A large, metallic, spherical shell has no net charge. It is supported on an insulating stand and has a small hole at the top. A small tack with charge Q is lowered on a silk thread through the hole into the interior of the shell.
1) What is the charge on the inner surface of the shell?
A) Q
B) Q/2
C) 0
D) -Q/2
E) -Q
2) What is the charge on the outer surface of the shell?
A) Q
B) Q/2
C) 0
D) -Q/2
E) -Q
3) The tack is now allowed to touch the interior surface of the shell. After this contact, what is the charge on the tack?
A) Q
B) Q/2
C) 0
D) -Q/2
E) -Q
4) What is the charge on the inner surface of the shell now?
A) Q
B) Q/2
C) 0
D) -Q/2
E) -Q
5) What is the charge on the outer surface of the shell now?
A) Q
B) Q/2
C) 0
D) -Q/2
E) -Q
The charge on the inner surface of the shell is -Q
The charge on the outer surface of the shell is Q
After this contact, the charge on the tack is 0
The charge on the inner surface of the shell now is 0
The charge on the outer surface of the shell now is Q
What is the charge on a shell ?The charge on a shell depends on the situation and the conditions of the shell. If the shell is an electrically neutral object, such as a metallic spherical shell, it has no net charge, meaning that the total positive charge is equal to the total negative charge. However, if the shell has an excess or deficit of electrons, it will have a net charge, either positive or negative, depending on whether it has an excess of electrons or a deficit of electrons.
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A 35.8 kg box initially at rest is pushed 2.38 m along a rough, horizontal floor with a constant applied horizontal force of 108.915 N. If the coefficient of friction between box and floor is 0.256, find the work done by the applied force. The acceleration of gravity is 9.8 m/s 2 . Answer in units of J.
Answer:
The work done by the applied force is 259.22 J.
Explanation:
The work done by the applied force is given by:
[tex] W = F*d [/tex]
Where:
F: is the applied horizontal force = 108.915 N
d: is the distance = 2.38 m
Hence, the work is:
[tex] W = F*d = 108.915 N*2.38 m = 259.22 J [/tex]
Therefore, the work done by the applied force is 259.22 J.
I hope it helps you!
A 6.11-g bullet is moving horizontally with a velocity of 366 m/s, where the sign indicates that it is moving to the right (see part a of the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part b. Note that both blocks are moving after the collision with the bullet. The mass of the first block is 1206 g, and its velocity is 0.662 m/s after the bullet passes through it. The mass of the second block is 1550 g. (a) What is the velocity of the second block after the bullet imbeds itself
Answer:
[tex]V=1.86m/s[/tex]
Explanation:
Mass of bullet [tex]M_B=6.11g[/tex]
Velocity of bullet [tex]V_B=366m/s[/tex]
Mass of first block [tex]M_b_1=1206g[/tex]
Velocity of block [tex]V_b=0.662m/s[/tex]
Mass of second block [tex]M_b_2=1550g[/tex]
Generally the total momentum before collision is mathematically given as
[tex]P_1=0.006kg*366+0+0[/tex]
[tex]P_1=2.196kg\ m/s[/tex]
Generally the total momentum after collision is mathematically given as
[tex]P_2=(1.206kg*0.633)+(1.550+0.00611)V[/tex]
[tex]P_2=0.763398+1.55611V[/tex]
Generally the total momentum is mathematically given as
[tex]P_1=P_2[/tex]
[tex]2.196=0.763398+1.55611V[/tex]
[tex]V=\frac{2.196+0.763398}{1.55611}[/tex]
[tex]V=1.86m/s[/tex]
If there is "waste" energy, does the Law of Conservation of Energy still apply?
Explanation:
Yes, the law of conservation of energy still applies even if there is waste energy.
The waste energy are the transformation products of energy from one form to another.
According to the law of conservation of energy "energy is neither created nor destroyed by transformed from one form to another in a system".
But of then times, energy is lost as heat or sound within a system.
If we take into account these waste energy, we can see that energy is indeed conserved. The sum total of the energy generated and those produced will be the same if we factor in other forms in which the energy has been transformed into.7. If the impact of the golf club on the ball in the previous question occurs over a time of 2 x 10 seconds, what
force does the ball experience to accelerate from rest to 73 m/s?
Answer:
3.65 x mass
Explanation:
Given parameters:
Time = 20s
Initial velocity = 0m/s
Final velocity = 73m/s
Unknown:
Force the ball experience = ?
Solution:
To solve this problem, we apply the equation from newton's second law of motion:
F = m [tex]\frac{v - u}{t}[/tex]
m is the mass
v is the final velocity
u is the initial velocity
t is the time taken
So;
F = m ([tex]\frac{73 - 0}{20}[/tex] ) = 3.65 x mass
How heavier elements formed during stellar nucleosynthesis and evolution?
Answer:
i honestly think its 21
Explanation:
da memes
10 + 10 =21
A rocket burns fuel to create hot gases that explode violently out of the rocket engine. This explosion creates thrust. Thrust is a force that pushes the rocket upward. What force must thrust overcome in order to send a rocket up into space?
Answer:
Thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.
Explanation:
From the concept of Escape Velocity, derived from Newton's Law of Gravitation, definition of Work, Work-Energy Theorem and Principle of Energy Conservation, which is the minimum speed such that rocket can overcome gravitational forces exerted by the Earth, and according to the Tsiolkovski's Rocket Equation, which states that thrust done by the rocket is equal to the change in linear momentum of the rocket itself, we conclude that thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.
two identical balls are rolling down a hill ball 2 is rolling faster than ball 1 which ball has more kinetic energy
A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m
Complete Question
A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?
Answer:
[tex]H_m=1.65m[/tex]
[tex]H_E=1.16307m[/tex]
Explanation:
From the question we are told that
Mass of ball [tex]M=2kg[/tex]
Length of string [tex]L= 2m[/tex]
Wind force [tex]F=13.2N[/tex]
Generally the equation for [tex]\angle \theta[/tex] is mathematically given as
[tex]tan\theta=\frac{F}{mg}[/tex]
[tex]\theta=tan^-^1\frac{F}{mg}[/tex]
[tex]\theta=tan^-^1\frac{13.2}{2*2}[/tex]
[tex]\theta=73.14\textdegree[/tex]
Max angle =[tex]2*\theta= 2*73.14=>146.28\textdegree[/tex]
Generally the equation for max Height [tex]H_m[/tex] is mathematically given as
[tex]H_m=L(1-cos146.28)[/tex]
[tex]H_m=0.9(1+0.8318)[/tex]
[tex]H_m=1.65m[/tex]
Generally the equation for Equilibrium Height [tex]H_E[/tex] is mathematically given as
[tex]H_E=L(1-cos73.14)[/tex]
[tex]H_E=0.9(1+0.2923)[/tex]
[tex]H_E=1.16307m[/tex]
Two point charges are placed on the x-axis as follows: charge q1 = 3.95 nC is located at x= 0.198 m , and charge q2 = 4.96 nC is at x= -0.297 m. What are the magnitude and direction of the total force exerted by these two charges on a negative point charge q3=6.00nCq that is placed at the origin?
Answer:
F = 2.40 × [tex]10^{-6}[/tex] N
Explanation:
given data
charge q1 = 3.95 nC
x= 0.198 m
charge q2 = 4.96 nC
x= -0.297 m
solution
force on a point charge kept in electric field F = E × q ................1
here E is the magnitude of electric field and q is the magnitude of charge
and
first we will get here electric field at origin
So net field at origin is
E = (Kq2÷r2²) - (kq1÷r1²) ...............2
put here value
E = 9[(4.96÷0.297²)-(3.95÷0.198²)]
E = 400.72 N/C ( negative x direction )
so that force will be
F = 6 × [tex]10^{-9}[/tex] × 400.72
F = 2.40 × [tex]10^{-6}[/tex] N
The net force on the third charge is 2.404 x 10⁻⁶ N.
The given parameters:
Position of first point charge, x1 = 0.198 mPosition of second point charge, x2 = -0.297 mFirst point charge, q1 = 3.95 nCSecond point charge, q2 = 4.96 nCThird point charge, q3 = 6 nC Position of the third charge, = 0The force on the third charge due to first charge is calculated as follows;
[tex]F_{13} = \frac{kq_1 q_3}{r^2} \\\\F_{13} = \frac{9\times 10^9 \times 3.95 \times 10^{-9} \times 6 \times 10^{-9} }{(0.198)^2} (+i)= 5.44 \times 10^{-6} \ N \ (+i)[/tex]
The force on the third charge due to second charge is calculated as follows;
[tex]F_{23} = \frac{kq_2q_3}{r^2} \\\\F_{23} = \frac{9\times 10^9 \times 4.96 \times 10^{-9}\times 6 \times 10^{-9} }{(0.297)^2} (-i)\\\\F_{23} = (3.036 \times 10^{-6} ) \ N \ (-i)[/tex]
The net force on the third charge is calculated as follows;
[tex]F_{net} = 5.44 \times 10^{-6} - 3.036 \times 10^{-6} \\\\F_{net} = 2.404 \times 10^{-6} \ N[/tex]
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, puck 1 of mass m1 ! 0.20 kg is sent sliding across a frictionless lab bench, to undergo a one-dimensional elastic collision with stationary puck 2. Puck 2 then slides off the bench and lands a distance d from the base of the bench. Puck 1 rebounds from the collision and slides off the opposite edge of the bench, landing a distance 2d from the base of the bench. What is the mass of puck 2
Answer:
1 kg
Explanation:
Assuming that,
Δx(2) = v(2)t, where Δx(2) = d and v(2) = 2m1 / (m1 + m2) v1i
On the other hand again, if we assume that
Δx(1) = v(1)t, where Δx(1) = -2d, and v(1)t = m1 - m2 / m1 + m2 v1i
From the above, we proceed to dividing Δx(2) by Δx(1), so that we have
d/-2d = [2m1 / (m1 + m2) v1i] / [m1 - m2 / m1 + m2 v1i], this is further simplified to
1/-2 = [2m1 / (m1 + m2)] / [m1 - m2 / m1 + m2]
1/-2 = 2m1 / (m1 + m2) * m1 + m2 / m1 - m2
1/-2 = 2m1 / m1 - m2, if we cross multiply, we have
m1 - m2 = -2 * 2m1
m1 - m2 = -4m1
m2 = 5m1
From the question, we're told that m1 = 0.2 kg, if we substitute for that, we have
m2 = 5 * 0.2
m2 = 1 kg
4. A ball is dropped from height of 45 m.
Then distance covered in last 0.6 sec of
its motion will be
Answer:
1.76 m
Explanation:
Height from which the object is dropped = 45 m
Time (t) remaining for the ball to land = 0.6 s
Height (h) in the remaining =?
The height to which the object falls in the remaining time can be obtained as follow:
Time (t) remaining for the ball to land = 0.6 s
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) in the remaining =?
h = ½gt²
h = ½ × 9.8 × 0.6²
h = 4.9 × 0.36
h = 1.76 m
Thus, the distance travelled in the last 0.6 s is 1.76 m
A ball is thrown straight upward and reaches the top of its path in 1.71 s (before it starts to come back down). A second ball is thrown at an angle of 34 degrees with the horizontal. At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically
Answer:
The second ball must be thrown at 30.01 m/s.
Explanation:
First, we need to find the maximum height (H) reached by the ball 1:
[tex] v_{f_{1}}^{2} = v_{0_{1}}^{2} - 2gH [/tex]
Where:
[tex]v_{f_{1}}[/tex]: is the final speed of ball 1 = 0 (at the maximum height)
[tex]v_{0_{1}}[/tex]: is the initial speed of ball 1
g: is the gravity = 9.81 m/s²
We need to find the initial speed, by using the following equation:
[tex] v_{f_{1}} = v_{0_{1}} - gt [/tex]
Where t is the time = 1.71 s (when it reaches the maximum height)
[tex] v_{0_{1}} = gt = 9.81 m/s^{2}*1.71 s = 16.78 m/s [/tex]
So, the maximum height is:
[tex] H = \frac{v_{0_{1}}^{2}}{2g} = \frac{(16.78 m/s)^{2}}{2*9.81 m/s^{2}} = 14.35 m [/tex]
Finally, the speed at which ball 2 must be thrown is:
[tex]v_{f_{2y}}^{2} = (v_{0_{2y}}sin(\theta)})^{2} - 2gH[/tex]
[tex]v_{0_{2y}}= \frac{\sqrt{2gH}}{sin(\theta)} = \frac{\sqrt{2*9.81 m/s^{2}*14.35 m}}{sin(34)} = 30.01 m/s[/tex]
Therefore, the second ball must be thrown at 30.01 m/s.
I hope it helps you!
PLEASE HELP PLEASEEEE
Answer:
How can I help you??? Plz insert some questions
Particle A of charge 2.76 10-4 C is at the origin, particle B of charge -6.54 10-4 C is at (4.00 m, 0), and particle C of charge 1.02 10-4 C is at (0, 3.00 m). We wish to find the net electric force on C. What is the x component of the electric force?
Answer:
a) F_net = 30.47 N , θ = 10.6º
b) Fₓ = 29.95 N
Explanation:
For this exercise we use coulomb's law
F₁₂ = k [tex]k \frac{ q_{1} \ q_{2} }{ r^{2} }[/tex]
the direction of the force is on the line between the two charges and the sense is repulsive if the charges are equal and attractive if the charges are different.
As we have several charges, the easiest way to solve the problem is to add the components of the force in each axis, see attached for a diagram of the forces
X axis
Fₓ = [tex]F_{bc x}[/tex]
Y axis
[tex]F_{y}[/tex]Fy = [tex]F_{ab} - F_{bc y}[/tex]
let's find the magnitude of each force
[tex]F_{ab}[/tex] = 9 10⁹ 2.76 10⁻⁴ 1.02 10⁻⁴ / 3²
F_{ab} = 2.82 10¹ N
F_{ab} = 28.2 N
[tex]F_{bc}[/tex] = 9 10⁹ 6.54 10⁻⁴ 1.02 10⁻⁴ / 4²
F_{bc} = 3.75 10¹ N
F_{bc} = 37.5 N
let's use trigonometry to decompose this force
tan θ = y / x
θ = tan⁻¹ and x
θ= tan⁻¹ ¾
θ = 37º
let's break down the force
sin 37 = F_{bcy} / F_{bc}
F_{bcy} = F_{bc} sin 37
F_{bcy} = 37.5 sin 37
F_{bcy} = 22.57 N
cos 37 = F_{bcx} /F_{bc}
F_{bcx} = F_{bc} cos 37
F_{bcx} = 37.5 cos 37
F_{bcx} = 29.95 N
let's do the sum to find the net force
X axis
Fₓ = 29.95 N
Axis y
Fy = 28.2 -22.57
Fy = 5.63 N
we can give the result in two ways
a) F_net = Fₓ i ^ + [tex]F_{y}[/tex] j ^
F_net = 29.95 i ^ + 5.63 j ^
b) in the form of module and angle
let's use the Pythagorean theorem
F_net = [tex]\sqrt{ F_{x}^2 + F_{y}^2 }[/tex]
F_net = √(29.95² + 5.63²)
F_net = 30.47 N
we use trigonometry for the direction
tan θ= [tex]\frac{ F_{y} }{ F_{x} }[/tex]
θ = tan⁻¹ \frac{ F_{y} }{ F_{x} }
θ = tan⁻¹ (5.63 / 29.95)
θ = 10.6º
A 35 kg box initially sliding at 10 m/s on a rough surface is brought to rest by 25 N
of friction. What distance does the box slide?
Answer:
the distance moved by the box is 70.03 m.
Explanation:
Given;
mass of the box, m = 35 kg
initial velocity of the box, u = 10 m/s
frictional force, F = 25 N
Apply Newton's second law of motion to determine the deceleration of the box;
-F = ma
a = -F / m
a = (-25 ) / 35
a = -0.714 m/s²
The distance moved by the box is calculated as follows;
v² = u² + 2ad
where;
v is the final velocity of the box when it comes to rest = 0
0 = 10² + (2 x - 0.714)d
0 = 100 - 1.428d
1.428d = 100
d = 100 / 1.428
d = 70.03 m
Therefore, the distance moved by the box is 70.03 m.
A neutral metal bob is hanging on the bottom of a pendulum that is 15 cm long. A charged balloon is held near the metal bob and the pendulum is pulled up to a vertical angle of 20-deg. If the mass of the metal bob is 0.025kg, what is the charge on the balloon.
Answer:
Explanation:
See the figure attached
F is electrostatic force .
T cos20 = mg
T sin20 = F
Tan20 = F / mg
F = mg tan 20 = .025 x 9.8 tan20
= .09 N
Distance between bob and balloon
= 15 sin20 = 5.1 cm = .051 m
If q be the charge on balloon
F = 9 x 10⁹ x q² / .051²
= 3460 x 10⁹ q² = .09
q² = 26 x 10⁻⁶ x 10⁻⁹
q = 16.12 x 10⁻⁸ C .
The 49-g arrow is launched so that it hits and embeds in a 1.45 kg block. The block hangs from strings. After the arrow joins the block, they swing up so that they are 0.44 m higher than the block's starting point.
Required:
How fast was the arrow moving before it joined the block?
Answer:
the initial speed of the arrow before joining the block is 89.85 m/s
Explanation:
Given;
mass of the arrow, m₁ = 49 g = 0.049 kg
mass of block, m₂ = 1.45 kg
height reached by the arrow and the block, h = 0.44 m
The gravitational potential energy of the block and arrow system;
P.E = mgh
P.E = (1.45 + 0.049) x 9.8 x 0.44
P.E = 6.464 J
The final velocity of the system after collision is calculated as;
K.E = ¹/₂mv²
6.464 = ¹/₂(1.45 + 0.049)v²
6.464 = 0.7495v²
v² = 6.464 / 0.7495
v² = 8.6244
v = √8.6244
v = 2.937 m/s
Apply principle of conservation of linear momentum to determine the initial speed of the arrow;
[tex]P_{initial} = P_{final}\\\\mv_{arrow} + mv_{block} = (m_1 + m_2)V\\\\0.049(v) + 1.45(0) = (0.049 + 1.45)2.937\\\\0.049v = 4.4026\\\\v = \frac{4.4026}{0.049} \\\\v = 89.85 \ m/s[/tex]
Therefore, the initial speed of the arrow before joining the block is 89.85 m/s
The arrow moving as the speed of "76.36 m/s".
According to the question,
By using the conservation of energy, we have
→ [tex]K.E=P.E[/tex]
→ [tex]\frac{1}{2} (m_1+m_2)v_2^2= (m_1+m_2)gh[/tex]
or,
→ [tex]v_2 = \sqrt{2mgh}[/tex]
By substituting the values, we have
→ [tex]= \sqrt{2\times 9.8\times 0.44}[/tex]
→ [tex]=2.469 \ m/s[/tex]
Now,
By using the conservation of momentum, we get
→ [tex]m_1 v_1 = (m_1+m_2) v_2[/tex]
or,
→ [tex]v_1 = \frac{(m_1+m_2)v_2}{m_1}[/tex]
[tex]= \frac{1.45+0.049}{0.049}\times 2.469[/tex]
[tex]= 30.6\times 2.496[/tex]
[tex]= 76.36 \ m/s[/tex]
Thus the above approach is correct.
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A rocket blasts off and moves straight upward from the launch pad with constant acceleration. After 2.7 s the rocket is at a height of 93 m.
What are the magnitude and direction of the rocket's acceleration?
What is its speed at this elevation?
Answer:
The magnitude and direction of the rocket acceleration is 68.89 m/s² upward.
The speed of the rocket at the given elevation is 186 m/s.
Explanation:
Given;
time to reach the given height, t = 2.7 s
height reached, h = 93 m
initial velocity of the rocket, u = 0
The magnitude and direction of the rocket acceleration is calculated as;
h = ut + ¹/₂at²
h = 0 + ¹/₂at²
h = ¹/₂at²
a = 2h / t²
a = (2 x 93) / 2.7
a = 68.89 m/s²
the direction of the acceleration is upward.
The speed at this elevation, V = u + at
V = at
V = 68.89 x 2.7
V = 186 m/s
If the nearest object in front of the detector is too far away, the echo will not get back before a second click is emitted. Once that happens, the computer has no way of knowing that the echo isn't an echo from the second click and the detector doesn't give correct results anymore. Once the speaker emits a click, how much time does the echo have to return to the microphone before the next click is emitted
Answer:
t = 2x / v ( time echo), t = 2.9 10⁻² s
Explanation:
In this case we can use the uniform motion relationships, since the sound wave has a constant speed. Let's start by calculating the time it takes for the click to reach the detector.
v = d / t
t = d / v
where d is the distance from the speaker to the detector and v the speed of sound (v = 340 m / s)
Now let's analyze the echo, it is produced by a reflection of the sound from a large obstacle in the direction of the sound, therefore if the distance to the obstacle is x, the echo travels a distance of 2x in this time (to)
2x = v to
2x = v (d / v)
d = 2x
if we substitute in the first equation
t = 2x / v ( time echo)
Let's analyze these results, if the distance relationship is fulfilled, the detector (microphone) is not able to distinguish between a click and the echo of the previous click
For a numerical result suppose that the distance from the loudspeaker to the detector is d = 10 m, we obtain that the obstacle must be at a distance from the loudspeaker of
x = 5 m
t = 2 5/ 340
t = 2.9 10⁻² s
This is the time the echo has to return in this speaker-microphone configuration
which of the following is used to answer scientific questions?
A. Experiments
B. Intuition
C. Opinion polls
D. Imagination
A box with a mass of 2 kg is pushed by a 10 N force. The acceleration
is
_m/s^2?
Answer:
a = 5 m/s^2
Explanation:
First, we look at Newton's 2nd Law:
F = ma
We now plug in the values,
10 N = 2 kg * a
10 N/2 kg = a
5 m/s^2 = a
This table shows the mass and volume of four different objects.
A two-column table with 4 rows. The first column titled objects has entries W, X, Y, Z. The second column titled Measurements has entries Mass: 16 grams Volume: 84 centimeters cubed in the first cell, Mass: 12 grams Volume: 5 centimeters cubed in the second cell, Mass: 4 grams Volume: 6 centimeters cubed in the third cell, Mass: 408 grams Volume: 216 centimeters cubed in the fourth cell.
Which ranks the objects from most to least dense?
Answer:
Here its right but its also better than Barney's response
Explanation:
W, Y, Z, X or C
Answer:
W, Y, Z, X
Explanation:
12
Select the correct answer.
What creates an electric force field that moves electrons through a circuit?
ОА.
energy source
B.
load
O c.
metal wires
OD.
resistance
Answer:
A
Explanation:
For a flourish at the end of her act, a juggler tosses a single ball high in the air. She catches the ball 3.2 s later at the same height from which it was thrown. What was the initial upward speed of the ball?
Answer:
15.68 m/s
Explanation:
Given that,
She catches the ball 3.2 s later at the same height from which it was thrown.
When it reaches the maximum height, its height is equal to 0.
It will move under the action of gravity.
[tex]t=\dfrac{2u}{g}[/tex]
2 here comes for the time of ascent and descent.
So,
[tex]u=\dfrac{tg}{2}\\\\u=\dfrac{3.2\times 9.8}{2}\\\\u=15.68\ m/s[/tex]
So, the initial upward speed of the ball is 15.68 m/s.
Give an example of mass making a difference in the amount of kinetic energy. Tell how you know the kinetic energy amount is different in your example
Please help due today!!
Answer:
An example would be
Explanation:
You have a ball with a mass of 10 kg swinging from a rope arond in a cirlce if we were to change the mass of the ball to 20 kg the kinetic energy would increase because we know the ball has more mass and more mass means ner force increases which is connected to kinetic energy. hope this answer helps!
Calculate the wavelength of the electromagnetic radiation required to excite an electron from the ground state to the level with in a one-dimensional box 34.0 pm in length.
The question is incomplete. The complete question is :
Calculate the wavelength of the electromagnetic radiation required to excite an electron from the ground state to the level with n = 6 in a one-dimensional box 34.0 pm in length.
Solution :
In an one dimensional box, energy of a particle is given by :
[tex]$E=\frac{n^2h^2}{8ma^2}$[/tex]
Here, h = Planck's constant
n = level of energy
= 6
m = mass of particle
a = box length
For n = 6, the energy associated is :
[tex]$\Delta E = E_6 - E_1 $[/tex]
[tex]$\Delta E = \left( \frac{n_6^2h_2}{8ma^2}\right) - \left( \frac{n_1^2h_2}{8ma^2}\right) $[/tex]
[tex]$=\frac{h^2(n_6^2 - n_1^2)}{8ma^2}$[/tex]
We know that,
[tex]$E = \frac{hc}{\lambda} $[/tex]
Here, λ = wavelength
h = Plank's constant
c = velocity of light
So the wavelength,
[tex]$= \frac{hc}{E}$[/tex]
[tex]$=\frac{hc}{\frac{h^2(n_6^2 - n_1^2)}{8ma^2}}$[/tex]
[tex]$=\frac{8ma^2c}{h(n_6^2 - n_1^2)}$[/tex]
[tex]$=\frac{8 \times 9.109 \times 10^{-31}(0.34 \times 10^{-10})^2 (3 \times 10^8)}{6.626 \times 10^{-34} \times (36-1)}$[/tex]
[tex]$= \frac{ 8 \times 9.109 \times 0.34 \times 0.34 \times 3 \times 10^{-43}}{6.626 \times 35 \times 10^{-34}}$[/tex]
[tex]$=\frac{25.27 \times 10^{-43}}{231.91 \times 10^{-34}}$[/tex]
[tex]$= 0.108 \times 10^{-9}$[/tex] m
= 108 pm
Which of the following is a mixture?
a air
biron
Chydrogen
d nickel
Many organisms on Earth exhibit similar ____________.
Question 3 options:
time
characteristics
nonliving
single-celled
k12 hurry and answer
Answer:
The correct answer is - Characteristics.
Explanation:
On Earth, there are many organisms that shared similar characteristics with other organisms in various ways. These similarities of the characteristics could result from similar habitat, common ancestor, similar function, genetics, and many other reasons.
The example of such shared characteristics are different kinds of birds that have wings and lay eggs, while mammals give birth to babies and many other traits and characteristics. On the basis of the traits and characteristics organisms shared they are grouped and classified.
HELP PLEASE!!!
Running at 3.0 m/s, Burce, the 50.0 kg quarterback, collides with Max, the 100.0 kg tackle, who is traveling at 6.0 m/s in the other direction. Upon collision, Max continues to travel forward at 2.0 m/s.If the collision between the players lasted for 0.04 s, determine the impact force on either during the collision
Answer:
10kN
Explanation:
Given data
m1= 50kg
u1= 3m/s
m2= 100kg
u2= 6m/s
v1= 2m/s
time= 0.04s
let us find the final velocity of Bruce v1
from the conservation of linear momentum
m1u1+m2u2=m1v1+m2v2
substitute
50*3+100*6= 50*v1+100*2
150+600=50v1+200
750-200=50v1
550= 50v1
divide both sides by 50
v1= 550/50
v1=11 m/s
From
F= mΔv/t
for Bruce
F=50*(11-3)/0.04
F=50*8/0.04
F=400/0.04
F=10000
F=10kN
for Max
F=100*(6-2)/0.04
F=100*4/0.04
F=400/0.04
F=10000
F=10kN