De los reptiles, anfibios; pájaros, peces; mamíferos y artrópodos, los que más se asemejan en características son los pájaros y los mamíferos porque ambos son de sangre caliente y no varia independientemente de en qué ambiente se encuentre
Answer: 361° C
Explanation:
Given
Initial pressure of the gas, P1 = 294 kPa
Final pressure of the gas, P2 = 500 kPa
Initial temperature of the gas, T1 = 100° C = 100 + 273 K = 373 K
Final temperature of the gas, T2 = ?
Let us assume that the gas is an ideal gas, then we use the equation below to solve
T2/T1 = P2/P1
T2 = T1 * (P2/P1)
T2 = (100 + 273) * (500 / 294)
T2 = 373 * (500 / 294)
T2 = 373 * 1.7
T2 = 634 K
T2 = 634 K - 273 K = 361° C
The coefficient of linear expansion, given that the length of the pipe increased by 1.5 cm is 1.67×10¯⁵ /°F
<h3>How to determine the coefficient of linear expansion</h3>
From the question given above, the following data were obtained
- Original diameter (L₁) = 10 m
- Change in length (∆L) = 1.5 cm = 1.5 / 100 = 0.015 m
- Change in temperature (∆T) = 90 °F
- Coefficient of linear expansion (α) =?
The coefficient of linear expansion can be obtained as illustrated below:
α = ∆L / L₁∆T
α = 0.015 / (10 × 90)
α = 0.015 / 900
α = 1.67×10¯⁵ /°F
Thus, we can conclude that the coefficient of linear expansion is 1.67×10¯⁵ /°F
Learn more about coefficient of linear expansion:
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Answer:
a. 79.1 N
b. 344 J
c. 344 J
d. 0 J
e. 0 J
Explanation:
a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force 
by the worker must be equal to the friction force 
on the crate, which is the product of friction coefficient μ and normal force N:
Let g = 9.81 m/s2
b. The work is done on the crate by this force is the product of its force and the distance traveled s = 4.35
c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35
This work is negative because the friction vector is in the opposite direction with the distance vector
d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.
e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction
In kynematics you describe the motion of particles using vectors and their change in time. You define a position vector r for a particle, and then define velocity v and acceleration a as
In dynamics Newton's laws predict the acceleration for a given force. Knowing the acceleration, and the kynematical relations defines above, you can solve for the position as a function of time: r(t)
