Answer:
The atmospheric pressure is 
.
Explanation:
Given that,
Atmospheric pressure 
drop height h'= 27.1 mm
Density of mercury 
We need to calculate the height
Using formula of pressure
Put the value into the formula
We need to calculate the new height
We need to calculate the atmospheric pressure
Using formula of atmospheric pressure
Put the value into the formula
Hence, The atmospheric pressure is .
Answer:
(C) deflected toward the top of the page.(
Explanation:
We can answer this problem by using Fleming's Left Hand Rule. By doing so, we have to place:
- The index finger of the left hand in the direction of the magnetic field
- The middle finger of the left hand in the direction of the particle's velocity
- The thumb will give the direction of the force, and therefore the deflection of the proton
In this problem, we have:
- Magnetic field direction: into the page --> index finger
- Proton's velocity: to the right --> middle finger
By doing so, we observe that the thumb points towards the top of the page: therefore, the correct answer is
(C) deflected toward the top of the page.
Answer:
sttouyietETwe2e664yrwtwwteuwtrwruwuuwwuwtwuw7w7w5w7w772253536464647
Answer:
the balls reached a height of 4.9985 m
Explanation:
Given the data in the question;
mass one m = 3.8 kg
mass two M = 2.1 kg
Initial velocities
u = 22 m/s
U = { moving downward} = 12 m/s
Now, using the law conservation of linear moment;
mu + MU = v( m + M )
we solve for "v" which is the velocity of the ball s after collision;
v = (mu + MU) / ( m + M )
so we substitute our given values into the equation
v = ( ( 3.8 × 22 ) + ( 2.1 × -12) ) / ( 3.8 + 2.1 )
v = ( 83.6 - 25.2 ) / 5.9
v = 58.4 / 5.9
v = 9.898 m/s
Now, we determine required height using the following relation;
v"² - v² = 2gh
where v" is the velocity at the top which is 0 m/s and g = -9.8 m/s²
0 - v² = 2gh
v² = -2gh
so we substitute
( 9.898 )² = -2 × -9.8 × h
97.97 = 19.6 × h
h = 97.97 / 19.6
h = 4.9985 m
Therefore, the balls reached a height of 4.9985 m
