A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 5.12 g coins stacked over the 24.2 cm mark, th
1 answer:
Answer:
The mass of the meter stick is
Explanation:
Here since the meter stick is in equilibrium position its net torque and net force should be equal to zero
Since at the begging the meter stick is balanced at center of the meter stick that means its center of mass should be present at
Now lets consider the later case where stick is balanced by two 5.12 g coins .
Here torque due to two coins =
Torque due to weight of meter stick = =
where m = mass of the meter stick
Here .
Upon equating we will be getting mass of the meter stick =
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We can solve the problem by using Ohm's law, which states that an Ohmic conductor the following relationship holds:
where
is the potential difference applied to the resistor
I is the current flowing through it
R is the resistance
In our problem, I=4.00 A and
, so the potential difference is
where
I is the current flowing through it
R is the resistance
In our problem, I=4.00 A and
Answer:
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:
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:
The initial potential energy of the spring is given by the equation:
the Kinetic energy of the block is then given by the equation:
so we can now set them both equal to each other, so we get:
This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:
so now we can solve this for the final velocity, so we get:
