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2 years ago

18. A swimmer steps out of a swimming pool into a light breeze. Why does

1 answer:

OverLord2011 [107]2 years ago

6 0

When the swimmer gets out, the water on her skin begins to evaporate. This process requires takes a lot of energy and that energy comes from the heat of her body. This is similar to when someone sweats. Hope this helps!

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Distance = 6 km south 60 minutes What was the average velocity

blsea [12.9K]

v = x/t

v = average velocity, x = displacement, t = elapsed time

Given values:

x = 6km south, t = 60min

Plug in and solve for v:

v = 6/60

v = 0.1km/min south

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3 years ago

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O operate this generator, a human applies a force by turning the crank. This action generates an electric current, which lights

Shtirlitz [24]

Answer A I’m very positive about it

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2 years ago

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Steam at 100°C is condensed into a 38.0 g aluminum calorimeter cup containing 280 g of water at 25.0°C. Determine the amount of

KonstantinChe [14]

Answer:

7.2g

Explanation:

From the expression of latent heat of steam, we have

Heat supplied by steam = Heat gain water + Heat gain by calorimeter

mathematically,

$m_{s}c_{w} \alpha _{w}$ + $m_{s}l$ = $m_{w}c_{w} \alpha _{w}$ + $m_{c}c_{c} \alpha c_{c}$

L=specific latent heat of water(steam)=2268J/g

$c_{w}$ =specific heat capacity=4.2J/gK

$c_{c}$ =specific heat capacity of calorimeter =0.9J/gk

$m_{w}$ =280g

$m_{c}$ =38g

α=change in temperature

$\alpha _{c}$ =(40-25)=15

$\alpha _{w}$ =(40-25)=15

$\alpha _{s}$ =(100-40)=60

Note: the temperature of the calorimeter is the temperature of it content.

From the equation, we can make $m_{s}$ the subject of formula

$m_{s}=\frac{m_{w}c_{w} \alpha +m_{c}c_{c}\alpha}{c_{w}\alpha +l}$

Hence

$m_{s}=\frac{(280*4.2*15) +(38*0.9*15)}{(4.2*60) +2268} \\m_{s}=\frac{18153}{2520}\\ m_{s}=7.2g$

Hence the amount of steam needed is 7.2g

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2 years ago

Evan drew a diagram to illustrate radiation

labwork [276]

The answer would be D, electromagnetic waves

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2 years ago

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A straight segment of a current-carrying wire has a current element IL where I = 2.70 A and L = 3.20 cm i + 4.30 cm j. The segme

myrzilka [38]

The component of the force in negative z-direction is -0.144 N.

The given parameters;

current in the wire, I = 2.7 A
length of the wire, L = (3.2 i + 4.3j) cm
magnetic filed, B = 1.24 i

The force on the segment of the wire is calculated as follows;

$F = ILBsin(\theta)$

where;

θ is the angle wire and magnetic field

The force on the wire segment will be perpendicular in negative z-direction (applying right hand rule), so there won't be any x and y component of the force.

The angle between the wire and the magnetic field is calculated as follows;

$\theta = tan^{-1} (\frac{y}{x} )\\\\\theta = tan^{-1} (\frac{4.3}{3.2} )\\\\\theta = 53.3 \ ^0$

The magnitude of the wire length is calculated as follows;

$|l | = \sqrt{3.2^2 + 4.3^2} = 5.36 \ cm = 0.0536 \ m$

The component of the force in negative z-direction is calculated as;

$F_z = -ILB sin(\theta)\\\\F_z = -2.7 \times 0.0536 \times 1.24 \times sin(53.3)\\\\F_z = -0.144 \ N$

Thus, the component of the force in negative z-direction is -0.144 N.

Learn more here:brainly.com/question/22719779

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2 years ago