Because they have different measurements and weight and mass and some measurements are the same
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Which nuclei is NOT radioactive?
A) Am-241 B) Mg-24 C) Pu-241 D) U-238
Solving:
It is noteworthy that chemical elements located on the periodic table in the lanthanide and actinide groups are radioactive.
Am-241 (americium) belongs to the group of actinides and is a heavy and radioactive metal.
Mg-24 (magnesium) is an essential element for the body, mainly for the nervous system, in addition to synthesizing proteins and serves for hormonal control, belongs to the group of alkaline earth metals and is a non-radioactive nucleus.
Pu-241 (plutonium) is an element that is isotope of fission by plutonium, belongs to the group of actinides and is a heavy and radioactive metal.
U-238 (uranium) is an element that is isotope of non-fission uranium, belongs to the group of actinides and is a heavy and radioactive metal.
Answer:
B) Mg-24
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Answer:

Explanation:
Change in velocity considering the x component will be
Final velocity-Initial velocity

Change in velocity considering the y component will be
Final velocity-Initial velocity

Resultant change in velocity
Acceleration= change in velocity per unit time hence

Answer:
17.6 N
Explanation:
The force exerted by the punter on the football is equal to the rate of change of momentum of the football:

where
is the change in momentum of the football
is the time elapsed
The change in momentum can be written as

where
m = 0.55 kg is the mass of the football
u = 0 is the initial velocity (the ball starts from rest)
v = 8.0 m/s is the final velocity
Combining the two equations and substituting the values, we find the force exerted on the ball:

The magnetic field at center of circular loops of wire is 3.78 x 10¯⁵ T.
We need to know about the magnetic field at the center of circular loops of wire to solve this problem. The magnetic field at the center can be determined as
B = μ₀ . I / 2r
where B is magnetic field, μ₀ is vacuum permeability (4π×10¯⁷ H/m), I is the current and r is radius.
From the question above, we know that:
r = 4 cm = 0.04 m
I = 1.7 A
By substituting the parameter, we get
B = μ₀ . I / 2r
B = 4π×10¯⁷ . 1.7 / (2.0.04)
B = 2.67 x 10¯⁵ T
Due to the perpendicular plane of loops, the total magnetic field at center will be
Btotal = √(2(B²))
Btotal = √(2(2.67 x 10¯⁵²))
Btotal = 3.78 x 10¯⁵ T
Find more on magnetic field at: brainly.com/question/7802337
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