Answer:
The variation rate is 5.42 10⁻⁵ cm²/ºC
Explanation:
When we have a thermal expansion problem we must have the relationship of the change in length as a function of the temperature, which are given in this problem, so we can write the expression for the area of a rectangle
a = L W
They ask us to find the rate of variation of this area depending on the temperature, so we can derive this expression with respect to the temperature
da / dT = d(LW) / dt
We use the derivative of a product since the two magnitudes change
da / dT = W dL/dT + L dW/dT
The values they give us are
= 1.9 10⁻⁵ cm/ºC
= 8.5 10⁻⁶ cm/ºC
W = 1.6 cm
L= 2.8 cm
Substituting the values and calculating
= 1.6 1.9 10⁻⁵ + 2.8 8.5 10⁻⁶
= 3.04 10⁻⁵ + 2.38 10⁻⁵
= 5.42 10⁻⁵ cm²/ºC
The variation rate is 5.42 10⁻⁵ cm²/ºC
Conservation of momentum: total momentum before = total momentum after
Momentum = mass x velocity
So before the collision:
4kg x 8m/s = 32
1kg x 0m/s = 0
32+0=32
Therefore after the collision
4kg x 4.8m/s = 19.2
1kg x βm/s = β
19.2 + β = 32
Therefore β = 12.8 m/s
Answer:
Answer is 98 joules
Explanation:
P. E=2*9.8*3=98
S.i unit of potential energy ⚡ is joules
So answer is 98 joules
Answer:
Umm...I guess mainly the percentage of successful ejaculations into the womb would be decreased. It also would suggest a malfunction in the testicles. The fertilizing process would also be less succesfull.
Explanation:
Not sure if this was a serious question or not.....
