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zvonat [6]
3 years ago
15

Pump A can fill a tank of water in 5 hours. Pump B can fill the same tank in 8 hours. How long does it take the two pumps workin

g together to fill the tank

Physics
2 answers:
valina [46]3 years ago
8 0

Answer:

It will take both pumps 3.08 hours to fill the tank working together.

Explanation:

Pump A can fill the tank in 5 hours. Assuming that the pump gives out a steady flow of water, in one hour, pump A can fill 1/5th of the tank. Similarly, pump B in an hour, fills up 1/8th of the tank.

We must add up these two values, in order to find how much of the tank the two pumps can fill up together in one hour.

1/5 +1/8 =13/40

So 13/40 of the tank is filled in an hour. We need to find how many hours it will take for the entire tank to be filled. To do so, divide 40 by 13. This gives:

3.08 hours to fill up the tank.

Dmitry [639]3 years ago
4 0

Explanation:

Below is an attachment containing the solution

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Physical activity is the most important thing you can do to improve and maintain health-related physical fitness. Which list con
katovenus [111]

Answer:

I guess it's d because age environment heredity Nd maturation r more obvious factors if we compre with only physical fitness

8 0
3 years ago
A freight car with a mass of 10 metric tons is rolling at 108 km/h along a level track when it collides with another freight car
Rufina [12.5K]

Answer:

20 metric tons

Explanation:

Using the Principle of Conservation of Linear Momentum which states that the total momentum of a system is constant in any direction in which no external force acts.

momentum before collision= momentum after collision

mathematically expressed as

m1u1 + m2u2 = m1v1+m2v2 =0

where:

m1= mass of rolling freight car = 10 metric tons = 10Mg

1 ton = 1000kg= 1Mg

u1 =initial velocity of freight car = 108km/h,

m2 = mass of stationary freight car = ?, this the unknown we are finding

u2 = initial velocity of

m2 = 0, because it is at rest,

v1= final velocity of m1,

v2= final velocity of m2

since they move together in the same direction they share a common final velocity as such

v1=v2= 36km/h

substituting valves in expression give

m1u1 + m2u2 = m1v1 + m2v2= 0

10Mgx 108km/hr + m2 x 0 = 10Mg x 36km/h + m2x36km/h

1080( Mg x km/hr) + 0 = 360 (Mg x Km/h) + 36m2 (km/h)

1080 (Mgx km/hr) - 360(Mg x km/h) = 36m2 (km/h)

720(Mg x km/h) = 36m2 (km/h)

divide both side by  36 (km/h) gives

m2 = 20Mg =20 metric tons

3 0
3 years ago
A projectile is launched with an initial velocity of (40 m/s), at an angle of (30°) above
zlopas [31]

Answer:

330.5  m

Explanation:

In this case, the object is launched horizontally at 30° with an initial velocity of 40 m/s .

The maximum height will be calculated as;

h=\frac{v^2_isin^2\alpha }{2g}

where ∝ is the angle of launch = 30°

vi= initial launch velocity = 40 m/s

g= 10 m/s²

h= 40²*sin²40° / 2*10

h={1600*0.4132 }/ 20

h= 661.1/2 = 330.5  m

8 0
3 years ago
Robert has just bought a new model rocket, and is trying to measure its flight characteristics. The rocket engine package claims
777dan777 [17]

Answer:

The work done by the drag force is given by 29.96 J

Explanation:

Given :

Thrust force F = 12.3 N

Displacement d = 10.2 m

Mass of rocket m = 0.663 Kg

From work energy theorem,

  W = \Delta K

 W_{t} - Wd - W_{g} = KE

Where W_{t} = thrust work W_{g} = gravitational work

KE = 12.3 \times 10.2 -Wd - 0.663 \times 9.8 \times 10.2

KE = 59.2 -Wd

After cutoff kinetic energy is converted into potential energy,

KE = Wd' + mg\Delta h

Put value of KE

59.2 -Wd = Wd' + 0.663 \times 9.8 \times 4.5

Work done by drag force is given by,

Wd'+Wd =  59.2 -29.23

                 = 29.96 J

Therefore, the work done by the drag force is given by 29.96 J

5 0
3 years ago
Find electric field at point p which is a distance l away from the both +q and -q
denis-greek [22]

Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

  • q is the charge
  • r is the distance
  • E is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

8 0
3 years ago
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