Answer:
See attached file pls
Explanation:
The strengths of the fields in the velocity selector of a Bainbridge mass spectrometer are B=0.500 T and E=1.2x105 V/m. The strength of the magnetic field that seperates the ions is Bo=0.750 T. A stream of single charged Li ions is found to bend in a circular arc of radius 2.32 cm. What is the mass of the Li ions?
Answer:
[tex]m = 1.16 \times 10^{-26}\ Kg[/tex]
Explanation:
Given,
Magnetic field, B = 0.5 T
Electric field, E = 1.2 x 10⁵ V/m
strength of the magnetic field that separates the ions, Bo=0.750 T
Radius, r = 2.32 cm
Relation of charge to mass ratio is given by
[tex]\dfrac{q}{m}=\dfrac{E}{BB_0R}[/tex]
[tex]m=\dfrac{qBB_0R}{E}[/tex]
Substituting all the values
[tex]m=\dfrac{1.6\times 10^{-19}\times 0.5\times 0.75\times 02.0232}{1.2\times 10^5}[/tex]
[tex]m = 1.16 \times 10^{-26}\ Kg[/tex]
Mass of Li ions is equal to [tex]m = 1.16 \times 10^{-26}\ Kg[/tex]
The mass of the Li ions is approximately [tex]\( 1.64 \times 10^{-26} \)[/tex] kg.
To find the mass of the Li ions, we can use the formula for the radius of the circular path of a charged particle moving in a magnetic field:
[tex]\[ r = \frac{m \cdot v}{q \cdot B} \][/tex]
Where:
- r is the radius of the circular path
- m is the mass of the ion
- v is the velocity of the ion
- q is the charge of the ion
- B is the magnetic field strength
We can also use the formula for the force experienced by a charged particle moving in both electric and magnetic fields:
[tex]\[ F = q \cdot (E + v \times B) \][/tex]
Since the particle moves in a circular path, the electric force F must be equal to the magnetic force [tex]\( q \cdot v \cdot B \)[/tex], where v is the speed of the particle.
We can rearrange the formulas to solve for m:
[tex]\[ m = \frac{q \cdot r \cdot B}{v} \][/tex]
Now, we need to find \( v \). Since the particle passes through both the electric and magnetic fields without deviation, the forces acting on it must be balanced. Therefore:
[tex]\[ F_{electric} = F_{magnetic} \]\[ q \cdot E = q \cdot v \cdot B \][/tex]
Solving for v:
[tex]\[ v = \frac{E}{B} \][/tex]
Now, substituting v into the equation for m :
[tex]\[ m = \frac{q \cdot r \cdot B}{\frac{E}{B}} \]\[ m = \frac{q \cdot r \cdot B^2}{E} \][/tex]
Now we can calculate \( m \) using the given values:
[tex]\[ m = \frac{(1.6 \times 10^{-19} C) \times (0.0232 m) \times (0.75 T)^2}{1.2 \times 10^5 V/m} \]\[ m= 1.64 \times 10^{-26} kg \][/tex]
Block B is attached to a massless string of length L = 1 m and is free to rotate as a pendulum. The speed of block A after the collision is half its speed before the collision. Block B was at rest before the collision. The mass of block A is 7 kg and the mass of block B is 2 kg. What is the minimum initial speed (in m/s) that block A must have for block B to swing through a complete vertical circle?
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The minimum velocity of A is [tex]v_A= 4m/s[/tex]
Explanation:
From the question we are told that
The length of the string is [tex]L = 1m[/tex]
The initial speed of block A is [tex]u_A[/tex]
The final speed of block A is [tex]v_A = \frac{1}{2}u_A[/tex]
The initial speed of block B is [tex]u_B = 0[/tex]
The mass of block A is [tex]m_A = 7kg[/tex] gh
The mass of block B is [tex]m_B = 2 kg[/tex]
According to the principle of conservation of momentum
[tex]m_A u_A + m_B u_B = m_Bv_B + m_A \frac{u_A}{2}[/tex]
Since block B at initial is at rest
[tex]m_A u_A = m_Bv_B + m_A \frac{u_A}{2}[/tex]
[tex]m_A u_A - m_A \frac{u_A}{2} = m_Bv_B[/tex]
[tex]m_A \frac{u_A}{2} = m_Bv_B[/tex]
making [tex]v_B[/tex] the subject of the formula
[tex]v_B =m_A \frac{u_A}{2 m_B}[/tex]
Substituting values
[tex]v_B =\frac{7 u_A}{4}[/tex]
This [tex]v__B[/tex] is the velocity at bottom of the vertical circle just at the collision with mass A
Assuming that block B is swing through the vertical circle(shown on the second uploaded image ) with an angular velocity of [tex]v__B'[/tex] at the top of the vertical circle
The angular centripetal acceleration would be mathematically represented
[tex]a= \frac{v^2_{B}'}{L}[/tex]
Note that this acceleration would be toward the center of the circle
Now the forces acting at the top of the circle can be represented mathematically as
[tex]T + mg = m \frac{v^2_{B}'}{L}[/tex]
Where T is the tension on the string
According to the law of energy conservation
The energy at bottom of the vertical circle = The energy at the top of
the vertical circle
This can be mathematically represented as
[tex]\frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L[/tex]
From above
[tex](T + mg) L = m v^2_{B}'[/tex]
Substitute this into above equation
[tex]\frac{1}{2} m(\frac{7 v_A}{4} )^2 = \frac{1}{2} (T + mg) L + mg 2L[/tex]
[tex]\frac{49 mv_A^2}{16} = \frac{1}{2} (T + mg) L + mg 2L[/tex]
[tex]\frac{49 mv_A^2}{16} = T + 5mgL[/tex]
The value of velocity of block A needed to cause B be to swing through a complete vertical circle is would be minimum when tension on the string due to the weight of B is zero
This is mathematically represented as
[tex]\frac{49 mv_A^2}{16} = 5mgL[/tex]
making [tex]v_A[/tex] the subject
[tex]v_A = \sqrt{\frac{80mgL}{49m} }[/tex]
substituting values
[tex]v_A = \sqrt{\frac{80* 9.8 *1}{49} }[/tex]
[tex]v_A= 4m/s[/tex]
Young's experiment is performed with light of wavelength 502 nm from excited helium atoms. Fringes are measured carefully on a screen 1.20 m away from the double slit, and the center of the 20th fringe (not counting the central bright fringe) is found to be 10.2 mm from the center of the central bright fringe. What is the separation of the two slits?
Answer:
Explanation:
wave length of light λ = 502 nm
screen distance D = 1.2 m
width of one fringe = 10.2 mm / 20
= .51 mm
fringe width = λ D / a , a is separation of slits
Puting the values given
.51 x 10⁻³ = 502 x 10⁻⁹ x 1.2 / a
a = 502 x 10⁻⁹ x 1.2 / .51 x 10⁻³
= 1181.17 x 10⁻⁶ m
1.18 x 10⁻³ m
= 1.18 mm .
Having straight ears (E) is dominant over droopy ears
(e). A parent with straight ears is crossed with a parent
with droopy ears. The Punnett square will show the
possible genotypes of the offspring.
What is the genotype of each square?
A:
B:
C:
D:
that is... A. B
................c. D
The Punnett square for the cross between a parent with straight ears and a parent with droopy ears would result in offspring with the genotypes EE, EE, Ee, and Ee.
Explanation:The Punnett square is a tool used in genetics to predict the possible genotypes of offspring based on the genotypes of the parents. In this case, the Punnett square for the cross between a parent with straight ears (genotype EE) and a parent with droopy ears (genotype ee) would look like:
EEeEEEEeEeEe
The genotypes of each square are as follows:
A: EE
B: EE
C: Ee
D: Ee
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An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the earth with a speed of 0.99543c relative to the earth. A scientist at rest on the earth's surface measures that the particle is created at an altitude of 43.0 km .
(a) As measured by the scientist, how much time does it take the particle to travel the 45.0 km to the surface of the earth?
(b) Use the length-contraction formula to calculate the distance from where the particle is created to the surface of the earth as measured in the particle’s frame.
(c) In the particle’s frame, how much time does it take the particle to travel from where it is created to the surface of the earth? Calculate this time both by the time dilation formula and from the distance calculated in part (b). Do the two results agree?
Answer:
Check attachment
The question have two distance
I decided to use the one in the question "a" in attachment and I will use the other one here
Explanation:
Given that,
Speed of particle relative to the earth is
V = 0.99543c
Where c is speed of light
c = 3 × 10^8 m/s
Particle height as detected by scientist is 43km
The initial length is 43km
Lo = 43km
Lo = 43,000m
A. Time taken for the particle to reach the earth surface?
Speed = distance / Time
Time = distance / speed
t = L / V
t = 43,000 / 0.99543c
t = 43,000 / (0.99543 × 3 × 10^8)
t = 1.4399 × 10^-4 seconds
b. Initial Lenght is given as Lo = 43km
Using length contraction formula
L = Lo√(1 — u² / c²)
L = 43√[1 — (0.99543c)² / c²]
L = 43√[1 — 0.990881c² / c²]
L = 43√[1 — 0.990881]
L = 43 × √(9.1191 × 10^-3)
L = 43 × 0.095494
L = 4.1062 km
c. Using time dilation formula
∆to = ∆t√(1 — u² / c²)
∆t is gotten from question a
∆t = 1.4399 × 10^-4 seconds
∆to = ∆t√[1 — (0.99543c)² / c²]
∆to = ∆t√[1 — 0.990881c² / c²]
∆to = ∆t√[1 — 0.990881]
∆to = ∆t × √(9.1191 × 10^-3)
∆to = 1.4399 × 10^-4 × 0.095494
∆to = 1.375 × 10^-5 seconds
To check if the time dilation agree
t = L / V
t = 4.1062 × 1000 / 0.99543c
t = 4.1062 × 1000 / 0.99543 × 3 × 10^8
t = 1.375 × 10^-5 seconds
The time dilation agreed
Suppose your telescope has perfect mirrors and is limited only by diffraction. If you are viewing with light of 400 nm wavelength, what diameter mirror in a space telescope would have 1.0-km resolution of objects on Mars when viewed from above the earth at a time when the Earth-Mars distance of 80 million kilometers
Answer:
39 m
Explanation:
We are given that
Wavelength=[tex]\lambda=400 nm=400\times 10^{-9} m[/tex]
1 nm=[tex]10^{-9} m[/tex]
y=1 km=[tex]1000 m[/tex]
1 km=1000 m
Earth mars distance =x=80 million Km=[tex]80\times 10^9 m[/tex]
1million km=[tex]10^9 m[/tex]
[tex]sin\theta=\frac{1.22\lambda}{d}[/tex]
[tex]sin\theta=\frac{y}{x}[/tex]
[tex]\frac{y}{x}=\frac{1.22\lambda}{d}[/tex]
[tex]\frac{1000}{80\times 10^9}=\frac{1.22\times 400\times 10^{-9}}{d}[/tex]
[tex]d=\frac{1.22\times 400\times 10^{-9}\times 80\times 10^9}{1000}[/tex]
[tex]d=39.04 m\approx 39 m[/tex]
Diameter of mirror =39 m
To achieve a 1.0-km resolution on Mars with light of a 400 nm wavelength, the size of the space telescope's mirror would need to be approximately 6 mm. This calculation is based on diffraction limit theory and may not be precisely accurate in practical application due to other potential limiting factors.
Explanation:This question is related to the resolution of a telescope which is primarily influenced by the aperture size and wavelength of light being observed. The bigger the telescope aperture (the diameter of the primary mirror), the greater the resolution as more light can be collected. The formula for the resolution limit due to diffraction (the smallest distinguishable detail) can be given by R = 1.22λ/D where λ is the wavelength of light and D is the diameter of the telescope's main mirror.
Here the wavelength given is 400 nm or 400 x 10^-9 m, and we need to calculate the mirror diameter required for a resolution of 1.0-km on Mars from Earth. Let's convert the resolution limit from kilometers to meters (which gives us 1000m), and then to radians (using the Earth-Mars distance), which results in an angle of 1.25x10^-11 radians approximately.
Substituting these values in the formula, we can solve for D and find that the telescope mirror size needed would be approximately 0.006 m or 6 mm for this resolution. Do note, that this is a theoretical value, in reality, the size might need to be larger due to factors like the diffraction limit, non-uniformities in mirrors, or aberrations in lenses.
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You are using a hand-held sound level meter to measure the intensity level of the roars produced by a lion prowling in the high grass. To decrease the measured sound intensity level by 36.2 dB requires the lion move away from you until its distance from you has increased by what factor? g
Answer:
64.57
Explanation:
We are given that
For decreasing measured sound intensity level=36.2 dB
We have to find the factor by which the distance increase.
Let initial distance=x
Final distance=x'
According to question
[tex]36.2=10log(\frac{x'^2}{x^2})[/tex]
[tex]36.2=10log(\frac{x'}{x})^2[/tex]
[tex]36.2=10\times 2log\frac{x'}{x}[/tex]
[tex]log\frac{x'}{x}=\frac{36.2}{20}=1.81[/tex]
[tex]\frac{x'}{x}=10^{1.81}[/tex]
[tex]x'=10^{1.81}x=64.57x[/tex]
Hence, the distance is increases by factor of 64.57
Imagine that an incident light wave encounters a thin film. Some of that light wave will be reflected off of the top surface of the thin film (call this Wave 1), while the rest will be transmitted through the film. Some of this transmitted light (Wave 2) will reflect off the back surface of the film and move back through the thin film and out into the original medium of the incident light wave. If the thin film has a thickness t, then what is the path length difference between Wave 1 and Wave 2?
a) t/4
b) t/2
c) t
d) 2t
Answer
D, 2t
Explanation:
See attached file
"2t" will be the path length difference between Wave 1 and Wave 2. A complete solution is below.
According to the question,
The path difference between Wave 1 and 2 will be:
When [tex]\Theta = 0^{\circ}[/tex],
→ [tex]P.d= \Delta x[/tex]
[tex]= AB+BC[/tex]
hence,
Path difference,
= [tex]t+t[/tex]
= [tex]2t[/tex]
Thus the above answer i.e., "option d" is appropriate.
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Suppose that a pendulum clock keeps time exactly correctly on Earth. Now suppose that you take this clock to planet X and find that the clock’s hour hand makes α revolutions every hour. Find a formula for the magnitude of the acceleration due to gravity near the surface of planet X. What would α be on the moon?
Answer:
The gravity at planet X is [tex]g= 3.05*10^{-6} \alpha^2 m/s^2[/tex]
The value of [tex]\alpha[/tex] on the moon is [tex]\alpha = 730.38 \ revolutions[/tex]
Explanation:
From the question we are told that
The clocks hour hand makes [tex]\alpha[/tex] revolution every 1 hour which is 3600 sec
this implies that the time peroid for 1 revolution would be [tex]= \frac{3600}{\alpha }sec[/tex]
The peroid for a pendulum is mathematically represented as
[tex]T = 2 \pi\sqrt{\frac{L}{g} }[/tex]
Where L is the pendulum length
g is the acceleration due to gravity
Let assume that we have a pendulum that counts in second on earth
This implies that its peroid would be = 2 second
i.e one second to swimg forward and one second to swing back to its original position
Now the length of this pendulum on earth is
[tex]L = \frac{gT^2}{4 \pi^2}[/tex] [Making L the subject in above equation]
Substituting values
[tex]L = \frac{9.8 * (2)^2}{4 * (3.142)^2}[/tex]
[tex]= 1[/tex]
When the same pendulum is taken to planet X the peroid would be
[tex]T = \frac{3600}{\alpha }[/tex]
Recall this value was obtained above for 1 revolution (from start point to end point back to start point)
So the acceleration due to gravity on this planet would be mathematically represented as
[tex]g = \frac{4 \pi L }{T^2}[/tex] [making g the subject in the above equation]
substituting values
[tex]g = \frac{4 * 3.142^2 * 1}{[ \frac{3600}{\alpha } ]^2}[/tex]
[tex]g= 3.05*10^{-6} \alpha^2 m/s^2[/tex]
On moon the acceleration due to gravity has a constant value of
[tex]g = 1.625 m/s^2[/tex]
The period of this pendulum on the moon can be mathematically evaluated as
[tex]T = 2\pi \sqrt{\frac{L}{g} }[/tex]
substituting value
[tex]T = 2 *3.142 \sqrt{\frac{1}{1.625} }[/tex]
[tex]= 4.929s[/tex]
given that
[tex]1 \ revolution ----> 4.929s\\ \\ \alpha \ revolution -------> 3600 \ s[/tex] {Note 1 revolution takes a peroid }
Making [tex]\alpha[/tex] the subject of the formula
[tex]\alpha =\frac{3600}{4.929}[/tex]
[tex]\alpha = 730.38 \ revolutions[/tex]
If it is fixed at C and subjected to the horizontal 60-lblb force acting on the handle of the pipe wrench at its end, determine the principal stresses in the pipe at point A, which is located on the outer surface of the pipe. Express your answers, separated by a comma, to three significant figures.
The principal stress in the pipe at point A, which is located on the outer surface of the pipe is 132.773 psi
Finding the principal stress:It is given that l = 12 in
The force F = 60 lb
The radiuses are as follows:
outer radius, R = 3.90 in/2
R = 1.95 in
Inner radius, r = 3.65 in/2
r = 1.825 in
The angle between the force applied and the distanced from the axis is 30°
So we get the torque:
T = l×Fsin 30°
T = 12 × 60 × (0.5)
T = 360 lb-in
Now, the angle of rotation or angular displacement ωt is given as
ωt = π(R⁴ - r⁴)/(2R)
ωt = π((1.95 in)⁴ - (1.825 in)⁴)/(2×1.95 in)
ωt = 2.7114 in³
Then the principal stress in the pipe at point A is:
principal stress = T/ωt
principal stress = (360 lb-in)/(2.7114 in³)
principal stress = 132.773 lb/in²
principal stress = 132.773 psi
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A 0.0140-kg bullet is fired straight up at a falling wooden block that has a mass of 2.42 kg. The bullet has a speed of 555 m/s when it strikes the block. The block originally was dropped from rest from the top of a building and had been falling for a time t when the collision with the bullet occurs. As a result of the collision, the block (with the bullet in it) reverses direction, rises, and comes to a momentary halt at the top of the building. Find the time t.
Answer:
0.16 s
Explanation:
• Falling from rest (V_block= 0 m/s) the block attains a final velocity V_block before colliding with the bullet. This velocity is given by Equation 2.4 as
V_block(final velocity of block just before hitting) =V_0,block +at
where a is the acceleration due to gravity (a = —9.8 m/s2) and t is the time of fall. The upward direction is assumed to be positive. Therefore, the final velocity of the falling block is
V_block = at
• During the collision with the bullet, the total linear momentum of the bullet/block system is conserved, so we have that
(M_bullet+M_block)V_f = M_bullet*V_bullet+ M_block*V_block
Total linear momentum after collision = Total linear momentum before collision
Here V_f is the final velocity of the bullet/block system after the collision, and V_bullet and V_block are the initial velocities of the bullet and block just before the collision. We note that the bullet/block system reverses direction, rises, and comes to a momentary halt at the top of the building. This means that V_f, the final velocity of the bullet/block system after the collision must have the same magnitude as V_block, the velocity of the falling block just before the bullet hits it. Since the two velocities have opposite directions, it follows that of V_f =-V_block, Substituting this relation and Equation (1) into Equation (2) gives
(M_bullet + M_block)(-at) = M_bullet*V_bullet + M_block(at)
t = -M_bullet*V_bullet/a(M_bullet +2M_block)
=-(0.0140-kg)*555 m/s/-9.8(0.0140-kg+2(2.42 kg)
=0.16 s
Final answer:
The student's physics question involves applying the conservation of momentum and kinematic equations to determine the time the block was falling before being struck by a bullet. By considering the bullet's speed, the mass of both the bullet and block, and the motions involved, one can calculate the time of fall.
Explanation:
The student's question involves a physics problem related to conservation of momentum and projectile motion. The bullet's speed when it strikes the block, and the subsequent motion of the block and bullet system provide key information. To solve for the time t that the block was falling, we need to use the principles of physics that dictate how objects move under the influence of gravity and how they interact in collisions.
To find the time t, we'll take the following steps:
Use the conservation of momentum to find the velocity of the block and bullet immediately after the collision.
Use kinematic equations to relate this velocity to the maximum height reached by the block and bullet, taking into account the direction reversal.
Use the kinematic equations again to find the time t during which the block was falling before the collision.
Assuming the collision is perfectly inelastic, the bullet embeds itself in the block, so we have a combined mass moving upward after the collision. This combined mass will move up to the original height due to the conservation of energy principle, as the initial kinetic energy is converted entirely to gravitational potential energy at the peak of the ascent after which it momentarily comes to a halt.
A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at 1200C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 6 10-11 m2/s, and the diffusion flux is found to be 1.2 10-7 kg/m2 -s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4 kg/m3. How far into the sheet from this high-pressure side will the concentration be 2.0 kg/m3? Assume a linear concentration profile.
Answer:
The answer is 0.001 m
Explanation:
Solution
Recall that,
A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at = 1200°C
The diffusion coefficient for nitrogen in steel at this temperature is =6 * 10-11 m2/s
The diffusion flux is = 1.2 *10^-7 kg/m2s
Concentration of nitrogen in the steel at the high-pressure surface is= 4 kg/m3.
The high-pressure side will the concentration is estimated to be = 2.0 kg/m3
Now,
The flux = -D dC/dx
1.2 x 10-7 kg/m2s = - 6 x 10-11 m2/s dC/dx
∫ˣ₀ dx = -5x^10-4 ∫²₄ dC
so,
x = (2-4) kg/m3 (-5x10-4 m4/kg)
where x = .001 m
Therefore x = 0.001 m
The distance from this high-pressure side will the concentration is 0.001 m
Calculation of the distance:Since the diffusion coefficient for nitrogen in steel at this temperature is 6 x 10-11 m2/s, the diffusion flux is found to be 1.2x 10-7 kg/m2 -s.
We know that
The flux =[tex]-D\ dC\div dx[/tex]
So,
1.2 x 10-7 kg/m2s = - 6 x 10-11 m2/s dC/dx
dC/dx = -2000
Now
-2000 = 4 - 2/0-x_B
x_B = 2/2000
x_B = 1*10^-3m
x_B = 0.001m
Hence, The distance from this high-pressure side will the concentration is 0.001 m
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A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 Joules (with the potential energy equal to zero at ground level) and is moving upward with a kinetic energy of 50 Joules. What is the maximum height h reached by the ball? Consider air friction to be negligible.
A. h ≈ 30 m
B. h ≈ 40 m
C. h ≈ 50 m
D. h ≈ 10 m
E. h ≈ 20 m
Given that,
A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 Joules. It is moving upward with a kinetic energy of 50 Joules.
We need to find the maximum height h reached by the ball. Let at a height of 10 meters, it has a potential energy of 50 Joules. So,
[tex]mgH=50\\\\mg=\dfrac{50}{h}\\\\mg=\dfrac{50}{10}\\\\mg=5\ N[/tex] ........(1)
Let at a height of h m, it reaches to a maximum height. at this point, it has a total of 100 J of energy. So,
[tex]mgh=50+50\\\\mgh=100\\\\h=\dfrac{100}{5}\\\\h=20\ m[/tex]
So, the correct option is (E) "h = 20 m".
The maximum height reached by the ball is about 20 meters, as determined by using the conservation of mechanical energy principle, considering that the total mechanical energy at the height of 10 meters was 100 Joules.
Explanation:To solve for the maximum height reached by the ball, we can use the conservation of mechanical energy principle, which states that the total mechanical energy (potential energy + kinetic energy) of the ball remains constant in the absence of air friction.
At 10 meters above the ground, the ball has a potential energy (PE) of 50 Joules and a kinetic energy (KE) of 50 Joules. Therefore, the total mechanical energy at that height is:
PE + KE = 50 J + 50 J = 100 J
As the ball rises, its kinetic energy is converted into potential energy until the kinetic energy becomes zero at the maximum height. The total mechanical energy at maximum height will be equal to the potential energy:
PE at maximum height = total mechanical energy = 100 J
Using the formula for gravitational potential energy, PE = mgh (where m is mass, g is the acceleration due to gravity (9.81 m/s²), and h is the height), and knowing that the PE at 10 meters is 50 J, we can find the mass of the ball:
50 J = m * 9.81 m/s² * 10 m
m = 50 J / (9.81 m/s² * 10 m) = 0.51 kg
With the mass of the ball, we can now calculate the maximum height using the total mechanical energy:
100 J = 0.51 kg * 9.81 m/s² * h
h = 100 J / (0.51 kg * 9.81 m/s²) ≈ 20 meters
Therefore, the maximum height h reached by the ball is about 20 meters.
The volume of a gas varies inversely as the pressure and directly as the temperature (in degrees Kelvin). If acertain gas occupies a volume of 2.2 liters at a temperature of 340 K and a pressure of 16 newtons per squarecentimeter, find the volume when the temperature is 408 K and the pressure is 24 newtons per squarecentimeter.
Final answer:
The volume of the gas at 408 K and 24 N/cm² will be 1.76 liters, calculated using the combined gas law that integrates Boyle's law and Charles's law for the given changes in pressure and temperature.
Explanation:
The volume of a gas is dependent on both its pressure and temperature, following Boyle's law and Charles's law, respectively. The relationship is such that volume varies inversely with pressure and directly with temperature (in kelvins) when the amount of gas is constant. To find the new volume when both the pressure and temperature change, we use the combined gas law, which integrates both Boyle's and Charles's laws:
V₁/T₁ * P₁ = V₂/T₂ * P₂
Given that at P₁ = 16 N/cm², T₁ = 340 K, and V₁= 2.2 liters, we can find V₂ when P₂ = 24 N/cm² and T₂ = 408 K using the formula:
V₂ = (V₁ * T₂/T₁) * (P₁/P₂)
Plugging in the values:
V₂ = (2.2 liters * 408 K / 340 K) * (16 N/cm² / 24 N/cm²)
V₂ = (2.2 liters * 1.2) * (2 / 3)
V₂ = 2.64 liters * (2 / 3)
V₂ = 1.76 liters
So, the volume of the gas when the temperature is 408 K and the pressure is 24 N/cm² would be 1.76 liters.
"One possible explanation for a galaxy's type invokes the angular momentum of the protogalactic cloud from which it formed. Suppose a galaxy forms from a protogalactic cloud with a lot of angular momentum. Assuming its type has not changed as a result of other interactions, we'd expect this galaxy to be ______."
Answer:
Spiral galaxy
Explanation:
If a galaxy forms from a protogalactic cloud with a lot of angular momentum, we would expect this galaxy to be a spiral galaxy. This is due to the conservation of angular momentum, where greater angular momentum usually results in a flatter, disk-like shape seen in spiral galaxies. Other factors also affect galaxy formation.
Explanation:In the subject of astrophysics, the angular momentum of the protogalactic cloud has significant implications for the type of galaxy formed. Specifically, if a galaxy forms from a protogalactic cloud with a high angular momentum, we'd expect this galaxy to be a spiral galaxy.
Angular momentum is a property of rotating bodies, determined by the mass, shape, and spin rate of the object. In the context of galaxies, a protogalactic cloud with higher angular momentum is more likely to flatten out into a spinning disk shape, leading to the formation of a spiral galaxy. This is due to the conservation of angular momentum. In contrast, lower angular momentum could lead to the formation of elliptical galaxies, which are less flat and more randomly arranged.
In the realm of galaxy formation and evolution, these are widely accepted ideas, though it's important to note that other factors, like the gas content of the protogalactic cloud and interactions with other galaxies, can also play a major role.
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During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 0.148 kg baseball crashing through the pane of a second-floor window in a nearby building. The ball strikes the glass at 14.5 m/s , shatters the glass as it passes through, and leaves the window at 11.5 m/s with no change of direction.
What is the direction of the impulse that the glass imparts to the baseball? Calculate the magnitude of this impulse (a positive number)?
Explanation:
Mass of baseball, m = 0.148 kg
Initial speed of the ball, u = 14.5 m/s
Final speed of the ball, v = 11.5 m/s
After crashing through the pane of a second-floor window, the ball shatters the glass as it passes through, and leaves the window at 11.5 m/s with no change of direction. So, the direction of the impulse that the glass imparts to the baseball is in opposite direction to the direction of the balls path.
The change in momentum of the ball is called impulse. It is given by :
[tex]J=m(v-u)\\\\J=0.148\times (11.5-14.5) \\\\J=-0.444\ kg-m/s\\\\|J|=0.444\ kg-m/s[/tex]
Hence, this is the required solution.
A proton is released from rest in a uniform electric field that has a magnitude of 8.0 x 104 V/m. The proton undergoes a displacement of 0.50 m in the direction of E Find the change in electric potential, the voltage, between point A and B. Find the change in potential energy of the proton-field system for this displacement. Hint: U
Answer:
Explanation:
The magnitude of electric field = 8 x 10⁴ V /m
there is a potential difference of 8 x 10⁴ V on a separation of 1 m
so on a separation of .5 m , potential drop or change in potential will be equal to
.5 x 8 x 10⁴
= 4 x 10⁴ V .
The increase in kinetic energy of proton = V X Q
V is potential drop x Q is charge on proton
= 4 x 10⁴ x 1.6 x 10⁻¹⁹
= 6.4 x 10⁻¹⁵ J
potential energy of the proton-field system will be correspondingly decreased by the same amount or by an amount of
- 6.4 x 10⁻¹⁵ J .
The change in electric potential is 4.0 x 10^4 volts. The change in potential energy of the proton-field system, given the charge of a proton, is calculated to be 6.4 x 10^-15 joules.
Explanation:The change in electric potential (or voltage) is calculated by multiplying the electric field strength by the displacement, so in this case it is (8.0 x 10^4 V/m) * (0.50 m) = 4.0 x 10^4 volts.
The change in potential energy of the proton-field system can be found by multiplying the change in electric potential by the charge of a proton. The charge of a proton is 1.6 x 10^-19 C, so the change in potential energy is (4.0 x 10^4 volts) * (1.6 x 10^-19 C), which equals to 6.4 x 10^-15 joules.
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Suppose a car approaches a hill and has an initial speed of
106
k
m
/
h
at the bottom of the hill. The driver takes her foot off the gas pedal and allows the car to coast up the hill.
A. If in actuality a
780
k
g
car with an initial speed of
106
k
m
/
h
is observed to coast up a hill and stops at a height
21.5
m
above its starting point, how much thermal energy was generated by friction in
J
?
B. What is the magnitude of the average force in newtons of friction if the hill has a slope of
27
∘
above the horizontal.
Answer:
a) 1.73*10^5 J
b) 3645 N
Explanation:
106 km/h = 106 * 1000/3600 = 29.4 m/s
If KE = PE, then
mgh = 1/2mv²
gh = 1/2v²
h = v²/2g
h = 29.4² / 2 * 9.81
h = 864.36 / 19.62
h = 44.06 m
Loss of energy = mgΔh
E = 780 * 9.81 * (44.06 - 21.5)
E = 7651.8 * 22.56
E = 172624.6 J
Thus, the amount if energy lost is 1.73*10^5 J
Work done = Force * distance
Force = work done / distance
Force = 172624.6 / (21.5/sin27°)
Force = 172624.6 / 47.36
Force = 3645 N
Suppose you had a collection of a large number of hypothetical quantum objects, each of whose individual energy levels were -4.6 eV, -3.0 eV, -2.1 eV, and -1.6 eV. If nearly all of these identical objects were in the ground state, what would be the energies of dark spectral lines in an absorption spectrum if visible white light (1.8 to 3.1 eV) passes through the material? Enter the energies in order of increasing energy, followed by entering 0 in any later boxes for which there is no dark line within the visible spectrum. (That is, if your answers were 1, 2, and 3 eV, you would enter 1 in the first box, 2 in the second box, and 3 in the third box. If your answers were 1 and 2 eV, you would enter 1 in the first box, 2 in the second box, and 0 in the third box. If your answer is just 1 eV, you would enter 1 in the first box, 0 in the second box, and 0 in the third box.) Smallest energy of a dark line: eV Next larger energy of a dark line (or 0): eV Next larger energy of a dark line (or 0): eV Additional Materials
The energy differences between the levels of the quantum objects and the photons in the visible spectrum determine the dark spectral lines in an absorption spectrum.
Explanation:In order to determine the energies of the dark spectral lines in an absorption spectrum, we need to consider the energy differences between the energy levels of the quantum objects and the energy levels of the photons in the visible spectrum. The energy levels of the quantum objects are -4.6 eV, -3.0 eV, -2.1 eV, and -1.6 eV. The range of photon energies for visible light is 1.63 to 3.26 eV. Only photons with energies matching the energy differences between the levels of the quantum objects will be absorbed, resulting in dark lines in the spectrum.
Based on the given information, the dark spectral lines within the visible spectrum would be:
Smallest energy of a dark line: -4.6 eVNext larger energy of a dark line (or 0): -3.0 eVNext larger energy of a dark line (or 0): -2.1 eVLearn more about Energy differences and absorption spectrum here:https://brainly.com/question/31230618
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Two boxes, mA = 18 kg and mB = 14 kg, are attached by a string under tension T1 . The rightmost box is being pulled horizontally across the floor by a different string under tension T2 . The coefficient of kinetic friction between the boxes and the floor is µK = 0.240. If the boxes are accelerating at 3.5 m/s2 to the right, what is the tensions T1 and T2 ?
Answer:
[tex]T_{1}=105.38 N[/tex]
[tex]T_{2}=187.34 N[/tex]
Explanation:
Applying the second Newton's law for the first box we have.
[tex]-f_{1f}+T_{1}=m_{A}a[/tex]
[tex]-\mu_{k}N_{1}+T_{1}=m_{A}a[/tex]
We know that the normal force is the product between the weight and the kinetic friction, so we have:
[tex]-\mu_{k}m_{A}g+T_{1}=m_{A}a[/tex]
Now we can find T₁:
[tex]\mu_{k}m_{A}g+m_{A}a=T_{1}[/tex]
The acceleration is the same for both boxes.
[tex]T_{1}=m_{1}(\mu_{k}g+a)[/tex]
[tex]T_{1}=18*(0.240*9.81+3.5)[/tex]
[tex]T_{1}=105.38 N[/tex]
Now let's analyze the forces of the second box.
[tex]-f_{2f}-T_{1}+T_{2}=m_{B}a[/tex]
[tex]-\mu_{k}m_{B}g-T_{1}+T_{2}=m_{B}a[/tex]
Let's solve it for T₂.
[tex]T_{2}=m_{B}a+T_{1}+\mu_{k}m_{B}g[/tex]
[tex]T_{2}=m_{B}a+T_{1}+\mu_{k}m_{B}g[/tex]
[tex]T_{2}=m_{B}(a+\mu_{k}g)+T_{1}[/tex]
[tex]T_{2}=14(3.5+0.240*9.81)+105.38[/tex]
[tex]T_{2}=187.34 N[/tex]
I hope it helps you!
The radius RH of a black hole, also known as the event horizon, marks the location where not even light itself can escape from the black hole. That is, no information about the interior of the black hole may escape to any observer located outside of the black hole. According to general relativity, RH = (2GM / c^2), where M is the mass of the black hole and c is the speed of light. you want to study a black hole by getting near it ith a radial distance of 50 RH. However, you don't want the difference in gravitational acceleration between your head and your feet to exceed 10 m/s^2.
a) As a multiple of the Sun's mass, approximate what is the limit to the mass of the black hole you can tolerate at the given radial distance.
b) Is the limit an upper limit(you can tolerate smaller masses) or a lower limit (you can tolerate larger masses)?
Solution:
a) We know acceleration due to gravity, g = GM/r²
Differential change, dg/dr = -2GM/r³
Here, r = 50*Rh = 50*2GM/c² = 100GM/c ²
My height, h=dr = 1.7 m
Difference in gravitational acceleration between my head and my feet, dg = -10 m/s²
or, dg/dr = -10/1.7 = -2GM/(100GM/c²)³
or, 5.9*100³*G²*M² = 2c⁶
or, M = 0.59*c³/(1000G) = 2.39*1032 kg = [(2.39*1032)/(1.99*1030 )]Ms = 120*Ms
Mass of black hole which we can tolerate at the given distance is 120 time the mass of Sun.
b) This limit an upper limit ,we can tolerate smaller masses only.
A particular gun (10kg) is able to fire 20 gram bullets at a speed of 350 m/s. From this information, calculate roughly how much energy contained in the powder in the bullet (Hint: you will need to calculate the recoil speed of the gun.) A. 2455 J B. 620 J C. Cannot be determined. D. 1230 J
Answer:
The answer is
D. 1230j
Explanation:
When a bullet is shot out of a gun the person firing experiences a backward impact, which is the recoil force, while the force propelling the bullet out of the gun is the propulsive force
given data
Mass of gun M=10kg
Mass of bullet m=20g----kg=20/1000 =0.02kg
Propulsive speed of bullet = 350m/s
Hence the moment of the bullet will be equal and opposite to that of the gun
mv=MV
where V is the recoil velocity which we are solving for
V=mv/M
V=0.02*350/10
V=7/10
V=0.7m/s
The energy contained in the bullet can be gotten using
KE=1/2m(v-V)²
KE=1/2*0.02(350-0.7)²
KE=1/2*0.02(349.3)²
KE=1/2*0.02*122010.49
KE=1/2*2440.20
KE=1220.1J
roughly the energy is 1230J
A solid block of mass m is suspended in a liquid by a thread. The density of the block is greater than that of the liquid. Initially, the fluid level is such that the block is at a depth d and the tension in the thread is T. Then, the fluid level is decreased such that the depth is 0.5d. What is the tension in the thread when the block is at the new depth?
Answer:
(C) T
The tension T at equilibrium will be equal to the Buoyant force.
The Buoyant force is given by:
Fb = density x acceleration due to gravity x volume displaced
The change in height doesn't affect the Buoyant force and hence the tension.
Note: The figure of question is added in the attachment
The tension in the thread when the block is at the new depth is :
(C) T
Buoyant forceThe definition of buoyancy refers to whether something can drift in water or discuss, or the control of water or other fluids to keep water above water, or an idealistic disposition.
The tension T at equilibrium will be equal to the Buoyant force.
The Buoyant force is given by:
Fb = density x acceleration due to gravity x volume displaced
The change in height doesn't affect the Buoyant force and hence, the tension.
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A man is traveling on a bicycle at 14 m/s along a straight road that runs parallel to some railroad tracks. He hears the whistle of a train that is behind him. The frequency emitted by the train is 840 Hz, but the frequency the man hears is 778 Hz. Take the velocity of sound to be 340 m/s.
a) What frequency is heard by a stationary observer located between the train and the bicycle?
b) What is the speed of the train, and is the train traveling away from or towardthe bicycle?
Answer:
Explanation:
b ) The problem is based on Doppler's effect of sound
f = f₀ x (V - v₀) /( [tex]V+v_s[/tex])
f is apparent frequency ,f₀ is real frequency , V is velocity of sound , v₀ is velocity of observer going away , [tex]v_s[/tex] is velocity of source going away
778 = 840 x (340 - 14)/ (340 + [tex]v_s[/tex])
340 + [tex]v_s[/tex] = 341.18
[tex]v_s[/tex] = 1.18 m /s
it will go away from the observer or the cyclist.
speed of train = 1.18 m /s
a )
For a stationary observer v₀ = 0
f = f₀ x V /( [tex]V+v_s[/tex])
= 840 x 340 / (340 + 1.180)
= 837 Hz
Final answer:
The stationary observer between the train and the bicycle hears a frequency of 809 Hz, and the train is moving away from the bicycle at 42 m/s.
Explanation:
A stationary observer located between the train and the bicycle would hear a frequency of 809 Hz as the train approaches and passes by.
The speed of the train can be calculated to be 42 m/s moving away from the bicycle.
n 38 g rifle bullet traveling at 410 m/s buries itself in a 4.2 kg pendulum hanging on a 2.8 m long string, which makes the pendulum swing upward in an arc. Determine the maximum vertical displacement of the pendulum (height).(g = 9.8 m/s2)
Answer:
68cm
Explanation:
You can solve this problem by using the momentum conservation and energy conservation. By using the conservation of the momentum you get
[tex]p_f=p_i\\mv_1+Mv_2=(m+M)v[/tex]
m: mass of the bullet
M: mass of the pendulum
v1: velocity of the bullet = 410m/s
v2: velocity of the pendulum =0m/s
v: velocity of both bullet ad pendulum joint
By replacing you can find v:
[tex](0.038kg)(410m/s)+0=(0.038kg+4.2kg)v\\\\v=3.67\frac{m}{s}[/tex]
this value of v is used as the velocity of the total kinetic energy of the block of pendulum and bullet. This energy equals the potential energy for the maximum height reached by the block:
[tex]E_{fp}=E_{ki}\\\\(m+M)gh=\frac{1}{2}mv^2[/tex]
g: 9.8/s^2
h: height
By doing h the subject of the equation and replacing you obtain:
[tex](0.038kg+4.2kg)(9.8m/s^2)h=\frac{1}{2}(0.038kg+4.2kg)(3.67m/s)^2\\\\h=0.68m[/tex]
hence, the heigth is 68cm
The coil in a 60-Hz ac generator has 125 turns, each having an area of 3.0 × 10-2 m2 and is rotated in a uniform 0.12-T magnetic field. What is the peak output voltage of this generator?
Answer:
Explanation:
Given that, .
Frequency
f = 60Hz
Number of turns
N = 125turns
Surface area of coil
A = 3 × 10^-2 m²
Magnetic field
B = 0.12T
Voltage peak to peak? I.e the EMF
EMF is given as
ε = —dΦ/dt
Where Φ is magnetic flux and it is given as
Φ = NBA Cosθ
Where N is number of turns
B is magnetic field
A is the cross sectional area
And θ is the resulting angle from the dot product of area and magnetic field
Where θ =ωt and ω = 2πf
Then, θ = 2πft
So, your magnetic flux becomes
Φ = NBA Cos(2πft)
Now, dΦ / dt = —NBA•2πf Sin(2πft)
dΦ / dt = —2πf • NBA Sin(2πft)
So, ε = —dΦ/dt
Then,
ε = 2πf • NBA Sin(2πft)
So, the maximum peak to peak emf will occur when the sine function is 1
I.e Sin(2πft) = 1
So, the required peak to peak emf is
ε = 2πf • NBA
Substituting all the given parameters
ε = 2π × 60 × 125 × 0.12 × 3 × 10^-2
ε = 169.65 Volts
The peak to peak voltage is 169.65 V
The required value of peak output voltage is 169.65 Volts.
Given data:
The frequency of ac generator is, f = 60 Hz.
The number of turns of coil is, n = 125 turns.
The area of each coil is, [tex]A =3.0 \times 10^{-2} \;\rm m^{2}[/tex].
The strength of magnetic field is, B = 0.12 T.
To start with this problem, we need to find the peak emf first. The peak output voltage is nothing but the value of this peak emf only. The expression for the peak EMF is given as,
ε = —dΦ/dt
here,
Φ is magnetic flux and it is given as
Φ = NBA Cosθ
Here,
θ is the resulting angle from the dot product of area and magnetic field
Where θ =ωt and ω = 2πf
Then, θ = 2πft
So, the expression for the magnetic flux becomes,
Φ = NBA Cos(2πft)
Now, dΦ / dt = —NBA•2πf Sin(2πft)
dΦ / dt = —2πf • NBA Sin(2πft)
So, ε = —dΦ/dt
Then,
ε = 2πf • NBA Sin(2πft)
So, the maximum peak to peak emf will occur when the sine function is 1
Sin(2πft) = 1
So, the required peak to peak emf is
ε = 2πf • NBA
Substituting all the given parameters
ε = 2π × 60 × 125 × 0.12 × 3 × 10^-2
ε = 169.65 Volts
Thus, we can conclude that the required value of peak output voltage is 169.65 Volts.
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STATE THE LAWS OF REFLECTION?
Answer:
hii bro.... from India. ....i am from Bilaspur cg
A glass lens, nglass=1.52, has a 127 nm thick antireflective film coating one side, nfilm=1.39. White light, moving through the air, is perpendicularly incident on the coated side of the lens. What is the largest wavelength of the reflected light that is totally removed by the coating? Assume that nair=1.00.
Answer:
The wavelength is [tex]\lambda = 706nm[/tex]
Explanation:
From the question we are told that
The refractive index of the glass is [tex]n__{glass}} = 1.52[/tex]
The thickness of film is [tex]D = 127nm = 127*10^{-9}m[/tex]
The refractive index of film [tex]n__{film}} = 1.39[/tex]
The refractive index of air is [tex]n__{air}} = 1.00[/tex]
Generally the thickness of the film can be obtained mathematically from this expression
[tex]D = \frac{\lambda}{4 * n__{film}} }[/tex]
Where [tex]\lambda[/tex] is the wavelength
Making the wavelength the subject of the formula
[tex]\lambda = 4 * n__{film}} * D[/tex]
Substituting values
[tex]\lambda = 4 *1.39 * 127 *10^{-9}[/tex]
[tex]\lambda =7.06 *10^{-7}m = 706nm[/tex]
Answer:
[tex]\lambda = 706.12 nm[/tex]
Explanation:
The optical path length for the reflection of light = [tex]2 n_{film} t[/tex]
For destructive interference, [tex]2 n_{film} t = \frac{\lambda}{2}[/tex]
The thickness of the anti-reflective film = 127 nm
The largest wavelength of the reflected light, [tex]\lambda = 4n_{film} t[/tex]
[tex]\lambda = 4 * 1.39 *127 * 10^{-9}[/tex]
[tex]\lambda = 706.12 * 10^{-9} m\\\lambda = 706.12 nm[/tex]
1200 N-m of torque is used to drive a gear (A) of diameter 25 cm, which in turn drives another gear (B) of diameter 52 cm. What torque is transmitted by gear B to its shaft? Give your answer in kN-m.
Answer:
2.5kN.m
Explanation:
Torque is directly proportional to pitch diameter
= Ta/Tb= Da/Db
=120/Tb= 0.25/0.5
Tb= 2.469kN.m approx 2.5kN.m
Answer:
Explanation:
answer explain below
Gasoline flows in a long, underground pipeline at a constant temperature of 15o C (rho = 680 kg/m3 ; ν = 4.6 × 10-7 m2 /s). Two pumping stations are located at the same elevation 13 km apart. The pressure drop between the stations is 1.4 MPa. The diameter of the pipe is 0.6 m and has a roughness of ε = 0.12 mm. Determine the volume flow rate.
Answer:
1.0416 m∧3/sec
Explanation:
check the pictures below for the solution