A 20kg mass approaches a spring at a speed of 30 m/s. The mass compresses the spring 12cm before coming to a stop. Calculate the
1 answer:
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The field lines spread apart as we move away from the charge, and they point away from the charge
Explanation:
The electric field produced by a single-point positive charge is a radial field, whose strength is given by the equation
where
k is the Coulomb's constant
Q is the magnitude of the charge
r is the distance from the charge at which the field is calculated
There are two pieces of information given by the field lines shown in the graph:
- The spacing between the lines gives an indication of the strength of the field: the closer to each other they are, the stronger the field. In this case, as we move away from the charge, the spacing between the lines increases, and this means that the field becomes weaker (in fact, it follows an inverse square law,
- The direction of the lines gives the direction of the electric field, which points away from the central charge. This is because the direction of the electric field corresponds to the direction of the force that a positive test charge would feel when immersed in the electric field: in this case, if we place a positive test charge in this field, then it would get repelled away from the central charge (remember that the electric force between two positive charges is repulsive), and therefore, the direction of the electric field is away from the central charge.
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Answer:
e% = 3.4%
Explanation:
This is a calorimetry problem where the heat released equals the heat absorbed
m (T₀ - T_f) = M c_{e2} (T₁ - T_f)
Index 1 refers to water and index 2 to metal, in this case it asks for the specific heat of the metal (c_{e2})
c_{e2} = m / M c_{e1} (T_f -T₀) / (T₁ - T_f)
Let's calculate
c_{e} = 60/100 4.19 (24-20) / (100-24)
c_{e2} = 0.1323 j / gC
This metal is possibly lead, which is its specific heat is 0.128 J / gC
The percentage error is
e% = (c_{e2} - 0.128) /0.128 100
e% = 3.4%