Answer:
D
Explanation:
<em>The most suitable testable question. in this case, would be that 'are there more home runs during the more humid months of the summer?'</em>
Since the aim of the investigation is to find the relationship between humidity and the number of home runs, measuring the number of home runs during the more humid months in the summer and comparing the data to the number of home runs during the less humid months in the same summer would provide the answer.
<u>Only option D raises a valid question that is relevant to the aim of the investigation.</u>
The gravitational constant was experimentally measured by W Cavendish using the attraction between big and small lead balls. is true
The correct answer is true
<h3>How do you define gravitational constant?</h3>
the strength of gravity. a factor in use in Newton's gravity law to relate the strength of the gravitational pull between two bodies with their masses and distance from one another. 6.67259 X 10-11 newtons per square kilogram is roughly the gravitational constant. G is its identifier.
<h3> where is the strongest gravity is?</h3>
The gravitational pull of the earth is greatest near sea level, normally, and weakens as you get further from the center, such as to the summit of Mt. Everest. Because the obloid earth was slightly wider, but only by a minor ratio, the gravity just at poles is stronger than that at the equator.
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Answer:
0.027648 kgm²
Explanation:
M = Mass of disc = 1.2 kg
r = Radius of disc = 0.16 m
m = Mass of rod = 0.16 kg
R = Rod distance = 0.16 m
Moment of inertia of disk is given by

Moment of inertia of the three rods

The total moment of inertia is given by

The moment of inertia of the stool with respect to an axis that is perpendicular to the plane of the disk at its center is 0.027648 kgm²
Answer:
a) Revolutions per minute = 2.33
b) Centripetal acceleration = 11649.44 m/s²
Explanation:
a) Angular velocity is the ratio of linear velocity and radius.
Here linear velocity = 72 m/s
Radius, r = 0.89 x 0. 5 = 0.445 m
Angular velocity

Frequency

Revolutions per minute = 2.33
b) Centripetal acceleration

Here linear velocity = 72 m/s
Radius, r = 0.445 m
Substituting

Centripetal acceleration = 11649.44m/s²
To solve the problem, use Kepler's 3rd law :
T² = 4π²r³ / GM
Solved for r :
r = [GMT² / 4π²]⅓
but first covert 6.00 years to seconds :
6.00years = 6.00years(365days/year)(24.0hours/day)(6...
= 1.89 x 10^8s
The radius of the orbit then is :
r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.89 x 10^8s)² / 4π²]⅓
= 6.23 x 10^11m