<span>A measurement must include both a number and an unit.
</span>
Answer:
1.603 s
Explanation:
Given that
Initial mass, = 0.45 kg
Initial period, = 1.45 s
Initial radius, = 0.14 m
Final mass, = 0.55 kg
Final period, = ?
Final radios, = 0.14 m
Since we are finding the rotation period of two masses of same radius, we can assume that the outward force is the same in both cases. This means that
m₁r₁ω₁² = m₂r₂ω2²
Where, ω = 2π/T, on substituting, we have
0.45 * 0.14 * (2π / 1.45)² = 0.550 * 0.14 * (2π / T₂)²
0.45 / 1.45² = 0.550 / T₂²
T₂² = 0.550 * 1.45² / 0.45
T₂² = 2.56972
T₂ = √2.56972
T₂ = 1.603 sec
Answer:
A
Explanation:
A nonpolar molecule has no separation of charge, so no positive or negative poles are formed. In other words, the electrical charges of nonpolar molecules are evenly distributed across the molecule. Nonpolar molecules tend to dissolve well in nonpolar solvents, which are frequently organic solvents.
The number of pulse beats elapsed before the rubber ball hits the ground can be obtained when you carry out the experiment yourself. However, the pulse beat method of timing used by Galileo is not reliable because it varies from time to time.
Galileo was interested in studying how objects fall. His discovery was that all objects had the same acceleration irrespective of their mass. This observation was in direct contrast to Aristotle's assertion that the velocity of objects is proportional to their mass.
However, he used his pulse beats as timer during the experiment. This method is unreliable because the pulse beats of a person changes depending on the person's state of mind. A stop clock could have been a more reliable timer than pulse beats.
Learn more: brainly.com/question/7201885
Answer:
The exit temperature of cold water is 37.6°C
Explanation:
Mass flow rate of water, 
Mass flow rate of air, 
Applying the conservation of heat law between air, water and the surrounding:
![[\dot{m}c(T_{in} - T_{out})]_{air} = [\dot{m}c(T_{in} - T_{out})]_{water} + \dot{Q}_{lost}](https://tex.z-dn.net/?f=%5B%5Cdot%7Bm%7Dc%28T_%7Bin%7D%20-%20T_%7Bout%7D%29%5D_%7Bair%7D%20%3D%20%5B%5Cdot%7Bm%7Dc%28T_%7Bin%7D%20-%20T_%7Bout%7D%29%5D_%7Bwater%7D%20%2B%20%5Cdot%7BQ%7D_%7Blost%7D)
We get the exit temperature of cold water by making 
the subject of the formula above:
![(T_{out})_{water} =\frac{ [\dot{m}c(T_{in} - T_{out})]_{air} + [\dot{m} c T_{in}]_{water} - \dot{Q}_{lost}}{(\dot{m}c)_{water}} \\\\(T_{out})_{water} = \frac{(2.5*1010*(90-20) + [1.2*4186*(8+273)] - 28000}{1.2*4186} \\\\(T_{out})_{water} = 310.6 K\\\\(T_{out})_{water} = 37.6^0 C](https://tex.z-dn.net/?f=%28T_%7Bout%7D%29_%7Bwater%7D%20%3D%5Cfrac%7B%20%5B%5Cdot%7Bm%7Dc%28T_%7Bin%7D%20-%20T_%7Bout%7D%29%5D_%7Bair%7D%20%2B%20%5B%5Cdot%7Bm%7D%20c%20T_%7Bin%7D%5D_%7Bwater%7D%20-%20%5Cdot%7BQ%7D_%7Blost%7D%7D%7B%28%5Cdot%7Bm%7Dc%29_%7Bwater%7D%7D%20%5C%5C%5C%5C%28T_%7Bout%7D%29_%7Bwater%7D%20%3D%20%5Cfrac%7B%282.5%2A1010%2A%2890-20%29%20%2B%20%5B1.2%2A4186%2A%288%2B273%29%5D%20-%2028000%7D%7B1.2%2A4186%7D%20%5C%5C%5C%5C%28T_%7Bout%7D%29_%7Bwater%7D%20%3D%20310.6%20K%5C%5C%5C%5C%28T_%7Bout%7D%29_%7Bwater%7D%20%3D%2037.6%5E0%20C)
