It is customary to work in SI units.
Calculate the volume of the concrete.
V = 3.7*2.1*5.8 cm³ = 45.066 cm³ = 45.066 x 10 ⁻⁶ m³
The mass is 43.8 g = 43.8 x 10⁻³ kg
The density is mass/volume.
Density = (43.8 x 10⁻³ kg)/(45.066 x 10⁻⁶ m³) = 971.9 kg/m³
Answer: 971.9 kg/m³
Distance traveled by the ball is given by
![distance = speed \times time](https://tex.z-dn.net/?f=%20distance%20%3D%20speed%20%5Ctimes%20time)
here we know that
speed = 20 m/s
times = 0.25 s
now we have
![distance = 20 \times 0.25](https://tex.z-dn.net/?f=distance%20%3D%2020%20%5Ctimes%200.25)
![distance = 5 m](https://tex.z-dn.net/?f=distance%20%3D%205%20m)
so ball will travel 5 m distance in the given interval of time
Answer:
a) ![v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s](https://tex.z-dn.net/?f=v_%7B1%7D%3D%5Cfrac%7B%2862.5-66%29ft%7D%7B%282.5-2%29s%7D%3D-7ft%2Fs)
![v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s](https://tex.z-dn.net/?f=v_%7B2%7D%3D%5Cfrac%7B%2865.94-66%29ft%7D%7B%282.1-2%29s%7D%3D-0.6ft%2Fs)
![v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s](https://tex.z-dn.net/?f=v_%7B3%7D%3D%5Cfrac%7B%2866.0084-66%29ft%7D%7B%282.01-2%29s%7D%3D0.84ft%2Fs)
![v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s](https://tex.z-dn.net/?f=v_%7B4%7D%3D%5Cfrac%7B%2866.001-66%29ft%7D%7B%282.001-2%29s%7D%3D1ft%2Fs)
b) ![v=65-32(2)=1ft/s](https://tex.z-dn.net/?f=v%3D65-32%282%29%3D1ft%2Fs)
Explanation:
From the exercise we got the ball's equation of position:
![y=65t-16t^{2}](https://tex.z-dn.net/?f=y%3D65t-16t%5E%7B2%7D)
a) To find the average velocity at the given time we need to use the following formula:
![v=\frac{y_{2}-y_{1} }{t_{2}-t_{1} }](https://tex.z-dn.net/?f=v%3D%5Cfrac%7By_%7B2%7D-y_%7B1%7D%20%20%7D%7Bt_%7B2%7D-t_%7B1%7D%20%20%7D)
Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001
![y_{t=2}=65(2)-16(2)^{2} =66ft](https://tex.z-dn.net/?f=y_%7Bt%3D2%7D%3D65%282%29-16%282%29%5E%7B2%7D%20%3D66ft)
![y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft](https://tex.z-dn.net/?f=y_%7Bt%3D2.5%7D%3D65%282.5%29-16%282.5%29%5E%7B2%7D%20%3D62.5ft)
![v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s](https://tex.z-dn.net/?f=v_%7B1%7D%3D%5Cfrac%7B%2862.5-66%29ft%7D%7B%282.5-2%29s%7D%3D-7ft%2Fs)
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![y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft](https://tex.z-dn.net/?f=y_%7Bt%3D2.1%7D%3D65%282.1%29-16%282.1%29%5E%7B2%7D%20%3D65.94ft)
![v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s](https://tex.z-dn.net/?f=v_%7B2%7D%3D%5Cfrac%7B%2865.94-66%29ft%7D%7B%282.1-2%29s%7D%3D-0.6ft%2Fs)
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![y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft](https://tex.z-dn.net/?f=y_%7Bt%3D2.01%7D%3D65%282.01%29-16%282.01%29%5E%7B2%7D%20%3D66.0084ft)
![v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s](https://tex.z-dn.net/?f=v_%7B3%7D%3D%5Cfrac%7B%2866.0084-66%29ft%7D%7B%282.01-2%29s%7D%3D0.84ft%2Fs)
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![y_{t=2.001}=65(2.001)-16(2.001)^{2} =66.001ft](https://tex.z-dn.net/?f=y_%7Bt%3D2.001%7D%3D65%282.001%29-16%282.001%29%5E%7B2%7D%20%3D66.001ft)
![v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s](https://tex.z-dn.net/?f=v_%7B4%7D%3D%5Cfrac%7B%2866.001-66%29ft%7D%7B%282.001-2%29s%7D%3D1ft%2Fs)
b) To find the instantaneous velocity we need to derivate the equation
![v=\frac{df}{dt}=65-32t](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bdf%7D%7Bdt%7D%3D65-32t)
![v=65-32(2)=1ft/s](https://tex.z-dn.net/?f=v%3D65-32%282%29%3D1ft%2Fs)
Answer:
A flat, horizontal line
Explanation:
A flat, horizontal line indicates a phase change.
The temperature does not increase because the added heat goes into converting one phase into another.
A is wrong. A downward-sloping line indicates that the temperature is decreasing with time.
C is wrong. An upward-sloping line indicates that the temperature is increasing with time.
Answer:
pizza is a heterogeneous mixture.
Explanation:
A heterogeneous mixture is a mixture in which the composition is not uniform throughout the mixture. Vegetable soup is a heterogeneous mixture. Any given spoonful of soup will contain varying amounts of the different vegetables and other components of the soup