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Marizza181 [45]
3 years ago
14

60 point!!!

Physics
1 answer:
erastova [34]3 years ago
7 0

Answer:

Playing hockey, driving a car, and even simply taking a walk are all everyday examples of Newton's laws of motion.

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A rigid container holds 0.30g of hydrogen gas.
Strike441 [17]

Answer:

Part A:    \mathbf{Q =94 \ J} to two significant figures

Part B:    \mathbf{Q =160  \ J} to two significant figures

Part C:    \mathbf{Q =220  \ J} to two significant figures

Explanation:

Given that :

mass of the hydrogen = 0.30 g

the molar mass of hydrogen gas molecule = 2 g/mol

we all know that:

number of moles = mass/molar mass

number of moles = 0.30 g /2 g/mol

number of moles = 0.15 mol

For low temperature between the range of 50 K to 100 K, the specific heat at constant volume for a diatomic gas molecule = C_v=\dfrac{3}{2}R

For Part A:

Q = mC_v\Delta T

Q= 0.15 \ mol (\dfrac{3}{2})(8.314 \ J/mole.K )(100-50)K

Q= 0.15 \times (\dfrac{3}{2}) \times (8.314 \ J )\times (50)

Q=93.5325 \ J

\mathbf{Q =94 \ J} to two significant figures

Part B. For hot temperature, C_v=\dfrac{5}{2}R

Q = mC_v\Delta T

Q= 0.15 \ mol (\dfrac{5}{2})(8.314 \ J/mole.K )(300-250)K

Q= 0.15 \times (\dfrac{5}{2}) \times (8.314 \ J )\times (50)

Q=155.8875 \ J

\mathbf{Q =160  \ J} to two significant figures

Part C. For an extremely hot temperature, C_v=\dfrac{7}{2}R

Q = mC_v\Delta T

Q= 0.15 \ mol (\dfrac{7}{2})(8.314 \ J/mole.K )(2300-2250)K

Q= 0.15 \times (\dfrac{7}{2}) \times (8.314 \ J )\times (50)

Q=218.2425 \ J

\mathbf{Q =220  \ J} to two significant figures

6 0
3 years ago
What are the characteristics and ph level of acids?
12345 [234]

General Characteristics of Acids & Bases. Chemists measure the strength of an acid or base by its pH, which is a term that refers to the "power of hydrogen.". The midpoint of the pH scale is neutral. Compounds with a pH lower than the midpoint value are acidic while those with a higher value are basic or alkaline.

7 0
3 years ago
Two identical satellites orbit the earth in stable orbits. one satellite orbits with a speed v at a distance r from the center o
wariber [46]
The available options are: (found the complete text on internet)
A- at a distance less than r
B- at a distance equal to r
C- at a distance greater than r

Solution:
The correct answer is C) at a distance greater than r.

In fact, the gravitational attraction between the satellite and the Earth provides the centripetal force that keeps the satellite in circular orbit, so we can write
G \frac{Mm}{r^2}=m \frac{v^2}{r}
where the term on the left is the gravitational force, while the term on the right is the centripetal force, and where
G is the gravitational constant
M is the Earth mass
m is the satellite mass
r is the distance of the satellite from the Earth's center
v is the satellite speed

Re-arranging the equation, we get
r= \frac{GM}{v^2}
and we see from this formula that, if the second satellite has a speed less than the speed v of the first satellite, it means that the denominator of the fraction is smaller, and so r is larger for the second satellite.
7 0
3 years ago
An empty 2,500 kg train car is headed northbound at a velocity of 5 m/s. Ahead of the first car, an empty 1,500 kg car is headed
Tema [17]

Let the mass of 2500 kg car be m_1 and it's velocity be v_1 and the mass of 1500 kg car be m_2 and it's velocity be v_2 .

After the bumping the mass be M and it's velocity be V.

     By law of conservation of momentum we have

                   m_1v_1+m_2v_2 = MV

                    2500 * 5 + 1500 * 1=4000 * V

                    V = 14000/4000 = 7/2 = 3.5 m/s

So the velocity of the two-car train = 3.5 m/s

9 0
3 years ago
I would love to stretch a wire from our house to the Shop so I can 'call' my husband in for meals. The wire could be tightened t
dezoksy [38]
Note: I'm not sure what do you mean by "weight 0.05 kg/L". I assume it means the mass per unit of length, so it should be "0.05 kg/m".

Solution:
The fundamental frequency in a standing wave is given by
f= \frac{1}{2L} \sqrt{ \frac{T}{m/L} }
where L is the length of the string, T the tension and m its mass. If  we plug the data of the problem into the equation, we find
f= \frac{1}{2 \cdot 24 m} \sqrt{ \frac{240 N}{0.05 kg/m} }=1.44 Hz

The wavelength of the standing wave is instead twice the length of the string:
\lambda=2 L= 2 \cdot 24 m=48 m

So the speed of the wave is
v=\lambda f = (48 m)(1.44 Hz)=69.1 m/s

And the time the pulse takes to reach the shop is the distance covered divided by the speed:
t= \frac{L}{v}= \frac{24 m}{69.1 m/s}=0.35 s
7 0
3 years ago
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