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Arturiano [62]
3 years ago
11

One group of students made a local watershed model. The watershed model was most likely used by the students to

Physics
2 answers:
hoa [83]3 years ago
4 0
It should be C
hope this helps you
scoray [572]3 years ago
3 0
The correct option is B.
A water shed refers to an area of land which serves as channel for rainfall and supply other bodies of water connected with it with water. The amount of water in the watershed determines the quantity of water that will be passed to the water bodies that are connected to the watershed.
A model is usually made in order to use it to forecast or predict a certain thing. A watershed model can be used to predict the amount of floods and drought at a particular location.<span />
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What are fun facts about lysosomes??
Ahat [919]
Lysosomes are used by the cell to digest or breakdown multifaceted organic molecules
8 0
3 years ago
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A cylinder of gas is at room temperature (20°C). The air conditioner breaks down, and the temperature rises
noname [10]

Answer: we divide this equation by pV and use {C}_{p}={C}_{V}+ . ... The temperature, pressure, and volume of the resulting gas-air mixture

Explanation:

4 0
3 years ago
87) Determine the equivalent resistances for the following circuits.
stepladder [879]

Answer:

15    and  11 ohms

Explanation:

First one =   For the parallel resistors 1 / (1/6 + 1/6 + 1/6) =   1/ (3/6 ) = 6/3 = 2 ohms   then add the 3 and the 10  = 15 ohms

second one    for the parallel portion   equiv =    (10+2)*24 / ( (10+2 + 24) = 8

  then add the  3 in series = 11 ohms

8 0
2 years ago
A string is stretched to a length of 308 cm and both ends are fixed. If the density of the string is 0.023 g/cm, and its tension
soldi70 [24.7K]

Answer:

Frquency=3,994Hz

Explanation:

Tension =967N

Density of string (μ)=0.023g/cm

Length of the stretched spring=308cm

Fundamental frequency for nth harmonic :

Fn=n/2L(√T/μ)

Substituting the given values to find the frequency :

f1=1/2(308cm) *(0.01m/1cm)[(√967N)/(0.023g/cm)(0.1kg)/(0.1kg/m)/(1g/cm)]

=6.16m[(√967N)/0.0023kg/m)]

=3,994.20Hz

Approximately,

The frequency will be =3,994Hz

7 0
2 years ago
The coefficients of static and kinetic friction between a 3.0-kg box and a horizontal desktop are 0.40 and 0.30, respectively. w
luda_lava [24]

The force of friction on the box is 8.82 N.

The force of friction is given by the formula

F=μN

Here N is the normal force of the box=mg=3*9.8=29.4 N

Now the static friction force on the box is=0.4*29.4=11.76 N

Since the box is applied with the horizontal pull of 15 N, which is greater than the static friction, therefore the box must be in the moving condition, in that case there is dynamic friction force is applied on the box=0.3*29.4=8.8 N

Therefore the force of friction on the box is 8.8 N.

6 0
3 years ago
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