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melamori03 [73]
3 years ago
15

When an atom releases gamma radiation____________.

Physics
1 answer:
wel3 years ago
3 0

Answer:

d. the atomic number remains the same.

Explanation:

First at all it's important to know how to read nuclear information

_{Z}^{A}X

X is the atomic symbol, A the mass number and Z the atomic number of the element.

Gamma rays emitted on gamma decay are characterized as _{0}^{0}\gamma

If we write the nuclear equation for the decay, we have that:

_{Z}^{A}X\,\rightarrow{}_{0}^{0}\gamma\,+\,{}_{Z}^{A}Y

The sum of the mass numbers and atomic numbers on the right side has to be equal to the left side numbers of the equation, that means the mass number and the atomic number remains the same for the resulting atom to preserve the equality.

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A 50.0 g toy car is released from rest on a frictionless track with a vertical loop of radius R (loop-the-loop). The initial hei
Mariana [72]

Answer:

the speed of the car at the top of the vertical loop  v_{top} = 2.0 \sqrt{gR \ \ }

the magnitude of the normal force acting on the car at the top of the vertical loop   F_{N} = 1.47 \ \ N

Explanation:

Using the law of conservation of energy ;

mgh = mg (2R) + \frac{1}{2}mv^2_{top}\\\\mg ( 4.00 \ R) = mg (2R) + \frac{1}{2}mv^2_{top}\\\\g(4.00 \ R) = g (2R) + \frac{1}{2}v^2 _{top}\\\\v_{top} = \sqrt{2g(4.00R - 2R)}\\\\v_{top} = \sqrt{2g(4.00-2)R

v_{top} = 2.0 \sqrt{gR \ \ }

The  magnitude of the normal force acting on the car at the top of the vertical loop can be calculated as:

F_{N} = \frac{mv^2_{top}}{R} \ - mg\\\\F_{N} = \frac{m(2.0 \sqrt{gR})^2}{R} \ - mg\\\\F_{N} = [(2.0^2-1]mg\\\\F_{N} = [(2.0)^2 -1) (50*10^{-3} \ kg)(9.8 \ m/s^2]\\\\

F_{N} = 1.47 \ \ N

4 0
3 years ago
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