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Xelga [282]
1 year 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]1 year 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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Which points are most efficient for the utilization of resources on a production possibilities diagram?
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Answer: most effective way is to practice reduce reuse and recycle for utilisation of resources

3 0
1 year ago
A circular loop of diameter 10 cm, carrying a current of 0.20 A, is placed inside a magnetic field B⃗ =0.30 Tk^. The normal to t
arlik [135]

Answer:

The magnitude of the torque on the loop due to the magnetic field is 4.7\times10^{-4}\ N-m.

Explanation:

Given that,

Diameter = 10 cm

Current = 0.20 A

Magnetic field = 0.30 T

Unit vectorn=-0.60\hat{i}-0.080\hat{j}

We need to calculate the torque on the loop

Using formula of torque

\tau=NIAB\sin\theta

Where, N = number of turns

A = area

I = current

B = magnetic field

Put the value into the formula

\tau=1\times0.20\times\pi\times(5\times10^{-2})^2\times0.30\times\sin90^{\circ}

\tau=4.7\times10^{-4}\ N-m

Hence, The magnitude of the torque on the loop due to the magnetic field is 4.7\times10^{-4}\ N-m.

5 0
1 year ago
What is the chemical formula for aluminum nitride?<br> A) Al3 <br> B) AlN3 <br> C) Al3N3 <br> D) AlN
OleMash [197]
The answer is c Ai3n3 yea that’s the right answer idk
6 0
2 years ago
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Derive an algebraic expression for the tension in the segment of the cord between the post and block A (assumed massless). Expre
lubasha [3.4K]
T=m1a+m2a 

a=v^2/r v=r*w w=2πF--->a=r(2f)^2 <r1 for m1 and r2 for m2> 

T=(m1r1+m2r2)(2πf)^2 <i factored out (2πf)^2 
3 0
2 years ago
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A rigid container equipped with a stirring device contains 1.5 kg of motor oil. Determine the rate of specific energy increase w
jenyasd209 [6]

To solve this problem we will apply the first law of thermodynamics which details the relationship of energy conservation and the states that the system's energy has. Energy can be transformed but cannot be created or destroyed.

Accordingly, the rate of work done in one cycle and the heat transferred can be expressed under the function,

\dot{U} = \dot{Q}-\dot{W}

Substitute 1W for \dot{Q} and 1.5 W for \dot{W}

\dot{U} = 1-1(1.5)

\dot{U} = 2.5W

Now calculcate the rate of specific internal energy increase,

\dot{u} = \frac{\dot{U}}{m}

\dot{u} = \frac{2.5}{1.5}

\do{u} = 1.6667W/kg

The rate of specific internal energy increase is 1.6667W/kg

8 0
2 years ago
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