Answer:
the second time there is a gas between you and the star,
Explanation:
When you observe the star for the first time you do not have a given between you and the star, therefore you observe the emission spectrum of the same that is formed by lines of different intensity and position that indicate the type and percentage of the atoms that make up the star.
When you observe the same phenomenon for the second time there is a gas between you and the star, this gas absorbs the wavelengths of the star that has the same energies and the atomisms and molecular gas, therefore these lines are not observed by seeing a series of dark bands,
The information obtained from the two spectra is the same, the type of atoms that make up the star
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
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
Which of the following is a mixture?
a air
biron
Chydrogen
d nickel
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:
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!
Which of the physical variables listed below will change when you change the area of the capacitor plates (while keeping the battery connected).
a. Capacitance
b. Charge on the plates
c. Voltage across the plates
d. Net electric field between the plates
e. Energy stored in the capacitor
Answer:
a. Capacitance
b. Charge on the plates
e. Energy stored in the capacitor
Explanation:
Let A be the area of the capacitor plate
The capacitance of a capacitor is given as;
[tex]C = \frac{Q}{V} = \frac{\epsilon _0 A}{d} \\\\[/tex]
where;
V is the potential difference between the plates
The charge on the plates is given as;
[tex]Q = \frac{V\epsilon _0 A}{d}[/tex]
The energy stored in the capacitor is given as;
[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} (\frac{\epsilon _0 A}{d} )V^2[/tex]
Thus, the physical variables listed that will change include;
a. Capacitance
b. Charge on the plates
e. Energy stored in the capacitor
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.
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 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.
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!
When a moving object collides with an object that isn't moving, what happens to the kinetic energy of each object?
All the objects are motionless, so kinetic energy of each object is zero after the collision.
What is Kinetic Energy?The kinetic energy of an object is defined as the energy which is possesses due to its motion. It is the work required to accelerate a body of a given mass from rest to its stated velocity. This energy is gained during its acceleration, the body maintains the kinetic energy as long as its momentum does not change.
Kinetic Energy can be expressed as
[tex]K.E.=[/tex] [tex]1/2 mv^2[/tex]
Where, m is the mass of the object
v is the velocity.
It is expressed in joules (J).
After the collision all the objects are at rest, therefore, the final kinetic energy is also zero which shows maximum loss of kinetic energy. Such collisions are called perfectly inelastic.
Thus, all the objects are motionless, so kinetic energy of each object is zero after the collision.
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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
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:
A physics student spends part of her day walking between classes or for recreation, during which time she expends energy at an average rate of 280 W. The remainder of the day she is sitting in class, studying or resting; during these activities, she expends energy at a rate of 100 W. If she expends a total of 1.1 x 10^7 J of energy in a 24 hour day, how much of the day did she spend walking
The time of the day she spent walking is equal to 3.70 hrs.
What is power?Power can be explained as the rate of doing work in unit time. The SI unit of measurement of power is J/s or Watt (W). Power can be described as a time based quantity. The mathematical expression for power can be represented as mentioned below.
Power = work/time
P = W/t
Given, the energy spends part of her day walking, Ew = 280 W
The energy is spent by sitting in the class, Es = 100 W
The total energy spends, Et = 1.1 × 10⁷J
[tex]E_w \times t + E_s(24\times 60\times 60-t)= 1.1 \times 10^7J[/tex]
[tex]280 \times t + 100(24\times 60\times 60-t)= 1.1 \times 10^7[/tex]
280 t + 0.86 × 10⁷ - 100 t = 1.1 × 10⁷
180 t = 0.24 × 10⁷
t = 0.24 × 10⁷/180 × 3600
t = 3.70 hr
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The moons phases are caused by
A. Eclipse of sun
B. Planets moving across the face of the moon
C. The alignment of the Earth, moon, and sun
D. The alignment of the planets.
Please help me!!!!
which of the following is used to answer scientific questions?
A. Experiments
B. Intuition
C. Opinion polls
D. Imagination
How heavier elements formed during stellar nucleosynthesis and evolution?
Answer:
i honestly think its 21
Explanation:
da memes
10 + 10 =21
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.
PLEASE HELP PLEASEEEE
Answer:
How can I help you??? Plz insert some questions
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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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!
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 fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.30 s for the boat to travel from its highest point to its lowest, a total distance of 0.660 m . The fisherman sees that the wave crests are spaced a horizontal distance of 5.50 m apart. How fast are the waves traveling
Answer:
v = 1.2 m/s
Explanation:
The wavelength of the waves is given as the horizontal distance between the crests:
λ = wavelength = 5.5 m
Now, the time period is given as the time taken by boat to move from the highest point again to the highest point. So it will be equal to twice the time taken by the boat to travel from highest to the lowest point:
T = Time Period = 2(2.3 s) = 4.6 s
Now, the speed of the wave is given as:
[tex]v = f\lambda[/tex]
where,
v= speed of wave = ?
f = frequency of wave = [tex]\frac{1}{T} = \frac{1}{4.6\ s} = 0.217\ Hz[/tex]
Therefore,
[tex]v = (0.217\ Hz)(5.5\ m)\\[/tex]
v = 1.2 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
a 250.0 g snowball of radius 4.00 cm starts from rest at the top of the peak of a roof and rolls down a section angled at 30.0 degrees
Answer:
The response to this question is as follows:
Explanation:
The whole question and answer can be identified in the file attached, please find it.
The force diagram of all the forces acting on the snowball include the normal force acting upwards, the weight of the snowball acting downwards and the frictional force acting horizontal.
The given parameters;
mass of the snow ball, m = 250 gradius of the snow ball, r = 4 cmangle of inclination of the plane, θ = 30 ⁰The force diagram of all the forces acting on the snowball is calculated as follows;
↑ N
⊕ → F
↓ W
Where;
N is the normal force on the snowballF is the frictional force on the snowballW is the weight of the ballThus, the force diagram of all the forces acting on the snowball include the normal force acting upwards, the weight of the snowball acting downwards and the frictional force acting horizontal.
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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
A stone is dropped from the top of a high cliff with zero initial velocity. In which system is the net momentum zero as the stone falls freely
Answer:
A system that includes the stone and the earth.
Explanation:
If the system of being dropped from the height of the cliff consists of just the stone alone, then it means that its momentum will certainly undergo changes as it falls freely. However, If the system is now expanded to include not only the stone but also the Earth, then it implies that the momentum of the stone which is in the downward direction will be equal and opposite to the momentum of the Earth in the upwards direction towards the stone. Therefore, the momentum will cancel out and net momentum will be zero.
A system of stone and earth can result to a net zero momentum.
Conservation of linear momentum
The principle of conservation of linear momentum states that the sum of the initial momentum is equal to the sum of final momentum.
[tex]m_1u_1 + m_2 u_2 = m_1v_1 + m_2 v_2[/tex]
A system that consists a linear system of stone and earth can result to a net zero momentum.
Thus, a system of stone and earth can result to a net zero momentum.
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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
two identical balls are rolling down a hill ball 2 is rolling faster than ball 1 which ball has more kinetic energy
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]