Answer:
724.3J/Kg.K
Explanation:
CHECK THE COMPLETE QUESTION BELOW
Compute the specific heat capacity at constant volume of nitrogen (N2) gas.and compare with specific heat of liquid water. The molar mass of N2 is 28.0 g/mol.
The specific heat capacity can be computed by using expression below
c= CV/M
Where c= specific heat capacity
M= molar mass
CV= molar hear capacity
Nitrogen is a diatomic element, the Cv can be related to gas constant with 5/2R
Where R= 8.314J/mol.k
Molar mass= 28 ×10^-3Kg/mol
If we substitute to the expression, we have
c= (5R/2)/(M)
=5R/2 × 1/M
=(5×8.314) /(2×28 ×10^-3)
=724.3J/Kg.K
Hence, the specific heat capacity at constant volume of nitrogen (N2) gas is
724.3J/Kg.K
The specific heat of liquid water is about 4182 J/(K kg) which is among substance with high specific heat, therefore specific heat of Nitrogen gas is 724.3J/Kg.K which is low compare to that of liquid water.
As a bicycle is ridden west in a straight line with decreasing speed,the acceleration of the bicycle must be
Answer:
Decreasing
Hope this helps! :)
A statement of the second law of thermodynamics is that:__________.
a) spontaneous reactions are always exothermic.
b) energy is conserved in a chemical reaction that has a decrease in entropy.
c) spontaneous reactions are always endothermic.
d) in a spontaneous process, the entropy of the universe increases.
Answer:
in a spontaneous process, the entropy of the universe increases.
Explanation:
Entropy is a measure of of the degree of randomness or disorderliness in a system.
The second law of thermodynamics can be stated as follows; "in any spontaneous process, the entropy of the universe increases."
The universe here refers to the system's disorder and the disorder of the surroundings. Therefore, a spontaneous process can occur, in which the entropy of the system decreases, only if the entropy increases in the surroundings.
For instance, when ice freezes, the entropy of liquid water decreases, that is, the entropy of the system decreases. However, heat is given off to the surroundings and the entropy of the surroundings increases. This is an obvious expression of this law.
A projectile is fired horizontally from a height of 10 m above level ground. The projectile lands a horizontal distance of 15 m from where it was launched.
-Find the hang time for the projectile.
-Find the initial speed of a projectile.
-What are the x and y components of the projectile’s velocity the moment before it strikes the ground?
-At what speed will the projectile strike the ground?
Answer:
a)t = 1,43 s
b) V = 10,49 m/s
c) V₀ₓ = 10,49 m/s ; V₀y = 14,01 m/s
d) Vf = 17,5 m/s
Explanation:
According to the problem statement
V₀ = V₀ₓ and V₀y = 0
And at the end of the movement t = ? the distance y = 10 m
Therefore as
h = V₀y - (1/2)*g*t²
Vertical distance y = h = 10 = V₀y*t - 0,5 (-9,8)*t²
10 = 4,9*t²
t² = 10/4,9 ⇒ t² = 2,04 s
t = 1,43 s
a) 1,43 s is the time of movement
b) V₀ = V₀ₓ V₀y = 0 and V₀ₓ = Vₓ ( constant )
Just before touching the ground, the horizontal distance is
hd = 15 = Vₓ * t
Then 15 /1,43 = Vₓ = V₀ₓ
Vₓ = 10,49 m/s
Then initial speed is V = 10,49 m/s since V₀y = 0
Vf² = Vₓ² + Vy²
Vyf = V₀y - g*t
Vyf = 0 - 9,8 *1,43
Vyf = - 14,01 m/s
And finally the speed when the projectile strike the ground is:
Vf² = Vₓ² + Vy²
Vf = √ (10,49)² + (14,01)²
Vf = 17,50 m/s
convert 100 Newton into dyne
Answer:10000000
Explanation:
What is the acceleration of the the object during the first 4 seconds?
Answer:
Velocity (m/s) over time (s) graph
Velocity (m/s) over time (s) graph
We could write out our average acceleration as:
a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t
a = (15 m/s - 0 m/s) / 0.2 seconds
a = 15 m/s / 0.2 seconds
a = 75 m/s / second
Explanation:
What this formula is telling us is that if we know the acceleration of an object, and the ... we can plug in our acceleration of 12.5 m/s2 for a, and 4 seconds for t.
Velocity (m/s) over time (s) graph
Velocity (m/s) over time (s) graph
We could write out our average acceleration as:
a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t
a = (15 m/s - 0 m/s) / 0.2 seconds
a = 15 m/s / 0.2 seconds
a = 75 m/s / second
A certain heat engine does 30.2 kJ of work and dissipates 9.14 kJ of waste heat in a cyclical process.
A) What was the heat input to this engine?
B) What was its efficiency?
Answer:
a) [tex]H_{in}=39.34 kJ[/tex]
b) Efficiency=76.77%
Explanation:
a)
In order to solve this problem, we can use the following formula:
[tex]H_{in}=H_{out}+W[/tex]
the problem provides us with all the necessary information so we can directly use the formula:
[tex]H_{in}=9.14kJ+30.2kJ[/tex]
[tex]H_{in}=39.34 kJ[/tex]
b) In order to find the efficiency, we can use the following formula:
[tex]Efficiency=\frac{W}{H_{in}}*100\%[/tex]
so we get:
[tex]Efficiency=\frac{30.2kJ}{39.34kJ}*100\%[/tex]
Efficiency=76.77%
What would its weight be on Jupiter?
24.9N
Answer:
1.898 × 10^27 kg
Explanation:
thats how much it ways
A sled is pulled with a force of 540 N at an angle of 40° with the horizontal. What are the horizontal and vertical components of this force?
Answer:
Fx = 467.65N
Fy = 270N
Explanation:
Given
Force = 540N
angle of inclination = 40 degree
Horizontal component Fx = Fcos 30
Fx = 540cos30
Fx = 540(0.8660)
Fx = 467.65N
Hence the horizontal component is 467.65N
Vertical component Fy = Fsin 30
Fy = 540sin30
Fy = 540(0.5)
Fy = 270N
Hence the vertical component is 270N
Galaxy B moves away from galaxy A at 0.577 times the speed of light. Galaxy C moves away from galaxy B in the same direction at 0.731 times the speed of light. How fast does galaxy C recede from galaxy A?
Answer:
The value is [tex]p = 0.7556 c[/tex]
Explanation:
From the question we are told that
The speed at which galaxy B moves away from galaxy A is [tex]v = 0.577c[/tex]
Here c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
The speed at which galaxy C moves away from galaxy B is [tex]u = 0.731 c[/tex]
Generally from the equation of relative speed we have that
[tex]u = \frac{p - v}{ 1 - \frac{ p * v}{c^2} }[/tex]
Here p is the velocity at which galaxy C recede from galaxy A so
[tex]0.731c = \frac{p - 0.577c }{ 1 - \frac{ p * 0.577c}{c^2} }[/tex]
=> [tex]0.731c [1 - \frac{ p * 0.577}{c}] = p - 0.577c[/tex]
=> [tex]0.731c - 0.4218 p = p - 0.577c[/tex]
=> [tex]0.731c + 0.577c = p + 0.4218 p[/tex]
=> [tex]1.308 c = 1.731 p[/tex]
=> [tex]p = 0.7556 c[/tex]
An element or compound used to enhance a semiconductor is called a(n) ____.
The element named boron can be used to enhance the properties of semiconductors.
What is a semiconductor?A semiconductor is a material that has electronic properties and has the value that falls in between a conductor. It can be a metallic copper or an insulator.
The rise in temperatures leads to a fall in resistivity. The element named boron can improve the electrical properties of the semiconductor as they form the impurities.
Find out more information about the element.
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