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sertanlavr [38]
3 years ago
10

13. A diver swims to a depth of 3.2 m in a freshwater lake. What is the increase in the force pushing in on her eardrum, compare

d to what it was at the lake surface? The area of the eardrum is 0.60 cm².
Physics
1 answer:
Gnoma [55]3 years ago
6 0

Answer:

1.88 N

Explanation:

h = 3.2 m, A = 0.6 cm^2 = 0.6 x 10^-4 m^2

density of water, d = 1000 kg/m^3, g = 9.8 m/s^2

Pressure at depth h, P = h x d x g = 3.2 x 1000 x 9.8 = 31.36 x 10^3 Pa

Force = pressure x Area = 31.36 x 10^3 x 0.6 x 10^-4 = 1.88 N

Thus, the force on ear drum is 1.88 N.

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Two spherical objects at the same altitude move with identical velocities and experience the same drag force at a time t. If Obj
Anna35 [415]

Answer:

<em>The object with the twice the area of the other object, will have the larger drag coefficient.</em>

<em></em>

Explanation:

The equation for drag force is given as

F_{D} = \frac{1}{2}pu^{2}  C_{D} A

where F_{D} IS the drag force on the object

p = density of the fluid through which the object moves

u = relative velocity of the object through the fluid

p = density of the fluid

C_{D} = coefficient of drag

A = area of the object

Note that C_{D} is a dimensionless coefficient related to the object's geometry and taking into account both skin friction and form drag. The most interesting things is that it is dependent on the linear dimension, which means that it will vary directly with the change in diameter of the fluid

The above equation can also be broken down as

F_{D} ∝ P_{D} A

where P_{D} is the pressure exerted by the fluid on the area A

Also note that P_{D} = \frac{1}{2}pu^{2}

which also clarifies that the drag force is approximately proportional to the abject's area.

<em>In this case, the object with the twice the area of the other object, will have the larger drag coefficient.</em>

5 0
3 years ago
Use the simulation to compare the masses of the three colored and unlabeled weights of different sizes. To do so, set the spring
neonofarm [45]

Answer:

The answer is given as follows,

Explanation:

Gold large-sized weight 100 g < M < 250g

50 g < Magenta small-sized weight < 100g

100g < Blue medium-sized weight < 250g

Hence,

100g < Blue medium-sized weight < 250g

50 g < Magenta small-sized weight < 100g

100 g < Gold large-sized weight < 250g.

8 0
3 years ago
Consider a space shuttle which has a mass of about 1.0 x 105 kg and circles the Earth at an altitude of about 200.0 km. Calculat
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Answer:

1.6675×10^-16N

Explanation:

The force of gravity that the space shuttle experiences is expressed as;

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G is the gravitational constant

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Substitute into the formula

g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²

g = 6.67×10^-6/4×10^10

g = 1.6675×10^{-6-10}

g = 1.6675×10^-16N

Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N

7 0
3 years ago
How does the scientific process generally begin?
Murljashka [212]
There are several approaches. The most favourable one (in my opinion) is this one:
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4. Experimenting (to prove the hypothesis)
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