Answer:
A cylindrical specimen of aluminum having a diameter of 0.505 in. (12.8 mm) and a gauge length of 2.0 in. (50.8 mm) is pulled in tension. Use the load-elongation characteristics tabulated below to complete parts (a) through (f).
A uniform 1500-kg beam, 20.0 m long, supports a 15,000-kg printing press
5.0 m from the right support column (Figure slide 8). Calculate the force
on each of the vertical support columns.
Answer:
[tex]\mathbf{F_1=4.41*10^4\ N}[/tex]
[tex]\mathbf{F_2 = 1.176*10^5 \ N}[/tex]
Explanation:
The missing image of the figure slide is attached in below.
However, from the model, it is obvious that it is in equilibrium.
As a result, the relation of the force and the torque is said to be zero.
i.e.
[tex]\sum F = 0[/tex] and [tex]\sum \tau = 0[/tex]
From the image, expressing the forces through the y-axis, we have:
[tex]F_1+F_2 = W_B + W_P \\ \\ \implies 9.8(1500+15000) \\ \\ \implies \mathtt{1.617\times 10^5 \ N}[/tex]
Also, let the force [tex]F_1[/tex] be the pivot and computing the torque to determine [tex]F_2[/tex]:
Then:
[tex]F_1(0)+F_2(20.0) = 10.0W_B + 15.0W_P[/tex]
[tex]F_2 = \dfrac{((10*1500)+(15*15000))*9.8}{20.0}[/tex]
[tex]F_2 = 117600 \ N[/tex]
[tex]\mathbf{F_2 = 1.176*10^5 \ N}[/tex]
For the force equation:
[tex]F_1+F_2=1.617*10^5 \ N;[/tex]
where:
[tex]F_2 = 1.176*10^5 \ N[/tex]
Then:
[tex]F_1+1.176*10^5 \ N=1.617*10^5 \ N[/tex]
[tex]F_1=1.617*10^5 \ N-1.176*10^5 \ N[/tex]
[tex]F_1=44100\ N[/tex]
[tex]\mathbf{F_1=4.41*10^4\ N}[/tex]
Harmonics a.are components of a complex waveform. b.have frequencies that are integer multiples of the frequency of the complex waveform. c.are pure tones. d.have sinusoidal waveforms. e.all of the above
Answer:
b.have frequencies that are integer multiples of the frequency of the complex waveform
Explanation:
Please correct me if I am wrong
A satellite of mass m, originally on the surface of the Earth, is placed into Earth orbit at an altitude h. (a) Assuming a circular orbit, how long does the satellite take to complete one orbit
Answer:
T = 5.45 10⁻¹⁰ [tex]\sqrt{(R_e + h)^3}[/tex]
Explanation:
Let's use Newton's second law
F = ma
force is the universal force of attraction and acceleration is centripetal
G m M / r² = m v² / r
G M / r = v²
as the orbit is circular, the speed of the satellite is constant, so we can use the kinematic relations of uniform motion
v = d / T
the length of a circle is
d = 2π r
we substitute
G M / r = 4π² r² / T²
T² = [tex]\frac{4\pi ^2 }{GM} \ r^3[/tex]
the distance r is measured from the center of the Earth (Re), therefore
r = Re + h
where h is the height from the planet's surface
let's calculate
T² = [tex]\frac{4\pi ^2}{ 6.67 \ 10^{-11} \ 1.991 \ 10^{30}}[/tex] (Re + h) ³
T = [tex]\sqrt{29.72779 \ 10^{-20}} \ \sqrt[2]{R_e+h)^3}[/tex]
T = 5.45 10⁻¹⁰ [tex]\sqrt{(R_e + h)^3}[/tex]
In the following calculations, be sure to express the answer in standard scientific notation with the appropriate number of
significant figures.
3.88 x 1079 - 4.701 x 1059
x 10
g
Answer:
-45,597.07
Explanation:
if not in scientific calculator and yung answer nung sa scientific sa comment na lang dinadownload ko ka eh
A light source radiates 60.0 W of single-wavelength sinusoidal light uniformly in all directions. What is the average intensity of the light from this bulb at a distance of 0.400 m from the bulb
Answer: [tex]29.85\ W/m^2[/tex]
Explanation:
Given
Power [tex]P=60\ W[/tex]
Distance from the light source [tex]r=0.4\ m[/tex]
Intensity is given by
[tex]I=\dfrac{P}{4\pi r^2}[/tex]
Inserting values
[tex]\Rightarrow I=\dfrac{60}{4\pi (0.4)^2}\\\\\Rightarrow I=\dfrac{60}{2.010}\\\\\Rightarrow I=29.85\ W/m^2[/tex]
Answer:
29.85 W/ m^2
Explanation:
Find the amount og work done
Answer:
100j
Explanation:
What is the pH of a solution with a hydrogen ion concentration of 2.0x10^3.(Use 3 digits)
Answer:
2.70
Explanation:
pH = -log[H+]
pH = -log[2.0x10^-3]
pH = 2.70
The energy truck travelling at 10 km/h has kinetic energy. How much kinetic energy does it have when it is loaded so its mass is twice and its speed is increased to twice?
Explanation:
The initial kinetic energy [tex]KE_0[/tex] is
[tex]KE_0 = \frac{1}{2}m_0v_0^2[/tex]
When its mass and velocity are doubled, its new kinetic energy KE is
[tex]KE = \frac{1}{2}(2m_0)(2v_0)^2 = \frac{1}{2}(2m_0)(4v_0^2)[/tex]
[tex]\:\:\:\:\:\:\:= 8 \left(\frac{1}{2}m_0v_0^2 \right)= 8KE_0[/tex]
Therefore the kinetic energy will increase by a factor of 8.
Drag the titles to the correct boxes to complete the pairs.
What is the temperature of a system in thermal equilibrium with another system made up of water and steam at one atmosphere of pressure
Full Question:
What is the temperature of a system in thermal equilibrium with another system made up of water and steam at one atmosphere of pressure?
A) 0°F
B) 273 K
C) 0 K
D) 100°C
E) 273°C
Answer:
The correction Option is D) 100°C
Explanation:
The temperature above is referred to as the critical point.
it is the highest temperature and pressure at which water (which has three phases - liquid, solid, and gas) can exist in vapor/liquid equilibrium. If the temperature goes higher than 100 degrees celsius, it cannot remain is liquid form regardless of what the pressure is at that point.
There is also a condition under which water can exist in its three forms: that is
- Ice (solid)
- Liquid (fluid)
- Gas (vapor)
That state is called triple point. The conditions necessary for that to occur are:
273.1600 K (0.0100 °C; 32.0180 °F) as temperature and611.657 pascals (6.11657 mbar; 0.00603659 atm) as pressureCheers
Cheers
Two positive charges, 91 = 5 x 10-'[C] and q2 =1 x 10-9 [C], are
separated by a distance of d=0.05 m. At location 'P' between the
two charges, the net electric field is found to be zero.
b. [10 points] The distance between charge qı and location 'P' is
considered to be 'x'. Find the value of 'x' in [cm]
Answer:
wareffctgggyyggghhhh
A conducting sphere of radius 5.0 cm carries a net charge of 7.5 µC. What is the surface charge density on the sphere?
Answer:
[tex]\sigma=0.014\ C/m^2[/tex]
Explanation:
Given that,
The radius of sphere, r = 5 cm = 0.05 m
Net charge carries, q = 7.5 µC = 7.5 × 10⁻⁶ C
We need to find the surface charge density on the sphere. Net charge per unit area is called the surface charge density. So,
[tex]\sigma=\dfrac{7.5\times 10^{-6}}{\dfrac{4}{3}\pi \times (0.05)^3}\\\\=0.014\ C/m^2[/tex]
So, the surface charge density on the sphere is [tex]0.014\ C/m^2[/tex].
When an object is in free fall, ____________________.
Answer:
Objects that are said to be undergoing free fall, are not encountering a significant force of air resistance; they are falling under the sole influence of gravity.
Explanation:
Under such conditions, all objects will fall with the same rate of acceleration, regardless of their mass.
An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m. (a) What is his deceleration
Answer:
a= -80.357 m/s
Explanation:
use the formula
vf^2=vi^2+2a(xf-xi)
Plug in givens
0=(7.50)^2+2a(0.350m)
solve for acceleration
a= -80.357 m/s
Displacement of a body moving in circular motion is
Explanation:
Displacement of a body moving in circular motion is called uniform circular motion.
hope it is helpful to you
Answer:
A constantly moving object with consistent circular movement. However, for its change in direction, it is accelerating
Explanation:
Uniform circular motion in a circle at constant rate can be described as the motion of the object. When an object moves in a circle, it changes its direction constantly. The object moves tangently to the circle at all times. As the velocity vector direction is the same as the object motion direction, the velocity vector is tangent to the circle. This is shown in the animation on the right by a vector arrow.
An item is accelerating that moves in a circle. Objects that accelerate are subjects that change their speed – either the velocity (i.e. the vecteur magnitude) or the direction. An object with consistent circular movement moves at a constant speed. However, because of its change in direction, it is accelerating. The acceleration direction is inside. The animation on the right shows this through a vector arrow
For an object with only a uniform circular movement, the final motion is the net force. The The net force acting on this object is directed to the middle of the circle. The net force is an inner or centripetal force. Without such a deepest force, an object would continue in a straight direction, never deviating. Regrettably, with the inward net force, perpendicular to the vector, the object changes the direction and is accelerated internally.
A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass first passes through the equilibrium position, and the position of the mass at any time is described by
The question is incomplete. The complete question is :
A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass first passes through the equilibrium position, and the position of the mass at any time is described by x = (4.7 cm)sin[(7.9 rad/s)πt].
Determine the following:
(a) frequency of the motion
(b) period of the motion
(c) amplitude of the motion
(d) first time after t = 0 that the object reaches the position x = 2.6 cm
Solution :
Given equation : x = (4.7 cm)sin[(7.9 rad/s)πt].
Comparing it with the general equation of simple harmonic motion,
x = A sin (ωt + Φ)
A = 4.7 cm
ω = 7.9 π
a). Therefore, frequency, [tex]$f=\frac{\omega}{2 \pi}$[/tex]
[tex]$=\frac{7.9 \pi}{2 \pi}$[/tex]
= 3.95 Hz
b). The period, [tex]$T=\frac{1}{f}$[/tex]
[tex]$T=\frac{1}{3.95}[/tex]
= 0.253 seconds
c). Amplitude is A = 4.7 cm
d). We have,
x = A sin (ωt + Φ)
[tex]$x_t=4.7 \sin (7.9 \pi t)$[/tex]
[tex]$2.6 = 4.7 \sin (7.9 \pi t)$[/tex]
[tex]$\sin (7.9 \pi t) = \frac{26}{47}$[/tex]
[tex]$7.9 \pi t = \sin^{-1}\left(\frac{26}{47}\right)$[/tex]
Hence, t = 0.0236 seconds.
Physics help please
Answer:
i think the answer is 0.001m³
(a) What is the maximum frictional force (in N) in the knee joint of a person who supports 45.0 kg of her mass on that knee if the coefficient of static friction is 0.016
Answer:
f = 7.06 N
Explanation:
The maximum frictional force on the knee joint of the person can be given by the following formula:
[tex]f = \mu R = \mu W \\[/tex]
where,
f = maximum frictional force = ?
μ = static friction coefficient = 0.016
W = Weight load on knee = mg
m = mass supported by knee = 45 kg
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]f = \mu mg\\f = (0.016)(45\ kg)(9.81\ m/s^2)\\[/tex]
f = 7.06 N
When a golfer tees off, the head of her golf club which has a mass of 158 g is traveling 48.2 m/s just before it strikes a 46.0 g golf ball at rest on a tee. Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s. Neglect the mass of the club handle and determine the speed of the golf ball just after impact.
Answer:
v₂ = 53.23 m/s
Explanation:
Given that,
The mass of a golf club, m₁ = 158 g = 0.158 kg
The initial speed of a golf club, u₁ = 48.2 m/s
The mass of a golf ball, m₂ = 46 g = 0.046 kg
It was at rest, u₂ = 0
Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s, v₁ = 32.7 m/s
We use the conservation of energy to find the speed of the golf ball just after impact as follows :
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.158(48.2)-0.158(32.7)}{0.046}\\\\=53.23\ m/s[/tex]
So, the speed of the golf ball just after the impact is equal to 53.23 m/s.
g Three masses are located in the x- y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m, 0 m), and a mass of 2 kg is located at (0 m, 3 m). Where is the center of mass of the system
Answer:
Xcm = (6 * 0 + 4 & 3 + 2 * 0) / 12 = 1
Ycm = (6 * 0 + 4 * 0 + 2 * 3) / 12 = 1/2
(Xcm , Ycm) = (1 , 1/2)
Using definition of center of mass
A 0.160kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.820m/s . It has a head-on collision with a 0.300kg glider that is moving to the left with a speed of 2.27m/s . Suppose the collision is elastic.
Part A
Find the magnitude of the final velocity of the 0.160kg glider. m/s
Part B
Find the direction of the final velocity of the 0.160kg glider.
i. to the right
ii. to the left
Part C
Find the magnitude of the final velocity of the 0.300kg glider. m/s
Part D
Find the direction of the final velocity of the 0.300kg glider.
Answer:
A) v_{f1} = -3.2 m / s, B) LEFT , C) v_{f2} = -0.12 m / s, D) LEFT
Explanation:
This is a collision exercise that can be solved using momentum conservation, for this we define a system formed by gliders, so that the forces during the collision are internal and the moment is conserved.
Let's use the subscript 1 for the lightest glider m1 = 0.160 kg and vo1 = 0.820 m / s
subscript 2 for the heaviest glider me² = 0.820 kg and vo2 = -2.27 m / s
Initial instant. Before the crash
p₀ = m₁ v₀₁ + m₂ v₀₂
Final moment. After the crash
p_f = m₁ v_{f1} + m₂ v_{f2}
p₀ = p_f
m₁ v₀₁ + m₂ v₀₂ = m₁ v_{f1} + m₂ v_{f2}
as the shock is elastic, energy is conserved
K₀ = K_f
½ m₁ v₀₁² + ½ m₂ v₀₂² = ½ m₁ [tex]v_{f1}^2[/tex] + ½ m₂ [tex]v_{f2}^2[/tex]
m₁ (v₀₁² - v_{f1}²) = m₂ (v_{f2}² -v₀₂²)
let's make the relationship
(a + b) (a-b) = a² -b²
m₁ (v₀₁ + v_{f1}) (v₀₁-v+{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)
let's write our two equations
m₁ (v₀₁ -v_{f1}) = m₂ (v_(f2) - v₀₂) (1)
m₁ (v₀₁ + v_{f1}) (v₀₁-v_{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)
we solve
v₀₁ + v_{f2} = v_{f2} + v₀₂
we substitute in equation 1 and obtain
M = m₁ + m₂
[tex]v_{f1} = \frac{m_1-m_2}{M} v_o_1 + 2 \frac{m_2}{M} v_f_2[/tex]
[tex]v_f_2 = \frac{2m_1}{M} v_o_1 + \frac{m_2-m_1}{M} v_o_2[/tex]vf2 = 2m1 / mm vo1 + m2-m1 / mm vo2
we calculate the values
m₁ + m₂ = 0.160 +0.3000 = 0.46 kg
v_{f1} = [tex]\frac{ 0.160 -0.300} {0.460} \ 0.820 + \frac{2 \ 0300}{0.460} \ (-2.27)[/tex]
v_{f1} = -0,250 - 2,961
v_{f1} = - 3,211 m / s
v_{f2} = [tex]\frac{2 \ 0.160}{0.460} \ 0.820 + \frac{0.300 - 0.160}{0.460 } \ (-2.27)[/tex]
v_{f2} = 0.570 - 0.6909
v_{f2} = -0.12 m / s
now we can answer the different questions
A) v_{f1} = -3.2 m / s
B) the negative sign indicates that it moves to the left
C) v_{f2} = -0.12 m / s
D) the negative sign indicates that it moves to the LEFT
A proton has been accelerated from rest through a potential difference of -1350 V. What is the proton's kinetic energy, in electron volts? What is the proton's kinetic energy, in joules? What is the proton's speed?
Answer:
1 eV = 1.60 * 10^-19 J work done in accelerating electron thru 1 V
KE (total energy) = 1350 ^ 1 eV (note proton goes from + to -)
KE = 1.60 * 10^-19 * 1350 = 2.16 * 10^-16 Joules
1/2 m v^2 = KE = 2.16 * 10^-16 J
v^2 = 4.32 * 10E-16 / 1.67 * 10-27 = 2.59 * 10^11
v = 5.09 * 10^5 m/s
The proton's kinetic energy, in joules is 2.16 *[tex]10^{-16}[/tex] J. The proton's velocity is 5.09 * [tex]10^{5}[/tex]m/s.
What is velocity?
When an item is moving, its velocity is the rate at which its direction is changing as seen from a certain point of view and as measured by a specific unit of time.
Uniform motion an object is said to have uniform motion when object cover equal distance in equal interval of time within exact fixed direction. For a body in uniform motion, the magnitude of its velocity remains constant over time.
1 eV = 1.60 * [tex]10^{-19} J[/tex] work done in accelerating electron throw 1 V
K.E (total energy) = 1350 ^ 1 eV (note proton goes from + to -)
K.E = 1.60 * [tex]10^{-19}[/tex]J * 1350 = 2.16 * [tex]10^{-16}[/tex] Joules
1/2 m v² = KE = 2.16 *[tex]10^{-16}[/tex] J
Velocity of proton is,
v² = 4.32 * 10[tex]e^{-16}[/tex] / 1.67 * [tex]10{-27}[/tex] = 2.59 * [tex]10^{11}[/tex]
v = 5.09 * [tex]10^{5}[/tex]m/s
The proton's kinetic energy, in joules is 2.16 *[tex]10^{-16}[/tex] J. The proton's velocity is 5.09 * [tex]10^{5}[/tex]m/s.
To learn more about velocity refer the link:
brainly.com/question/18084516
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A television tube can accelerate electrons to 2.00 · 104 ev. Calculate the wavelength of emitted X-rays with the highest energy.
λ = _____ m
9.9 x 10 -30
6.2 x 10 -11
1.6 x 10 10
7.1 x 10 -57
Answer:
6.2 × 10^-11 m
Explanation:
1 eV = 1.602 × 10-19 joule
2.00 × 104 ev. = 2.00 × 10^4 eV × 1.602 × 10^-19 joule/1eV
= 3.2 × 10^-15 J
From;
E= hc/λ
λ = hc/E
λ = 6.6 × 10^-34 × 3 × 10^8/3.2 × 10^-15
λ = 6.2 × 10^-11 m
brainly A person's eye lens is 2.9 cm away from the retina. This lens has a near point of 25 cm and a far point at infinity. What must the focal length of this lens be in order for an object placed at the near point of the eye to focus on the retina
Answer: The focal length of the lens is 2.60 cm
Explanation:
The equation for lens formula follows:
[tex]\frac{1}{f}=\frac{1}{v}-\frac{1}{u}[/tex]
where,
f = focal length = ? cm
v = image distance = 2.9 cm
u = Object distance = -25 cm
Putting values in above equation, we get:
[tex]\frac{1}{f}=\frac{1}{2.9}-\frac{1}{(-25)}\\\\\frac{1}{f}=\frac{1}{2.9}+\frac{1}{(25)}\\\\\frac{1}{f}=\frac{25+2.9}{2.9\times 25}\\\\f=\frac{72.5}{27.9}=2.60cm[/tex]
Hence, the focal length of the lens is 2.60 cm
What power (in kW) is supplied to the starter motor of a large truck that draws 260 A of current from a 25.5 V battery hookup
Answer:
P = 6.63 kW
Explanation:
Given that,
Current, I = 260 A
Voltage of the battery, V = 25.5 V
We need to find the power supplied to the starter motor. We know that,
P = VI
Put all the values,
P = 25.5 × 260
P = 6630 W
or
P = 6.63 kW
So, the power supplied to the motor is 6.63 kW.
Answer:
The power is 6.63 kW.
Explanation:
Current, I = 260 A
Voltage, V = 25.5 V
Power of an electrical appliance is given by
P = V I
P = 25.5 x 260
P = 6630 W
1 kW = 1000 W
So, the power is
P = 6.63 kW
An electric lamp consumes 60W at 220 volts. How many dry cells of 1.5 V and internal resistance 1 Ohm are required to glow the lamp?
Answer:
1. Number of dry cells of 1.5 V required is 40.
2. Number of internal resistance of 1 ohm required is 807
Explanation:
We'll begin by calculating the resistance. This can be obtained as follow:
Power (P) = 60 W
Voltage (V) = 220 V
Resistance (R) =?
P = V²/R
60 = 220² / R
Cross multiply
60 × R = 220²
60 × R = 48400
Divide both side by 60
R = 48400 / 60
R ≈ 807 Ohm
1. Determination of the number of dry cells of 1.5 V required.
Voltage (V) = 220
Dry Cells = 1.5 V
Number of dry cells (n) =?
n = Voltage / Dry cells
n = 60 / 1.5
n = 40
2. Determination of the number of internal resistance of 1 ohm required.
Resistance (R) = 807 Ohm
Internal resistance (r) = 1 ohm
Number of internal resistance (n) =?
n = R/r
n = 807 / 1
n = 807
SUMMARY:
1. Number of dry cells of 1.5 V required is 40.
2. Number of internal resistance of 1 ohm required is 807
The block in the drawing has dimensions L0×2L0×3L0,where L0 =0.2 m. The block has a thermal conductivity of 150 J/(s·m·C˚). In drawings A, B, and C, heat is conducted through the block in three different directions; in each case the temperature of the warmer surface is 35 ˚C and that of the cooler surface is 16 ˚C Determine the heat that flows in 6 s for each case.
Answer:
1140 J, 6840 J, 10260 J
Explanation:
Lo x 2 Lo x 3 Lo, Lo = 0.2 m, K = 150 J/(s · m · C˚) , T = 35 ˚C, T' = 16 ˚C,
time, t = 6 s
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 1140 J[/tex]
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 6840 J[/tex]
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{2\times 0.2}\\\\H = 10260 J[/tex]
A hot air balloon is a sphere of volume 2210 m3. The density of the hot air inside is 1.13 kg/m3, while the air outside has a density of 1.29 kg/m3. The balloon itself has a mass of 240 kg. What is the TOTAL NET force acting on the balloon?
[?]N
The total net force acting on the balloon will be 24498 Newtons
Given that
Volume of the balloon = 2210 cubic meter
Density of the air inside the balloon = 1.13 kg/m3
What will be the net force exerted on the balloon ?Here force on the balloon will be equal to the weight of the air displaced by balloon
[tex]F= mass of air displaced\times gravity[/tex]
[tex]F= Density \times volume \times gravity[/tex]
[tex]F=1.13 \times 2210 \times 9.81[/tex]
[tex]F=24498 N[/tex]
The total net force acting on the balloon will be 24498 Newtons
To know more about buoyancy force follow
https://brainly.com/question/117714
An electron in a hydrogen atom is in a p state. Which of the following statements is true?
a.
The electron’s wavefunction has at least one node (i.e., at least one place in space where it goes to zero).
b.
The electron has an energy of -13.6 eV.
c.
The electron has a total angular momentum of ħ.
d.
The electron has a z-component of angular momentum equal to sqrt(2)* ħ.
Answer:
The electron’s wavefunction has at least one node (i.e., at least one place in space where it goes to zero).
Explanation:
We know that the p-orbitals have nodes. A node is a region where the probability of finding an electron goes down to zero.
P orbitals are oriented along the x,y,z Cartesian axes and are known to have angular nodes along the axes.
Hence, if an electron in a hydrogen atom is in a p state, the electron’s wavefunction has at least one node
if the tin is made of a metal which has a density of 7800 kg per metre cubic calculate the volume of the metal used to make tin and lead
Answer:
XL sleep usual Addison officer at home and ear is not a short time to be be free and ear is a short time to make a short time
Explanation:
so that I can take the class on Monday and ear is not a short time to be be free and ear is not a short time to be be free and ear is not a short time to be be free and ear is not a short time to time for a day or night and ear buds is Anshu and duster and duster fgor a day or night is Anshu and duster for a day or not a week of computer science from your computer and I am in the same as I am a short of ti and you can be the first time I will be be