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Verizon [17]
1 year ago
5

How many air molecules are in a 13.0×12.0×10.0 ft room (28.2 L=1 ft3)? Assume atmospheric pressure of 1.00 atm, a room temperatu

re of 20.0 ∘C, and ideal behavior.
Physics
1 answer:
aksik [14]1 year ago
4 0

Answer:

1963.93 Moles

Explanation:

-We know the standard conversion ratio for the volume of a mole is 1 Mole=22.4L

Given volume of rooms as 13.0ft\times12.0ft\times10ft=1560 ft^3

Convert the volume into liters:1560\times28.2l=43992l

#From our conversion ratio above, we get the volume of air molecules in moles as:

V_m=\frac{43992L}{22.4L}\times 1\ mole\\\\=1963.93\ Moles

Hence, the volume of air molecules is 1963.93 Moles

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A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?
ankoles [38]
Given required solution

M=10kg W=? W=Fd
v=5.0m/s F=mg
t=2.40s =10*10=100N
S=VT
=5m/s*2.4s
=12m
so W=12*100
W=1200J
4 0
2 years ago
The atomic number of beryllium (Be) is 4, and the atomic number of barium (Ba) is 56. Which comparison is best supported by this
Thepotemich [5.8K]

The answer is:

B)They are in the same group because they have similar chemical properties, but they are in different periods because they have very different atomic numbers.

The explanation:

The vertical columns on the periodic table are called groups or families because of their similar chemical behavior. All the members of a family of elements have the same number of valence electrons and similar chemical properties.

A period is a horizontal row of elements on the periodic table. For example, the elements sodium ( Na ) and magnesium ( Mg ) are both in period 3. The elements astatine ( At ) and radon ( Rn ) are both in period 6.

6 0
2 years ago
Read 2 more answers
The atmosphere pressure can support mercury in a tube, which the upper end is closed, up to 0.76 meter. If the mercury is replac
Leni [432]

Answer:

Maximum height the atmosphere pressure can support the

water=10.336 m

Explanation:

We know that ,

Pressure = h\cdot\rho\cdot g

Case 1 - Mercury in the tube

Density\ of\ mercury =\rho_1\\and\ height\ attained\ for\ mercury\ column = h_1

Case 2 - Water in the tube

Density\ of\ water =\rho_2\\and\ height\ attained\ for\ water\ column = h_2

Since atmospheric pressure is same

.P=h_1\cdot\rho_1\cdot g = h_2\cdot\rho_2\cdot g

or,  h_2=\frac{h_1\rho_1}{\rho_2}

Given\ h_1= 0.76\  m,\rho_1=13.6\cdot\rho_2

∴ h_2=0.76\cdot13.6=10.336\ m

Hence height of the water column =10.336 m

6 0
2 years ago
A particle leaves the origin with an initial velocity v → = (3.00iˆ) m/s and a constant acceleration a → = (−1.00iˆ − 0.500jˆ) m
tatiyna

Answer:

the position vector (x,y) will be (1.5 m,-2.25 m) and the velocity vector (vx,vy) will be ( 0 m/s , -1.5 m/s) when x reaches its maximum x coordinate

Explanation:

Since the velocity is related with the acceleration and coordinates through

vx²=v₀x²+2*ax*x

where

vx = velocity in the x direction

v₀x = initial velocity in the x direction = 3 m/s

ax = acceleration in the x direction = −1.00 m/s²

x= coordinates in the x-axis

when x reaches its maximum coordinate , then vx=0

thus

vx²=v₀x²+2*ax*x

0 = (3 m/s)² + 2* (−1.00 m/s²)*x

x= 1.5 m

also for the time t

vx = v₀x + ax*t → t= (vx-v₀x)/ax = (0- 3 m/s)/  (−1.00 m/s²) = 3 seconds

for the y coordinates

y = y₀+v₀y*t + 1/2 ay*t²

where

v₀y = initial velocity in the y direction = 0 m/s

ay = acceleration in the x direction = −0.5 m/s²

y= coordinates in the y-axis

y₀= initial coordinate in the y-axis =0

then since y₀=0 and v₀y=0

y = 1/2*ay*t²

y = 1/2*ay*t² = 1/2*(−0.5 m/s²)*(3 s)² = -2.25 m

and

vy=v₀y+ ay*t= 0+(−0.5 m/s²)*(3 s)= (-1.5 m/s)

therefore the position vector (x,y) will be (1.5 m,-2.25 m)

and the velocity vector (vx,vy) will be ( 0 m/s , -1.5 m/s)

7 0
1 year ago
At room temperature, a typical person loses energy to the surroundings at the rate of 62 W. If this energy loss has to be made u
Alex_Xolod [135]

To solve this problem it is necessary to use the given proportions of power and energy, as well as the energy conversion factor in Jules to Calories.

The power is defined as the amount of energy lost per second and whose unit is Watt. Therefore the energy loss rate given in seconds was

P = \frac{E}{t} \rightarrow E= Energy, t = time

P = 62W = 62 \frac{J}{s}

The rate of energy loss per day would then be,

P = 62\frac{J}{s} (\frac{86400s}{1day})

P = 5356800 \frac{J}{day}

That is to say that Energy in Jules per lost day is 5356800J

By definition we know that 1KCal = 4.184*10^{6}J

In this way the energy in Cal is,

E = 5356800J \frac{1KCal}{4.184*10^{6}J}

E = 1279.694 KCal

The number of kilocalories (food calories) must be 1279.694 KCal

4 0
2 years ago
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