Answer:
He is known as the first microbiologist and also “the Father of Microbiology” because he was the first to observe bacteria underneath a microscope. He made many other significant discoveries in the field of biology and also made important changes to the microscope.
Explanation:
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The magnitude of the downward acceleration of the hollow cylinder is 6m/s^2.
Z = I α
T.R =1/2 M ( 
+ 
)α
T.R = 1/2M 5/4 α
T = 5Ma/8
Mg - T = Ma
Mg - 5Ma/8 = Ma
Mg= 5Ma/8 + Ma = 13Ma / 8
acceleration = 8g/13 = 6 m/s^2
The rate at which an object's velocity with respect to time changes is called its acceleration. The direction of the net force imposed on an item determines its acceleration in relation to that force. According to Newton's Second Law, the magnitude of an object's acceleration is the result of two factors working together
The size of the net balance of all external forces acting on that item is directly proportional to the magnitude of this net resultant force; the magnitude of that object's mass, depending on the materials from which it is built, is inversely related to its mass.
Learn more about acceleration here:
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Air and water have a good day
<span>3598 seconds
The orbital period of a satellite is
u=GM
p = sqrt((4*pi/u)*a^3)
Where
p = period
u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits.
a = semi-major axis of orbit.
Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So
u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2
The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So
150000 m + 3.396x10^6 m = 3.546x10^6 m
Substitute the known values into the equation for the period. So
p = sqrt((4 * pi / u) * a^3)
p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3)
p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3)
p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3)
p = sqrt(1.2945785x10^7 s^2)
p = 3598.025212 s
Rounding to 4 significant figures, gives us 3598 seconds.</span>
