When placing the piece of aluminium in water, the level of water will rise by an amount equal to the volume of the piece of aluminum.
Therefore, we need to find the volume of that piece.
Density can be calculated using the following rule:
Density = mass / volume
Therefore:
volume = mass / density
we are given that:
the density = 2.7 g / cm^3
the mass = 16 grams
Substitute in the equation to get the volume of the piece of aluminum as follows:
volume = 16 / 2.7 = 5.9259 cm^3
Since the water level will rise to an amount equal to the volume of aluminum, therefore, the water level will rise by 5.9259 cm^3
Answer:
Plug in the given values and solve for the final velocity. Remember, when the ball is on the ground it has a height of zero.
Explanation:
<span>The answer is C: water is drawn up a straw by cohesion and adhesion. Water molecules stick to one another and the walls of the straw, just like in a capillary.
Cohesion is the attractive force between like materials (between water
molecules).
Adhesion is the attractive force between twounlike materials (such as between
water and a solid container).
Capillary action is the tendency of a liquid to rise innarrow tubes or small openings as a result of adhesion and cohesion.
The liquid water molecules bind to the straw—a process known as adhesion. In the narrow space of the straw, the interaction of cohesion and adhesion causes theliquid to be drawn upward in the straw.</span>
So power is considered as the rate of doing work. Base on the problem given, my analysis is that the machine who finish the work faster is machine C. Therefore, in order to finish the same amount of work in a short period of time you are going to expend the most power. My answer is Machine C.
Answer:
(a) W= 44N
(b)W= 31.65 N
Explanation:
Data
T=44 N : Maximum force that the rope can withstand without breaking
Newton's second law:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
(a) We apply the formula (1) at constant speed , then, a=0
W: heaviest fish that can be pulled up vertically
∑F = 0
T-W =0
W = T
W= 44N
(b) We apply the formula (1) , a= 1.26 m/s²
W: heaviest fish that can be pulled up vertically
W= m*g
m= W/g
g= 9.8 m/s² : acceleration due to gravity
∑F = 0
T-W = m*a
T= W+(W/g)*a
44=W*(1+1/9.8)* (1.26 )
44= W* 1.39
W= 44/1.39
W= 31.65 N