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:
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so now we can solve this for the final velocity, so we get:
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Answer:
E = 0 r <R₁
Explanation:
If we use Gauss's law
Ф = ∫ E. dA =
/ ε₀
in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.
Consequently by Gauss's law the electric field is ZERO
E = 0 r <R₁