Whay is the other name of negative acceleration.

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

3 years ago

object carries a charge of -8.1 ÂµC, while another carries a charge of -2.0 ÂµC. How many electrons must be transferred from the

LenKa [72]

Number of electrons transferred: $1.91\cdot 10^{13}$

Explanation:

The charge on the first object is

$Q_1 = -8.1\mu C$

while the charge on the 2nd object is

$Q_2=-2.0 \mu C$

When they are in contact, the final charge on each object will be

$Q=\frac{Q_1+Q_2}{2}=\frac{-8.1+(-2.0)}{2}=-5.05 \mu C$

So, the amount of charge (electrons) transferred from the 1st object to the 2nd object is

$\Delta Q = Q_1 - Q = -8.1 -(5.05)=-3.05 \mu C = -3.05\cdot 10^{-6}C$

The charge of one electron is

$e=-1.6\cdot 10^{-19}C$

Therefore, the number of electrons transferred is

$N=\frac{Q}{e}=\frac{-3.05\cdot 10^{-6}}{-1.6\cdot 10^{-19}}=1.91\cdot 10^{13}$

Learn more about electrons:

brainly.com/question/2757829

#LearnwithBrainly

8 0

3 years ago

A satellite is in a circular Earth orbit of radius r. The area A enclosed by the orbit depends on r2 because A = πr2. Determine

Salsk061 [2.6K]

Answer:

a. $T=r^{3/2}$

b. $K=\frac{1}{r}$

c. $v=\frac{1}{\sqrt{r}}$

d. $v=\sqrt{r}$

Explanation:

To make analysis about the satellite circular earth the depends or r and A

$T^2=\frac{4\pi}{GM}*r^3$

$K=\frac{GM*m}{2*r}$

a.

$T^2=r^3$

$T=r^{3/2}$

b.

$K=\frac{GM*m}{2}*\frac{1}{r}$

$K=\frac{1}{r}$

c.

$K=\frac{1}{2}*m*v^2$

$v^2=\frac{2*K}{m}=\frac{2*Gm*m}{2*m*r}$

$v=\frac{1}{\sqrt{r}}$

d.

$v=\sqrt{2*GMr}$

$v=\sqrt{r}$

3 0

3 years ago

You push a 50 kg wooden box across a wooden floor at a constant speed of 1.0 m/s. The coefficient of kinetic friction is 0.15. N

WARRIOR [948]

Answer:

0.68 seconds

Explanation:

Data provided in the question:

Mass of the box = 50 kg

Speed of the box = 1.0 m/s

Coefficient of friction, μ = 0.15

Now,

Force applied = μmg

Here,

g is the acceleration due to gravity = 9.8 m/s²

Thus,

F = 0.15 × 50 × 9.8

= 73.5 N

Also,

Force = Mass × Acceleration

thus,

73.5 N = 50 × a

or

a = 1.47 m/s²

After doubling the speed

Final speed = 2 × Initial speed

= 2 × 1 m/s

= 2 m/s

Also,

Acceleration = [change in speed] ÷ Time

or

1.47 = [ 2 - 1 ] ÷ Time

or

Time = 1 ÷ 1.47

or

Time = 0.68 seconds

6 0

4 years ago

Consider two copper wires. one has twice the length and twice the cross-sectional area of the other. how do the resistances of t

Anestetic [448]

Answer:

R = ρ L / A where R is the resistance and ρ the resistivity:

R2 / R1 = (L2 / A2) / (L1 / A1) = L2 * A1 / (L1 * A2)

R2 / R1 = (2 L * A) / (L * 2 A) = 1

Their resistances should be the same.

5 0

2 years ago

Whay is the other name of negative acceleration.​

Whay is the other name of negative acceleration.