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3 years ago

The colour of the shadow of coloured objects is not same as the colour of the objects

Physics

1 answer:

Tom [10]3 years ago

7 0

<h3>The colour of the shadow of coloured objects is not same as the colour of the objects</h3>

When an opaque item absorbs all of the light falling from a light source, no light is reflected, and the thing appears black. The color of an object's <u>shadow is independent of its hue; it can be red, yellow, green, or black.</u>

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Consider two circular metal wire loops each carrying the same current I as shown below. In what region(s) could the net magnetic

vova2212 [387]

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3 years ago

Hole filling fasteners (for example, MS20470 rivets) should not be used in composite structures primarily because of the?

DIA [1.3K]

Answer:

We can cause delamination.

Explanation:

The reason why is because the probability of causing delamination increase considerably when we use Hole-filling fasteners. If we use a typical rivet, these tends to expands in order to fill the hole.

If we analyze the force applied by the expanded rod will cause that the matrial will be deteriorated and will cause that the material to delaminate around the edges of the hole and we can cause possible control and no protection to the material.

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3 years ago

Calculate the work done by a 50.0 N force on an object as it moves 9.00 m, if the force is oriented at an angle of 135° from the

aivan3 [116]

Answer:

Work done, W = -318.19 Joules

Explanation:

It is given that,

Force acting on the object, F = 50 N

Distance covered by the force, d = 9 m

Angle between the force and the distance traveled, $\theta=135^{\circ}$

The work done by an object is equal to the product of force and distance traveled. It is equal to the dot product of force and the distance. Mathematically, it is given by :

$W=Fd\ cos\theta$

$W=50\times 9\times \ cos(135)$

W = -318.19 Joules

So, the work done by the force is 318.19 Joules. The work is done in opposite to the direction of motion. Hence, this is the required solution.

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3 years ago

A practical rule is that a radioactive nuclide is essentially gone after 10 half-lives. What percentage of the original radioact

ArbitrLikvidat [17]

Answer:

0.09 % of the original radioactive nucllde its left after 10 half-lives
It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

Explanation:

The equation for radioactive decay its:

$N ( t) \ = \ N_0 \ e^{ \ - \frac{t}{\tau}}$ ,

where N(t) its quantity of material at time t, $N_0$ its the initial quantity of material and $\tau$ its the mean lifetime of the radioactive element.

The half-life $t_{\frac{1}{2}}$ its the time at which the quantity of material its the half of the initial value, so, we can find:

$N (t_{\frac{1}{2} }) \ = \ N_0 \ e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}$

so:

$\ N_0 \ e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}$

$e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{1}{2}$

$- \frac{t_{\frac{1}{2}}}{\tau}} \ = - \ ln( 2 )$

$t_{\frac{1}{2}}\ = \tau ln( 2 )$

So, after 10 half-lives, we got:

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - \frac{10 \ t_{\frac{1}{2}}}{\tau}}$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - \frac{10 \ \tau \ ln( 2 ) }{\tau}}$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - 10 \ \ ln( 2 ) }$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ * \ 9.76 * 10^{-4}$

So, we got that a 0.09 % of the original radioactive nucllde its left.

Putonioum-239 has a half-life of 24,110 years. So, 10 half-life will take to pass

$10 \ * \ 24,110 \ years \ = \ 241,100 \ years$

It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

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3 years ago

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A bob of mass m is suspended from a fixed point with a massless string of length L (i.e., it is a pendulum). You are to investig

Brrunno [24]

Answer:

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2 years ago

The colour of the shadow of coloured objects is not same as the colour of the objects​

The colour of the shadow of coloured objects is not same as the colour of the objects