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avanturin [10]
3 years ago
9

Suppose that the height (in centimeters) of a candle is a linear function of the amount of time (in hours) it has been burning.

After 11 hours of burning, a candle has a height of 27.9 centimeters. After 30 hours of burning, its height is 26 centimeters. What is the height of the candle after 20 hours?
Physics
1 answer:
iragen [17]3 years ago
3 0

Answer:

y = 27 cm

Explanation:

given,

After 11 hours of burning height of candle is 27.9 cm

After 30 hours of burning height of candle is 26 cm

height after 20 hour = ?

the candle height decreases linearly

using linear equation

y = m x+ c

27.9 = m (11 ) + C.......(1)

26 = m (30) + C..............(2)

on solving equation (1) and (2)

1.9 = -19 m

m = -0.1

from equation 2

26 = -3 + C

C = 29

y = m x + 29

at 20 hour

y = -0.1× 20 + 29

y = 27

height of candle after 20 hours is 27 cm.

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A weather balloon is filled to a volume of 6000 L while it is on the ground, at a pressure of 1 atm and a temperature of 273 K.
avanturin [10]

Answer : The final volume of the balloon at this temperature and pressure is, 17582.4 L

Solution :

Using combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 1 atm

P_2 = final pressure of gas = 0.3 atm

V_1 = initial volume of gas = 6000 L

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 273 K

T_2 = final temperature of gas = 240 K

Now put all the given values in the above equation, we get the final pressure of gas.

\frac{1atm\times 6000L}{273K}=\frac{0.3atm\times V_2}{240K}

V_2=17582.4L

Therefore, the final volume of the balloon at this temperature and pressure is, 17582.4 L

4 0
3 years ago
Read 2 more answers
A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. T
Serga [27]

Answer:

a)  k_{e} = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m,  e)  v = 15.23 m / s  

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

      k_{e} = ½ k x²

      k_{e} = ½ 2900 0.80²

      k_{e} = 928 J

b) place the origin at the point of the uncompressed spring, the spider's potential energy

     U = m h and

     U = 8 9.8 (-0.80)

     U = -62.7 J

c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

      K = ½ m v²

      K = 0

d) write the energy at two points, maximum compression and maximum height

     Em₀ = ke = ½ m x²

     E_{mf} = mg y

     Emo = E_{mf}

     ½ k x² = m g y

     y = ½ k x² / m g

     y = ½ 2900 0.8² / (8 9.8)

     y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

     Y = y- 0.80

     Y = 11.8367 -0.80

     Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

     Emo = ½ k x² = 928 J

     E_{mf}= ½ m v²

     Emo = E_{mf}

     Emo = ½ m v²

      v =√ 2Emo / m

     v = √ (2 928/8)

     v = 15.23 m / s

8 0
3 years ago
A car is traveling at a speed of 10m/s. A 0.5kg clump of mudis
CaHeK987 [17]

Answer:B

Explanation:

Given

speed of car v=10 m/s

mass of clump m=0.5 kg

Radius of car tire r=0.2 m

Since the tire is rotating about axle so a centripetal force is acting constantly on each particle towards the center of tire.

Centripetal force is given by

F_c=\frac{mv^2}{r}

where m=mass\ of\ element

v=speed

r=distance\ from\ center

F_c=\frac{0.5\times 10^2}{0.2}

F_c=250\ N (inward)

           

3 0
3 years ago
Two auto technicians are discussing the stages of engine operation. Auto Technician A says during the intake stage of engine ope
pav-90 [236]

Neither technician is correct.

Please don't touch my car.

7 0
3 years ago
A closed-end organ pipe is used to produce a mixture of sounds. The third and fifth harmonics in the mixture have frequencies of
nexus9112 [7]

Answer:

F_1=366.67Hz

Explanation:

From the question we are told that:

Frequency of 3rd harmonics F_3=1100

Frequency of 5th harmonics F_3=1833

Generally the equation for Wavelength at 3rd Harmonics is mathematically given by

 \lambda_3=\frac{4}{3}l

Therefore

 F_3=\frac{3v}{4l}

Generally the equation for Wavelength at 1st Harmonics is mathematically given by

 \lambda_1=\frac{4}{1}l

Therefore

 F_1=\frac{v}{4l}

Generally the equation for the frequency of the first harmonic is mathematically given by

 F_1=\frac{F_3}{3}

 F_1=\frac{1100}{3}

 F_1=366.67Hz

7 0
3 years ago
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