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lora16 [44]
2 years ago
9

Is 72.3 L of an ideal gas is cooled from 36 Celsius to -156 Celsius what is the volume of the gas become

Physics
1 answer:
Furkat [3]2 years ago
7 0

Answer:

72300cm^3

Explanation:

1 liter = 1000cm^3

72.3 liter = x

Cross multiply

72.3 x 1000 = x

x = 72300cm^3

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You're driving down the highway late one night at 20 m/s when a deer steps onto the road 49 m in front of you. You reaction time
Leno4ka [110]

Answer:

v = 26.7 m/s

Explanation:

Given,

speed of the car = 20 m/s

distance between the car and the deer = 49 m

time taken to press the brake = 0.50 s

maximum deceleration of the car = 10 m/s²

Now,

distance travel by the car in 0.5 s = u x t = 20 x 0.5 = 10 m

distance travel by the car after the break is pressed

Using equation of motion

v² = u² + 2 a s

0² = 20² - 2 x 10 x s

s = 20 m

Total distance travel by the car = 20 + 10 = 30 m

Distance between deer and car = 49-30 = 19 m.

b. Maximum speed a car could have

Distance travel by the car in reaction time = v' x 0.5

v' is the maximum speed of the car.

maximum distance car can cover = 49 - 0.5 v'

Now, Using equation of motion

v² = u² + 2 a s

0² =v'² - 2 x 10 x (49- 0.5 x v')

v'² +10 v' -980 = 0

By solving

v = 26.7 m/s

Hence, maximum speed of the car can be 26.7 m/s

 

4 0
3 years ago
Calculate the unit cell edge length for an 79 wt% Ag- 21 wt% Pd alloy. All of the palladium is in solid solution, and the crysta
ankoles [38]

Answer:

The edge length is 0.4036 nm

Solution:

As per the question:

Density of Ag, \rho = 10.49 g/cm^{3}

Density of Pd, \rho = 12.02 g/cm^{3}

Atomic weight of Ag, A = 107.87 g/mol

Atomic weight of Pd, A' = 106.4 g/mol

Now,

The average density, \rho_{a} = \frac{n A_{avg}} {V_{c}\times N_{A}}

where

V_{c} = a^{3}  = Volume of crystal lattice

a = edge length

n = 4 = no. of atoms in FCC

Therefore,

\rho_{a} = = \frac{n A_{avg}} {V_{c}\times N_{A}}

Therefore, the length of the unit cell is given as:

a = (\frac{nA_{avg}}{\rho_{a}\times N_{a}})^{1/3}            (1)

Average atomic weight is given as:

A_{avg} = \frac{100}{\frac{C_{Ag}}{A_{Ag}} + \frac{C_{Pd}}{A_{Pd}}}

where

C_{Ag} = 79 %

A_{Ag} = 107

C_{Pd} = 21%

A_{Pd} = 106

Therefore,

A_{avg} = \frac{100}{\frac{79}{107} + \frac{21}{106}} = 106.78

In the similar way, average density is given as:

\rho_{a} = \frac{100}{\frac{C_{Ag}}{\rho_{Ag}} + \frac{C_{Pd}}{\rho_{Pd}}}

\rho_{a} = \frac{100}{\frac{79}{10.49} + \frac{21}{12.02}} = 10.78 g/cm^{3}

Therefore, edge length is given by eqn (1) as:

a = (\frac{4\times 106.78}{10.78\times 6.023 X 10^23})^{1/3} = 4.036\times 10^{- 8} cm = 0.4036\times 10^{- 9} m = 0.4036 nm

5 0
3 years ago
According to Newton's 3rd law of motion, if you hit a baseball with a bat it will _____.
Anna [14]
Fly in a straight line unless an outside force changes its course because i tried it once in a baseball game that my mommy rekt me in.
4 0
3 years ago
Read 2 more answers
The position vector of a particle of mass 1.65 kg as a function of time is given by = (6.00 î + 4.15 t ĵ), where is in meters an
SashulF [63]

Answer:

 L = 41.09 Kg m2 / s      The angular momentum does not depend on the time

Explanation:

The definition of angular momentum is

        L = r x p

Where blacks indicate vectors

Let's apply this definition our case. Linear momentum

      p = m v

Let's replace

      L = m r x v

The given function is

      x = 6.00 i ^ + 4.15 t j ^

We look for speed

     v = dx / dt

     v = 0 + 4.15 j ^

To evaluate the angular momentum one of the best ways is to use determinants

     L = m \left[\begin{array}{ccc}i&j&k\\6&4.15t&0\\0&4.15&0\end{array}\right]

      L = m 6 4.15 k ^

The other products give zero

Let's calculate

      L = 1.65 6 4.15 k ^

      L = 41.09 Kg m2 / s

The angular momentum does not depend on the time

7 0
3 years ago
A Smart Car, which has a mass of 1000 kg, is going 20 m/s. When it hits the barrier, it stops with a time of 0.5 seconds. What i
antiseptic1488 [7]

Answer:

The change in momentum = -20000 kg m/s.

Explanation:

Mass m = 1000 kg

speed v₁ = 20 m/s

speed v₂ = 0 m/s

We know that,

The change in momentum

ΔP = m (Δv)

ΔP = m (v₂ - v₁)

     = 1000 (0 - 20)

     = 1000 (-20)

     = -20000 kg m/s

Thus, the change in momentum = -20000 kg m/s.

Note: negative sign indicates that the velocity is reducing when it hits the barrier.

4 0
2 years ago
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