Seasonal rainstorms because there is a lot of water vapor in the air causing it to always be moist.
Initial volume of water is 155 mL. Convert the mass of iron and lead into volume using density values as follows:
d=\frac{m}{V}
Here, d is density, m is mass and V is volume of objects
volume of piece of iron can be calculated as:
V=\frac{m}{d}=\frac{15 g}{7.86 g/mL}=1.90 mL
Similarly, volume of lead can be calculated as:
V=\frac{m}{d}=\frac{20 g}{11.3g/mL}=1.77 mL
To calculate the final volume of water, volume of both iron and lead should be added to the initial volume of water.
V_{final}=V_{initial}+V_{iron}+V_{lead}=155 mL+1.90 mL+1.77mL=158.67 mL
Thus, new water level in milliliters will be 158.67 mL.
Answer:
Current needed = 704A
Explanation:
Using the fomula; torque(τ) = (I)(A)(B)Sinθ
Where B = uniform magnetic field
I = current and A = Area
Diameter = 19cm = 0.19m so, radius = 0.19/2 = 0.095m
Area(A) = πr^(2) = πr^(2)
= π(0.095)^(2) = 0.0284 m^(2)
Now, B(earth)= 5x10^-5 T
While, we can ignore the angle because it's insignificant since the angle of the wire is oriented for maximum torque in the earth's field.
Now, if we arrange the formula to solve for charge (I):
I = (τ)/(A)(B)
I = (1.0x10^-3) / (0.0284)(5x10^-5)
I = 704A
Mario lioedrama is the season to fall asleep in the front and back down to my school laptop to