A screw-jack used to lift a bus is a
1 answer:
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Answer:
a. q2 = 16.4μC, positive charge
b. F = 0.900N
c. downward
Explanation:
a. In order to calculate the charge of the unknown charge you use the following formula, for the electric force between two charges:
(1)
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
r: distance between the charges = 0.300m
q1: charge 1 = -0.550 μC = 0.550*10^-6C
q2: charge 2 = ?
Fe: electric force = 0.900N
The force exerted in the second charge points upward, then, the sign of the second charge is positive because this charge is getting closer to the first one.
You solve the equation (1) for the second charge ans replace the values of the other parameters:
The values of the second charge is 1.64 μC
b. By the third Newton Law, you have that the force exerted in the second charge is equal to the force exerted by the first charge on the second one.
The force exerted on the first charge is 0.900N
c. The charges are attracting between them, then, the force exerted on the first charge is pointing downward.
The electrical potential energy of a charge q located at a point at potential V is given by
Therefore, if the charge must move between two points at potential V1 and V2, the difference in potential energy of the charge will be
In our problem, the electron (charge e) must travel across a potential difference V. So the energy it will lose traveling from the metal to the detector will be equal to
Therefore, if we want the electron to reach the detector, the minimum energy the electron must have is exactly equal to the energy it loses moving from the metal to the detector:
Therefore, if the charge must move between two points at potential V1 and V2, the difference in potential energy of the charge will be
In our problem, the electron (charge e) must travel across a potential difference V. So the energy it will lose traveling from the metal to the detector will be equal to
Therefore, if we want the electron to reach the detector, the minimum energy the electron must have is exactly equal to the energy it loses moving from the metal to the detector:
Answer:
a) m = 993 g
b) E = 6.50 × 10¹⁴ J
Explanation:
atomic mass of hydrogen = 1.00794
4 hydrogen atom will make a helium atom = 4 × 1.00794 = 4.03176
we know atomic mass of helium = 4.002602
difference in the atomic mass of helium = 4.03176-4.002602 = 0.029158
fraction of mass lost = = 0.00723
loss of mass for 1000 g = 1000 × 0.00723 = 7.23
a) mass of helium produced = 1000-7.23 = 993 g (approx.)
b) energy released in the process
E = m c²
E = 0.00723 × (3× 10⁸)²
E = 6.50 × 10¹⁴ J