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olga2289 [7]
2 years ago
11

Water, of density 1000 kg/m3, is flowing in a drainage channel of rectangular cross-section. The width of the channel is 15 m, t

he depth of the water is 8.0 m and the speed of the flow is 2.5 m/s. At what rate is water flowing in this channel?
Physics
1 answer:
Rom4ik [11]2 years ago
8 0

Answer:

Water flowing rate=  (300000kg/s) = (300000l/s)

Explanation:

First with the section of the channel, the depth of the water and the speed of the fluid we can determine the volume of fluid that circulates per second through the channel:

Volume per time= 15m × 8m × (2.5m/s)= 300 m³/s

With this volume of circulating fluid per second elapsed, we multiply it by the density of the water to determine the kilograms or liters of water that circulate through the channel per second elapsed:

Water flowing rate= (300m³/s) × (1000kg/m³)= (300000kg/s) = (300000l/s)

Taking into account that 1kg of water is approximately equal to 1 liter of water.

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Answer:

Magnitude of impulse, |J| = 4 kg-m/s                                                                                

Explanation:

It is given that,

Mass of cart 1, m_1=2\ kg

Mass of cart 2, m_2=4\ kg  

Initial speed of cart 1, u_1=3\ m/s          

Initial speed of cart 2, u_2=0 (stationary)

The carts stick together. It is the case of inelastic collision. Let V is the combined speed of both carts. The momentum remains conserved.

m_1u_1+m_2u_2=(m_1+m_2)V

V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}        

V=\dfrac{2\times 3}{(2+4)}

V = 1 m/s

The magnitude of the impulse exerted by one cart on the other is given by:

J=F\times t=m(V-u)

J=m(V-u)

J=2\times (1-3)    

J = -4 kg-m/s

or

|J| = 4 kg-m/s

So, the magnitude of the impulse exerted by one cart on the other 4 kg-m/s. Hence, this is required solution.

8 0
2 years ago
(Another tomato/skyscraper problem.) You are looking out your window in a skyscraper, and again your window is at a height of 45
Ivan

Answer:

1027.2 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.2 ft/s

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{450-\frac{1}{2}\times 32.2\times 2^2}{2}\\\Rightarrow u=192.8\ ft/s

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{192.8^2-0^2}{2\times 32.2}\\\Rightarrow s=577.20\ m

The height the tomato would fall is 450+577.2 = 1027.2 m

6 0
3 years ago
A 900-kg car traveling east at 15.0 m/s collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together. Assume
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Explanation:

It is given that,

Mass of the car 1, m_1=900\ kg

Initial speed of car 1, u_1=15i\ m/s (east)

Mass of the car 2, m_2=750\ kg

Initial speed of car 2, u_1=20j\ m/s (north)

(b) As the cars stick together. It is a case of inelastic collision. Let V is the common speed after the collision. Using the conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)V

900\times 15i +750\times 20j=(900+750)V

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|V|=\sqrt{8.18^2+9.09^2}

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(b) Let \theta is the direction the wreckage move just after the collision. It is given by :

tan\theta=\dfrac{v_y}{v_x}

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Hence, this is the required solution.

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Explanation:

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