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SSSSS [86.1K]
3 years ago
6

An inflatable backyard swimming pool is being filled from a garden hose with a flowrate of 0.12 gal/s. (a) If the pool is 8 ft i

n diameter, determine the time rate of change of the depth of the water in the pool.

Physics
2 answers:
Over [174]3 years ago
7 0

Explanation:

Below is an attachment containing the solution.

zzz [600]3 years ago
6 0

Answer:

0.00318 ft/s

Explanation:

Volume of the inflatables swimming pool, V = πr²h

V = πr²h

a) (dh/dt) = (dh/dV) × (dV/dt)

V = πr²h

(dV/dh) = πr²

(dh/dV) = (1/πr²)

So, we require the rate of change of depth when diameter = 8 ft, radius = (diameter)/2 = 4 ft

(dh/dt) = (dh/dV) × (dV/dt)

(dh/dV) = (1/πr²) = (1/π(4²)) = 0.0199 ft⁻²

(dV/dt) = 0.12 gal/s = 0.016 ft³/s

(dh/dt) = (0.0199 × 0.016) = 0.00318 ft/s

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Tema [17]

Answer:

v = 8.57 m/s

Explanation:

As we know that the wagon is pulled up by string system

So the net force on the wagon along the inclined is due to tension in the rope and component of weight along the inclined plane

So as per work energy theorem we know that

work done by tension force + work done by force of gravity = change in kinetic energy

F_t . d - (mgsin\theta)(d) = \frac{1}{2}mv^2 - 0

so we have

F_t = 129 N

\theta = 17.5^o

m = 38.2 kg

d = 85.4 m

so now we have

129(85.4) - (38.2)9.8sin17.5 (85.4) = \frac{1}{2}(38.2) v^2

v = 8.57 m/s

7 0
3 years ago
An insulated beaker with negligible mass contains 0.250 kg of water at 75.0C. How many kilograms of ice at -20.0C must be droppe
kkurt [141]

Answer:

The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg

Explanation:

Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water

Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C

To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.

Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C

Latent heat of ice = L = 334000 J/kg

Specific heat capacity of water = C = 4186 J/kg.°C

Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m

Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J

543600 m = 36627.5

m = 0.0674 kg = 67.4 g of ice.

3 0
3 years ago
P*V = n*R*T where P = pressure V = volume n = number of moles R = the universal gas constant T = temperature in degrees Kelvin T
seraphim [82]

The Gay-Lussac's law or Amonton's law states that the pressure of a given amount of a gas is directly propotional to its temperature if its volume is kept constant  .

            P∝T

and  

The Charles Law states that volume of given amount of gas at constant pressure  is directly propotional to temperature.

              V∝T

So, by Gay-Lussac's law if we increase the temperature the Pressure will increase and by Charles Law, if we increase the temperature the volume will increase.

Therefore, if the temperature of gas increases either the pressure of the gas, the volume of the gas, or both, will increase.

Hence,

Answer is option C

3 0
3 years ago
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Why is the moon’s surface cratered, but the Earth’s is not
Scorpion4ik [409]

The comets hit the moon's surface because there is no atmosphere on the moon to protect it. The earth has an atmosphere so it is protected.

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Stella [2.4K]

Answer:

Explanation:

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