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sergey [27]
3 years ago
11

Which type of line is shown on the map

Physics
1 answer:
emmainna [20.7K]3 years ago
3 0

Answer:

A

Explanation:

completed the test

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The Gulf Stream off the east coast of the United States can flow at a rapid 3.9 m/s to the north. A ship in this current has a c
Alex Ar [27]

Answer:

72.54 degree west of south

Explanation:

flow = 3.9 m/s north

speed = 11 m/s

to find out

point due west from the current position

solution

we know here water is flowing north and ship must go south at an equal rate so that the velocities cancel and the ship just goes west

so it become like triangle with 3.3 point down and the hypotenuse is 11

so by triangle

hypotenuse ×cos(angle) = adjacent side

11 ×cos(angle) = 3.3

cos(angle) = 0.3

angle = 72.54 degree west of south

3 0
3 years ago
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Rheumatoid arthritis is a disorder in which the joints swell and cause pain. Rheumatoid arthritis causes the body’s own immune c
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This would be the musculoskeletal 
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3 years ago
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A student has a displacement of 739 m north in 162 s. What was the student’s average velocity?A. 0.22 m/sB. 119,718 m/sC. 162 m/
user100 [1]

Answer:

answer below

Explanation:

Displacement of the student is 739 m due North and it takes 162 s.

We need to find the student's average velocity. Using formula of velocity.

Velocity = displacement/time

v= 739/162

v= 4.56

4 0
2 years ago
At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m2. A fan is 30 m from the loudspeak
aliina [53]

Explanation:

Below is an attachment containing the solution.

5 0
3 years ago
Since the aluminum bar is not an isolated system, the second law of thermodynamics cannot be applied to the bar alone. Rather, i
max2010maxim [7]

Answer:

ΔS total ≥ 0 (ΔS total = 0 if the process is carried out reversibly in the surroundings)

Explanation:

Assuming that the entropy change in the aluminium bar is due to heat exchange with the surroundings ( the lake) , then the entropy change of the aluminium bar is, according to the second law of thermodynamics, :

ΔS al ≥ ∫dQ/T

if the heat transfer is carried out reversibly

ΔS al =∫dQ/T  

in the surroundings

ΔS surr ≥ -∫dQ/T = -ΔS al → ΔS surr ≥ -ΔS al = - (-1238 J/K) = 1238 J/K

the total entropy change will be

ΔS total = ΔS al + ΔS surr

ΔS total ≥ ΔS al + (-ΔS al) =

ΔS total ≥ 0

the total entropy change will be ΔS total = 0 if the process is carried out reversibly in the surroundings

4 0
3 years ago
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