Answer:
0.4112 m
Explanation:
The mass of the 1st ball = 3.6 kg
The height of the 1st ball =3.5 m
The mass of the 2nd ball = 3.6 kg
Mass of the bar M = 9.9 kg
Length of the bar L = 4.2 m
The velocity of the ball when it dropped from the height is calculated by using the formula:
[tex]\dfrac{1}{2}mv_1^2 = mgh_1 \\ \\ v = \sqrt{2gh_1} \\ \\ v =\sqrt{2\times9.8 \times 3.5} \\ \\ v = 8.283 \ m/s\\\\[/tex]
Provided that the bar is pivoted at the center and the ball is placed at the two ends, the moment of inertia for the bar is:
[tex]I = \dfrac{1}{12}ML^2 + m_1 (\dfrac{L}{2})^2 + m_2(\dfrac{L}{2})^2 \\ \\ =\dfrac{1}{12}(9.9kg)(4.2m)^2 + [3.6 kg+3.6kg](\dfrac{4.2}{2 \ m})^2 \\ \\ = 46.305 \ kg.m^2[/tex]
The angular momentum of the system due to the ball can be determined by using the formula:
L = mvr
L = (3.6 kg) (8.283 m/s) (2.1 m)
L = 62.61948 kg. m²
Now, Using the law of conservation:
[tex]L_i = L_f \\ \\ 62.61948 \ kg.m^2/s = I \omega \\ \\[/tex]
[tex]\omega = \dfrac{62.6198 \ kg.m^2/s}{46.305 \ kg.m^2}[/tex]
[tex]\omega =1.352 \ rad/s[/tex]
The linear angular velocity is deduced to be:
[tex]v = r \omega \\ \\ v = (2.1 \ m) ( 1.352 \ rad/s) = 2.839 \ m/s[/tex]
∴
the height raised by the second ball is:
[tex]h_2 = \dfrac{v^2}{2g} \\ \\ h_2 = \dfrac{(2.839)^2}{2(9.8 \ m/s^2)} \\ \\ h_2 =0.4112 \ m[/tex]
Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.370 mm wide. The diffraction pattern is observed on a screen 3.70 m away. Define the width of a bright fringe as the distance between the minima on either side. You may want to review (Page) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Single-slit diffraction. Part A What is the width of the central bright fringe
Answer:
Δx = 6.33 x 10⁻³ m = 6.33 mm
Explanation:
We can use the Young's Double Slit Experiment Formula here:
[tex]\Delta x = \frac{\lambda L}{d}\\\\[/tex]
where,
Δx = distance between consecutive dark fringes = width of central bright fringe = ?
λ = wavelength of light = 633 nm = 6.33 x 10⁻⁷ m
L = distance between screen and slit = 3.7 m
d = slit width = 0.37 mm = 3.7 x 10⁻⁴ m
Therefore,
[tex]\Delta x = \frac{(6.33\ x\ 10^{-7}\ m)(3.7\ m)}{3.7\ x \ 10^{-4}\ m}[/tex]
Δx = 6.33 x 10⁻³ m = 6.33 mm
when two capacitor 3muF and 6muF are connected in a parallel and combination is charged to a potential of 120 volt the potential difference across the 3muF capacitor is
Answer:
V₁ = V = 120 V
Explanation:
Such a combination of capacitors in which;
1- Potential difference across each capacitor is the same
2- Total charge is distributed amongst the capacitors
; is called Parallel Combination.
Therefore, in this case, the potential difference across each capacitor will also be the same. Because the capacitors are connected in parallel here. So the voltage across 3 μF capacitor will be the same as the voltage across the 6 μF capacitor and they both will be equal to the total potential difference.
V₁ = V = 120 V
Electroconvulsive therapy would be done under the
supervision of a counseling psychologist, where high level
of electric shock would be admistered.
Select one:
True
False
Answer:
the answer of this question is true
Which of the following best defines climate? PLEASE HELP!!
Do all the cells look the same in a filament? If there is a difference, describe
how they are different.
Answer:
All cells in filamentous algae look the same except for the spores. Explanation:
An 800 kg charging bull rams through a wooden fence. It was travelling at
5 m/s, now it's travelling at 3 m/s. How much impulse did the bull
experience by smashing the fence?
Answer:
J = 1600 kg-m/s
Explanation:
Given that,
The mass of charging bull rams, m = 800 kg
Initial speed, u = 5 m/s
Final speed, v = 3 m/s
We need to find the impulse the bull experience by smashing the fence. Let it is J. We know that, impulse is equal to the change in momentum such that,
J = m(v-u)
Put all the values,
J = 800(3-5)
= 800(-2)
= -1600 kg-m/s
Hence, the magnitude of impulse is equal to 1600 kg-m/s.
Please help if you can!
Answer:
C, Red has the longest one
c red
red has longest wavelength
amnh dot org
PLEASE HELP What are some ways to increase momentum?
In the Zeeman effect, the energy levels of hydrogen are split by a magnetic field. Each state with a different value of mlml has a different energy. The differences in energy between adjacent values of mlml are all equal. The Zeeman effect is observed as a splitting in the spectral lines of hydrogen in the presence of a magnetic field. What is the greatest number of lines that a single spectral line can be split into by a magnetic field
Answer:
the Zeeman effect without spin is three lines
Explanation:
The Zeeman effect is the result of the interaction of the magnetic field with the orbital angular momentum of the electrons, if we do not take the spin into account it is called the Normal Zeeman effect.
Therefore we only take into account the orbital moments (m_l) of the transition, from the selection rules of the refreshed harmonics, only the transition with
[tex]\Delta m_l[/tex] = 0, ± 1
ΔE [tex]\DeltaE = \mu_B \ \Delta m_l \ B[/tex]
Let's analyze for the case of the Hydrogen atom
For a transition between levels n = 1 and n = 2 the values of m_l are n_f = 1 m_l = 0 and for n₀ = 2 m_l = 0, 1
so we only have two lines.
For transition n_f = 2 and n₀ = 3
n_f = 2 m_l = 0, 1
n₀ = 3 m_l = -1, 0, 1
There are only two lines plus the central line, so there are three spectral lines
for n_f = 3 and n_o = 4
n_f = 3 ml = -1, 0, 1
n₀= 4 ml = -2, -1, 0, 1, 2
Only transitions with Δm_l = ±1 are allowed, so there are only two transitions plus the central transition (Δm_l = 0), so there are only 3 spectral lines.
In summary, due to the selection rule of spherical harmonics, the greatest number of lines in the Zeeman effect without spin is three lines.
An astronaut is a distance L from her spaceship, and is at rest with respect to the ship, when she discovers that her tether has broken. She tosses a wrench with a speed Vw in the opposite direction of the ship to propel herself back to the ship. The astronaut has mass MA, and the wrench has mass Mw.
Required:
a. Draw a sketch, showing the subsequent motion of the astronaut and the wrench.
b. What is the initial momentum (before toss) of the astronaut plus wrench system? What is the final momentum?
c. Use conservation of momentum to solve for the speed of the astronaut VA, relative to the ship, in terms of MA, Mw and Vw.
d. How long does it take her to reach the ship in terms of L, MA, Mw and Vw?
e. How far has the wrench traveled from its original position when the astronaut reaches the ship? Express your answer in terms of L, MA and Mw.
Answer:
B) I₀ = I_f= 0, C) vₐ = [tex]\frac{m_w}{m_a} \ v_w[/tex] , D) t = [tex]\frac{m_a}{m_w} \ \frac{L}{v_w}[/tex]
Explanation:
A) in the attachment you can see a diagram of the movement of the key and the astronaut that is in the opposite direction to each other.
B) Momentum equals the change in momentum in the system
I = ∫ F dt = Δp
since the astronaut has not thrown the key, the force is zero, so the initial impulse is zero
I₀ = 0
The final impulse of the two is still zero, since it is a vector quantity, subtracting the impulse of the two gives zero, since it is an isolated system
I_f = 0
C) We define the system formed by the astronaut and the key, for which the forces during the separation are internal and the moment is conserved
initial instant.
p₀ = 0
final instant
p_f = [tex]m_a v_a - m_w v_w[/tex]
We used the subscript “a” for the astronaut and the subscript “w” for the key
the moment is preserved
po = p_f
0 = mₐ vₐ - m_w v_w
vₐ = [tex]\frac{m_w}{m_a} \ v_w[/tex]
D) as the astronaut goes at constant speed we can use the uniform motion relationships
vₐ = x / t
t = x / vₐ
t = [tex]\frac{m_a}{m_w} \ \frac{L}{v_w}[/tex]
You are playing in a volley ball game Your team has 12 and the other team has 18.
How many points does your team needs to win?
How many points does the other team needs to win?
Answer:
you need 7 points and the other team just needs to stop you from scoring
Explanation:
The period of a sound wave coming from an instrument is 2 seconds. What 1 point
is the frequency of the sound? (f = 1/T) *
5 Hz
50 Hz
ОО
0.5 Hz
Answer:
b
Explanation:
The length of the slope of a mountain is 2780 m, and it makes
its base?
angle of 14.1° with the horizontal. What is the height of the mountain, relative to
Additional Materials
Reading
Answer:
677 m
Explanation:
Using the definition of the sine of an angle, we can write
sin 14.1 = (height of mountain) / (slope length of mountain)
sin 14.1 = H / (2780 m) ---> H = (2780 m) x sin 14.1
= 677 m
The height of the mountain is 677.21 m
The given parameters;
length of the slope, L = 2780 m
angle of inclination, Ф = 14.1°
let the height of the mountain, = h
A simple sketch of the problem is given below;
↓P
↓
↓ h
↓ 14.1°
↓------------------------------------------------Q
A straight line joining PQ is the hypotenuse of the right triangle.
The height of the right triangle is calculated as follows;
[tex]sin(14.1) = \frac{h}{PQ} \\\\\h = PQ \times sin(14.1)\\\\h = 2780 \times sin(14.1)\\\\h = 677.21 \ m[/tex]
Thus, the height of the mountain is 677.21 m
Learn more here: https://brainly.com/question/4326804
Two band members are marching toward each other beginning from opposite end zone goal lines (100 yards apart). One band member is playing a picolo and marching at 5 ft/s and the other is playing a sousaphone and marching at 1 ft/s. At what yard line do they meet?
Answer:
17
Explanation:
answer is 16.6666 which is equal to 17
Man-made climate change is an example
of...
True or false. When a girl walks the action of pushing and the equal amd opposite reaction is being projected forward
This is true I think
It applies to Newton's Laws
it's true because it's a part of newtons law
Light of intensity I0 passes through 4 ideal polarizing sheets. Unpolarized light enters the 1st sheet that has a horizontal transmission axis. Light continues to the 2nd sheet that has its transmission axis at 25 degrees with respect to the 1st sheet, then to the 3rd sheet that has its transmission axis at 47 degrees with respect to the 1st sheet then to the 4th sheet that has its transmission axis at 10 degrees with respect to the 3d sheet. The intensity of the emerging light as percentage of I0 is close to:
Answer:
34.24 %
Explanation:
Since I₀ is the intensity of the un-polarized light, the intensity I₁ of the light polarized by the 1st sheet is (by the one-half rule) I₁ = I₀/2.
The intensity of polarized light I from a polarized source I' is I = I'cos²Ф where Ф is the angle between the direction of I' and I. Since the second sheet has its transmission axis at 25° with respect °o the 1st sheet, the intensity of light I₂ from the second sheet is I₂ = I₁cos²25°.
Also, the 3rd sheet has its transmission axis 47° with respect to the 1st sheet. So, the angle between the transmission axis of the 2nd sheet and 3rd sheet is 47° - 25° = 22°. So, the intensity I₃ from the 3rd sheet is I₃ = I₂cos²22°
Finally, the 4th sheet has its transmission axis 10° with respect to the 3rd sheet. So, the intensity I₄ from the 4th sheet is I₄ = I₃cos²10°.
So, I₄ = I₃cos²10°
I₄ = I₂cos²22°cos²10°
I₄ = I₁cos²25°cos²22°cos²10°
I₄ = (I₀/2)cos²25°cos²22°cos²10°
I₄/I₀ = cos²25°cos²22°cos²10°/2
I₄/I₀ = (cos25°cos22°cos10°)²/2
I₄/I₀ = (0.9063 × 0.9272 × 0.9848)²/2
I₄/I₀ = 0.8275²/2
I₄/I₀ = 0.6848/2
I₄/I₀ = 0.3424
So, as a percentage,
I₄/I₀ × 100% = 0.3424 × 100% = 34.24 %
Write about
a time you had to ride a bicycle on a difficult
surface. What did you have to do to adjust your
riding?
Hans Full is pulling on a rope to drag his backpack to school across the ice. He pulls upwards and rightwards with a force of 22.9 Newtons at an angle of 35 degrees above the horizontal to drag his backpack a horizontal distance of 129 meters to the right. Determine the work (in Joules) done upon the backpack.
Answer:
2420 J
Explanation:
From the question given above, the following data were obtained:
Force (F) = 22.9 N
Angle (θ) = 35°
Distance (d) = 129 m
Workdone (Wd) =?
The work done can be obtained by using the following formula:
Wd = Fd × Cos θ
Wd = 22.9 × 129 × Cos 35
Wd = 22.9 × 129 × 0.8192
Wd ≈ 2420 J
Thus, the workdone is 2420 J.
Which of the following is a category of mechanical wave?
O A. Transverse
B. Frictional
C. Parallel
D. Perpendicular
Answer:
a
because the mechanical wave is when it goes over and over again
Answer:
The answer is a like i said 3hrs ago i dont know if this guy copied me tbh
Explanation:
A man applies a force of 315 N to push the block of 225 kg onto 10 m ramp. Calculate the efficiency of the person if the mass gains a height of 1.2 m.
Answer:
η = 0.84 = 84%
Explanation:
The efficiency of the man can be given by the following formula:
η = output/input
where,
η = efficiency of man = ?
output = potential energy gain of the block = mgh
input = work done by man = Fd
Therefore,
[tex]\eta = \frac{mgh}{Fd}[/tex]
where,
m = mass of block = 225 kg
g = acceleration due to gravity = 9.81 m/s²
h = height gained by block = 1.2 m
F = force exerted by man = 315 N
d = distance covered by man = 10 m
Therefore,
[tex]\eta = \frac{(225\ kg)(9.81\ m/s^2)(1.2\ m)}{(315\ N)(10\ m)}[/tex]
η = 0.84 = 84%
An object of mass 45 kg is observed to accelerate at the rate of 6 m/s2. Calculate the force required to produce this acceleration
find the total electric field along the line of two of the two charges shown in figure at the point midway between them
help its due in 10 minutes lol
PLS HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
A) pass the ball to a teammate
B) Smash the shuttlecock downward in your opponents court
C)Do a fake hit . . .
D) Do a fake hit . . .
(Best guess)
1. Amplitude is the maximum distance that the particles in a medium _____
2. A Wavelenght of a wave is the distance from ____
3. As the frequency of a wave increases, the wavelengths ___
4. The frequency of a wave is the number of ___
5. Changing the frequency of waves also changes ___
Answer:T or F: Mechanical waves need a medium through which to transport energy. ... T or F: Wave A has a greater frequency than wave B, but the amplitude of the two waves is the ... past the raft increases, you also find that the distance between each crest ___. ... For a given wave, the frequency doubles, the wavelength is ____.
Explanation:
Rays of light radiating from Sun are
radiations
Explanation:
All of the energy from the Sun that reaches the Earth arrives as solar radiation, part of a large collection of energy called the electromagnetic radiation spectrum. Solar radiation includes visible light, ultraviolet light, infrared, radio waves, X-rays, and gamma rays. Radiation is one way to transfer heat.
Answer:
All of the energy from the Sun that reaches the Earth arrives as solar radiation, part of a large collection of energy called the electromagnetic radiation spectrum. Solar radiation includes visible light, ultraviolet light, infrared, radio waves, X-rays, and gamma rays. Radiation is one way to transfer heat
(121 m/s, or 43
4. The pilot of an airplane, which has been diving at a speed of
540 km/h, pulls out of the dive at constant speed.
(a) What is the minimum radius of the plane's circular path in
order that the acceleration of the pilot at the lowest point
will not exceed 7 g?
(b) What force is applied on an 80 kg pilot by the plane seat at
the lowest point of the pull-out? (328 m, 6.3 X 10N)
Answer:
(a) r = 328 m
(b) F = 6271.8 N
Explanation:
speed, v = 540 km/h = 150 m/s
Acceleration, a = 7 g
(a) Let the radius is r.
[tex]a =\frac{v^{2}}{r}\\7\times 9.8 = \frac{150\times 150}{r}\\r = 328 m[/tex]
(b) The force on 80 kg pilot at the lowest point is
[tex]F =m \times g + m\frac{v^{2}}{r}\\F = 80\times9.8 +80\times \frac{150\times 150}{328}\\\\F =784 + 5487.8 = 6271.8 N[/tex]
What does it mean when work is positive?
Answer:
When force and displacement are in the same direction, the work performed on an object is said to be positive work. Example: When a body moves on the horizontal surface, force and displacement act in the forward path. The work is done in this case known as Positive work.
Explanation:
Hope this helps you
A 25.0 kg mass is traveling to the right with a speed of 2.80 m/s on a smooth horizontal surface when it collides with and sticks to a second 25.0 kg mass that is initially at rest but is attached to one end of a light, horizontal spring with force constant 170.0 N/m. The other end of the spring is fixed to a wall to the right of the second mass.
1) Find the frequency of the subsequent oscillations.2) Find the amplitude of the subsequent oscillations.3) Find the period of the subsequent oscillations.4) How long does it take the system to return the first time to the position it had immediately after the collision?
Answer:
Explanation:
1 ) angular frequency ω = √ ( k / m )
=√ ( 170 / 50 )
= 1.844 rad /s
2πn = 1.844 where n is frequency of oscillation
n = 1.844 / (2 x 3.14 )
= .294 per sec
= .294 x 60 = 18 approx. per minute .
Velocity just after collision of composite mass ( using law of conservation of momentum )
= 25 x 2.8 / 50
v = 1.4 m/s
If new amplitude be A
1/2 k A² = 1/2 m v²
m = 25 + 25 = 50 kg
170 x A² = 50 x 1.4²
A = 0.76 m
3 ) period of oscillation = 1 /n
= 1 / .294
= 3.4 s
4 ) It will take complete one period of oscillation ie 3.4 s to come to its original position.