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lidiya [134]
3 years ago
11

An undersea research chamber is spherical with an external diameter of 5.50 m . The mass of the chamber, when occupied, is 87600

kg . It is anchored to the sea bottom by a cable. The density of seawater is 1025 kg/m3. Part A What is the buoyant force on the chamber?
Physics
1 answer:
dmitriy555 [2]3 years ago
4 0

Answer:

the buoyant force on the chamber is F = 7000460 N

Explanation:

the buoyant force on the chamber is equal to the weight of the displaced volume of sea water due to the presence of the chamber.

Since the chamber is completely covered by water, it displaces a volume equal to its spherical volume

mass of water displaced = density of seawater * volume displaced

m= d * V , V = 4/3π* Rext³

the buoyant force is the weight of this volume of seawater

F = m * g = d * 4/3π* Rext³ * g

replacing values

F = 1025 kg/m³ * 4/3π * (5.5m)³ * 9.8m/s² = 7000460 N

Note:

when occupied the tension force on the cable is

T = F buoyant - F weight of chamber  = 7000460 N - 87600 kg*9.8 m/s² = 6141980 N

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Rina8888 [55]

Answer:

E. greater than the angle of incidence.

Explanation:

Snell's law states that:

n_i sin \theta_i = n_r sin \theta_r (1)

where

n_i, n_r are the refractive index of the first and second medium

\theta_i, \theta_r are the angle of incidence and refraction, respectively

For light moving from water to air, we have:

n_i = 1.33 (index of refraction of water)

n_r = 1.00 (index of refraction of air)

Substituting into (1) and re-arranging the equation, we get

\sin \theta_r = \frac{n_i}{n_r} sin \theta_i = 1.33 sin \theta_i

which means that

\theta_r > \theta_i

so, the correct answer is

E. greater than the angle of incidence.

4 0
3 years ago
Read 2 more answers
a hot piece of copper is placed in an insulated cup. what is the final temperature of the water and copper?
slavikrds [6]

Answer:

Option C. 30°C.

Explanation:

The following data were obtained from the question:

Mass of water (Mw) = 0.5 Kg

Specific heat capacity of water (Cw) = 4.18 KJ/Kg°C

Initial temperature of water (Tw1) =

22°C

Change in temperature (ΔT) = T2 – Tw1 = T2 – 22°C

Mass of copper (Mc) = 0.5 Kg

Specific heat capacity of copper (Cc) = 0.386 KJ/kg°C

Initial temperature of copper (Tc1) = 115°C

Change in temperature (ΔT) = T2 – Tc1 = T2 – 115°C

Final temperature (T2) =..?

Note: Both the water and the piece copper will have the same final temperature and the heat will be zero since the water will cool the piece of copper.

Thus, we can determine the final temperature of the water and copper as follow:

Q = MwCwΔT + McCcΔT

0 = 0.5 x 4.18 x (T2 – 22) + 0.5 x 0.386 x (T2 – 115)

0 = 2.09 (T2 – 22) + 0.193 (T2 – 115)

0 = 2.09T2 – 45.98 + 0.193T2 – 22.195

Collect like terms

2.09T2 + 0.193T2 = 45.98 + 22.195

2.283T2 = 68.175

Divide both side by the coefficient of T2 i.e 2.283

T2 = 68.175/2.283

T2 = 29.8 ≈ 30°C

Therefore, the final temperature of water and copper is 30°C.

8 0
2 years ago
An object 2.2 kg object accelerates at 4.0 m/s^2. What is the mass of the object?
KatRina [158]

Answer:

2.2kg

Explanation:

kilograms is a measurement of mass so the answer is 2.2kg

5 0
2 years ago
Which are characteristics of light? Check all that apply
Schach [20]

Answer:

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Explanation:

7 0
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Consider f(x) = -4x2 + 24x + 3. Determine whether the function has a maximum or minimum value. Then find the
murzikaleks [220]

Answer:

The function has a maximum in x=3

The maximum is:

f(3) = 39

Explanation:

Find the first derivative of the function for the inflection point, then equal to zero and solve for x

f(x)' = -4*2x + 24=0

-4*2x + 24=0

8x=24

x=3

Now find the second derivative of the function and evaluate at x = 3.

If f (3) ''< 0 the function has a maximum

If f (3) '' >0 the function has a minimum

f(x)''= 8

Note that:

f(3)''= -8

the function has a maximum in x=3

The maximum is:

f(3)=-4(3)^2+24(3) + 3\\\\f(3) = 39

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3 years ago
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