Electric power is given by:
Power = voltage x current
10000 = 15.8V
V = 633 Volts
Answer:
We can cause delamination.
Explanation:
The reason why is because the probability of causing delamination increase considerably when we use Hole-filling fasteners. If we use a typical rivet, these tends to expands in order to fill the hole.
If we analyze the force applied by the expanded rod will cause that the matrial will be deteriorated and will cause that the material to delaminate around the edges of the hole and we can cause possible control and no protection to the material.
Given :
The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12.
To Find :
How many grains are there in the ball?
Solution :
Volume of ball of the ballpoint is :
Now, grain size of 12 has about 520000 grains/mm³.
Therefore, number of grains are :
<h2>Hello!</h2>
The answer is: There are 3.42 moles of MgCl2 in 326g of the compound.
<h2>Why?</h2>
Assuming that the compound is MgCl2, we can find how many moles of the compound are in 326 g of the same compound, calculating the molar mass of the compound, so:
Then,
Therefore, to calculate how many moles are in 326 of the compound, we can use the following equation
Hence,
There are 3.42 moles of MgCl2 in 326g of the compound.
Have a nice day!
Answer:
a) When its length is 23 cm, the elastic potential energy of the spring is
0.18 J
b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
Explanation:
Hi there!
a) The elastic potential energy (EPE) is calculated using the following equation:
EPE = 1/2 · k · x²
Where:
k = spring constant.
x = stretched lenght.
Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).
First, let´s convert the spring constant units into N/m:
4 N/cm · 100 cm/m = 400 N/m
EPE = 1/2 · 400 N/m · (0.03 m)²
EPE = 0.18 J
When its length is 23 cm, the elastic potential energy of the spring is 0.18 J
b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:
EPE = 1/2 · 400 N/m · (0.06 m)²
EPE = 0.72 J
When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
