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uranmaximum [27]
1 year ago
12

Armand is monitoring a large sealed tank by way of a sensor that records the liquid level over time on a graph. He looks at the

graph and claims that the sensor indicates there are waves in the liquid in the tank.
In the space provided, answer each of the following.

Part A: Explain how Armand knows that there is a wave in the tank.
Part B: Find the amplitude, in centimeters (cm), and frequency, in number of waves every second/cycle per second (Hz), of the wave.
Physics
2 answers:
Brums [2.3K]1 year ago
5 0

Answer:

A.) armand probably looked at the graph that the liquid sensor sends information too

Explanation:

timofeeve [1]1 year ago
4 0

Answer:

i need ppoints

Explanation:

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A raft is made of a plastic block with a density of 650 kg/m 3 , and its dimensions are 2.00 m à 3.00 m à 5.00 m. 1. what is the
cupoosta [38]
1) The volume of the raft is the product between the lenghts of its three dimensions:
V = (2.00 m)(3.00m)(5.00m)=30 m^3

2) The mass of the raft is the product between its density, d, and its volume, V:
m=dV=(650 kg/m^3)(30 m^3)=19500 kg

3) The weight of the raft is the product between its mass m and the gravitational acceleration, g=9.81 m/s^2:
W=mg=(19500 kg)(9.81 m/s^2)=1.91 \cdot 10^5 N

4) The apparent weight is equal to the difference between the weight of the raft and the buoyancy (the weight of the displaced fluid):
W_a = W- \rho_W V_{disp} g
where \rho _W = 1000 kg/m^3 is the water density and V_{disp} is the volume of displaced fluid.
The density of the raft (650 kg/m^3) is smaller than the water density (1000 kg/m^3), this means that initially the buoyancy (which has upward direction) is larger than the weight (downward direction) and so the raft is pushed upward, until it reaches a condition of equilibrium and it floats. At equilibrium, the weight and the buoyancy are equal and opposite in sign:
W=B=\rho _W V_{disp} g
and therefore, the apparent weight will be zero:
W_a = W-B=W-W=0

5) The buoyant force B is the weight of the displaced fluid, as said in step 4):
B=\rho_W V_{disp} g
When the raft is completely immersed in the water, the volume of fluid displaced V_{disp} is equal to the volume of the raft, V_{disp}=V. Therefore the buoyancy in this situation is
B= \rho_W V g = (1000 kg/m^3)(30 m^3)(9.81 m/s^2)=2.94 \cdot 10^5 N
However, as we said in point 4), the raft is pushed upward until it reaches equilibrium and it floats. At equilibrium, the buoyancy will be equal to the weight of the raft (because the raft is in equilibrium), so:
B=W=1.91 \cdot 10^5 N

6) At equilibrium, the mass of the displaced water is equal to the mass of the object. In fact, at equilibrium we have W=B, and this can be rewritten as
mg = m_{disp} g
where m_{disp}= \rho_W V_{disp} is the mass of the displaced water. From the previous equation, we obtain that m_{disp}=m=19500 kg.

7) Since we know that the mass of displaced water is equal to the mass of the raft, using the relationship m=dV we can rewrite m=m_{disp} as:
d V =d_W V_{disp}
and so
V_{disp}= \frac{d V}{d_W}= \frac{(650 kg/m^3)(30m^3)}{1000kg/m^3}= 19.5 m^3

8) The volume of water displaced is (point 7) 19.5 m^3. This volume is now "filled" with part of the volume of the raft, therefore 19.5 m^3 is also the volume of the raft below the water level. We can calculate the fraction of raft's volume below water level, with respect to the total volume of the raft, 30 m^3:
\frac{19.5 m^3}{30 m^3}\cdot 100= 65 \%
Viceversa, the volume of raft above the water level is 30 m^3-19.5 m^3 = 10.5 m^3. Therefore, the fraction of volume of the raft above water level is
\frac{10.5 m^3}{30 m^3}\cdot 100 = 35 \%

9) Let's repeat steps 5-8 replacing \rho _W, the water density, with \rho_E=806 kg/m^3, the ethanol density.

9-5) The buoyant force is given by:
B=\rho _E V_{disp} g = (806 kg/m^3)(30 m^3)(9.81 m/s^2)=2.37 \cdot 10^5 N
when the raft is completely submerged. Then it goes upward until it reaches equilibrium and it floats: in this condition, B=W, so the buoyancy is equal to the weight of the raft.

9-6) Similarly as in point 6), the mass of the displaced ethanol is equal to the mass of the raft: 
m_E = m = 19500 kg

9-7) Using the relationship d= \frac{m}{V}, we can find the volume of displaced ethanol:
V_E =  \frac{m}{d_E} = \frac{19500 kg}{806 kg/m^3}=24.2 m^3

9-8) The volume of raft below the ethanol level is equal to the volume of ethanol displaced: 24.2 m^3. Therefore, the fraction of raft's volume below the ethanol level is 
\frac{24.2 m^3}{30 m^3}\cdot 100 = 81 \%
Consequently, the raft's volume above the ethanol level is 
30 m^3 - 24.2 m^3 = 5.8 m^3
and the fraction of volume above the ethanol level is
\frac{5.8 m^3}{30 m^3}\cdot 100 = 19 \%
8 0
2 years ago
Water is a colorless and odorless liquid. It can exist in solid, liquid, and gas states. It boils at 100 degrees C and melts at
BARSIC [14]

Answer: Option (c) is the correct answer.

Explanation:

Physical properties are the properties in which there is no change in chemical composition of a substance. On the other hand, chemical properties are the properties which change the chemical composition of a substance.

For example, when water boils at 100 ^{o}C then it changes into vapor state whereas when water freezes at 0^{0}C then it changes state from liquid to solid.

This means only physical state of water is changing and there is no change in chemical composition of water.

Hence, we can conclude that best option describing given information is that these are the physical changes water undergoes.

4 0
2 years ago
If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed?
sweet [91]

Answer:

Speed of the electron will be v=5.896\times 10^7m/sec

Explanation:

We have given that charge on electron e=1.6\times 10^{-19}C

Mass of electron m=9.11\times 10^{-31}kg

Potential difference = V=9.9KV=9.9\times 10^3volt

Now according to energy conservation eV=\frac{1}{2}mv^2

1.6\times 10^{-19}\times 9.9\times 10^3=\frac{1}{2}\times 9.11\times 10^{-31}v^2

v=5.896\times 10^7m/sec

4 0
1 year ago
The resistivity of a semiconductor can be modified by adding different amounts of impurities. A rod of semiconducting material o
zavuch27 [327]

Answer:

pp

Explanation:

7 0
1 year ago
A lightning bolt transfers 6.0 coulombs of charge from a cloud to the ground in 2.0 x 10-3 second. what is the average current d
AlladinOne [14]
The current is defined as the amount of charge transferred through a certain point in a certain time interval:
I= \frac{Q}{\Delta t}
where
I is the current
Q is the charge
\Delta t is the time interval

For the lightning bolt in our problem, Q=6.0 C and \Delta t= 2.0 \cdot 10^{-3}s, so the average current during the event is
I= \frac{Q}{\Delta t} = \frac{6.0 C}{2.0 \cdot 10^{-3} s}=3000 A
4 0
2 years ago
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