Why would you wanna know a cure for all sickness.?

Explain the method to measure the external diameter of a sphere

Anton [14]

Answer:

The sphere is that the circular objects in the two-dimensional space (1) circle

(2) disk. Two-dimensional space is a set of points and the distance of that point, The two points of Sphere that length and center.

The sphere can be constructed as the name of surface form circle about any diameter. The circle is the special type of the revolution replacing the circle,

the sphere is the distance r is the radius of the ball and the circle is the center of the mathematical ball, as the center and the radius of the sphere is to respectively.

The ball and sphere have not to be maintained mathematical references as solid references. A sphere of any radius is centered at the number of zero.

Explanation: Hope this helps and good luck :)

3 0

3 years ago

A straight fin is made from copper (k = 388 W/m-K) and is 0.5 cm in diameter and 30 cm long. The temperature at the base of the

erastovalidia [21]

Answer:

The rate of transfer of heat is 0.119 W

Solution:

As per the question:

Diameter of the fin, D = 0.5 cm = 0.005 m

Length of the fin, l =30 cm = 0.3 m

Base temperature, $T_{b} = 75^{\circ}C$

Air temperature, $T_{infty} = 20^{\circ}$

k = 388 W/mK

h = $20\ W/m^{2}K$

Now,

Perimeter of the fin, p = $\pi D = 0.005\pi \ m$

Cross-sectional area of the fin, A = $\frac{\pi}{4}D^{2}$

A = $\frac{\pi}{4}(0.5\times 10^{-2})^{2} = 6.25\times 10^{- 6}\pi \ m^{2}$

To calculate the heat transfer rate:

$Q_{f} = \sqrt{hkpA}tanh(ml)(T_{b} - T_{infty})$

where

$m = \sqrt{\frac{hp}{kA}} = \sqrt{\frac{20\times 0.005\pi}{388\times 6.25\times 10^{- 6}\pi}} = 41.237$

Now,

$Q_{f} = \sqrt{20\times 388\times 0.005\pi\times 6.25\times 10^{- 6}\pi}tanh(41.237\times 0.3)(75 - 20) = 0.119\ W$

5 0

3 years ago

Speakers A and B are vibrating in phase. They are directly facing each other, are 8.2 m apart, and are each playing a 78.0 Hz to

Stels [109]

Answer:6.298,4.1,1.9015

Explanation:

Wavelength= $\frac{velocity of sound }{frequency}$

= $\frac{343}{78}$ =4.397 m

Distance of 3rd speaker from speaker A is x

From B 78-x

Difference between the distances must be a whole number of wavelengths

First

$x-\left ( 8.2-x\right )=4.397$ for 1 st wavelength

2x=8.2+4.397=12.597

x=6.298m

second

For zero wavelength

$x-\left ( 8.2-x\right )=0$

2x=8.2

x=4.1m

Third

$\left ( 8.2-x\right )-x=4.397$

x=1.9015 m

6 0

3 years ago

The magnitude of a force vector is 89.6 newtons (N). The x component of this vector is directed along the +x axis and has a magn

insens350 [35]

Answer:

(a) θ = 33.86°

(b) Ay = 49.92 N

Explanation:

You have that the magnitude of a vector is A = 89.6 N

The x component of such a vector is Ax = 74.4 N

(a) To find the angle between the vector and the x axis you use the following formula for the calculation of the x component of a vector:

$A_x=Acos\theta$ (1)

Ax: x component of vector A

A: magnitude of vector A

θ: angle between vector A and the x axis

You solve the equation (1) for θ, by using the inverse of cosine function:

$\theta=cos^{-1}(\frac{A_x}{A})=cos{-1}(\frac{74.4N}{89.6N})\\\\\theta=33.86\°$

the angle between the A vector and the x axis is 33.86°

(b) The y component of the vector is given by:

$A_y=Asin\theta\\\\A_y=(89.6N)sin(33.86\°)=49.92N$

the y comonent of the vecor is Ay = 49.92 N

3 0

3 years ago

A delivery truck travels with a constant velocity up an 8 slope. A 60 kg box sits on the floor of the truck and, because of sta

Blababa [14]

Explanation:

Given that,

The slope of the ramp, $\theta=8^{\circ}$

Mass of the box, m = 60 kg

(a) Distance covered by the truck up the slope, d = 300 m

Initially the truck moves with a constant velocity. We know that the net work done on the box is equal to 0 as per work energy theorem as :

$W=\dfrac{1}{2}m(v^2-u^2)=0$

u and v are the initial and the final velocity of the truck

(b) The work done on the box by the force of gravity is given by :

$W=Fd\ cos\theta$

Here, $\theta=90+8=98^{\circ}$

$W=mgd\ cos\theta$

$W=60\times 9.8\times 300\ cos(98)$

W = -24550.13 J

(c) What is the work done on the box by the normal force is equal to 0 as the angle between the force and the displacement is 90 degrees.

(d) The work done by friction is given by :

$W_f=-W$

$W_f=24550.13\ J$

Hence, this is the required solution.

6 0

3 years ago