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3 years ago

The first modern atomic scientist was:

Physics

2 answers:

Rina8888 [55]3 years ago

8 0

Answer:

John Dalton was the first modern atomic scientist.

Explanation:

kykrilka [37]3 years ago

6 0

Answer:

John Dalton

Explanation:

John Dalton was the first to adapt Democritus' theory into the first modern atomic model.

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A bob attached to a string of length L = 1.25 m, initially found at the equilibrium

malfutka [58]

The maximum displacement angle of the bob is 13⁰.

The given parameters;

<em>Length of the pendulum, L = 1.25 m</em>
<em>Initial velocity of the bob, v = 0.8 m/s</em>

The maximum displacement of the bob is calculated by applying the principle of conservation of energy;

$P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\h = \frac{v^2}{2g} \\\\h = \frac{0.8^2}{2\times 10} \\\\h = 0.032 \ m$

The maximum displacement angle is calculated as follows;

$cos \theta = \frac{L-h}{L} \\\\cos \theta = \frac{1.25 - 0.032}{1.25} \\\\\cos \theta = \frac{1.218}{1.25} \\\\cos \theta = 0.9744\\\\\theta = cos^{-1}(0.9744)\\\\\theta = 13\ ^0$

Thus, the maximum displacement angle of the bob is 13⁰.

Learn more here:brainly.com/question/13981780

4 0

3 years ago

A square loop of wire, with sides of length a, lies in the first quadrant of the xy plane, with one corner at the origin. In thi

Ainat [17]

Answer:

$emf=-\dfrac{1}{2}kta^5$

Explanation:

Given that

B(y, t) = k y ³t²

To find the total flux over the loop we have to integrate over the loop

$\phi =\int B.dS$

Given that loop is square,so

$\phi =\int B.dS$

B(y, t) = k y ³t²

$\phi =kt^2\int_{0}^{a}dx\int_{0}^{a}y^3dy$

$\phi =\dfrac{1}{4}kt^2a^5$

We know that emf given as

$emf=-\dfrac{d\phi }{dt}$

$\phi =\dfrac{1}{4}kt^2a^5$

$emf=-\dfrac{1}{2}kta^5$

5 0

3 years ago

If the specific surface energy for magnesium oxide is 1.0 J/m2 and its modulus of elasticity is (225 GPa), compute the critical

alexdok [17]

Answer:

The critical stress required for the propagation of an initial crack $\sigma_{c}$ = 21.84 M pa

Explanation:

Given data

Modulus of elasticity E = 225 × $10^{9}$ $\frac{N}{m^{2} }$

Specific surface energy for magnesium oxide is $\gamma_{s}$ = 1 $\frac{J}{m^{2} }$

Crack length (a) = 0.3 mm = 0.0003 m

Critical stress is given by $\sigma_{c}^{2} }$ = $\frac{2 E \gamma}{\pi a}$ -------- (1)

⇒ 2 E $\gamma_{s}$ = 2 × 225 × $10^{9}$ × 1 = 450 × $10^{9}$

⇒ $\pi$ a = 3.14 × 0.0003 = 0.000942

⇒ Put these values in equation 1 we get

⇒ $\sigma_{c}^{2} }$ = $\frac{450 }{0.000942} 10^{9}$

⇒ $\sigma_{c}^{2} }$ = 4.77 × $10^{14}$

⇒ $\sigma_{c}$ = 2.184 × $10^{7}$ $\frac{N}{m^{2} }$

⇒ $\sigma_{c}$ = 21.84 $\frac{N}{mm^{2} }$

⇒ $\sigma_{c}$ = 21.84 M pa

This is the critical stress required for the propagation of an initial crack.

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3 years ago

When traveling from air to a solid, what would you expect a sound wave to do?

sdas [7]

A. Speed up is the answer. Hope it is right.

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3 years ago

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in 10 minutes a heart can beat 700 times at this rate in how many minutes will a heart beat 140 times at what rate can a heart b

Andru [333]

Answer: Heart can beat 140 times in = 2 minutes

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3 years ago