What did Thomson’s model of the atom include that Dalton’s model did not have? A)a nucleus
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The component of the force in negative z-direction is -0.144 N.
The given parameters;
- <em>current in the wire, I = 2.7 A</em>
- <em>length of the wire, L = (3.2 i + 4.3j) cm</em>
- <em>magnetic filed, B = 1.24 i</em>
The force on the segment of the wire is calculated as follows;
where;
- <em>θ is the angle wire and magnetic field</em>
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The force on the wire segment will be perpendicular in negative z-direction (applying right hand rule), so there won't be any x and y component of the force.
The angle between the wire and the magnetic field is calculated as follows;
The magnitude of the wire length is calculated as follows;
The component of the force in negative z-direction is calculated as;
Thus, the component of the force in negative z-direction is -0.144 N.
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The change in the speed of the space capsule will be -0.189 m/s.
The average force exerted by each on the other will be 567 N.
The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.
<h3>Given:</h3>Mass of the astronaut, = 126 kg
Speed he acquires, = 2.70 m/s
Mass of the space capsule, = 1800kg
The initial momentum of the astronaut-capsule system is zero due to rest.
= 0
m/s
Therefore,
According, to the impulse-momentum theorem;
FΔt = ΔP
ΔP = m Δv
ΔP = 126×2.70
= 340.2 kgm/sec
t is time interval = 0.600s
F = ΔP/Δt
F = 340.2/0.600
= 567 N
Therefore, the average force exerted by each on the other will be 567 N.
The Kinetic Energy of the astronaut;
K.E =
× 126 ×
= 459.27 J
The Kinetic Energy of the capsule;
K.E =
= ×1800×
= 32.14 J
Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.
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