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Lesechka [4]
3 years ago
8

60m27Co → 6027Co Predict the type of radioactive emission produced from the decay of metastable cobalt-60 to cobalt-60. Describe

this type of emission and its reaction to an electric field. A) During the radioactive decay, alpha particles are released. These positive particles are attracted to the negative plate in the electric field and represent a ground energy level. B) Beta particles are released during the radioactive decay. These negative particles are attracted to the positive plate in the electric field and represent an excited energy state. C) Radioactive gamma decay is produced by the reaction. This neutral electromagnetic radiation allows the isotope to return to its ground state and is not attracted to the electric field. D) Both types of radioactive emissions, particles and electromagnetic radiation, are produced during this decay. None of these are attracted to the electric field and both present an intermediate level. Eliminate
Physics
1 answer:
lara [203]3 years ago
4 0
Some one already asked this question and you can copy paste and google it but I believe it is c you may want to double check
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Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field
Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength  at the midpoint between the two rings is given as;

E_{mid} = E_{right} +E_{left}\\\\E_{right}  = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt}  = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }  - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0
3 years ago
The instantaneous velocity of an object is the blank of the object with a blank
777dan777 [17]
The instantaneous velocity of the object is its speed and direction at that instant.
7 0
3 years ago
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Texture hope this helps! :)

5 0
3 years ago
Cole is riding a sled with initial speed of 5 m/s from west to east. the frictional force of 50 n exists due west. the mass of t
stepan [7]
We can calculate the acceleration of Cole due to friction using Newton's second law of motion:
F=ma
where F=-50 N is the frictional force (with a negative sign, since the force acts against the direction of motion) and m=100 kg is the mass of Cole and the sled. By rearranging the equation, we find
a= \frac{F}{m}= \frac{-50 N}{100 kg}=-0.5 m/s^2

Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
a= \frac{v_f^2-v_i^2}{2d}
where
v_f=0 is the final speed of the sled
v_i=5 m/s is the initial speed
d is the distance covered

By rearranging the equation, we find d:
d= \frac{v_f^2-v_i^2}{2a}= \frac{-(5 m/s)^2}{2 \cdot (-0.5 m/s^2)}=25 m
3 0
3 years ago
Puck 1 (1 kg) travels with velocity 20 m/s to the right when it collides with puck 2 (1 kg) which is initially at rest. After th
Burka [1]

Answer:

Explanation:

Parameters given:

Mass of Puck 1, m = 1 kg

Mass of Puck 2, M = 1 kg

Initial velocity of Puck 1, u = 20 m/s

Initial velocity of Puck 2, U = 0 m/s

Final velocity of Puck 1, v = 5 m/s

Since we are told that momentum is conserved, we apply the principle of conservation of momentum:

Total initial momentum of the system = Total final momentum of the system

mu + MU = mv + MV

(1 * 20) + (1 * 0) = (1 * 5) + (1 * V)

20 = 5 + V

V = 20 - 5 = 15 m/s

Puck 2 moves with a velocity of 15 m/s

7 0
3 years ago
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