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3 years ago

9

Calculating Net Force

1 answer:

VARVARA [1.3K]3 years ago

6 0

Explanation:

225=m (2.20m/s2 which give m=16kg I used newtons second law to fins the required average force

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Which explains why more energy is released in nuclear reactions than in chemical reactions? Chemical reactions are always endoth

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More energy is released in nuclear reactions than in chemical reactions; this is because in nuclear reactions, mass is converted to energy. Nuclear energy released in nuclear fission and fusion is several 100 million times as large as an ordinary chemical reaction like the combustion process. The reason why nuclear energy release so much energy is because tremendous amounts of energy is released at one time. The nuclei in a nuclear reaction undergo a chain reaction, causing the neutrons to move extremely fast and release high amounts of energy.

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a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How f

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Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for $\Delta$ DAC:

$tan\theta = \frac{d}{x}$

⇒ $\theta = tan^{-1} \frac{d}{x}$

Now for the $\Delta$ BAC:

$tan\theta = \frac{d + h}{x}$

⇒ $\theta = tan^{-1} \frac{d + h}{x}$

Now, differentiating w.r.t x:

$\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}]$

For maximum angle, $\frac{d\theta }{dx}$ = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}][/tex]

0 = $\frac{-(d + h)}{(d + h)^{2} + x^{2}} -\frac{-d}{x^{2} + d^{2}}$

$\frac{-(d + h)}{(d + h)^{2} + x^{2}} = \frac{{d}{x^{2} + d^{2}}$

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Answer:

Current, I = 0.0011 A

Explanation:

It is given that,

Diameter of rod, d = 2.56 cm

Radius of rod, r = 1.28 cm = 0.0128 m

The resistivity of the pure silicon, $\rho=2300\ \Omega-m$

Length of rod, l = 20 cm = 0.2 m

Voltage, $V=1\times 10^3\ V$

The resistivity of the rod is given by :

$R=\rho\dfrac{L}{A}$

$R=2300\ \Omega-m\dfrac{0.2\ m}{\pi (0.0128\ m)^2}$

R = 893692.30 ohms

Current flowing in the rod is calculated using Ohm's law as :

V = I R

$I=\dfrac{V}{R}$

$I=\dfrac{10^3\ V}{893692.30\ \Omega}$

I = 0.0011 A

So, the current flowing in the rod is 0.0011 A. Hence, this is the required solution.

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D. all of these

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