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3 years ago

14

The wave functions for states of the hydrogen atom with orbital quantum number l=0 are much simpler than for most other states,

because the angular part of the wave.
a. True
b. False

1 answer:

Thepotemich [5.8K]3 years ago

4 0

True. Fun fact. Hope this helps

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Two manned satellites approach one another at a relative velocity of v=0.190 m/s, intending to dock. The first has a mass of m1=

drek231 [11]

Answer:

Their final relative velocity is 0.190 m/s

Explanation:

The relative velocity of the satellites, v = 0.190 m/s

The mass of the first satellite, m₁ = 4.00 × 10³ kg

The mass of the second satellite, m₂ = 7.50 × 10³ kg

Given that the satellites have elastic collision, we have;

$v_2 = \dfrac{2 \cdot m_1}{m_1 + m_2} \cdot u_1 - \dfrac{m_1 - m_2}{m_1 + m_2} \cdot u_2$

$v_2 = \dfrac{ m_1 - m_2}{m_1 + m_2} \cdot u_1 + \dfrac{2 \cdot m_2}{m_1 + m_2} \cdot u_2$

Given that the initial velocities are equal in magnitude, we have;

u₁ = u₂ = v/2

u₁ = u₂ = 0.190 m/s/2 = 0.095 m/s

v₁ and v₂ = The final velocities of the satellites

We get;

$v_1 = \dfrac{2 \times 4.0 \times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 - \dfrac{4.0 \times 10^3- 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095$

$v_2 = \dfrac{ 4.0 \times 10^3 - 7.50\times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 + \dfrac{2 \times 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095$

The final relative velocity of the satellite, $v_f$ = v₁ + v₂

∴ $v_f$ = 0.095 + 0.095 = 0.190

The final relative velocity of the satellite, $v_f$ = 0.190 m/s

4 0

2 years ago

A cyclist rides in a circle with speed 8.1 m/s. What is his centripetal

Alina [70]

Explanation:

We know that the tangent velocity is 8.1 m/s. We also know that the tangent velocity can be written in the following way:

Vt = ωr with ω being the angular velocity.

We now calculate ω:

ω = Vt/r = 8.1 m/s / 27m = 0.3 rad/s

Now that we have ω we can calculate the centripetal aceleration:

a = ω^2 * r = ( 0.3 )^2 * 27 = 2.43 m/s^2

5 0

2 years ago

Sam threw a stone horizontally at a speed of 5m/s from a 9.8 m tall building. how much time does the stone take to hit the groun

taurus [48]

The given is a projectile motion with angle equal to 90° at which the motion would become horizontal. The projectile motion has two components for the distance. The horizontal component would progress at a constant rate each second while the vertical component would become greater as time progress.

In this item, we are asked to determine the amount of time that would be needed for the stone to hit the ground. This can be calculated through the equation,

d = V₀t + (1/2)(gt²)
It is to be noted that V₀ has a y-component equal to zero.

Substituting this value to the equation,
9.8 m = (0) + (1/2)(9.8 m/s²)(t²)
t = sqrt (2) = 1.41 s

Hence, the time would be equal to 1.41 s.

7 0

2 years ago

A skier starts from rest at the top of a 20 degree incline and skis in a straight line to the bottom of the slope, a distance d

SSSSS [86.1K]

Answer:

Explanation:

Loss of potential energy = mgh.

h = d sin 20

= 400 sin20 = 136.8 m

Loss of potential energy = m x 9.8 x 136.8

= 1340.64 m

negative work done by friction = μ mg cosθ x d

= .2 x m x 9.8 x cos 20 x 400

= 736.72 m

Net loss of potential energy = 1340.64 m - 736.72 m

= 603.92 m

= gain of kinetic energy = 1/2 m v²

1/2 m v² = 603.92 m

v² = 1207.84

v = 34.75 m /s .

6 0

2 years ago

Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 400°C.

Jet001 [13]

Answer:

q=471.19kj.kg

Explanation:water constant volume = C

First state

Quantity X1=1

Pressure=1 bar

Second state

T2=400C

Required heat transferred q in kj.kg

Assumption:

System is at equilibrium

Kinetic and Potential energy are neglected

Therefore, from 1st law of thermodynamics q - w=∆u. . . . 1

From steam table A

U1=Ug=2553.6kj.kg

U2=Vg=1625m³.kg

V1=V2=4625m³.kg

From steam Table B

Px= 7 bar

Vx=4667kj.kg

Ux=3026.6kj.kg

Py= 10 bar

Vy= 3066kj.kg

Uy=2957.5kj.kg

Assuming a Linear interpolation

Ux-V2/Ux-Vy=Vx-U2/Ux-Vy

U2 =Vx-{vx-vy}× vx-v2/Ux-Vy

U2 = 3026.6 - (3026.6-2957.3)× 4667-4625/4667-3066

=3024.79kj.kg

Therefore, from q-w=∆u

q=3024.79-253.6

=471.19kj.kg

Where u = internal energy

V= specific volume

6 0

3 years ago