A neutron changes into a proton during a radioactive decay process called beta decay (positron), which is option D.
What is beta decay?A beta decay in physics is a nuclear reaction in which a beta particle (electron or positron) is emitted.
A positron is an electron with a positive charge.
During a beta decay, a neutron in the nucleus of the radioactive material suddenly changes into a proton, causing an increase in the atomic number of an element.
Learn more about beta decay at: https://brainly.com/question/25455333
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If 1.02 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved through it?
Answer:
Explanation:
one electron has [tex]1.60217662*10^{-19}~coulombs~then\\\\1.02*10^{20}~electrons------->1.02*10^{20}*1.60217662*10^{-19}~coulombs= 16.3422~coulombs[/tex]
A ship is flying away from Earth at 0.9c (where c is the speed of light). A missile is fired that moves toward the Earth at a speed of 0.5c relative to the ship. How fast does the missile move relative to the Earth
Answer:
the required speed with which the missile move relative to the Earth is -0.727c
Explanation:
Given the data in the question;
relative velocity relation;
u' = u-v / 1 - [tex]\frac{uv}{c^2}[/tex]
so let V[tex]_B[/tex] represent the velocity as seen by an external reference frame; u=V[tex]_B[/tex]
and let V[tex]_A[/tex] represent the speed of the secondary reference frame; v=V[tex]_A[/tex]
hence, u' is the speed of B as seen by A
so
u' = V[tex]_B[/tex]-V[tex]_A[/tex] / 1 - [tex]\frac{V_BV_A}{c^2}[/tex]
now, given that; V[tex]_A[/tex] = 0.9c and V[tex]_B[/tex] = 0.5c
we substitute
u' = ( 0.5c - 0.9c ) / 1 - [tex]\frac{(0.5c)(0.9c)}{c^2}[/tex]
u' = ( 0.5c - 0.9c ) / 1 - [tex]\frac{c^2(0.5)(0.9)}{c^2}[/tex]
u' = ( 0.5c - 0.9c ) / 1 - (0.5 × 0.9)
u' = ( -0.4c ) / 1 - 0.45
u' = -0.4c / 0.55
u' = -0.727c
Therefore, the required speed with which the missile move relative to the Earth is -0.727c
A scenario where reaction time is important is when driving on the highway. During the delay between seeing an obstacle and reacting to avoid it (or to slam on the brakes!) you are still moving at full highway speed. Calculate how much distance you cover in meters before you start to put your foot on the brakes if you are travelling 65 miles per hour.
Answer:
66.83 meters
Explanation:
After a quick online search, it seems that scientists calculate the average reaction time of individuals as 2.3 seconds between seeing an obstacle and putting their foot on the brakes. Now that we have this reaction time we need to turn the miles/hour into meters/second.
1 mile = 1609.34 meters (multiply these meters by 65)
65 miles = 104,607 meters
1 hour = 3600 seconds
Therefore the car was going 104,607 meters every 3600 seconds. Let's divide these to find the meters per second.
[tex]\frac{104,607}{3600} = \frac{29.0575 meters}{1 second}[/tex]
Now we simply multiply these meters by 2.3 seconds to find out the distance covered before the driver puts his/her foot on the brakes...
29.0575m * 2.3s = 66.83 meters
A 2 kg object traveling at 5 m s on a frictionless horizontal surface collides head-on with and sticks to a 3 kg object initially at rest. Which of the following correctly identifies the change in total kinetic energy and the resulting speed of the objects after the collision? Kinetic Energy Speed
(A) Increases 2 m/s 3.2 m/s
(B) Increases Soold 2 m/s
(C) Decreases 3.2 m/s
(D) Decreases
Answer: (d)
Explanation:
Given
Mass of object [tex]m=2\ kg[/tex]
Speed of object [tex]u=5\ m/s[/tex]
Mass of object at rest [tex]M=3\ kg[/tex]
Suppose after collision, speed is v
conserving momentum
[tex]\Rightarrow mu+0=(m+M)v\\\\\Rightarrow v=\dfrac{2\times 5}{2+3}\\\\\Rightarrow v=2\ m/s[/tex]
Initial kinetic energy
[tex]k_1=\dfrac{1}{2}\times 2\times 5^2\\\\k_1=25\ J[/tex]
Final kinetic energy
[tex]k_2=\dfrac{1}{2}\times (2+3)\times 2^2\\\\k_2=10\ J[/tex]
So, it is clear there is decrease in kinetic energy . Thus, energy decreases and velocity becomes 2 m/s.
What is 3*10^-6 divided by 2.5*10^6 expressed in standard notation?
Answer:
1.2 x 10^-12
Explanation:
3/2.5 x 10^-6/10^6
1.2 x 10^-6 x 10^-6
1.2 x 10^-12
A strong trough in a Rossby wave occurs when the jet stream A. bends towards the Equator. B. bends toward the poles. C. does not bend but maintains an east to west flow. D. does not bend but maintains a west to east flow.
Answer:
A. bends towards the Equator.
Explanation:
Rossby waves are inertial waves that are naturally occurring in a rotating fluids. These waves are also called as the planetary waves.
The Rossby waves are undulated that occur in the polar front jet stream when there is a significant differences in the temperatures between the polar and the tropical air masses.
It occurs when the polar air masses moves towards the equator and when the tropical air masses moves towards the pole. It is formed when the air bends away from the poles and bends towards the equator.
Hence the correct option is (A).
