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3 years ago

All of the orbitals in a given subshell have the same value of the __________ quantum number.

1 answer:

3 0

All of the orbitals in a given subshell have the same value of the "magnetic and principal" quantum number

Hope this helps!

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Consult Multiple Concept Example 10 in preparation for this problem. Traveling at a speed of 18.2 m/s, the driver of an automobi

Anettt [7]

Answer:

The speed of the automobile after 1.43s is 10 $\frac{m}{s}$

Explanation:

$a= \frac{-f}{m}= \frac{-u_{k}*m*g}{m}$

$a= -u_{k}*g=- 0.590* 9.8 \frac{m}{s^{2} }= -5.782 \frac{m}{s^{2} }$

$V_{f} = V_{i} + a*t$

$V_{f} = 18.2 \frac{m}{s} - (5.782 \frac{m}{s^{2} }* 1.43 s)$

$V_{f} = 9.93174 \frac{m}{s}$

$V_{f}$ ≅ 10 $\frac{m}{s}$

7 0

3 years ago

Which of the following would most likely happen if water did not form hydrogen bonds?

Trava [24]

Option C is the correct answer

6 0

3 years ago

What is another unit for momentum besides kg-m/s? a. N b. N-s c. N-s2 d. N/s

lara [203]

'Newton-second' is dimensionally equivalent to 'kilogram-meter/second'.

8 0

2 years ago

During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with large catapu

ella [17]

Answer:

The work done on the hose by the time the hose reaches its relaxed length is 776.16 Joules

Explanation:

The given spring constant of the of the spring, k = 88.0 N/m

The length by which the hose is stretched, x = 4.20 m

For the hose that obeys Hooke's law, and the principle of conservation of energy, the work done by the force from the hose is equal to the potential energy given to the hose

The elastic potential energy, P.E., of a compressed spring is given as follows;

P.E. = 1/2·k·x²

∴ The potential energy given to hose, P.E. = 1/2 × 88.0 N/m × (4.20 m)²

1/2 × 88.0 N/m × (4.20 m)² = 776.16 J

The work done on the hose = The potential energy given to hose, P.E. = 776.16 J

5 0

3 years ago

A string along which waves can travel is 4.36 m long and has a mass of 222 g. The tension in the string is 60.0 N. What must be

lora16 [44]

Answer:

frequency is 195.467 Hz

Explanation:

given data

length L = 4.36 m

mass m = 222 g = 0.222 kg

tension T = 60 N

amplitude A = 6.43 mm = 6.43 × $10^{-3}$ m

power P = 54 W

to find out

frequency f

solution

first we find here density of string that is

density ( μ )= m/L ................1

μ = 0.222 / 4.36

density μ is 0.050 kg/m

and speed of travelling wave

speed v = √(T/μ) ...............2

speed v = √(60/0.050)

speed v = 34.64 m/s

and we find wavelength by power that is

power = μ×A²×ω²×v / 2 ....................3

here ω is wavelength put value

54 = ( 0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 ) / 2

0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 = 108

ω² = 108 / 7.160 × $10^{-5}$

ω = 1228.16 rad/s

so frequency will be

frequency = ω / 2π

frequency = 1228.16 / 2π

frequency is 195.467 Hz

7 0

3 years ago