Is the text in red correct (0n0')

MakcuM [25]

<h2>Hello there, to answer your question...</h2><h2 /><h2>1,893,9379 x 0 = 0</h2><h2 /><h2>Anything times zero is always 0</h2><h2 /><h2>For example</h2><h2>1x0=0</h2><h2>4x0=0</h2><h2>12x0=0</h2><h2>48x0=0</h2><h3>I hope this helps :)</h3>

4 0

3 years ago

A 50 mm diameter steel shaft and a 100 mm long steel cylindrical bushing with an outer diameter of 70 mm have been incorrectly s

BARSIC [14]

Answer:

The axial force is $P = 15.93 k N$

Explanation:

From the question we are told that

The diameter of the shaft steel is $d = 50mm$

The length of the cylindrical bushing $L =100mm$

The outer diameter of the cylindrical bushing is $D = 70 \ mm$

The diametral interference is $\delta _d = 0.005 mm$

The coefficient of friction is $\mu = 0.2$

The Young modulus of steel is $207 *10^{3} MPa (N/mm^2)$

The diametral interference is mathematically represented as

$\delta_d = \frac{2 *d * P_B * D^2}{E (D^2 -d^2)}$

Where $P_B$ is the pressure (stress) on the two object held together

So making $P_B$ the subject

$P_B = \frac{\delta _d E (D^2 - d^2)}{2 * d* D^2}$

Substituting values

$P_B = \frac{(0.005) (207 *10^{3} ) * (70^2 - 50^2))}{2 * (50) (70) ^2 }$

$P_B = 5.069 MPa$

Now he axial force required is

$P = \mu * P_B * A$

Where A is the area which is mathematically evaluated as

$\pi d l$

So $P = \mu P_B \pi d l$

Substituting values

$P = 0.2 * 5.069 * 3.142 * 50 * 100$

$P = 15.93 k N$

8 0

3 years ago

Differentiate conductors and insulators with the help of suitable examples

andrew-mc [135]

Answer:

Any material that keeps energy such as electricity, heat, or cold from easily transferring through is an insulator

A conductor is a material which contains movable electric charges.

3 0

2 years ago

Read 2 more answers

The eye of the Atlantic giant squid has a diameter of 3.50 × 10^2 mm. If the eye

Lunna [17]

Answer:

q = 224 mm, h ’= - 98 mm, real imagen

Explanation:

For this exercise let's use the constructor equation

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image respectively.

In a mirror the focal length is

f = R / 2

indicate us radius of curvature is equal to the diameter of the eye

R = 3,50 10² mm

f = 3.50 10² /2 = 1.75 10² mm

they also say that the distance to the object is p = 0.800 10³ mm

1 / q = 1 / f - 1 / p

1 / q = 1 / 175 - 1 /800

1 / q = 0.004464

q = 224 mm

to calculate the size let's use the magnification ratio

m = $\frac{h'}{h} = - \frac{q}{p}$

h '= $- \frac{q}{p} \ h$

h ’= - 224 350 / 800

h ’= - 98 mm

in concave mirrors the image is real.

3 0

3 years ago

onsidering light at the two ends of the visible light spectrum, violet light has a _____ wavelength and a _____ photon energy th

Alenkasestr [34]

Answer:

Shorter, higher

Explanation:

6 0

1 year ago