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2 years ago

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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?

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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As per the question Bob drops the bag full with feathers from the top of the building.

The mass of the bag(m)= 1.0 lb

Let the air resistance is neglected.As the bag is under free fall ,hence the only force that acts on the bag is the force of gravity which is in vertical downward direction.

Here the acceleration produced on bag due to the free fall will be nothing else except the acceleration due to gravity i.e g =9.8 m/s^2

Here we are asked to calculate the distance travelled by the bag at the instant 1.5 s

Hence time t= 1.5 s

From equation of kinematics we know that -

S=ut + 0.5at^2 [ here S is the distance travelled]

For motion under free fall initial velocity (u)=0.

Hence S= 0×1.5+{0.5×(-9.8)×(1.5)^2}

⇒ -S =0-11.025 m

⇒ S= 11.025 m

=11 m

Here the negative sign is taken only due to the vertical downward motion of the body .we may take is positive depending on our frame of reference .

Hence the correct option is B.

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2 years ago

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A girl swings on a playground swing in such a way that at her highest point she is 4.1 m from the ground, while at her lowest po

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Answer:

V1 =8.1 m/s

Explanation:

height at highest point (h2) = 4.1 m

height at lowest point (h1) = 0.8 m

acceleration due to gravity (g) = 9.8 m/s^{2}

from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore

mgh1 + $0.5mV1^{2}$ = mgh2 + $0.5mV2^{2}$

where

speed at highest point = V2

speed at lowest point = V1

mass of the girl and swing = m

at the highest point, the speed is minimum (V1 = 0)

at the lowest point the speed is maximum (V2 is the maximum speed)

therefore the equation becomes mgh1 + $0.5mV1^{2}$ = mgh2

m(gh1 + $0.5V1^{2}$ ) = m(gh2)

gh1 + $0.5V1^{2}$ = gh2

V1 = $\sqrt{\frac{gh2 - gh1}{0.5}}$

now we can substitute all required values into the equation above.

V1 = $\sqrt{\frac{(9.8x4.1) - (9.8x0.8)}{0.5}}$

V1 = $\sqrt{\frac{32.34}{0.5}}$

V1 =8.1 m/s

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2 years ago

If the volume of an object is reported as 5.0 ft3 what is the volume in cubic meters

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The problem statement is simply asking us to convert units. We convert from units of ft^3 to units of m^3. To do this, we need a conversion factor. For this case, we use 1 m is equal to 3.28084 ft. We do as follows:

5.0 ft^3 ( 1 m / 3.28084 ft )^3 = 0.1416 m^3

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2 years ago

Dane is standing on the moon holding an 8 kilogram brick 2 metres above the ground. How much energy is in the brick's gravitatio

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The gravitational potential energy of the brick is 25.6 J

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its position in a gravitational field.

Near the surface of a planet, the gravitational potential energy is given by

$PE=mgh$

where

m is the mass of the object

g is the strength of the gravitational field

h is the height of the object relative to the ground

For the brick in this problem, we have:

m = 8 kg is its mass

g = 1.6 N/kg is the strenght of the gravitational field on the moon

h = 2 m is the height above the ground

Substituting, we find:

$PE=(8)(1.6)(2)=25.6 J$

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Answer:

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Solution:

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Mass density of dust, $\rho_{d} = 3\times 10^{- 18}\ kg/m^{3}$

Now, to calculate the average decelerating force exerted by the particle:

$Mass,\ m = \rho_{d}V$ (1)

Volume, $V = A\times v\times t$

Thus substituting the value of volume, V in eqn (1):

$m = \rho_{d}(Avt)$

where

A = Area

v = velocity

t = time

$m = \rho_{d}(A\times v\times t)$ (2)

$Momentum,\ p = \rho_{d}(Avt)v = \rho_{d}Av^{2}t$

From Newton's second law of motion:

$F = \frac{dp}{dt}$

Thus differentiating w.r.t time 't':

$F_{avg} = \frac{d}{dt}(\rho_{d}Av^{2}t) = \rho_{d}Av^{2}$

where

$F_{avg}$ = average decelerating force of the particle

Now, substituting suitable values in the above eqn:

$F_{avg} = 3\times 10^{- 18}\times 10\times 1\times 10^{6} = 3\times 10^{- 11}\ N$

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