The total momentum of the block and putty prior to their collision is
(0.454 kg) (0 m/s) + (5.70 × 10⁻² kg) (8.99 m/s) ≈ 0.512 kg•m/s
and the total momentum after the collision is
(0.454 kg) v + (5.70 × 10⁻² kg) (-8.99 m/s)
where v is the velocity of the block. Momentum is conserved, so
(0.454 kg) v + (5.70 × 10⁻² kg) (-8.99 m/s) = 0.512 kg•m/s
==> v ≈ 2.26 m/s
The total work done on the block by the spring as it gets compressed by a distance x is equal to the change in the block's kinetic energy:
1/2 (25.0 N/m) x ² = 1/2 (0.454 kg) (2.26 m/s) - 0
==> x ≈ 0.202 m ≈ 20.2 cm
Students have 4 different pairs of sunglasses, each with lenses where the polarization filter is oriented in different directions (indicated by the direction of lines drawn on the lenses below). They are planning to ride rafts down a river on a sunny day where there is lots of glare on the water. Which pair would they want to wear to eliminate the glare from the sun
Answer:
the lens you must select has an angle of 143º measured with respect to the horizontal, this angle is 53º with respect to the vertical.
Explanation:
The glare is caused by the reflection of light in the water, the polarization of the reflected light is polarized in a direction parallel to the surface of the water, the polarization is total for the angles
n = tan [tex]\theta_{p}[/tex]
\theta_{p} = tan⁻¹ n
the refractive index for seawater is 1.33
\theta_{p}= tan⁻¹ 1.33
\theta_{p} = 53º
for this angle the light is totally polarized, for the other angles the polarization is partial.
Based on this, the lenses must eliminate this polarization, so its polarization direction must have 90º with respect to this polarization,
\theta_{lens} = 53 +90
\theta_{lens}= 143º
Therefore, the lens you must select has an angle of 143º measured with respect to the horizontal, this angle is 53º with respect to the vertical.
A lens that could work is one that is polarized 45º with respect to the vertical.
The glass that is worn to eliminate glare must be able to remove glare away from the user.
What are sunglasses?The term sunglasses refers to the kind of glasses that are worn to prevent glare from the sun. This is because the glasses are made of polaroid materials that eliminate glare.
The question is incomplete as the sunglasses are not shown. However, the glass that is worn to eliminate glare must be able to remove glare away from the user.
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if a car travels 200 m to the east in 8.0s what is the car's average velocity.
Answer:
25mph
Explanation:
you do distance divided by time. 200/8=25
For a damped oscillator with a mass of 360 g, a spring constant 130 N/m and a damping coefficient of 86 g/s, what is the ratio of the amplitude of the damped oscillations to the initial amplitude at the end of 11 cycles
Answer:
A/A₀ = 0.5106
Explanation:
To do this, we need to use several formulas and expressions. First, we need to know the period of time of the oscillator. This can be calculated using the following expression:
ω = 2π/T (1)
But angular frequency (ω) can be also be calculated using:
ω = √(k/m) (2)
Using (1) and (2), we can solve for the period T:
2π/T = √(k/m) (mass in kg)
2π/T = √(130/0.360)
2π/T = √361.11
2π/T = 19.003
T = 2π/19.003 = 0.331 s
Now, the amplitude A at a time t, is:
A = x exp(-bt/2m) (3)
At time 0, A = x. so A₀ = x
The problem states that we have 11 cycles respect to the initial amplitude, so expression (3) can be rewritten as:
A = x exp(-b(17t/2m)) using b as kg/s = 0.086 kg/s
Replacing the data we have:
A = x exp(-0.086(17*0.331)/2*0.36)
A = x exp(-0.086 * 7.815)
A = x exp(-0.6721)
A = 0.5106x (4)
Now, doing the ratio with the innitial we have:
A / A₀ = 0.5106x / x
The ratio is:
A/A₀ = 0.5106Hope this helps
use a trigonometric equation to determine the leg of this triangle
C=90°
A=30°
c=10m
What is a?
Answer: 5
Explanation: B is for sure 60°, c* cosB = 10*1/2 =5
Does the mass change how a bouncy ball bounces
Answer:
hiii what's up. and good morning hyd
Please help, confused on this whole lesson
Answer:
1,4,3and 2 would happen hope this helps
Three liquids that do not mix are poured into a cylindrical container with a diameter of 10.0 cm. The densities and volumes of the liquids are as follows.
Liquid 1: ????1 = 2.80 ✕ 103 kg/m3 and V1 = 2.00 ✕ 10−3 m3
Liquid 2: ????2 = 1.00 ✕ 103 kg/m3 and V2 = 1.50 ✕ 10−3 m3
Liquid 3: ????3 = 0.600 ✕ 103 kg/m3 and V3 = 1.00 ✕ 10−3 m3
Determine the pressure on the bottom of the container.
Answer:
P = 9622.9 Pa = 9.62 KPa
Explanation:
First, we will calculate the mass of all three liquids:
m = ρV
where,
m = mass of liquid
ρ = density of liquid
V = Volume of liquid
FOR LIQUID 1:
m₁ = (2.8 x 10³ kg/m³)(2 x 10⁻³ m³) = 5.6 kg
m₂ = (1 x 10³ kg/m³)(1.5 x 10⁻³ m³) = 1.5 kg
m₃ = (0.6 x 10³ kg/m³)(1 x 10⁻³ m³) = 0.6 kg
The total mass will be:
m = m₁ + m₂+ m₃ = 5.6 kg + 1.5 kg + 0.6 kg
m = 7.7 kg
Hence, the weight of the liquids will be:
W = mg = (7.7 kg)(9.81 m/s²) = 75.54 N
Now, we calculate the base area:
A = πr² = π(0.05 m)²
A = 7.85 x 10⁻³ m²
Now the pressure will be given as:
[tex]P = \frac{F}{A}\\\\P = \frac{75.54\ N}{7.85\ x\ 10^{-3}\ m^2}[/tex]
P = 9622.9 Pa = 9.62 KPa
Please help
Which of the following is an example of heat energy being transformed into mechanical energy
A. An electric mixer being used to beat eggs
B. A person turning a crank to sharpen a pencil.
C. A car battery being used to power a car radio
D. Boiling water producing steam used to power a boat
Answer:
D
Explanation:
because it is the only one that has something to do with heat keyword would be boiling
In a stunt, three people jump off a platform and fall 8.5 m onto a large air bag. A fourth person at the other end of the air bag, the "flier," is launched 16 m vertically into the air. Assume all four people have a mass of 74 kg. What is the momentum of the flier just after launch? Let upward be the positive direction.
Answer:
They Died Period NO MORE
Complete the sentence below using the words provided in parentheses (). For two universes that are the same size, the universe with the faster expansion rate must be _______ (younger/older) than the universe with the slower expansion rate. The slope of the line in the Hubble plot of the ______ (younger/older) universe will be ______ (steeper/flatter)
Answer:
For two universe that is the same size, the universe with the faster expansion rate must be younger than the universe with a slower expansion rate. The slope in the line of the Hubble plot of the older universe will be flatter.
Explanation:
Hubble's law establishes a direct relationship between a galaxy's distance and its redshift-determined recessional velocity. The redshifts of galaxy clusters and distant galaxies are equivalent and proportional to their distances from us, according to Hubble's theorem.
Thus, to copy and complete the given question.
For two universe that is the same size, the universe with the faster expansion rate must be younger than the universe with a slower expansion rate. The slope in the line of the Hubble plot of the older universe will be flatter.
3
Explain the working principles
of 아
has/ Constant volume
thermometer
a gas
Answer:
A constant volume gas thermometer works on a thermometric property of " change in pressure with temperature at a constant volume"
P Flag question
Matt, shown below, is able to move
the large truck because
8. Consider a capacitor that is made of two large conducting plates that are rectangular in shape (1 cm by 6 cm), aligned parallel to each other, and separated by an air-filled gap of 0.001 cm. This capacitor is included in a circuit where a battery provides 15,000 V of potential difference. When the capacitor is fully charged in this circuit, what is the c
Answer: [tex]7.96\ \mu C[/tex]
Explanation:
Given
The dimension of the plate is [tex]1\ cm\times 6\ cm[/tex]
The gap between the plate is [tex]0.001\ cm[/tex]
Voltage applied [tex]V=15,000\ V[/tex]
The capacitance of the capacitor is
[tex]C=\dfrac{\epsilon_o A}{d}\\\\C=\dfrac{8.85\times 10^{-12}\times 1\times 6\times 10^{-4}}{10^{-5}}\\\\C=53.1\times 10^{-11}\ F[/tex]
Charge acquired by the capacitor
[tex]\Rightarrow Q=CV\\\Rightarrow Q=53.1\times 10^{-11}\times 15,000\\\Rightarrow Q=796.5\times 10^{-8}\\\Rightarrow Q=7.96\times 10^{-6}\ C[/tex]
A soccer ball with mass 0.450 kg is initially moving with speed 2.20 m/s. A soccer player kicks the ball, exerting a constant force of magnitude 38.0 N in the same direction as the ball's motion. Over what distance must her foot be in contact with the ball to increase the ball's speed to 6.00m/s?
Answer:
0.187 m
Explanation:
We'll begin by calculating the acceleration of the ball. This can be obtained as follow:
Mass (m) = 0.450 Kg
Force (F) = 38 N
Acceleration (a) =?
F = m × a
38 = 0.450 × a
Divide both side by 0.450
a = 38 / 0.450
a = 84.44 m/s²
Finally, we shall determine the distance. This can be obtained as follow:
Initial velocity (u) = 2.20 m/s.
Final velocity (v) = 6 m/s
Acceleration (a) = 84.44 m/s²
Distance (s) =?
v² = u² + 2as
6² = 2.2² + (2 × 84.44 × s)
36 = 4.4 + 168.88s
Collect like terms
36 – 4.84 = 168.88s
31.52 = 168.88s
Divide both side by 168.88
s = 31.52 / 168.88
s = 0.187 m
Thus, the distance is 0.187 m
Help pls... 100points
1 Answer:
The pitch of a sound depends on
a. frequency
b. intensity
c. amplitude
d. quality
2 Answer:
The loudness of a sound depends on
a. frequency
b. intensity
c. overtones
d. timbre
3 Answer:
The number of sound waves per unit time is called
a. amplitude
b. wavelength
c. speed
d. frequency
4 Answer:
The blending of fundamental tone and overtones produces a sound's
a. intensity
b. pitch
c. timbre
d. loudness
5 Answer:
Sonar is used
a. to locate objects in water
b. to determine underwater distances
c. for both a and b
d. for neither a nor b
6 Answer:
Sounds of frequencies below 20 Hz are called
a. fundamentals
b. overtones
c. ultrasonic sounds
d. infrasonic sounds
7 Answer:
A sound is music if it has
a. a definite pitch
b. a pleasing quality
c. rhythm
d. all of these characteristics
8 Answer:
The stirrup is a part of the
a. outer ear
b. middle ear
c. inner ear
d. cochlea
9 Answer:
An example of a percussion instrument is a (an)
a. trumpet
b. harp
c. drum
d. flute
10 Answer:
The speed of sound depends on
a. temperature
b. density
c. elasticity
d. all of these factors
11 Answer:
Sounds with frequencies above 20,000 Hz are called _____.
12 Answer:
The science of sound is called _____.
13 Answer:
High-frequency waves are used in a sound navigation and ranging
system called _____.
14 Answer:
Sounds produced when sections of a string vibrate separately are
called _____.
15 Answer:
Sound without pleasing quality, identifiable pitch, and repeating
patterns is called _____.
16 Answer:
The speed of sound decreases with increasing temperature.
a. true
b. false
17 Answer:
Sound waves are transverse.
a. true
b. false
18 Answer:
The decibel is a unit of pitch.
a. true
b. false
19 Answer:
A change in the pitch of a sound due to motion of source or
receiver (observer) is called the Doppler effect.
a. true
b. false
20 Answer:
Guitars and cellos are examples of percussion instruments.
a. true
b. false
Answer:
Explanation:
1. a as pitch depends on frequency
2. b as loudness depends on intensity
3. d as that is definition of frequency
4. c as that is definition of timbre
5. c as sonar can do both
6. d as that is definition of infrasound
7. d
8. b as that is one of three bones in the middle ear
9. c as you hit the drum to make sound
10. d
11. ultrasound
12. acoustics
13. ultrasound
14. harmonics? or resonance?
15. noise
16. b
17. b
18. b
19. a
20. b
Which term refers to how often a person works out?
Answer: Frequency
Explanation: Frequency is term which best describes how often a person exercises.
Answer:
frequency
Explanation:
I took the. test
¿Cuál es la frecuencia de rotación en la tierra?
Answer:
frequency is approximately 11.5 µHz, or more exactly, 11.5740740e-6 Hz
Explanation:
An echo is heard from a building 0.4 s after you shout "hello." How many feet away is
the building if the air temperature is 27°C? (Velocity of sound in air at 0°C is 1087 ft/s)
Answer:
Circular motion: find period, find radius, find velocity, find centripetal acceleration 27 V= T a =vºlr=rw
Explanation:
Your _________ is simply your weight-to-height ratio which can be calculated using certain body measurements.
O health
O size
O weight loss
O BMI
Answer:
SiZe
Explanation:
bad bunny is the best!!!
The figure below shows regions of the electromagnetic
spectrum
Gamma
Radio Microwaves Infrared Visible Ultraviolet X-rays Rays
Which of the following waves has the highest frequency?
{GAMMA RAYS}
Answer:
The electromagnetic spectrum in order of increasing frequency is - radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays and gamma rays.
The frequency of the gamma rays is >3×10
19
m.
Hence, the gamma rays has the maximum frequency in the electromagnetic spectrum.
what heat transfer occurs when warm air rises
Suppose that you make a series RC circuit with a capacitor and a known resistor that has a 5% tolerance: R= 5.20 ± 0.26kΩ. You plan is to measure the time constant of the circuit in order to determine the value of the capacitor. You do ten trials of your experiment and measure the time constant each time, and then you calculate the average and standard error of the ten results to determine that τ= 2.150 ± 0.002s.
Which of the following is the best estimate of the uncertainty on your determination of the value of your capacitor?
a. 1%
b. 0.1%
c. 5%
Answer:
correct answer is C
Explanation:
The time constant of an RC circuit is
τ = RC
so to find the capacitance
C = τ/ R
C = 2.150 / 5.20 10³
C = 4.13 10⁻⁴ F
to find the error we use the worst case
ΔC = | [tex]|\frac{dC}{d \tau }| \ \Delta \tau + | \frac{dC}{dR} | \ \Delta R[/tex]
the absolute value guarantees that we find the worst case, we evaluate the derivatives
ΔC = 1 /R Δτ + τ/R² ΔR
the absolute values of the errors are
Δτ = 0.002 s
ΔR = 0.3 kΩ
we substitute
ΔC = 0.002 /5.20 10³ + 2.150/(5.20 10³)² 0.3 10³
ΔC = 3.8 10⁻⁷ + 1.74 10⁻⁵
ΔC = 1.77 10⁻⁵ F
the uncertainty or error must be expressed with a significant figure
ΔC = 2 10⁻⁵ F
the percentage error is
Er% =[tex]\frac{\Delta C}{C} \ 100[/tex]
Er% = [tex]\frac{2 \ 10^{-5} }{ 4.13 \ 10^{-4} } \ 100[/tex]
Er% = 4.8%
the correct answer is C
A 12 ohm resistor and a 24 ohm resistor are connected in parallel across a 6.0 V battery. The correct statement from the following is a. the current is the same in each resistor. b. the power dissipated is the same in each resistor. c. the larger current is in the larger resistor. d. the voltage difference between the ends of each resistor is the same.
Answer:
d. the voltage difference between the ends of each resistor is the same.
Explanation:
A resistor is a circuit component that offers opposition to the flow of current. Resistors may be connected in series or in parallel.
Resistors in parallel are connected at common junctions. The potential difference (voltage difference between the ends of each resistor) is the same for each resistor in a parallel connection while the current across each resistor is different.
In addition to producing images, ultrasound can be used to heat tissues of the body for therapeutic purposes. An emitter is placed against the surface of the skin; the amplitude of the ultrasound wave at this point is quite large. When a sound wave hits the boundary between soft tissue and bone, most of the energy is reflected. The boundary acts like the closed end of a tube which can lead to standing waves. Suppose 0.7 MHz ultrasound is directed through a layer of tissue at a bone 0.55 cm below the surface. Remember, sound waves in the body travel at 1540 m/s. Explain.
Answer:
ΔT = 4.058 10²⁰ [tex]\frac{S_o^2}{r^2 \ c_e}[/tex]
Explanation:
In this experiment the system can be approximated as a tube with one end open and the other closed.
The open end is where the ultrasound emitter is and the closed end where the limit between the tissue and the bone is, the length of the tube is L = 0.55 cm = 5.5 10⁻³ m
a node is formed at the closed end and a belly at the open end, so the resonance has the form
λ = 4L 1st harmonic
λ = 4/3 L third harmonic
λ = 4/5 L fifth harmonic
λ = 4L / (2n + 1) n = 0, 1, 2, (2n + 1)
This wave is a standing wave therefore energy density remains in place
P = 1/2 ρ v (w S₀)²
angular velocity is related to frequency
w = 2π f
we substitute
E = P = 2π² ρ v f² S₀²
if this energy per unit area is transformed into heat
E = m c_e DT
let's use the concept of density
ρ = m / V
m = ρ V
if there are no losses in the system
½ π² ρ v f² S₀² = ρ V c_e ΔT
ΔT = [tex]\frac{\pi ^2 \ v \f^2 S_o^2}{2V \ c_e}[/tex]
When analyzing this expression the temperature increase is
* quadratic at the frequency and maximum amplitude of the wave
* proportional to the speed of the wave in the tissue
* inversely proportional to tissue volume
we can approximate the volume of the tissue to the volume of a cylinder tube
V = π r² L
ΔT = [tex]\frac{\pi \ v \ f^2 S_o^2 }{r^2 \ L \ c_e}[/tex]
we calculate
ΔT = π 1450 (0.7 10⁶)² S₀² /( r² 5.5 10-3 c_e)
ΔT = 4.058 10²⁰ [tex]\frac{S_o^2}{r^2 \ c_e}[/tex]
Plz help w answer 1:/ confused ash
Answer:
I would say d I had the same question yesterday and I got it correct so hope that helps
The chart lists the masses of four planets.
Planetary Masses
Planet
Mass
Neptune
1.02 x 1026
Uranus
8.68 x 1025
Mars
6.42 x 1023
Venus
4.87 X 1024
According to evidence that supports Einstein's general
theory of relativity, which list shows the planets that
would cause curvature in space-time from the least
amount of curvature to the greatest?
O Mars, Venus, Uranus, Neptune
O Neptune, Uranus, Venus, Mars
O Neptune, Uranus, Mars, Venus
O Venus, Mars, Uranus, Neptune
Answer:
I think it's A
Explanation:
It's definitely not B on edge
The correct option for the given question about Einstein's general
theory of relativity is Option A) Mars, Venus, Uranus, Neptune.
What is the Einstein's general theory of relativity?Albert Einstein established that the rules of physics apply to all non-accelerating observers in his theory of special relativity.He also demonstrated that the speed of light in a vacuum remains constant regardless of the velocity of an observer.The theory may be used to anticipate everything and describes how objects behave in space and time. For instance: that if there are black holes or not, if Gravity can causes light to bend, The way Mercury behaves when it is in orbit and many more interesting things.As a Conclusion, we can state that the planets who would cause curvature in space time from the least amount of curvature to the greatest will be in order Mars, Venus, Uranus, Neptune.
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1. Light moving through air is incident on a piece of crown glass (1.61) Jt an angle of 45. What is the angle of refraction?
a test tube has a diameter of 3cm . how many turns would a piece of thread of length 90.42 make round test tube
Answer:
4.8 turns would be made around the tube
Explanation:
You need the circumference of the tube since it's just a lifted circle
2×pi×3=18.85
90.42/18.85=4.79
And round that to 4.8
Consider a composite cube made of epoxy with fibers aligned along one axis of the cube (the fibers are parallel to four of the twelve cube edges). If the cube can only be loaded in axial tension such that the force is uniformly applied over - and is normal to - a cube face, what is the lowest possible positive length change the cube can experience under this tension
The question is incomplete. The complete question is :
Consider a composite cube made of epoxy with fibers aligned along one axis of the cube (the fibers are parallel to four of the twelve cube edges). If the cube can only be loaded in axial tension such that the force is uniformly applied over - and is normal to - a cube face, what is the lowest possible positive length change the cube can experience under this tension? The applied tensile force is 102 KN. The unloaded cube edge length is 56 mm. The glass fibers have an elastic modulus of 200 GPa. The epoxy has an elastic modulus of 38 GPa. The cube is comprised of 18 vol% epoxy (the balancing vol % is glass fiber). Hint: The loading axis is intentionally unspecified. Answer Format: Lowest possible length increase (change of length) under tension.
Solution :
Given :
[tex]$E_{glass fibre}$[/tex] = 200 GPa
[tex]$V_{glass fibre} = 82\%$[/tex]
[tex]$E_{epoxy}$[/tex] = 38 GPa
[tex]$V_{epoxy} = 82\%$[/tex]
Edge length = 56 mm
Cube is loaded in axial tension such that the force is uniformly applied over a cube face.
[tex]$E_{\text{composite}}=\frac{E_{glass fibre} \times E_{epoxy}}{(E_{glass fibre .E_{epoxy}})+(E_{fibre}.V_{glass fibre})}$[/tex]
[tex]$E_{composite} = \frac{200 \times 38}{(200 \times 0.18)+(38\times 0.82)}$[/tex]
[tex]$=113.16 $[/tex] GPa
Applied stress [tex]$=\frac{\text{applied load}}{\text{area}}$[/tex]
[tex]$\sigma=\frac{102 \times 10^3 \ N}{56 \times 56 \times 10^{-6} \ m^2}$[/tex]
= 32.5 MPa
By Hooke's law
[tex]$\sigma = E . \epsilon$[/tex]
[tex]$\sigma = E. \frac{\Delta l}{l}$[/tex]
[tex]$\Delta l = \frac{\sigma}{E}\times l$[/tex]
Length change, [tex]$\Delta l =\frac{32.5 \times 10^6 \ Pa}{113.16 \times 10^9 \ Pa}\times 56 \times 10^{-2} \ m$[/tex]
[tex]$\Delta l = \frac{32.5 \times 56}{113.16} \times 10^{-3} \ mm$[/tex]
= 0.016 mm
NO LINKS: Which of the following best demonstrates Newton's Third Law?
Gravity is a force pulling down on you.
The ahrder you hit a volleyball the faster it will get over the net
Air resistance slows down a parachutist
Your foot pushing on the floor and the floor pushing back
The best demonstration that applies to Newton's Third Law of motion would be D) When you walk your foot pushes down on the ground while the ground pushes back on your foot.
------------------
Newton's Third Law states that for every action, there is an equal and opposite reaction. This is actually explains that forces come in pairs and forces are an interaction between two objects. As per the correct option given in the question explains Newton’s Third Law.
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When you walk your foot (say object A) pushes down on the ground while the ground (say object Q) pushes back on your foot with the same force but in the opposite direction.