When is a body said to be motion (movement )?

If the elevation in reservoir b is 100m, what must the elevation in reservoir a be if thevolume flow rate through the cast-iron

Zinaida [17]

The elevation in reservoir at the rate of flow using is 03m/s is 114m.

The Reynolds range is the ratio of inertial forces to viscous forces. The Reynolds variety is a dimensionless variety used to categorize the fluids structures in which the impact of viscosity is crucial in controlling the velocities or the flow sample of a fluid.

The reason of the Reynolds number is to get a few experience of the relationship in fluid glide between inertial forces (this is those that maintain going by using Newton's first law – an item in motion stays in movement) and viscous forces, this is people who cause the fluid to come back to a forestall because of the viscosity of the fluid.

calculation,

Let L = 100 m pipe

L1 = 150 m pipe

H f = friction losses

Using Reynolds number, relative roughness, friction co- effiicients and friction losses

Substitute the value in equation

Z = 110= 0.48= 3.54

Z = 114m

Therefore water surface elevation at reservoir is 114 meter.

Learn more about rate of flow here:-brainly.com/question/21630019

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6 0

1 year ago

A mass of 5kg starts from rest and pulls down vertically on a string wound around a disk-shaped, massive pulley. The mass of the

Paha777 [63]

Answer:

c. V = 2 m/s

Explanation:

Using the conservation of energy:

$E_i =E_f$

so:

Mgh = $\frac{1}{2}IW^2 +\frac{1}{2}MV^2$

where M is the mass, g the gravity, h the altitude, I the moment of inertia of the pulley, W the angular velocity of the pulley and V the velocity of the mass.

Also we know that:

V = WR

Where R is the radius of the disk, so:

W = V/R

Also, the moment of inertia of the disk is equal to:

I = $\frac{1}{2}MR^2$

I = $\frac{1}{2}(5kg)(2m)^2$

I = 10 kg*m^2

so, we can write the initial equation as:

Mgh = $\frac{1}{2}IV^2/R^2 +\frac{1}{2}MV^2$

Replacing the data:

(5kg)(9.8)(0.3m) = $\frac{1}{2}(10)V^2/(2)^2 +\frac{1}{2}(5kg)V^2$

solving for V:

(5kg)(9.8)(0.3m) = $V^2(\frac{1}{2}(10)1/4 +\frac{1}{2}(5kg))$

V = 2 m/s

8 0

3 years ago

If you wanted to find the area of the hot filament in a light bulb, you would have to know the temperature (determinable from th

Sladkaya [172]

Answer:

To find out the area of the hot filament of a light bulb, you would need to know the temperature, the power input, the Stefan-Boltzmann constant and <u>Emissivity of the Filament</u>.

Explanation:

The emissive power of a light bulb can be given by the following formula:

E = σεAT⁴

where,

E = Power Input or Emissive Power

σ = Stefan-Boltzmann constant

ε = Emissivity

A = Area

T = Absolute Temperature

Therefore,

A = E/σεT⁴

So, to find out the area of the hot filament of a light bulb, you would need to know the temperature, the power input, the Stefan-Boltzmann constant and <u>Emissivity of the Filament</u>.

3 0

3 years ago

If the sprinter accelerates at that rate for a distance of 15 m, and then maintains the velocity he has at that point for the re

ahrayia [7]

Answer:

The time for the entire race is 11.39 sec.

Explanation:

Given that,

Distance = 15 m

Remainder distance = 100 m

Suppose A sprinter begins a race with an acceleration of 3.4 m/s².

We need to calculate the time

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value in the equation

$15=0+\dfrac{1}{2}\times3.4\times t^2$

$t^2=\dfrac{30}{3.4}$

$t=2.97\ sec$

We need to calculate the final velocity of sprinter

Using equation of motion again

$v=u+at$

Put the value into the formula

$v=0+3.4\times2.97$

$v=10.09\ m/s$

We need to calculate the distance covers by sprinter

$d=100-15=85\ m$

The sprinter need to covers only 85 m.

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t'=\dfrac{85}{10.09}$

$t'=8.42\ sec$

We need to calculate the time for the entire race

$t''=t+t'$

Put the value into the formula

$t''=2.97+8.42$

$t''=11.39\ sec$

Hence, The time for the entire race is 11.39 sec.

4 0

2 years ago

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g

Morgarella [4.7K]

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:

$k=\sqrt{\frac{I}{M} }$

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration $k=\sqrt{\frac{I}{M} }$

So, moment of rotational inertia (I) of a cylinder about it axis = $\frac{MR^2}{2}$