Answer:
B) 2.7 g of aluminium has a volume of 1 cm^3
Explanation:
Density can be defined as mass all over the volume of an object.
Simply stated, density is mass per unit volume of an object.
Mathematically, density is given by the equation;
![Density = \frac{mass}{volume}](https://tex.z-dn.net/?f=Density%20%3D%20%5Cfrac%7Bmass%7D%7Bvolume%7D)
If the density of aluminum is 2.7 g/cm³, it simply means that 2.7 g of aluminium has a volume of 1 cm³
Check:
Given the following data;
Mass = 2.7 grams
Volume = 1 cm³
Substituting into the formula, we have;
![Density = \frac{2.7}{1}](https://tex.z-dn.net/?f=Density%20%3D%20%5Cfrac%7B2.7%7D%7B1%7D)
Density = 2.7 g/cm³
So, first you find your acceleration which is 3m/s^2, using the acceleration formula.
Now set up your equation, F=ma, so put in the stuff, F=80kg·3m/s^2. Then solve your equation by multiplying, and you get F=240N, since newtons are your measurement.
Hope this helps
Answer:
A. The wavelength doubles but the wave speed is unchanged
Explanation:
The relationship between the period and wavelength is direct. Doubling the period of the oscillator will correspondingly double the wavelength but the wave speed is unaffected
1. H2O- The water completely dissolved the salt.
Alcohol- The alcohol dissolved the salt slightly.
Glycerin-The salt has not dissolved at all.
2. Water
3. Like dissolve like basically works on the principle of polarity. It means the substances which possess similar chemical properties may dissolve in each other. For example, ethanol can be dissolved in water because both are polar in nature whereas non-polar molecules can be dissolved in non-polar solvents only.
4. Generally, a solute dissolves faster in a warmer solvent than it does in a cooler solvent because particles have more energy of movement. For example, if you add the same amount of sugar to a cup of hot tea and a cup of iced tea, the sugar will dissolve faster in the hot tea.
Learn more about The choice of solvent here:-brainly.com/question/14918321
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1. Velocity at which the packet reaches the ground: 121.2 m/s
The motion of the packet is a uniformly accelerated motion, with constant acceleration
directed downward, initial vertical position
, and initial vertical velocity
. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:
![v_f^2 -v_i^2 =2ad](https://tex.z-dn.net/?f=v_f%5E2%20-v_i%5E2%20%3D2ad)
substituting, we find
![v_f^2 = 2ad\\v_f = \sqrt{2ad}=\sqrt{2(9.8 m/s^2)(750 m)}=121.2 m/s](https://tex.z-dn.net/?f=v_f%5E2%20%3D%202ad%5C%5Cv_f%20%3D%20%5Csqrt%7B2ad%7D%3D%5Csqrt%7B2%289.8%20m%2Fs%5E2%29%28750%20m%29%7D%3D121.2%20m%2Fs)
2. height at which the packet has half this velocity: 562.6 m
We need to find the heigth at which the packet has a velocity of
![v_f=\frac{121.2 m/s}{2}=60.6 m/s](https://tex.z-dn.net/?f=v_f%3D%5Cfrac%7B121.2%20m%2Fs%7D%7B2%7D%3D60.6%20m%2Fs)
In order to do that, we use again the same SUVAT equation substituting
with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:
![v_f^2-v_i^2=2ad\\d=\frac{v_f^2}{2a}=\frac{(60.6 m/s)^2}{2(9.8 m/s^2)}=187.4 m](https://tex.z-dn.net/?f=v_f%5E2-v_i%5E2%3D2ad%5C%5Cd%3D%5Cfrac%7Bv_f%5E2%7D%7B2a%7D%3D%5Cfrac%7B%2860.6%20m%2Fs%29%5E2%7D%7B2%289.8%20m%2Fs%5E2%29%7D%3D187.4%20m)
Which means that the heigth of the packet was
![h=750 m-187.4 m=562.6 m](https://tex.z-dn.net/?f=h%3D750%20m-187.4%20m%3D562.6%20m)