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2 years ago

The chemists'_ is another name for the periodic table. Answer here

Physics

1 answer:

Alina [70]2 years ago

3 0

Answer: Calendar

Explanation: I'm pretty sure it is calendar.

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Are momentum and kinetic energy conserved during an inelastic collision?

nevsk [136]

Energy and momentum are always conserved. Kinetic energy is not conserved in an inelastic collision though. And that is because it is converted to another form of energy

8 0

3 years ago

1 Calculate the size of the quantum involved in the excitation of (a) an electronic motion of frequency 1.0 × 1015 Hz, (b) a mol

torisob [31]

Answer: a) E= 6.63x10^-19J

E= 3.97×10^2KJ/mol

b) E = 3.31×10^-19J

E= 18.8×10^4 KJ/mol

C) E = 1.32×10^-33J

E= 8.01×10^-10KJ/mol

Explanation:

a) E = h ×f

h= planks constant= 6.626×10^-34

E=(6.626×10^-34)×(1.0×10^15)

E=6.63×10^-19J

1mole =6.02×10^23

E=( 6.63×10^-19)×(6.02×10^23)

E=3.97×10^2KJ/mol

b) E =(6.626×10^-34)/(1.0×10^15)

E=3.13×10^-19J

E= 3.13×10^-19) ×(6.02×10^23)

E= 18.8×10^3KJ/MOL

c) E= (6.626×10^-34) /0.5

E= 1.33×10^-33J

E= (1.33×10^-33) ×(6.02×10^23)

E= 8.01×10^-10KJ/mol

8 0

3 years ago

A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves through the point

seropon [69]

since centripetal acceleration is always towards the center of the circle

so at the given position where speed and acceleration is given the center coordinate will be towards the center of circle

also we know that

$a_c = \frac{v^2}{r}$

$11.6 = \frac{2.5^2}{R}$

$R = 0.54 m$

so the coordinates of the center will be

$(x, y) = (3.20 , (3.5 + 0.54))$

$(x, y) = (3.20, 4.04)$

so the coordinate is (3.20 m, 4.04 m)

6 0

3 years ago

A 1.120 kg car is traveling with a speed of 40 m/s. find its energy

Aleonysh [2.5K]

Answer:

896 kJ

Explanation:

KInetic Energy = 1/2 m v^2

= 1/2 (1120)(40^2) = 896 000 J or 896 kJ

4 0

2 years ago

The opening to a cave is a tall, 30.0-cm-wide crack. A bat is preparing to leave the cave emits a 30.0 kHz ultrasonic chirp. How

Vlada [557]

Answer:

The value is $w = 7.54 \ m$

Explanation:

From the question we are told that

The length of the crack is $a = 0.3 \ m$

The frequency is $f = 30.0 \ kHz = 30 *10^{3} \ Hz$

The distance outside the cave that is being consider is $D = 100 \ m$

The speed of sound is $v_s = 340 \ m/s$

Generally the wavelength of the wave is mathematically represented as

$\lambda = \frac{v}f}$

=> $\lambda = \frac{340 }{30*10^{3}}$

=> $\lambda = 0.0113 \ m/s$

Generally for a single slit the path difference between the interference patterns of the sound wave and the center is mathematically represented as

$y = \frac{ n * \lambda * D}{a}$

=> $y = \frac{ 1 * 0.0113 * 100}{0.3}$

=> $y = 3.77 \ m$

Generally the width of the sound beam is mathematically represented as

$w = 2 * y$

=> $w = 2 * 3.77$

=> $w = 7.54 \ m$

4 0

3 years ago