Answer:
The value is
Explanation:
From the question we are told that
The mass of matter converted to energy on first test is
The mass of matter converted to energy on second test
Generally the amount of energy that was released by the explosion is mathematically represented as
=>
=>
Yes, the ions can exit slit P without being deflected, if the electric field strength is 170.6 N/C
Explanation:
When the ions are inside the container, they are subjected to two forces, with directions opposite to each other:
The ions will move straight and undeflected if the two forces are equal and opposite. By using Fleming Left Hand rule, we notice that the magnetic force on the (negative) ions point upward: this means that the electric field must be also upward (so that the electric force on the ions is downward). Then, the two forces are balanced if
which translates into
Therefore, if the speed of the ions is equal to this ratio, the ions will go undeflected.
We can even calculate the value of E at which this occurs. In fact, we know that the ions are earlier accelerated by a potential difference , so we have that their kinetic energy is given by the change in electric potential energy:
where
Solving for v, the speed,
And since the magnetic field strength is
B = 0.83 T
The strength of the electric field must be
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Answer:
s = 1.7 m
Explanation:
from the question we are given the following:
Mass of package (m) = 5 kg
mass of the asteriod (M) = 7.6 x 10^{20} kg
radius = 8 x 10^5 m
velocity of package (v) = 170 m/s
spring constant (k) = 2.8 N/m
compression (s) = ?
Assuming that no non conservative force is acting on the system here, the initial and final energies of the system will be the same. Therefore
• Ei = Ef
• Ei = energy in the spring + gravitational potential energy of the system
• Ei = \frac{1}{2}ks^{2} + \frac{GMm}{r}
• Ef = kinetic energy of the object
• Ef = \frac{1}{2}mv^{2}
• \frac{1}{2}ks^{2} + (-\frac{GMm}{r}) = \frac{1}{2}mv^{2}
• s =
s =
s = 1.7 m