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Tpy6a [65]
4 years ago
6

A severe thunderstorm dumped 2.0 in of rain in 30 min on a town of area 22 km2. what mass of water fell on the town? one cubic m

eter of water has a mass of 103 kg.
Physics
2 answers:
AlekseyPX4 years ago
7 0
For this case, the first thing you should know is that density is defined as the quotient between mass and volume.
 d = m / v
 We have then for this case that the density is:
 d = 103 Kg / m ^ 3
 The volume is given by:
 V = A * d
 Where
 A: area of the city
 d: height of the rain
 Also, for calculation purposes, you must know the following measure:
 1Km = 1000m
 1 inch = 0.0254 meters
 Calculating the volume we have:
 V = (22 * (1000/1) ^ 2) * (2 * (0.0254 / 1)) = 1117600 m ^ 3
 Then, calculating the mass we have:
 m = d * v
 m = (103) * (1117600) = 115112800 Kg
 Equivalently
 m = 1.15 * 10 ^ 8 Kg
 answer
 mass of water fell on the town was 1.15 * 10 ^ 8 Kg
Naily [24]4 years ago
5 0
We want to convert everything to meters so we can calculate cubic meter volume and use the given mass information.  
2 inches = 5.08 centimeters = .0508 meters.  
22 square kilometers = 22,000,000 sq meters.  
The product of these gives the number of cubic meters of water that rained. .0508 * 22,000,000 = 1,117,600 cubic meters.  
Multiply by the mass to arrive at final answer. 1,117,600 m^3 * 103kg/m^3 = 115,112,800 kg  
115,112,800 kg of water fell.
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a 2.0*10^3 kg car accelerates from rest under the actions of two forces. One is a forward force of 1140 N provided by traction b
BlackZzzverrR [31]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

<span>KE = 1/2mv^2 = (1/2)(2000)(2^2) = 4000 J This must equal the net work acting on the car. W=Fd The net force is 1140-950= 190N. so, d=W/F = 4000/190 = 21.05 m</span>
4 0
3 years ago
7. A 1.0 kg metal head of a geology hammer strikes a solid rock with a velocity of 5.0 m/s. Assuming all the energy is retained
marta [7]

The increase in temperature of the metal hammer is 0.028 ⁰C.

The given parameters:

  • <em>mass of the metal hammer, m = 1.0 kg</em>
  • <em>speed of the hammer, v = 5.0 m/s</em>
  • <em>specific heat capacity of iron, 450 J/kg⁰C</em>

The increase in temperature of the metal hammer is calculated as follows;

Q = K.E\\\\mc \Delta T = \frac{1}{2}  mv^2\\\\\Delta T = \frac{v^2}{2 c}

where;

<em>c is the </em><em>specific heat capacity</em><em> of the metal hammer</em>

<em />

Assuming the metal hammer is iron, c = 450 J/kg⁰C

\Delta T = \frac{5^2}{2 \times 450} \\\\\Delta T = 0.028 \ ^0C

Thus, the increase in temperature of the metal hammer is 0.028 ⁰C.

Learn more about heat capacity here: brainly.com/question/16559442

8 0
3 years ago
A quantity of 14.1 cm^3 of water at 8.4°C is placed in a freezer compartment and allowed to freeze to solid ice at -7.2°C. How m
serious [3.7K]

Answer:920.31 J

Explanation:

Given

Volume of water (V)=14.1 cm^3

mass(m)=\rho \times V=1000\times 14.1\times 10^{-6}=14.1 gm

Temperature =8.4^{\circ} C

Final Temperature =-7.2 ^{\circ}C

specific heat of water(c)=4.184 J/g-^{\circ}C

Therefore heat required to removed is

Q=mc(\Delta T)

Q=14.1\times 4.184\times (8.4-(-7.2))

Q=920.31 J

3 0
4 years ago
An unbalanced force of 20 N is applied to a 4.0 kg mass at rest. What is the
inna [77]

Answer:

<h2>5 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

We have

a =  \frac{20}{4}  = 5 \\

We have the final answer as

<h3>5 m/s²</h3>

Hope this helps you

5 0
3 years ago
A carnot heat engine receives 600 kj of heat from a source of unknown temperature and rejects 175 kj of it to a sink at 20°c. de
oee [108]

(b) 71%

The thermal efficiency of a Carnot heat engine is given by:

\eta = \frac{W}{Q_{in}}

where

W is the useful work done by the engine

Q_{in} is the heat in input to the machine

In this problem, we have:

Q_{in}=600 kJ is the heat absorbed

W=600 kJ-175 kJ=425 kJ is the work done (175 kJ is the heat released to the sink, therefore the work done is equal to the difference between the heat in input and the heat released)

So, the efficiency is

\eta = \frac{425 kJ}{600 kJ}=0.71 = 71\%

(a) 737^{\circ}C

The efficiency of an engine can also be rewritten as

\eta = 1-\frac{T_C}{T_H}

where

T_C is the absolute temperature of the cold sink

T_H is the temperature of the source

In this problem, the temperature of the sink is

T_C = 20^{\circ}C + 273=293 K

So we can re-arrange the equation to find the temperature of the source:

T_H = \frac{T_C}{1-\eta}=\frac{293 K}{1-0.71}=1010 K\\T_H = 1010 K - 273=737^{\circ}C

7 0
3 years ago
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