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Bas_tet [7]
3 years ago
9

A hiker, caught in a rainstorm might absorb 1 liter of water in her clothing. if it is windy so that this amount of water is eva

porated quickly at 20�c, how much heat would be required for this process?
Physics
1 answer:
NeTakaya3 years ago
4 0

Water evaporates at 100⁰C

So change in temperature = 100-20 = 80⁰C

Amount of water to be evaporated = 1 liter = 1L*1kg/liter = 1 kg

Specific heat of water is 1 calorie/gram ⁰C = 4.186 joule/gram =4186 J/kg

So heat required E = mcΔT = 1 * 4186 *80= 334880 J =334.88 kJ

So amount of heat require to evaporate water = 334.88 kJ

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Bro i need help omg.
sergey [27]

Answer:

25.

Explanation:

8 0
3 years ago
Read 2 more answers
What are the wavelengths of electromagnetic wave in free space that have the following frequencies?.
irina [24]

The wavelengths of the light are 4.3 * 10^-12 m and 0.2 m respectively.

<h3>What is wavelength?</h3>

The term wavelength has to do with the horizontal distance that is covered by a wave. We know that a long wavelength implies that the wave is able to travel a long distance from one point to another.

Given that;

c = λf

c = speed of light

λ = wavelength of ight

f = frequency of light

Thus;

λ = 3 * 10^8/ 7.00 x 10^19

λ = 4.3 * 10^-12 m

λ = 3 * 10^8/1.50 x 10^9

λ = 2 * 10^-1 or 0.2 m

Learn more about wavelength:brainly.com/question/13533093

#SPJ1

Missing parts:

What are the wavelengths of electromagnetic wave in free space that have the following frequencies? (a) 7.00 x 10^19 Hz______ pm (b) 1.50 x 10^9 Hz__________ cm

7 0
1 year ago
Read 2 more answers
What type of energy results from the burning of wood or gasoline?
Sonja [21]

Answer:

Chemical energy

Explanation:

Chemical energy is energy stored in the bonds of atoms and molecules. Batteries, biomass, petroleum, natural gas, and coal are examples of chemical energy. Chemical energy is converted to thermal energy when people burn wood in a fireplace or burn gasoline in a car's engine.

7 0
3 years ago
A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t
notsponge [240]

A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is 1.37 m/s^2

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

A)

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, F_f, backward

The frictional force can be written as

F_f = \mu mg

where

\mu=0.03 is the coefficient of kinetic friction

m=150 kg is the mass of the motorbike

g=9.8 m/s^2 is the acceleration of gravity

Therefore the net force is given by

\sum F = F - F_f = F - \mu mg

And substituting, we find

\sum F=250 - (0.03)(150)(9.8)=205.9 N

2)

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

\sum F = ma

where

m is the mass

a is the acceleration

In this problem, we have

\sum F = 205.9 N is the net force

m = 150 kg is the mass

Solving for a, we find the acceleration:

a=\frac{\sum F}{m}=\frac{205.9}{150}=1.37 m/s^2

C)

Since the motion of the motorbike is a uniformly accelerated motion, we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For this motorbike, we have:

u = 0 (it starts from rest)

a=1.37 m/s^2

s = 350 m

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(1.37)(9.8)}=5.2 m/s

4)

For this part of the problem, we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the elapsed time

Here we have:

v = 5.2 m/s

u = 0

a=1.37 m/s^2

Solving for t, we find

t=\frac{v-u}{a}=\frac{5.2-0}{1.37}=3.80 s

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6 0
3 years ago
Legacy issues $570,000 of 8.5%, four-year bonds dated January 1, 2019, that pay interest semiannually on June 30 and December 31
choli [55]

Answer:

1) Determine the total bond interest expense to be recognized.

Total bond interest expense over life of bonds:

Amount repaid:    

8 payments of $24,225:           $193,800    

Par value at maturity:                 $570,000    

Total repaid:                                   $763800 (193,800 + 570,000)  

Less amount borrowed:         $508050    

Total bond interest expense: $255750 (763800 - 508,050)

2)Prepare a straight-line amortization table for the bonds' first two years.

Semiannual Interest Period­ End; Unamortized Discount; Carrying Value

01/01/2019                                      61,950                           508,050  

06/30/2019                                      54,206                          515,794  

12/31/2019                                       46,462                         523,538  

06/30/2020                                       38,718                        531,282  

12/31/2020                                         30,974                          539,026

3) Record the interest payment and amortization on June 30:

June 30            Bond interest expense, dr                         31969  

                       Discount on bonds payable, Cr     (61950/8)  7743.75

                                        Cash, Cr                     ( 570000*8.5%/2)  24225  

4) Record the interest payment and amortization on December 31:

Dec 31                 Bond interest expense, Dr               31969  

                           Discount on bonds payable, Cr  7744  

                                    Cash, Cr                                24225

6 0
3 years ago
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