After impairing memory, focus, and reasoning, alcohol next causes a deterioration in __________.
1 answer:
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Explanation:
It is given that,
Magnetic field, B = 0.15 T
Charge on a proton,
Mass of a proton,
The cyclotron frequency is given by :
f = 2286785.40 Hz
or
Hence, this is the required solution.
Answer:
h '= [ ( \frac{ M-m}{M +m } )+ 2 (\frac{M}{M+m})]² h
Explanation:
Let's analyze this problem, first the two bodies travel together, second the superball bounces, third it collides with the marble and fourth the marble rises to a height h ’
let's start by finding the velocity of the two bodies just before the collision, let's use the concepts of energy
starting point. Starting point
Em₀ = U = m g h
final point. Just before the crash
Em_f = K = ½ m v²
as there is no friction the mechanical energy is conserved
Em₀ = Em_f
mg h = ½ m v²
v = √2gh
this speed is the same for the two bodies.
Second point. The superball collides with the ground, this process is very fast, so we will assume that the marble has not collided, let's use the concept of conservation of moment
initial instant. Just when the superball starts contacting the ground
p₀ = M v
this velocity is negative because it points down
final instant. Just as the superball comes up from the floor
p_f = M v '
the other body does not move
p₀ = p_f
- m v = M v '
v ’= -v
Therefore, the speed of the asuperbola is the same speed with which it arrived, but in the opposite direction, that is, upwards.
Let's use the subscript 1 for the marble and the subscript 2 for the superball
Third part. The superball and the marble collide
the system is formed by the two bodies, so that the forces during the collision are internal and the moment is conserved
initial instant. Moment of shock
p₀ = M + m v_2
final instant. When the marble shoots out.
P_f = Mv_{1f'}+ m v_{2f}
p₀ = p_
M v_{1'}+ m v_2 = M v_{1f'} + m v_{2f}
M (v_{1'} - v_{1f'}) = -m (v_2 - v_{2f})
in this expression we look for the exit velocity of the marble (v2f), as they indicate that the collision is elastic the kinetic nerve is also conserved
K₀ = K_f
½ M v_{1'}² + m v₂² = M v_{1f'}² + ½ m v_{2f}²
M (v_{1'}² - v_{1f'}²) = - m (v₂² - v_{2f}²)
Let's set the relation (a + b) (a-b) = a² - b²
M (v_{1'} + v_{1f'}) (v_{1'} - v_{1f'}) = -m (v₂ + v_{2f}) (v₂ - v_{2f})
let's write our two equations
M ( v_{1'} - v_{1f'}) = -m (v₂ - v_{2f}) (1)
M (v_{1'} + v_{1f'}) (v_{1'} - v_{1f'}) = -m (v₂ + v_{2f}) (v₂ - v_{2f})
if we divide these two expressions
(v_{1'}+ v_{1f'}) = (v₂ + v_{2f} )
we substitute this result in equation 1 and solve
v_{1f'}= (v₂ + v_{2f}) - v_{1'}
M (v_{1'} - [(v₂ + v_{2f}) - v_{1'}] = -m (v₂ - v_{2f})
-M v₂ - M v_{2f1'} + 2M v_{1'} = m v₂ - m v_{2f}
-M v_{2f} -m v_{2f} = m v₂ -M v₂ + 2M v_{1'}
v_{2f} (M + m) = - v₂ (M-m) + 2 M v_{1'}
v_{2f} = - ) v₂ + 2
v_{1'}
now we can substitute the velocity values found in the first two parts
= - ( \frac{ M-m}{M +m } ) √2gh + 2(\frac{M}{M+m}) √2gh
we simplify
v_{2f} = [( \frac{ M-m}{M +m } ) + 2 (\frac{M}{M+m})]
let's call the quantity in brackets that only depends on the masses
A = ( \frac{ M-m}{M +m } )+ 2 (\frac{M}{M+m})]
v_{2f}= A \sqrt{2gh}
in general, the marble is much lighter than the superball, so its speed is much higher than the speed of the superball
finally with the conservation of energy we find the height that the marble reaches
Starting point
Emo = K = ½ mv_{2f}²
Final point
Emf = U = m g h'
Em₀ = Em_f
½ m v_{2f}² = m g h ’
h ’= ½ v_{2f}² / g
h ’= ½ (A \sqrt{2gh})² / g
h ’= A² h
h '= [ ( \frac{ M-m}{M +m } )+ 2 (\frac{M}{M+m})]² h
The correct option is B.
There were 4 treatment conditions compared in the experiment.
F ratio plays an essential role in performing a particular dataset of ANOVA. F ratio is particularly a ratio which is obtained by the between-group variance or it is also called MSB and also within-group variance known as MSW. Any F-ratio which is computed is compared with the critical F-ratios from the table as it will check out if there are any variations available between the groups or not.
The researcher reports an F-ratio where the degrees of freedom = 3, 36.
F-ratio is obtained by the calculation of dividing the Mean squared errors because of the treatment by the Mean squared error which occurs due to error.
For the particular case, the researcher reports an F-ratio having degrees of freedom = 3, 36. It is indicating that the treatments are being distributed with degrees of freedom which is 3 and specified errors are distributed with degrees of freedom which is 36.
Let the treatments which were involved in the study can be denoted by k.
Let the total number of individuals involved in the study can be taken as N.
Then, the treatments will be having degrees of freedom as,
= k-1
3=k-1
k=3+1=4
Therefore, the required treatment condition number that was compared is 4.
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