How many significant figures are in 246.32

A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein

RSB [31]

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 $8m^{3}/s$ . 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:

$V_{cone}=\frac{1}{3} \pi r^{2}h$

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:

$\frac {r}{h}=\frac{4}{5}$

When solving for r, we get:

$r=\frac{4}{5}h$

so we can substitute this into our volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h$

which simplifies to:

$V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h$

$V_{cone}=\frac{16}{75} \pi h^{3}$

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

$\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}$

Which simplifies to:

$\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}$

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:

$\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}$

Now we can substitute the provided values into our equation. So we get:

$\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}$

so:

$\frac{dh}{dt}=0.25m/s$

3 0

2 years ago

A vehicle travelling at an initial velocity of 20km/hr,accelerates at 4m/s².calculate its final velocity after 10 seconds.

Nikitich [7]

acceleration = Velocity changes ÷ time of the velocity changes

4 m/s^2 =

4 × 10^(-3) × 3600 km / h =

4 × 3.6 =

14.4 km / h

Thus :

14.4 = V(2) - V(1) / t(2) - t(1)

14.4 = V(2) - 20 / 10

Multiply both sides by 10

10 × 14.4 = 10 × ( V(2) - 20 ) / 10

144 = V(2) - 20

Add both sides 20

144 + 20 = V(2) - 20 + 20

V(2) = 164 Km/h

Thus the final velocity after 10 seconds is 164 Km/h .

4 0

2 years ago

How long does it take a car to cross a 20m bridge if it starts from rest and accelerates at 5 m/s^2?

polet [3.4K]

The correct answer is 2.8s

5 0

3 years ago

2) A motorcycle is moving at a constant speed of 40 km/h. How long does it take the motorcycle to

docker41 [41]

Answer:

2 hours

Explanation:

The motorcycle travels 40 km per hour.

80km / 40km/h = 2 hours.

7 0

3 years ago

Read 2 more answers

If measurements of gas are 75 L and 300 kIlopascals and then the gas is measured is second time and found to be 50 L, describe w

Ann [662]

Answer:

The pressure must have increased in the process

Explanation:

The State Equation for gasses reads: $P*V=n*R*T$

where P is the gas' pressure, V its volume, n the number of moles of gas, R the gas constant and T the temperature in degrees Kelvin.

If the temperature of the gas doesn't change in the described process, the right hand side of the equation stays the same. If that is the case, given that when the Volume of the gas diminishes from 75 liters to 50 liters, then the pressure must have increased to keep that product "P * V" constant:

$P_i*V_i=P_f*V_f\\75 *300=50*X\\X=\frac{75*300}{50} =450$

So the pressure must have gone up to 450 kilopascals.

3 0

2 years ago