A golfer starts with the club over her head and swings it to reach maximum speed as it contacts the ball. Halfway through her sw
ing, when the golf club is parallel to the ground, does the acceleration vector of the club head point straight down, parallel to the ground, approximately toward the golfer's shoulders, approximately toward the golfer's feet, or toward a point above the golfer's head?
A. Parallel to the ground, approximately toward the golfer's shouldersB. Straight downC. Approximately toward the golfer's feetD. Toward a point above the golfer's head
A. Parallel to the ground, approximately toward the golfer's shouldersB. Straight downC. Approximately toward the golfer's feetD. Toward a point above the golfer's head
1 answer:
Answer:
A. Parallel to the ground, approximately toward the golfer's shoulders
Explanation:
Elliptical path represent the motion of the golf club.
From the figure we have the motion of the club. So the acceleration provided will be centripetal acceleration so halfway through the swing the direction of acceleration will be towards the golfer's shoulder.
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Answer:
(A). The resulting nucleus is Fr.
(B). The resulting nucleus is Po.
(C). The resulting nucleus is Ne
(D). The resulting nucleus is Tc.
Explanation:
Given that,
A nucleus undergoes a nuclear decay.
(A). In alpha decay,
We know that,
When the nucleus emit alpha particle then atomic mass of particle reduce by 4 and atomic number reduce by 2.
We need to calculate the resulting nucleus
Using given data
The resulting nucleus is Fr.
(B). In beta-minus decay,
We know that,
When the nucleus emit beta- minus particle then atomic mass of particle is same and atomic number increase by 1.
We need to calculate the resulting nucleus
Using given data
The resulting nucleus is Po.
(C). In beta-plus decay,
We know that,
When the nucleus emit beta- plus particle then atomic mass of particle is same and atomic number decrease by 1.
We need to calculate the resulting nucleus
Using given data
The resulting nucleus is Ne.
(D). In gamma decay,
We know that,
When the nucleus emit gamma particle then atomic mass and atomic number of particle is same.
We need to calculate the resulting nucleus
Using given data
The resulting nucleus is Tc.
Hence, (A). The resulting nucleus is Fr.
(B). The resulting nucleus is Po.
(C). The resulting nucleus is Ne
(D). The resulting nucleus is Tc.
Answer:
Given:
Thermal Kinetic Energy of an electron,
= Boltzmann's constant
Temperature, T = 1800 K
Solution:
Now, to calculate the de-Broglie wavelength of the electron, :
(1)
where
h = Planck's constant =
= momentum of an electron
= velocity of an electron
= mass of electon
Now,
Kinetic energy of an electron = thermal kinetic energy
(2)
Using eqn (2) in (1):
Now, to calculate the de-Broglie wavelength of proton, :
(3)
where
= mass of proton
= velocity of an proton
Now,
Kinetic energy of a proton = thermal kinetic energy
(4)
Using eqn (4) in (3):