What is the name of the device used to measure wind speed?

a cone is constructed by cutting a sector from a circular sheet of metal with radius . the cut sheet is then folded up and welde

Svet_ta [14]

The expression for the radius and height of the cone can be obtained from

the property of a function at the maximum point.

$The \ radius, \ of \ the \ base \ of \ the \ cone \ is \ \sqrt{ \dfrac{3}{4}} \times radius \ of \ circular \ sheet \ metal$

The height of the cone is half the length of the radius of the circular sheet metal.

Reasons:

The part used to form the cone = A sector of a circle

The length of the arc of the sector = The perimeter of the circle formed by the base of the cone.

$Volume \ of \ a \ cone = \dfrac{1}{3} \cdot \pi \cdot r^3 \cdot h$

$Volume \ of \ a \ cone, \, V = \dfrac{1}{3} \cdot \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}$

θ/360·2·π·s = 2·π·r

Where;

s = The radius of he circular sheet metal

h = s² - r²

$\dfrac{dV}{dr} = \dfrac{d}{dr} \left(\dfrac{1}{3} \cdot \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}\right) = \dfrac{\pi \cdot (3 \cdot r^2 \cdot s^2 - 4 \cdot r^4)}{\sqrt{(s^2- r^2)}} = 0$

3·r²·s² - 4·r⁴ = 0

3·r²·s² = 4·r⁴

3·s² = 4·r²

$\underline{\left \right. The \ radius, \, r =\sqrt{ \dfrac{3}{4}} \cdot s}$

$\underline{The \ height, \, h =\sqrt{s^2 - \dfrac{3}{4}\cdot s^2} = \dfrac{s}{2}}}$

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3 0

3 years ago

A steady electric current flows through a wire. If 9.0 C of charge passes a particular spot in the wire in a time period of 2.0

defon

1) Current: 4.5 A

2) Time taken: 4.7 s

Explanation:

1)

The electric current intensity is defined as the rate at which charge flows in a conductor; mathematically:

$I=\frac{q}{t}$

where

I is the current

q is the amount of charge passing a given point in a time t

For the wire in this problem, we have

q = 9.0 C is the amount of charge

t = 2.0 s is the time interval

Solving for I, we find the current:

$I=\frac{9.0}{2.0}=4.5 A$

2)

To solve this problem, we can use again the same formula

$I=\frac{q}{t}$

where

I is the current

q is the amount of charge passing a given point in a time t

In this problem, we have:

I = 3.0 A (current)

q = 14.0 C (charge)

Therefore, the time taken for the charge to move past a particular spot in the wire is

$t=\frac{q}{I}=\frac{14.0}{3.0}=4.7 s$

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8 0

3 years ago

If 175 kJ of heat caused a chunk of ice at 0°C to melt to liquid water at 0°C, how many moles were in the ice? How many grams of

elena55 [62]

Answer is 29.1 mol or 525 g

Explanation:

8 0

3 years ago

What force must the worker exert to get the box moving & what force must the worker exert to accelerate the box at 0.1 meter

photoshop1234 [79]

Since static friction is the minimum force required to just start the motion of a stationary object.

Here if we need to start an object from rest then we required F = 700 N

So for the first part of the above problem Force will be F = 700 N

Now if the box is already moving then we will have to use kinetic friction force between box and floor

now we can write the equation of net force as

$F - F_k = m*a$

here

$F_k = kinetic friction = 220 N$

$m = mass = 500 kg$

$a = acceleration = 0.1 m/s^2$

now we will have

$F - 220 = 500* 0.1$

$F = 220 + 50 = 270 N$

3 0

3 years ago

A ball is projected upward at time t = 0.0 s, from a point on a roof 70 m above the ground. The ball rises, then falls and strik

pentagon [3]

Answer:

The velocity of a ball will be "-70.13 m/s".

Explanation:

The given values are:

u = 70 m

t = 0.0 s

g = a = -9.8 m/s²

s = -1 m

v = ?

As we know,

The equation of motion will be:

⇒ $v^2-u^2=2as$

On substituting the estimated values, we get

⇒ $v^2-(70)^2=2\times (-9.8)\times (-1)$

⇒ $v^2-4900=19.6$

⇒ $v^2=19.6+4900$

⇒ $v^2=4919.6$

⇒ $v=\sqrt{4919.6}$

⇒ $v=70.13 \ m/s$

In downward direction, it will be:

⇒ $v=-70.13 \ m/s$

8 0

3 years ago