Answer:
a) The acceleration of the sprinter is 6.521 meters per square second, b) The acceleration of 84512.16 kilometers per square hour.
Explanation:
Note: Statement is incomplete, complete description of the problem is: <em>A sprinter accelerates from rest to </em>
<em> in 1.38 seconds. What is his acceleration in </em><em>a.</em><em> </em>
<em> and </em><em>b.</em><em> </em>
<em>?</em>
a) We assume that sprinter accelerates uniformly, so that acceleration (
), measured in meters per square second, can be obtained from this kinematic expression as function of initial and final velocities (
,
), measured in meters per second, and time (
), measured in seconds, as well.


If we know that
,
and
, the acceleration experimented by the sprinter is:


The acceleration of the sprinter is 6.521 meters per square second.
b) If we know that a hour is equivalent to 3600 seconds and a kilometer is equivalent to 1000 meters, the result of the previous item is converted by using two consecutive unit conversions:


The acceleration of 84512.16 kilometers per square hour.
(i) The direction of the current will be in opposite direction to the magnetic field.
(ii) The magnitude of the emf induced in the circuit is 0.359 V.
(iii) The induced current if the circular coil is 0.048 A.
b(i) The magnetic flux linkage through the coil is 1.41 x 10⁻⁴ Tm².
b(ii) The radius of the coil is r√3 m.
<h3>
Direction of the current</h3>
The direction of the current will be in opposite direction to the magnetic field.
<h3>Emf induced in the circuit</h3>
Initial area of the coil = (πd²)/4
Initial area of the coil = (π x 0.225²)/4 = 0.0398 m²
Final area of the coil = (π x 0.072²)/4 = 0.001296 m²

<h3>Induced current</h3>
I = emf/R
I = 0.359/7.5
I = 0.048 A
<h3>Magnetic flux linkage through the coil</h3>
Ф = LI
Ф = 0.015 x 0.0094
Ф = 1.41 x 10⁻⁴ Tm²
<h3>Radius of the coil</h3>

where;
Learn more about inductance of coil here: brainly.com/question/17086348
We use the Raoult's Law to answer this question.
Partial Pressure = Mole Fraction*Total Pressure
When concentration is expressed in ppm, it means 'parts per million'. So, there can be 0.67 mol of ozone per 10⁶ mol of mixture.
Mole fraction = 0.67/10⁶ = 0.67×10⁻⁶
Partial pressure = 0.67×10⁻⁶(735 torr) = <em>4.92×10⁻⁴ torr</em>
Choice-D is the correct statement.
Based on this electric field diagram, the statement which best compares the charge of A with B is "A is negatively charged and B is positively charged. The charge on A is greater than that on B".
<u>Answer:</u> Option A
<u>Explanation:</u>
The charge is quantized represented as elementary charge, about 1.602×10−19 coulombs. Their are two kinds of electric charging: positive and negative (usually transported, separately, by protons and electrons). Like charges repel each other, while attraction occurs among unlike charges. An entity without net charge is considered neutral. If a piece of matter comprises more electrons than protons, it has a negative charge, when there are fewer, it'll have a positive charge and when there are equal amounts, this will be neutral.