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2 years ago

What is dimension of volume flow rate.?

Physics

1 answer:

ella [17]2 years ago

6 0

is the volume of fluid which passes per unit time; usually represented by the symbol Q (sometimes V̇).

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A dairy farmer notices that a circular water trough near the barn has become rusty and now has a hole near the base. The hole is

ryzh [129]

Answer: 2.73m/s

Explanation:

Potential energy of the water is turned into kinetic energy

mgd = mv^2/2

Where m = mass of water

g = acceleration due to gravity = 9.81m/s^2

d = depth of the water = 0.38m

v = velocity of leaking water

m × 9.81 × 0.38 = mv^2/

eliminate m on both sides and multiply through by 2

7.4456 = v^2

v = sqrt(7.4456)

v = 2.73m/s

3 0

2 years ago

Two cars A and B are 100m apart moving towards each other with

maxonik [38]

Let car A's starting position be the origin, so that its position at time t is

A: x = (40 m/s) t

and car B has position at time t of

B: x = 100 m - (60 m/s) t

They meet when their positions are equal:

(40 m/s) t = 100 m - (60 m/s) t

(100 m/s) t = 100 m

t = (100 m) / (100 m/s) = 1 s

so the cars meet 1 second after they start moving.

They are 100 m apart when the difference in their positions is equal to 100 m:

(40 m/s) t - (100 m - (60 m/s) t) = 100 m

(subtract car B's position from car A's position because we take car A's direction to be positive)

(100 m/s) t = 200 m

t = (200 m) / (100 m/s) = 2 s

so the cars are 100 m apart after 2 seconds.

3 0

2 years ago

40 meters divided by 5

denis23 [38]

Answer:

8 meters

Explanation:

40/ 5 = 8

8 0

2 years ago

The demand equation for the Roland portable hairdryer is given as follows where x (measured in units of a hundred) is the quanti

lesya [120]

Answer:

Explanation:

The concept of elastic and inelastic demand is applied.

for an elastic demand, the elasticity must be greater than 1 and for an Inelastic demand, the elasticity must be less than 1.

The steps and appropriate calculation is as shown in the attached file.

6 0

3 years ago

A stuntman with a mass of 80.5 kg swings across a moat from a rope that is 11.5 m. At the bottom of the swing the stuntman's spe

goldenfox [79]

Answer:

No
5.49 m/s

Explanation:

The net force required to accelerate the stuntman in a circular arc of radius 11.5 m will be ...

F = mv²/r . . . . where this m is the mass being accelerated, v is the tangential velocity, and r is the radius.

Here, the net force needs to be ...

F = (80.5 kg)(8.45 m/s)²/(11.5 m) . . . . . where this m is meters

≈ 499.8175 kg·m/s² = 499.8 N

Gravity exerts a force on the stuntman of ...

F = mg = (80.5 kg)(9.8 m/s²) = 788.9 kg·m/s² = 788.9 N

Then the tension required in the rope/vine is ...

499.8 N+788.9 N= 1288.7 N

This is more than the capacity of the rope, so we do not expect the stuntman to make it across the moat.

_____

The allowed net force for centripetal acceleration is ...

1000 N -788.9 N = 211.1 N

Then the allowed velocity is ...

211.1 = 80.5v²/11.5

30.16 = v² . . . . multiply by 11.5/80.5

5.49 = v . . . . . . take the square root

The maximum speed the stuntman can have is 5.49 m/s.

_____

Comment on crossing the moat

The kinetic energy at the bottom of the swing translates to potential energy at the end of the swing. At the lower speed, the stuntman cannot rise as high, so will traverse a shorter arc. At 8.45 m/s, the moat could be about 16.8 m wide; at 5.49 m/s, it can only be about 11.5 m wide.

5 0

2 years ago