Answer:
a
[tex]m = 0.169 \ kg[/tex]
b
[tex]|v_{max} |= 0.5653 \ m/s[/tex]
Explanation:
From the question we are told that
The spring constant is [tex]k = 14 \ N/m[/tex]
The maximum extension of the spring is [tex]A = 6.0 \ cm = 0.06 \ m[/tex]
The number of oscillation is [tex]n = 30[/tex]
The time taken is [tex]t = 20 \ s[/tex]
Generally the the angular speed of this oscillations is mathematically represented as
[tex]w = \frac{2 \pi}{T}[/tex]
where T is the period which is mathematically represented as
[tex]T = \frac{t}{n}[/tex]
substituting values
[tex]T = \frac{20}{30 }[/tex]
[tex]T = 0.667 \ s[/tex]
Thus
[tex]w = \frac{2 * 3.142 }{ 0.667}[/tex]
[tex]w = 9.421 \ rad/s[/tex]
this angular speed can also be represented mathematically as
[tex]w = \sqrt{\frac{k}{m} }[/tex]
=> [tex]m =\frac{k }{w^2}[/tex]
substituting values
[tex]m =\frac{ 15 }{(9.421)^2}[/tex]
[tex]m = 0.169 \ kg[/tex]
In SHM (simple harmonic motion )the equation for velocity is mathematically represented as
[tex]v = - Awsin (wt)[/tex]
The velocity is maximum when [tex]wt = \(90^o) \ or \ 1.5708\ rad[/tex]
[tex]v_{max} = - A* w[/tex]
=> [tex]|v_{max} |= A* w[/tex]
=> [tex]|v_{max} |= 0.06 * 9.421[/tex]
=> [tex]|v_{max} |= 0.5653 \ m/s[/tex]
The frequency of light emitted from hydrogen present in the Andromeda galaxy has been found to be 0.10% higher than that from hydrogen measured on Earth.
Is this galaxy approaching or receding from the Earth, and at what speed?
Answer:
3x10^5m/s
Explanation:
See attached file
Explanation:
The speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].
Doppler's Effect
According to the Doppler effect, the difference between the frequency at which light wave leave a source and reaches an observer is caused by the relative motion of the observer and the wave source.
Given that the difference in the frequency is 0.10 %. The speed of light emitted from the galaxy can be calculated by the Doppler effect.
[tex]\dfrac {\Delta f}{f} = \dfrac {v}{c}[/tex]
Where f is the frequency of the light, v is the speed of light emitted from the galaxy and c is the speed of light emitted from the earth.
[tex]\dfrac {0.10 f}{100 f} = \dfrac {v}{3\times 10^8}[/tex]
[tex]v = 3\times 10^5\;\rm m/s[/tex]
Hence we can conclude that the speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].
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A velocity selector in a mass spectrometer uses a 0.100-T magnetic field. (a) What electric field strength is needed to select a speed of 4.00 . 106 m/s
Answer:
The electric field strength needed is 4 x 10⁵ N/C
Explanation:
Given;
magnitude of magnetic field, B = 0.1 T
velocity of the charge, v = 4 x 10⁶ m/s
The velocity of the charge when there is a balance in the magnetic and electric force is given by;
[tex]v = \frac{E}{B}[/tex]
where;
v is the velocity of the charge
E is the electric field strength
B is the magnetic field strength
The electric field strength needed is calculated as;
E = vB
E = 4 x 10⁶ x 0.1
E = 4 x 10⁵ N/C
Therefore, the electric field strength needed is 4 x 10⁵ N/C
Consider two parallel plate capacitors. The plates on Capacitor B have half the area as the plates on Capacitor A, and the plates in Capacitor B are separated by twice the separation of the plates of Capacitor A. If Capacitor A has a capacitance of CA-17.8nF, what is the capacitance of Capacitor? .
Answer:
CB = 4.45 x 10⁻⁹ F = 4.45 nF
Explanation:
The capacitance of a parallel plate capacitor is given by the following formula:
C = ε₀A/d
where,
C = Capacitance
ε₀ = Permeability of free space
A = Area of plates
d = Distance between plates
FOR CAPACITOR A:
C = CA = 17.8 nF = 17.8 x 10⁻⁹ F
A = A₁
d = d₁
Therefore,
CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F ----------------- equation 1
FOR CAPACITOR B:
C = CB = ?
A = A₁/2
d = 2 d₁
Therefore,
CB = ε₀(A₁/2)/2d₁
CB = (1/4)(ε₀A₁/d₁)
using equation 1:
CB = (1/4)(17.8 X 10⁻⁹ F)
CB = 4.45 x 10⁻⁹ F = 4.45 nF
A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3 degrees angle. What is the y-component of the velocity of the second ball?
Answer:
v_{1fy} = - 0.4549 m / s
Explanation:
This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved
initial. Before the crash
p₀ = m v₁₀
final. After the crash
[tex]p_{f}[/tex] = m [tex]v_{1f}[/tex] + m v_{2f}
Recall that velocities are a vector so it has x and y components
p₀ = p_{f}
we write this equation for each axis
X axis
m v₁₀ = m v_{1fx} + m v_{2fx}
Y Axis
0 = -m v_{1fy} + m v_{2fy}
the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components
sin 23.3 = v_{2fy} / v_{2f}
cos 23.3 = v_{2fx} / v_{2f}
v_{2fy} = v_{2f} sin 23.3
v_{2fx} = v_{2f} cos 23.3
we substitute in the momentum conservation equation
m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3
0 = - m v_{1f} sin θ + m v_{2f} sin 23.3
1.83 = v_{1f} cos θ + 1.15 cos 23.3
0 = - v_{1f} sin θ + 1.15 sin 23.3
1.83 = v_{1f} cos θ + 1.0562
0 = - v_{1f} sin θ + 0.4549
v_{1f} sin θ = 0.4549
v_{1f} cos θ = -0.7738
we divide these two equations
tan θ = - 0.5878
θ = tan-1 (-0.5878)
θ = -30.45º
we substitute in one of the two and find the final velocity of the incident ball
v_{1f} cos (-30.45) = - 0.7738
v_{1f} = -0.7738 / cos 30.45
v_{1f} = -0.8976 m / s
the component and this speed is
v_{1fy} = v1f sin θ
v_{1fy} = 0.8976 sin (30.45)
v_{1fy} = - 0.4549 m / s
How many turns of wire are needed in a circular coil 13 cmcm in diameter to produce an induced emf of 5.6 VV
Answer:
Number of turns of wire(N) = 3,036 turns (Approx)
Explanation:
Given:
Diameter = 13 Cm
emf = 5.6 v
Note:
The given question is incomplete, unknown information is as follow.
Magnetic field increases = 0.25 T in 1.8 (Second)
Find:
Number of turns of wire(N)
Computation:
radius (r) = 13 / 2 = 6.5 cm = 0.065 m
Area = πr²
Area = (22/7)(0.065)(0.065)
Area = 0.013278 m²
So,
emf = (N)(A)(dB / dt)
5.6 = (N)(0.013278)(0.25 / 1.8)
5.6 = (N)(0.013278)(0.1389)
N = 3,036.35899
Number of turns of wire(N) = 3,036 turns (Approx)
Air bags greatly reduces the chance og injury in a car accident.explain how they do si in terms of energy transfer
Answer:
in an accident, when the body collides with the air bags, the collision time of impact between the two bodies will increase due to the presence of air bags in the car. Larger is the impact time smaller is the transformation of energy between the body and air bag. That is why air bags greatly reduce the chance of injury in a car accident.
48. A patient presents with a thrombosis in
the popliteal vein. This thrombosis most likely
causes reduction of blood flow in which of the
following veins?
Answer:
the interation blood veins
Explanation:
If mirror M2 in a Michelson interferometer is moved through 0.233 mm, a shift of 792 bright fringes occurs. What is the wavelength of the light producing the fringe pattern?
Answer:
The wavelength is [tex]\lambda = 589 nm[/tex]
Explanation:
From the question we are told that
The distance of the mirror shift is [tex]k = 0.233 \ mm = 0.233*10^{-3} \ m[/tex]
The number of fringe shift is n = 792
Generally the wavelength producing this fringes is mathematically represented as
[tex]\lambda = \frac{ 2 * k }{ n }[/tex]
substituting values
[tex]\lambda = \frac{ 2 * 0.233*10^{-3} }{ 792 }[/tex]
[tex]\lambda = 5.885 *10^{-7} \ m[/tex]
[tex]\lambda = 589 nm[/tex]
Expectant mothers many times see their unborn child for the first time during an ultrasonic examination. In ultrasonic imaging, the blood flow and heartbeat of the child can be measured using an echolocation technique similar to that used by bats. For the purposes of these questions, please use 1500 m/s as the speed of sound in tissue. I need help with part B and C
To clearly see an image, the wavelength used must be at most 1/4 of the size of the object that is to be imaged. What frequency is needed to image a fetus at 8 weeks of gestation that is 1.6 cm long?
A. 380 kHz
B. 3.8 kHz
C. 85 kHz
D. 3.8 MHz
Answer:
380 kHz
Explanation:
The speed of sound is taken as 1500 m/s
The length of the fetus is 1.6 cm long
The condition is that the wavelength used must be at most 1/4 of the size of the object that is to be imaged.
For this 1.6 cm baby, the wavelength must not exceed
λ = [tex]\frac{1}{4}[/tex] of 1.6 cm = [tex]\frac{1}{4}[/tex] x 1.6 cm = 0.4 cm =
0.4 cm = 0.004 m this is the wavelength of the required ultrasonic sound.
we know that
v = λf
where v is the speed of a wave
λ is the wavelength of the wave
f is the frequency of the wave
f = v/λ
substituting values, we have
f = 1500/0.004 = 375000 Hz
==> 375000/1000 = 375 kHz ≅ 380 kHz
A current of 5 A is flowing in a 20 mH inductor. The energy stored in the magnetic field of this inductor is:_______
a. 1J.
b. 0.50J.
c. 0.25J.
d. 0.
e. dependent upon the resistance of the inductor.
Answer:
C. 0.25J
Explanation:
Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;
L is the inductance
I is the current flowing in the inductor
Given parameters
L = 20mH = 20×10^-3H
I = 5A
Required
Energy stored in the magnetic field.
E = 1/2 × 20×10^-3 × 5²
E = 1/2 × 20×10^-3 × 25
E = 10×10^-3 × 25
E = 0.01 × 25
E = 0.25Joules.
Hence the energy stored in the magnetic field of this inductor is 0.25Joules
The roller coaster car reaches point A of the loop with speed of 20 m/s, which is increasing at the rate of 5 m/s2. Determine the magnitude of the acceleration at A if pA
Answer and Explanation:
Data provided as per the question is as follows
Speed at point A = 20 m/s
Acceleration at point C = [tex]5 m/s^2[/tex]
[tex]r_A = 25 m[/tex]
The calculation of the magnitude of the acceleration at A is shown below:-
Centripetal acceleration is
[tex]a_c = \frac{v^2}{r}[/tex]
now we will put the values into the above formula
= [tex]\frac{20^2}{25}[/tex]
After solving the above equation we will get
[tex]= 16 m/s^2[/tex]
Tangential acceleration is
[tex]= \sqrt{ac^2 + at^2} \\\\ = \sqrt{16^2 + 5^2}\\\\ = 16.703 m/s^2[/tex]
You add 500 mL of water at 10°C to 100 mL of water at 70°C. What is the
most likely final temperature of the mixture?
O A. 80°C
OB. 10-C
OC. 20°C
O D. 60°C
Answer:
Option (c) : 20°C
Explanation:
[tex]t(final) = \frac{w1 \times t1 + w2 \times t2}{w1 + w2} [/tex]
T(final) = 500* 10 + 100*70/600 = 20°C
"A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if"
Answer:
A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if
the dispersion is great
Calculate the density of the following material.
1 kg helium with a volume of 5.587 m³
700 kg/m³
5.587 kg/m³
0.179 kg/m³
Answer:
[tex]density \: = \frac{mass}{volume} [/tex]
1 / 5.587 is equal to 0.179 kg/m³
Hope it helps:)
Answer:
The answer is
0.179 kg/m³Explanation:
Density of a substance is given by
[tex]Density \: = \frac{mass}{volume} [/tex]
From the
mass = 1 kg
volume = 5.583 m³
Substitute the values into the above formula
We have
[tex]Density \: = \frac{1 \: kg}{5.583 \: {m}^{3} } [/tex]
We have the final answer as
Density = 0.179 kg/m³Hope this helps you
Suppose you drop paperclips into an open cart rolling along a straight horizontal track with negligible friction. As a result of the accumulating paper clips, explain whether the momentum and kinetic energy increase, decrease, or stay the same.
Answer:
Stay the same
Explanation:
Since, friction is negligible:
Initial Momentum = Final Momentum
Initial KE = Final KE
m1 * v1 = m2 * v2
When m increases v decreases.
The momentum and kinetic energy remain the same if you drop paper clips into an open cart rolling along a straight horizontal track with negligible friction.
What is friction?Between two surfaces that are sliding or attempting to slide over one another, there is a force called friction. For instance, friction makes it challenging to push a book down the floor. Friction always moves an object in a direction that is counter to the direction that it is traveling or attempting to move.
Given:
The paperclips into an open cart rolling along a straight horizontal track with negligible friction,
Calculate the momentum, Since friction is negligible,
Initial Momentum = Final Momentum
Initial Kinetic Energy = Final Kinetic Energy
m₁ × v₁ = m₁ × v₂
When m increases, v decreases,
Thus, momentum will remain the same.
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Which notation is better to use? (Choose between 4,000,000,000,000,000 m and 4.0 × 1015 m)
Answer:
4 x 10¹⁵
Explanation:
Which unbalanced force accounts for the direction of the net force of the rocket?
a. Air resistance
b. Friction
c. Gravity
d. Thrust of rocket engine
It depends on what stage of the mission you're talking about.
==> While it's sitting on the pad before launch, the forces on the rocket are balanced, so there's no net force on it.
==> When the engines ignite, their thrust (d) is greater than the force of gravity. So the net force on the rocket is upward, and the spacecraft accelerates upward.
==> After the engines shut down, the net force acting on the rocket is due to Gravity (c).
. . . If the rocket has enough vertical speed, it escapes the Earth completely, and just keeps going.
. . . If it has enough horizontal speed, it enters Earth orbit.
. . . If it doesn't have enough vertical or horizontal speed, it falls back to Earth.
A rocket will preserve to speed up so long as there's a resultant pressure upwards resulting from the thrust of the rocket engine.
What unbalanced force bills for the course of the internet pressure of the rocket?A rocket launches whilst the pressure of thrust pushing it upwards is greater than the burden force because of gravity downwards. This unbalanced pressure reasons a rocket to accelerate upwards. A rocket will maintain to hurry up so long as there's a resultant force upwards resulting from the thrust of the rocket engine.
What's the net pressure of unbalanced?
If the forces on an item are balanced, the net pressure is zero. If the forces are unbalanced forces, the results do not cancel each difference. Any time the forces acting on an object are unbalanced, the net pressure is not 0, and the movement of the item modifications.
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A train on one track moves in the same direction as a second train on the adjacent track. The first train, which is ahead of the second train and moves with a speed of 36.4 m/s , blows a horn whose frequency is 123 Hz .what is its speed?
Answer:
51. 7m/s
Explanation:
Take speed of sound in air = 340 m/s
fp = fs (V + Vp)/(V + Vs)
128 = 123 (340 + Vp)/(340 + 36.4)
Vp = 51.7m/s
Explanation:
An electron is accelerated from rest through a potential difference. After acceleration the electron has a de Broglie wavelength of 880 nm. What is the potential difference though which this electron was accelerated
Answer:
3x10⁴v
Explanation:
Using
Wavelength= h/ √(2m.Ke)
880nm = 6.6E-34/√ 2.9.1E-31 x me
Ke= 6.6E-34/880nm x 18.2E -31.
5.6E-27/18.2E-31
= 3 x 10⁴ Volts
In a physics lab, Asha is given a 11.5 kg uniform rectangular plate with edge lengths 62.9 cm by 46.9 cm . Her lab instructor requires her to rotate the plate about an axis perpendicular to its plane and passing through one of its corners, and then prepare a report on the project. For her report, Asha needs the plate's moment of inertia ???? with respect to given rotation axis. Calculate ???? .
Answer:
6.9kgm²
Explanation:
For an axis through the center of the rectangle, I = m[(w²+L²)/12
Using the parallel axis theorem, the added value of I = mR² = m[(w²/4 + L²/4]
Adding the 2 expressions,
I = (m/3)*(w²+L²)
I =6.95 kg∙m²
A beam of light from a laser illuminates a glass how long will a short pulse of light beam take to travel the length of the glass.
Answer:
The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]
Explanation:
Given that,
A beam of light from a laser illuminates a glass.
Suppose, the length of piece is [tex]L=25.21\times10^{-2}\ m[/tex]
Index of refraction is 2.83.
We need to calculate the speed of light pulse in glass
Using formula of speed
[tex]v=\dfrac{c}{\mu}[/tex]
Put the value into the formula
[tex]v=\dfrac{3\times10^{8}}{2.83}[/tex]
[tex]v=1.06\times10^{8}\ m/s[/tex]
We need to calculate the time of short pulse of light beam
Using formula of velocity
[tex]v=\dfrac{d}{t}[/tex]
[tex]t=\dfrac{d}{v}[/tex]
Put the value into the formula
[tex]t=\dfrac{25.21\times10^{-2}}{1.06\times10^{8}}[/tex]
[tex]t=2.37\times10^{-9}\ sec[/tex]
Hence, The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]
Based on the passage, why is it important that different ethnic groups worked together on the strike? The groups needed to avoid speaking to one another because they wouldn’t understand. The different ethnic groups believed in being separate. The groups needed to trick the owners. They needed to be able to unite even though they spoke different languages.
Answer:D
Explanation:I got it right
Answer:
They needed to be able to unite even though they spoke different languages.
Explanation:
An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.165×105 km and the planet has a radius of 4175 km and a mass of 6.70×1022 kg , how long (in days) does it take the moon to make one revolution around the planet? The gravitational constant is 6.67×10−11N·m2/kg2 .
Answer:
364days
Explanation:
Pls see attached file
Explanation:
The moon will take 112.7 days to make one revolution around the planet.
What is Kepler's third law?The period of the satellite around any planet only depends upon the distance between the planet's center and satellite and also depends upon the planet's mass.
Given, the distance from the moon's center to the planet's surface,
h = 2.165 × 10⁵ km,
The radius of the planet, r = 4175 km
The mass of the planet = 6.70 × 10²² kg
The total distance between the moon's center to the planet's center:
a = r +h = 2.165 × 10⁵ + 4175
a = 216500 + 4175
a = 220675
a = 2.26750 × 10⁸ m
The period of the planet can be calculated as:
[tex]T =2\pi \sqrt{\frac{a^3}{Gm} }[/tex]
[tex]T =2\3\times 3.14 \sqrt{\frac{(2.20675 \times 10^8)^3}{(6.67\times 10^{-11}).(6.70\times 10^{22})} }[/tex]
T = 9738253.26 s
T = 112.7 days
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At what temperature (degrees Fahrenheit) is the Fahrenheit scale reading equal to:_____
(a) 3 times that of the Celsius and
(b) 1/5 times that of the Celsius
Answer:
C = 26.67° and F = 80°C = -20° and F = -4°Explanation:
Find:
3 times that of the Celsius and 1/5 times that of the CelsiusComputation:
F = (9/5)C + 32
3 times that of the Celsius
If C = x
So F = 3x
So,
3x = (9/5)x + 32
15x = 9x +160
6x = 160
x = 26.67
So, C = 26.67° and F = 80°
1/5 times that of the Celsius
If C = x
So F = x/5
So,
x/5 = (9/5)x + 32
x = 9x + 160
x = -20
So, C = -20° and F = -4°
An aluminum rod 17.400 cm long at 20°C is heated to 100°C. What is its new length? Aluminum has a linear expansion coefficient of 25 × 10-6 C-1.
Answer:
the new length is 17.435cm
Explanation:
the new length is 17.435cm
pls give brainliest
The new length of aluminum rod is 17.435 cm.
The linear expansion coefficient is given as,
[tex]\alpha=\frac{L_{1}-L_{0}}{L_{0}(T_{1}-T_{0})}[/tex]
Given that, An aluminum rod 17.400 cm long at 20°C is heated to 100°C.
and linear expansion coefficient is [tex]25*10^{-6}C^{-1}[/tex]
Substitute, [tex]L_{0}=17.400cm,T_{1}=100,T_{0}=20,\alpha=25*10^{-6}C^{-1}[/tex]
[tex]25*10^{-6}C^{-1} =\frac{L_{1}-17.400}{17.400(100-20)}\\\\25*10^{-6}C^{-1} = \frac{L_{1}-17.400}{1392} \\\\L_{1}=[25*10^{-6}C^{-1} *1392}]+17.400\\\\L_{1}=17.435cm[/tex]
Hence, The new length of aluminum rod is 17.435 cm.
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Which examination technique is the visualization of body parts in motion by projecting x-ray images on a luminous fluorescent screen?
Answer:
Fluoroscopy
Explanation:
A Fluoroscopy is an imaging technique that uses X-rays to obtain real-time moving images of the interior of an object. In its primary application of medical imaging, a fluoroscope allows a physician to see the internal structure and function of a patient, so that the pumping action of the heart or the motion of swallowing, for example, can be watched.
CAN SOMEONE HELP ME PLEASE ITS INTEGRATED SCIENCE AND I AM STUCK
Answer:
[tex]\huge \boxed{\mathrm{Option \ D}}[/tex]
Explanation:
Two forces are acting on the object.
Subtracting 2 N from both forces.
2 N → Object ← 5 N
- 2 N - 2N
0 N → Object ← 3 N
The force 3 N is pushing the object to the left side.
The mass of the object is 10 kg.
Applying formula for acceleration (Newton’s Second Law of Motion).
a = F/m
a = 3/10
a = 0.3
Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.
A. Which skater, if either, has the greater momentum after the push-off? Explain.
B. Which skater, if either, has the greater speed after the push-off? Explain.
Answer:
the two ice skater have the same momentum but the are in different directions.
Paula will have a greater speed than Ricardo after the push-off.
Explanation:
Given that:
Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.
A. Which skater, if either, has the greater momentum after the push-off? Explain.
The law of conservation of can be applied here in order to determine the skater that possess a greater momentum after the push -off
The law of conservation of momentum states that the total momentum of two or more objects acting upon one another will not change, provided there are no external forces acting on them.
So if two objects in motion collide, their total momentum before the collision will be the same as the total momentum after the collision.
Momentum is the product of mass and velocity.
SO, from the information given:
Let represent the mass of Paula with [tex]m_{Pa}[/tex] and its initial velocity with [tex]u_{Pa}[/tex]
Let represent the mass of Ricardo with [tex]m_{Ri}[/tex] and its initial velocity with [tex]u_{Ri}[/tex]
At rest ;
their velocities will be zero, i.e
[tex]u_{Pa}[/tex] = [tex]u_{Ri}[/tex] = 0
The initial momentum for this process can be represented as :
[tex]m_{Pa}[/tex][tex]u_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]u_{Ri}[/tex] = 0
after push off from each other then their final velocity will be [tex]v_{Pa}[/tex] and [tex]v_{Ri}[/tex]
The we can say their final momentum is:
[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex] = 0
Using the law of conservation of momentum as states earlier.
Initial momentum = final momentum = 0
[tex]m_{Pa}[/tex][tex]u_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]u_{Ri}[/tex] = [tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]
Since the initial velocities are stating at rest then ; u = 0
[tex]m_{Pa}[/tex](0) + [tex]m_{Pa}[/tex](0) = [tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]
[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex] = 0
[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] = - [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]
Hence, we can conclude that the two ice skater have the same momentum but the are in different directions.
B. Which skater, if either, has the greater speed after the push-off? Explain.
Given that Ricardo weighs more than Paula
So [tex]m_{Ri} > m_{Pa}[/tex] ;
Then [tex]\mathsf{\dfrac{{m_{Ri}}}{m_{Pa} }= 1}[/tex]
The magnitude of their momentum which is a product of mass and velocity can now be expressed as:
[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] = [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]
The ratio is
[tex]\dfrac{v_{Pa}}{v_{Ri}} =\dfrac{m_{Ri}}{m_{Pa}} = 1[/tex]
[tex]v_{Pa} >v_{Ri}[/tex]
Therefore, Paula will have a greater speed than Ricardo after the push-off.
(A) Both the skaters have the same magnitude of momentum.
(B) Paula has greater speed after push-off.
Conservation of momentum:Given that two skaters Paula and Ricardo are initially at rest.
Ricardo weighs more than Paula.
Let us assume that the mass of Ricardo is M, and the mass of Paula is m.
Let their final velocities be V and v respectively.
(A) Initially, both are at rest.
So the initial momentum of Paula and Ricardo is zero.
According to the law of conservation of momentum, the final momentum of the system must be equal to the initial momentum of the system.
Initial momentum = final momentum
0 = MV + mv
MV = -mv
So, both of them have the same magnitude of momentum, but in opposite directions.
(B) If we compare the magnitude of the momentum of Paula and Ricardo, then:
MV = mv
M/m = v/V
Now, we know that M>m
so, M/m > 1
therefore:
v/V > 1
v > V
So, Paula has greater speed.
Learn more about conservation of momentum:
https://brainly.com/question/2141713?referrer=searchResults
A radar installation operates at 9000 MHz with an antenna (dish) that is 15 meters across. Determine the maximum distance (in kilometers) for which this system can distinguish two aircraft 100 meters apart.
Answer:
R = 36.885 km
Explanation:
In order to distinguish the two planes we must use the Rayleigh criterion that establishes two distinguishable objects if in their diffraction the central maximum of one coincides with the first minimum of the other
The diffraction equation for slits is
a sin θ = m λ
the first minimum occurs for m = 1
sin θ = λ a
as the diffraction experiments the angles are very small, we approximate
sin θ = θ
θ = λ / a
This expression is for a slit, in the case of circular objects, when solving the system in polar coordinates, a numerical constant appears, leaving the expression of the form
θ = 1.22 λ / a
In this problem they give us the frequency, let's find the wavelength with the relation
c = λ f
λ = c / f
θ = 1.22 c/ f a
since they ask us for the distance between the planes, we can use the definition of radians
θ = s / R
if we assume that the distance is large, we can approximate the arc to the horizontal distance
s = x
we substitute
x / R = 1.22 c / fa
R = x f a / 1.22c
Let's reduce the magnitudes to the SI system
f = 9000 MHz = 9 109 Hz
a = 15 m
x = 100 m
let's calculate
R = 100 10⁹ 15 / (1.22 3 108)
R = 3.6885 10⁴ m
let's reduce to km
R = 3.6885 10¹ km
R = 36.885 km
A 28.0 kg child plays on a swing having support ropes that are 2.30 m long. A friend pulls her back until the ropes are 45.0 ∘ from the vertical and releases her from rest.
A: What is the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing?
B: How fast will she be moving at the bottom of the swing?
C: How much work does the tension in the ropes do as the child swings from the initial position to the bottom?
Answer
A)184.9J
B)=3.63m/s
C) Zero
Explanation:
A)potential energy of the child at the initial position, measured relative the her potential energy at the bottom of the motion, is
U=Mgh
Where m=28kg
g= 9.8m/s
h= difference in height between the initial position and the bottom position
We are told that the rope is L = 2.30 m long and inclined at 45.0° from the vertical
h=L-Lcos(x)= L(1-cosx)=2.30(1-cos45)
=0.674m
Her Potential Energy will now
= 28× 9.8×0.674
=184.9J
B)we can see that at the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:
E= 0.5mv^2
where
m = 28.0 kg is the mass of the child
v is the speed of the child at the bottom position
Solving the equation for v, we find
V=√2k/m
V=√(2×184.9/28
=3.63m/s
C)we can find work done by the tension in the rope is given using expresion below
W= Tdcosx
where W= work done
T is the tension
d = displacement of the child
x= angle between the directions of T and d
In this situation, we have that the tension in the rope, T, is always perpendicular to the displacement of the child, d. x= 90∘ and cos90∘=0 hence, the work done is zero.
