Can you be fluid overload and dehydrated at the same time

____ are solid while lines stretching across one or more lanes in the same direction, indicating the proper place to come to a s

Natalka [10]

Stop lines are solid white lines painted across the traffic lanes at intersections and pedestrian crosswalks, indicating the exact place to stop.

8 0

2 years ago

A bird sits on a high-voltage power line with its feet 3.87 cm apart. The wire is made from aluminum, is 2.11 cm in diameter, an

Svetlanka [38]

Answer:

ΔV=0.484mV

Explanation:

The potential difference across the end of conductor that obeys Ohms law:

ΔV=IR

Where I is current

R is resistance

The resistance of a cylindrical conductor is related to its resistivity p,Length L and cross section area A

R=(pL)/A

Given data

Length L=3.87 cm =0.0387m

Diameter d=2.11 cm =0.0211 m

Current I=165 A

Resistivity of aluminum p=2.65×10⁻⁸ ohms

So

ΔV=IR

$=I(\frac{pL}{A})\\ =I(\frac{pL}{\pi r^{2} } )\\=I(\frac{pL}{\pi (d/2)^{2} } )\\=165A((\frac{(2.65*10^{-8})(0.0387m)}{\pi (0.0211m/2)^{2} } ))\\=4.84*10^{-4}V$

ΔV=0.484mV

3 0

2 years ago

It takes 15 min to drive 6.0 mi in a straight line to the local hospital. It takes 10 min to go the last 3.0 mi, 2.0 min to go t

Gala2k [10]

Answer:

36.87 km/h

Explanation:

Convert all the units in SI system

1 mile = 1609.34 m

d1 = 6 mi = 9656.04 m

t1 = 15 min = 15 x 60 = 900 s

d2 = 3 mi = 4828.02 m

t2 = 10 min = 10 x 60 = 600 s

d3 = 1 mi = 1609.34 m

t3 = 2 min = 2 x 60 = 120 s

d4 = 0.5 mi = 804.67 m

t4 = 0.5 min = 0.5 x 60 = 30 s

Total distance, d = d1 + d2 + d3 + d4

d = 9656.04 + 4828.02 + 1609.34 + 804.67 = 16898.07 m = 16.898 km

total time, t = t1 + t2 + t3 + t4

t = 900 + 600 + 120 + 30 = 1650 s = 0.4583 h

The ratio of the total distance covered to the total time taken is called average speed.

Average speed = 16.898 / 0.4583 = 36.87 km/h

6 0

2 years ago

What causes an object to move, change direction or change speed?

Wittaler [7]

I believe it would be an unbalanced force. Because the forces are unbalanced, one side is stronger and, therefore, the object will move.

7 0

2 years ago

Read 2 more answers

The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.

Elena L [17]

Answer:

The magnitude of the acceleration is $a_r = 1.50 \ m/s^2$

The direction is $\theta = 32.5 6^o$ north of east

Explanation:

From the question we are told that

The force exerted by the wind is $F_{sail} = (330 ) \ N \ north$

The force exerted by water is $F_{keel} = (210 ) \ N \ east$

The mass of the boat(+ crew) is $m_b = 260 \ kg$

Now Force is mathematically represented as

$F = ma$

Now the acceleration towards the north is mathematically represented as

$a_n = \frac{F_{sail}}{m_b}$

substituting values

$a_n = \frac{330 }{260}$

$a_n = 1.269 \ m/s^2$

Now the acceleration towards the east is mathematically represented as

$a_e = \frac{F_{keel}}{m_b }$

substituting values

$a_e = \frac{210}{260}$

$a_e =0.808 \ m/s^2$

The resultant acceleration is

$a_r = \sqrt{a_e^2 + a_n^2}$

substituting values

$a_r = \sqrt{(0.808)^2 + (1.269)^2}$

$a_r = 1.50 \ m/s^2$

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

$tan \theta = \frac{a_e}{a_n}$

$\theta = tan ^{-1} [\frac{a_e}{a_n } ]$

substituting values

$\theta = tan ^{-1} [\frac{0.808}{1.269 } ]$

$\theta = tan ^{-1} [0.636 ]$

$\theta = 32.5 6^o$

5 0

3 years ago