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Xelga [282]
2 years ago
11

Weddell seals make holes in sea ice so that they can swim down to forage on the ocean floor below. Measurements for one seal sho

wed that it dived straight down from such an opening, reaching a depth of 0.30 km in a time of 5.0 min.what is the speed of the diving seal in m/s?
A. 0.60 m/s B. 1.0 m/s C. 1.6 m/s D. 6.0 m/s E. 10 m/s
Physics
1 answer:
Mariulka [41]2 years ago
3 0

Answer:

B. 1 m/s

Explanation:

Metric unit conversions:

0.3 km = 300m

5 minutes = 5*60 = 300 seconds

So if a seal can reach a depth of 300m in a time of 300 seconds, its diving speed is the distance divided by time duration

v = s/t = 300/300 = 1m/s

So B is the correct answer

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Considerable scientific work is currently under way to determine whether weak oscillating magnetic fields such as those found ne
slega [8]

Answer:

\epsilon = 2.96 \times 10^{-11} \ V

Explanation:

given,

magnetic field strength =  1.40 ✕ 10⁻³ T

frequency of oscillation = 60 Hz

diameter of RBC = 7.5 μm

EMF = ?

\epsilon = NBA\omega

\epsilon = NB(\pi\ r^2)\ (2\pi f)

\epsilon = NB(\pi\ (\dfrac{d}{2})^2)\ (2\pi f)

\epsilon = (1)\ 1.4 \times 10^{-3}(\pi\ (\dfrac{7.5 \times 10^{-6}}{2})^2)\ (2\pi\times 60)

\epsilon = 2.96 \times 10^{-11} \ V

maximum emf that can generate around the perimeter of the cell \epsilon = 2.96 \times 10^{-11} \ V

5 0
2 years ago
A boat moves through the sea.
sergiy2304 [10]

Answer:

dont you have to times it

Explanation:

4 0
2 years ago
A policeman kicks in a door with a force of 4500 N. What force does the door apply to the policeman’s leg?
Soloha48 [4]

Answer:

-4500 N

Source: Brainly

The police officer must be angry 0_0

4 0
2 years ago
If isomerization requires breaking the pi bond, what minimum energy is required for isomerization in j/mol?
aliina [53]
<span>The minimum energy required for isomerization is 267 000 J/mol 
</span>

The isomerization of cis-but-2-ene to trans-but-2-ene requires breaking of the π bond.

The bond energy of a C-C σ bond is 347 kJ/mol.

The bond energy of a C=C double bond (σ + π) is 614 kJ/mol.

So the bond energy of a π bond is (614 – 347) kJ/mol = 267 kJ/mol =
267 000 J/mol.

6 0
2 years ago
Tara is giving a speech on the how to get a U.S. passport for the first time. During her speech, she tells the audience the requ
Alchen [17]

Answer:

a. 1. she informed the audience when she was concluding

2. She simply outlined the steps tying her conclusion to the topic

3. her conclusion was concise and simple.

b. 13.1 grams

Explanation:

a. signalling the end of a speech prepares the mind of the audience and leaves them satisfied, and Kara correctly signaled the end of her speech when she said "In conclusion".

Also, Kara summarized the ideas in her speech by simply outlining the requirements needed to get a U.S passport.

Her conclusion was concise and simple which is very important especially for lengthy speeches because the audience at this point must be getting tired and distracted.

b. Amount of cholesterol in 100 mL of blood = 248 mg

Total Volume of blood = 5.3 Liters.

the question is asking us to find the total grams of cholesterol in the individual's body, that is the total grams of cholesterol in 5.3 Liters of the individuals blood.

To solve, let us use the following steps:

1. unify the units of measuring volume.

since the first volume is given in Milliliters, I will convert the second volume (5.3 Liters) to milliliters too. To do this, note that:

1 liter = 1000 Milliliter (mL)

∴ 5.3 Liters = 5.3 × 1000 = 5300 Milliliters (mL)

2. Calculate the amount of cholesterol in 5300 mL.

100 mL = 248 mg of cholesterol

∴ 1 mL = \frac{248}{100\\} = 2.48 mg

∴ 5300 mL = 2.48 × 5300 = 13,144 mg of cholesterol.

3. Convert the amount of cholesterol from Milligram to gram

Since the question asked for the gram of cholesterol, not the milligram of cholesterol, converting from milligram to gram, we use the following steps:

1000 mg = 1 g

∴ 1 mg = \frac{1}{1000} g

∴ 13144 mg = 13144 × \frac{1}{1000} = \frac{13144}{1000} = 13.144 ≅ 13.1 g of cholesterol ( 1 decimal place).

I suggest you go and study the conversion relationships between the sub units milli, micro, centi and deci.

7 0
2 years ago
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