Answer:
h'=0.25m/s
Explanation:
In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).
So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of 
. As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:
notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.
If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:
When solving for r, we get:
so we can substitute this into our volume of a cone formula:
which simplifies to:
So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:
Which simplifies to:
So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)
So we get:
Now we can substitute the provided values into our equation. So we get:
so:
Answer:
Option (D)
Explanation:
The velocity at which blood flows in the blood vessels is inversely proportional to the total cross-sectional area of the blood vessels present in the body. This means that if the cross sectional area of the vessels low, then there will be high rate of blood flow, and vice versa. This blood flow is minimum in the case of capillaries, where it gets enough time for the exchanging of essential nutrients as well as gases.
Thus, the correct answer is option (D).
By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s
Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:
mass m = 0.170 kg
initial speed u = 6 m/s.
Distance covered s = 61 m
To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V
To do this, let us first calculate the kinetic energy at which the ball move.
K.E = 1/2m
K.E = 1/2 x 0.17 x 
K.E = 3.06 J
The work done on the ball is equal to the kinetic energy. That is,
W = K.E
But work done = Force x distance
F x S = K.E
F x 61 = 3.06
F = 3.06/61
F = 0.05 N
From here, we can calculate the acceleration of the ball from Newton second law
F = ma
0.05 = 0.17a
a = 0.05/0.17
a = 0.3 m/
To calculate the final velocity, let us use third equation of motion.
= + 2as
= + 2 x 0.3 x 61
= 36 + 36
= 72
V = 
V = 8.485 m/s
Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.
Learn more about dynamics here: brainly.com/question/402617
Answer:
Option A
Carrots are cut into small pieces and mixed into a salad
Explanation:
When physical changes occur, the actual composition remain the same but the molecules are re-arranged. Therefore, when carrots are cut into smaller pieces and mixed into salad, there will be no chemical reaction hence the actual composition will remain the same despite being cut and molecules in it re-arranged. Considering the other options, new substances are formed hence they are deemed as chemical changes. Therefore, option A is correct.
