Given:
u = 10⁵ m/s, the entrance velocity
v = 2.5 x 10⁶ m/s, the exit velocity
s = 1.6 cm = 0.016 m, distance traveled
Let a = the acceleration.
Then
u² + 2as = v²
(10⁵ m/s)² + 2*(a m/s²)*(0.016 m) = (2.5 x 10⁶ m/s)²
0.032a = 6.25 x 10¹² - 10¹⁰ = 6.24 x 10¹²
a = 1.95 x 10¹⁴ m/s²
Answer: 1.95 x 10¹⁴ m/s²
TheThe relative velocity of Deigo will be 27km/h .
<h3>What is relative Velocity? </h3>
The relative velocity is the velocity of an object with respect to another observer. It is the time rate of relative position of one object with respect to another object .
Vab = Va-Vb
Where a is object and b is observer .
We have given here the velocity of ball 20km/ h
and the another velocity is 7km/ h
Let us assume that Diego is also moving in the direction where ball is moving so the relative velocity will be
V = 20+7=27km/ h
When both observer and object is moving in same direction than there sum of velocity will be relative velocity ,if they are moving opposite than subtraction.
to learn more about Relative velocity click Here brainly.com/question/6070547
#SPJ4
The trachea is a tube that carries air inside the lungs.
The calculated magnitude is 6.73 x 10³ V/m.
AMU is described as being one-twelfth the mass of a carbon-12 atom (12C). C makes up more than 98% of the carbon that can be found in nature, making it the most prevalent isotope. The magnitude of the field is the change in potential across a small distance in the indicated direction divided by that distance.
Potential difference = 8.20 kV= 8.20 x 10³ V
radius= 19.4/100=0.194 m
total distance that is circumference of the circle= 2πr =2 x 3.14 x 0.194
= 1.218 m
therefore Magnitude= 8.20 x 10³ / 1.218
=6.73 x 10³ V/m
Learn more about Magnitude here-
brainly.com/question/15681399
#SPJ9
