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Mrac [35]
3 years ago
8

Concrete is pumped from a cement mixer to the place it is being laid, instead of being carried in wheelbarrows. The flow rate is

200 L/min through a 50.0-m-long, 8.00-cm-diameter hose, and the pressure at the pump is
8.00×106N/m2.
(a) Calculate the resistance of the hose. (b) What is the viscosity of the concrete, assuming the flow is laminar? (c) How much power is being supplied, assuming the point of use is at the same level as the pump? You may neglect the power supplied to increase the concrete's velocity.

Physics
1 answer:
max2010maxim [7]3 years ago
3 0

Answer:

a. 2.4 ×109  N ⋅ m2/s

b. 48.3 N⋅s /m2  

c. 8.00×104W

Explanation:

See Attached file for explanation

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A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra
kkurt [141]

Answer:

V_{ft}= 317 cm/s

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

P_i = P_f

Where:

P_i=M_cV_{ic} + M_tV_{it}

P_f = M_cV_{fc} + M_tV_{ft}

Now:

M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}

Where M_c is the mass of the car, V_{ic} is the initial velocity of the car, M_t is the mass of train, V_{fc} is the final velocity of the car and V_{ft} is the final velocity of the train.

Replacing data:

(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}

Solving for V_{ft}:

V_{ft}= 3.17 m/s

Changed to cm/s, we get:

V_{ft}= 3.17*100 = 317 cm/s

b) The kinetic energy K is calculated as:

K = \frac{1}{2}MV^2

where M is the mass and V is the velocity.

So, the initial K is:

K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2

K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2

K_i = 22.06 J

And the final K is:

K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = 19.61 J

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

4 0
3 years ago
Strontium 3890Sr has a half-life of 28.5 yr. It is chemically similar to calcium, enters the body through the food chain, and co
patriot [66]

Answer:

Thus the time taken is calculated as 387.69 years

Solution:

As per the question:

Half life of ^{3890}Sr\, t_{\frac{1}{2}} = 28.5 yrs

Now,

To calculate the time, t in which the 99.99% of the release in the reactor:

By using the formula:

\frac{N}{N_{o}} = (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}

where

N = No. of nuclei left after time t

N_{o} = No. of nuclei initially started with

\frac{N}{N_{o}} = 1\times 10^{- 4}

(Since, 100% - 99.99% = 0.01%)

Thus

1\times 10^{- 4} = (\frac{1}{2})^{\frac{t}{28.5}}}

Taking log on both the sides:

- 4 = \frac{t}{28.5}log\frac{1}{2}

t = \frac{-4\times 28.5}{log\frac{1}{2}}

t = 387.69 yrs

5 0
3 years ago
When the temperature of water increases from room temperature to 90*C the process of heating the water is
polet [3.4K]
The process is called ENDOTHERMIC
8 0
3 years ago
What is Plancks constant (h)?
Tatiana [17]
Planck's constant is 4.39048042 × 10-67 m4 kg2 / s2.
6 0
3 years ago
Mention two ways to increase the strength of an electro magnet.
VikaD [51]

1.) Law of Induction.

2.) Increase Winding Count

6 0
3 years ago
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