Answer:
81,725 N/m^2
Explanation:
Given the following :
Height of water = 2m
Density of water = 1000kg/m^3
Atmospheric pressure (Pat) = 101325 N/m^2
Lowest pressure (Pl) can be obtained thus ;
Pat - Pl = density × height × acceleration due to gravity
101325 - Pl = 1000 × 2 × 9.8
101325 - Pl = 19600
101325 - 19600 = Pl
Lowest pressure (Pl) = 81,725 N/m^2
81,725 N/m^2 = (81,725/101325) atm
= 0.8065630 = 0.8065 atm
A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 193 m/s and a frequency of 235 Hz . The amplitude of the standing wave at an antinode is 0.380 cm.
A) Calculate the amplitude at a point on the string a distance of 18.0 cm from the left-hand end of the string.
B) How much time does it take the string to go from its largest upward displacement to its largest downward displacement at this point?
C) Calculate the maximum transverse velocity of the string at this point.
D) Calculate the maximum transverse acceleration of the string at this point.
Answer:
A. We know that amplitude at x is
Asin (kx)
But k= 2πf/v
k= 2*3.132*235/193= 7.65
So A = 0.35*sin( 7.65x 0.18)= 0.00841m
C
Vmax = Amplitude x angular velocity
= 0.0084 x 2πf
= 0.0084* 2*3.142* 235= 12.4m/s
D. Maximum acceleration = omega² x Amplitude
= (2πf)²* 0.00841= 183.40m/s²
12. A sprinter has an acceleration of 5m/s” during the first 2 seconds of the race. What
velocity does she reach after this time?
Answer:
v=10m/s
Explanation:
[tex]a=\frac{vf-vi}{t}\\ 5m/s^{2}=\frac{vf-0}{2}\\ vf=5*2\\vf=10 m/s[/tex]
We can model the human back as a pivoted rod?
Answer:
So the answer is yes, we can the back be shaped like a spinning rod
spinal column that is approximated by a long and narrow rod,
Explanation:
The bone system of the body is very well modeled in physics, the back has a spinal column that is approximated by a long and narrow rod, this rod is fixed in the lower part to the coccyx and has a weight in the upper part (head), this rod has longitudinal vertical movement and twisting movement around the lower part of the bar.
So the answer is yes, we can the back be shaped like a spinning rod
Speedy Sue, driving at 35.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 160 m ahead traveling at 5.20 m/s. Sue applies her brakes but can accelerate only at −1.90 m/s2 because the road is wet. Will there be a collision?
Answer:
Hence, there will be a collision
Explanation:
First we calculate total distance covered by the speedy sue's car before coming to rest:
2as = Vf² - Vi²
where,
a = deceleration = - 1.9 m/s²
s = distance covered = ?
Vf = Final Velocity = 0 m/s (since car finally stops)
Vi = Initial Velocity = 35 m/s
Therefore,
2(-1.9 m/s²)s = (0 m/s)² - (35 m/s)²
s = 322.37 m
Now, we calculate time taken by car to stop:
Vf = Vi + at
0 m/s = 35 m/s + (-1.9 m/s²)t
t = 18.42 s
Now, we calculate distance traveled by van in this time:
s₁ = V₁t
where,
s₁ = distance traveled by van = ?
V₁ = speed of van = 5.2 m/s
Therefore,
s₁ = (5.2 m/s)(18.42 s)
s₁ = 95.78 m
Now, for collision to occur, the following relation must be satisfied:
s ≥ 160 m + s₁
using values:
322.37 m > 160 m + 95.78 m
322.37 m > 255.78 m
Hence, there will be a collision
A sinusoidal transverse wave travels along a long, stretched string. The amplitude of this wave is 0.0957 m, its frequency is 3.75 Hz, and its wavelength is 1.97 m. What is the shortest transverse distance between a maximum and a minimum of the wave?
Answer:
Shortest distance = 0.1914 m
Explanation:
Given that,
Amplitude of the wave is 0.0957 m
Frequency of the wave is 3.75 Hz
Wavelength of the wave is 1.97 m
We need to find the shortest transverse distance between a maximum and a minimum of the wave.
The distance between maximum point in positive axis and the baseline is equal to amplitude.
Shortest distance = 2 A
D = 2 × 0.0957
D = 0.1914 m
So, the shortest transverse distance between a maximum and a minimum of the wave is 0.1914 m.
similarity of gravity and electromagnetic force
Answer:
Gravity is an attractive force as well as electromagnetic, but electromagnetic attracts and repels.
Explanation:
what was the average speed in km/h of a car that travels 490.0 km in 4.2 h?
Answer:
116.67 km/h
Explanation:
avarge speed = total distance / total time
Answer: 490 / 4.2 = 116.67 km/h
How could you record the number 4000 and report 2 significant figures?
Explanation:
Write in scientific notation.
4000 = 4.0×10³
URGENT!!! Which option(s) correctly define an electric circuit? (Select all that apply) a set-up where current flows without a voltage difference an open path of conductors a set-up where current flows due to a voltage difference a closed path of conductors
Answer:
a set up where current flows without a voltage difference
Explanation:
because a circuit is a set up of different components, and throughout the circuit the voltage is the same, even with more components
Answer:
a set-up where current flows due to a voltage difference
a closed path of conductors
Explanation:
In a liquid with a density of 1400 kg/m3, longitudinal waves with a frequency of 390 Hz are found to have a wavelength of 7.60 m. Calculate the bulk modulus of the liquid. Express your answer in pascals.
Answer:
The bulk modulus of the liquid is 1.229 x 10¹⁰ Pa
Explanation:
Given;
density of liquid, ρ = 1400 kg/m³
frequency of the wave, f = 390 Hz
wavelength, λ = 7.60 m
The speed of the sound is given by;
v = fλ
v = 390 x 7.6
v = 2964 m/s
The bulk modulus of the liquid is given by;
[tex]v = \sqrt{\frac{B}{\rho}}\\\\v^2 = \frac{B}{\rho}\\\\B = \rho v^2[/tex]
where;
B is bulk modulus
B = (1400)(2964)²
B = 1.229 x 10¹⁰ N/m²
B = 1.229 x 10¹⁰ Pa
Therefore, the bulk modulus of the liquid is 1.229 x 10¹⁰ Pa
Here are the positions at three different times for a bee in flight (a bee's top speed is about 7 m/s). Time 6.6 s 6.9 s 7.2 s Position 1.8, 5.0, 0 m 0.5, 6.9, 0 m −0.4, 9.5, 0 m (a) Between 6.6 s and 6.9 s, what was the bee's average velocity? Be careful with signs. vavg, a = 7.6739 (b) Between 6.6 s and 7.2 s, what was the bee's average velocity? Be careful with signs. vavg, b = 4.58557 (c) Of the two average velocities you calculated, which is the best estimate of the bee's instantaneous velocity at time 6.6 s? (d) Using the best information available, what was the displacement of the bee during the time interval from 6.6 s to 6.65 s? Δr = m
Answer:
(A.) (- 4.33, 6.33 , 0); (B.) (- 3.66, 7.5, 0); (C.) average at (A) (- 4.33, 6.33 , 0) ; (D.) (- 0.2165, 0.3165, 0)
Explanation:
Given the following :
Time - - - - - - - 6.6s - - - - - - - - - 6.9s - - - - - 7.2s
Position - (1.8,5.0,0) - (0.5,6.9,0) - - (−0.4,9.5,0)
(a) Between 6.6 s and 6.9 s, what was the bee's average velocity?
Vavg = Distance / time
[(0.5,6.9,0) - (1.8,5.0,0)] / 6.9 - 6.6
Vavg = [(0.5 - 1.8), (6.9 - 5.0), (0 - 0)] / 0.3
Vavg = - 1.3 / 0.3, 1.9/0.3, 0/3
Vavg = (- 4.33, 6.33 , 0)
b) Between 6.6 s and 7.2 s, what was the bee's average velocity?
Vavg = [(−0.4,9.5,0) - (1.8,5.0,0)] / 7.2 - 6.6
Vavg = - 2. 2/0.6, 4.5/0.6, 0/0.6
Vavg = (- 3.66, 7.5, 0)
c.) Of the two averages (- 4.3, 6.3 , 0) is closer to the instantaneous Velocity at 6.6s
D.) (d) Using the best information available, what was the displacement of the bee during the time interval from 6.6 s to 6.65 s?
Displacement = Velocity * time
Vavg between 6.6 to 6.9 ; time = (6.65 - 6.6) = 0.05 s
= (- 4.33, 6.33 , 0) * 0.05
= (- 0.2165, 0.3165, 0)
A bat emits a sound at a frequency of 30.0 kHz as it approaches a wall. The bat detects a beat frequency of 700 Hz. The speed of the bat is closest to
Answer:
3.948m/s
Explanation:
To solve this we need to apply Doppler effect theory
So
To find the frequency received by insect will be gotten when the Source and observer both are moving in same direction which is given by
f1 = f0 x (V - Vo)/(V - Vs)
f0 = 30.0 kHz
V = 344 m/s
Vs will now be the speed of the bat and
Vo will be the speed of the object which is = 0 m/s
So substituting we have
f1 = 30 x 10^3 x (344- 0)/(344- Vs)
Next to find the frequency reflected by wall we use
f2 = f1 x (V + Vs)/(V + Vo)
So substituting the value of f1 calculated above we have
f2 = 30 x 10^3 x (344 + Vs) x (344 - 0)/[(344 - Vs) x (344 + 0)]
f2 = 30 x 10^3 x (344 + Vs)/(344- Vs)
But the beat frequency detected by bat is 700 Hz,
So we say
f2 - f0 = 700 Hz
30 x 10^3 x (344+ Vs)/(344 - Vs) - 30x 10^3 = 700
(344 + Vs)/(344 - Vs) = 1 + 700/30000 = 1.023
344 + Vs = 344 x 1.023 - Vs x 1.0233
Vs = 344 x ( 1.023 - 1)/(1 + 1.023)
So finally
Vs = Speed of source that is the bat is = 3.949m/s
Describe a sound wave. Discuss what determines pitch and volume.
Answer:
The amplitude of a sound wave is a reflection of how much energy is carried, which contributes to the intensity of the sound. Intensity is measured in decibels and is perceived as sound volume. Thus, the volume is proportional to the amplitude of the sound wave. The frequency of a sound wave is perceived as pitch.
Explanation:
The amount of energy carried by a sound wave is reflected in its amplitude, which affects the sound's intensity. The perception of intensity as sound loudness is expressed in decibels. Hence, the relationship between the volume and the sound wave's amplitude is clear. Pitch is the perceived frequency of a sound wave.
What is wave ?Wave is is a disturbance in a medium that carries energy as well as momentum . wave is characterized by amplitude, wavelength and phase. Amplitude is the greatest distance that the particles are vibrating. especially a sound or radio wave, moves up and down. Amplitude is a measure of loudness of a sound wave. More amplitude means more loud is the sound wave.
Wavelength is the distance between two points on the wave which are in same phase.
Phase is the position of a wave at a point at time t on a waveform.
There are two types of the wave longitudinal wave and transverse wave.
Longitudinal wave : in which, vibration of the medium (particle) is parallel to propagation of the wave. Sound wave is a longitudinal wave.
Transverse wave : in which, vibration of the medium (particle) is perpendicular to propagation of the wave. Light wave is a Transverse wave.
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You're driving on a straight road (in the x direction) at a constant speed of 28 m/s. In 10 seconds, you speed up to 37 m/s to pass a truck. (a) Assuming your car speeds up at a constant rate (constant force by the road on the tires), what is your average x component of velocity vavg,x during this maneuver
Answer:
The average [tex]x[/tex] component velocity [tex]V_{avg,x}[/tex] is 32.5 m/s
Explanation:
v = u +at bhjklj kj h
x = (u + v / 2 )t
Average velocity is given by
[tex]V_{avg} = \frac{Displacement}{Time} \\V_{avg} = \frac{x_{2} - x_{1} }{t_{2} - t_{1} }[/tex]
From the question,
Initial speed, u = 28 m/s
Final speed, v = 37 m/s
Time, t = 10 secs
From the formula
[tex]x = (\frac{u + v}{2})t\\[/tex]
where [tex]x[/tex] is the displacement
Put the given values into the equation to find the displacement [tex]x[/tex]
[tex]x = (\frac{28 + 37}{2}) 10\\ x = (\frac{65}{2})10\\ x = (\frac{32.5}{10})\\ x = 325 m[/tex]
Now, for the average [tex]x[/tex] component velocity [tex]V_{avg,x}[/tex]
[tex]V_{avg} = \frac{Displacement}{Time} \\V_{avg} = \frac{x_{2} - x_{1} }{t_{2} - t_{1} }[/tex]
[tex]V_{avg,x} = \frac{325 - 0}{10 - 0}\\V_{avg,x} = \frac{325}{10}\\ V_{avg,x} = 32.5 m/s[/tex]
Hence, the average [tex]x[/tex] component velocity [tex]V_{avg,x}[/tex] is 32.5 m/s
The average x component of velocity during the maneuver of the truck is 32.5 m/s.
Velocity:
The term velocity of any object is defined as the ratio of displacement covered by any object to the time taken by the object to displace.
Given data:
The magnitude of initial speed is, u = 28 m/s.
The magnitude of final speed is, v = 37 m/s.
The time interval is, t = 10 s.
The displacement covered by the truck is calculated as,
[tex]d = \dfrac{v+u}{2} \times t\\\\ d = \dfrac{37+28}{2} \times 10\\\\ d = 325 \;\rm m[/tex]
Now, the expression for the x-component of the average velocity is given as,
[tex]v_{x}=\dfrac{d}{t}[/tex]
Solving as,
[tex]v_{x}=\dfrac{325}{10}\\\\ v_{x}=32.5 \;\rm m/s[/tex]
Thus, we can conclude that the average x component of velocity during the maneuver of truck is 32.5 m/s.
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What is the variable that is manipulated by the experimenter during an experiment called
34. JAnswer this: Pure (24 carat) gold has a density of 19 g/mL. If you bought"gold" ring and found it had a volume of 0.3 ml and that it weighed 5.7 grams, is it pure gold? Show your work to support your answer.
Explanation:
Density = mass / volume
ρ = 5.7 g / 0.3 mL
ρ = 19 g/mL
Yes, it's pure gold.
Yes, it's pure gold.
How do you calculate gold density?The mass of an object divided by its volume is the formula for determining density. This is expressed mathematically as d = m/v, where d denotes density, m denotes mass, and v denotes object volume. The common measurements are kg/m³.What is the formula of volume?The basic formula for volume is length, breadth, and height, as opposed to length, width, and height for the area of a rectangular shape.What is the volume of pure gold?Gold has a specific gravity of 19.3. In light of this, 19.3 grams take up 1 cubic centimeter. 31.1 grams make up a Troy ounce. As a result, we need 31.1/19.3 = 1.61 cubic centimeters to equal 1 Troy Oz.According to the question:
Density = mass / volume.
ρ = 5.7 g / 0.3 mL.
ρ = 19 g/mL.
Hence, Yes it's pure gold.
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In the drawing, what is the vector sum of forces A→+B→+C→ if each grid square is 7.00 N on a side? If the resultant is eastward, enter a positive value and if the resultant is westward, enter a negative value.
Answer:
resultant force = 14 N ( East direction)
Explanation:
A = √( (4*7)² + (4*7)² )
A = 39.6 N
B = 4 * 7
B = 28 N
C = 2 * 7
C = 14 N
∑ y forces = Ay - B = (4*7) - 28 = 0
∑ x forces = Ax - C = (4*7) - 14 = 14 N
so the resultant force = 14 N ( East direction)
A train which is traveling at 73mi/hr applies its brakes as it reaches point A and slows down with a constant deceleration. Its decreased velocity is observed to be 56mi/hr as it passes a point 1/2mi beyond A. A car moving at 45mi/hr passes point B at the same instant that the train reaches point A. In an unwise effort to beat the train to the crossing, the driver "steps on the gas." Calculate the constant acceleration a that the car must have in order to beat the train to the crossing by 3.9sec and find the velocity v of the car as it reaches the crossing.
Answer:
First to find deceleration of the train we use
v²= u²+ 2as
56²= 73²+ 2(0.5)a
a= -2193mi/hr²
Then we find time in which the train does the intersection
Using
S= ut+ 1/2 at²
1= 73t-1/2(1293)t²
t =68.5s
But since the train is to intersect in 3.9s the time will be the difference which is
65.68s
So finding acceleration
S= ut + 1/2at²
1.3mi= 45/3600mi/s(65.58s)+ 1/2a(65.5)²
So a= 1.179ft/s²
To find velocity we use
V= u + at
= 45/3600mi/s + (-2.33E-4mis²)(65.58s)
V= 0.0271mi/s
= 97.6ft/s
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3
This question is incomplete, the complete question is;
Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' separated by a distance d is
|FI = |QQ'I / d²
where K = 1/4π∈0, and
∈0 = 8.854 × 10⁻¹² C²/(N.m²) is the permittivity of free space.
Consider two point charges located on the x-axis:
one charge, q₁ = -18.5 nC, is located at
x₁ = -1.715m; the charge q₂ = 30.5 nC, is at the origin ( x₂=0 )
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q₃ = 51.0 nC placed between q₁ and q₂ at x₃ = -1.085 m ?
Answer: (Fnet3)x = -3.3287 × 10⁻⁵ N
Explanation:
Given that;
Q₁ = -18.5 nC Q₃ = 51 nC Q₂ = 30.5 nC
x₁ = - 1.715m x₃ = - 1.085m x₂ = 0
Now
x - component of Net force on charge Q₃ is
(Fnet3)x = -K|Q₁I|Q₃I / r₁3² - -K|Q₂I|Q₃I / r₂3²
(Fnet3)x = -(9×10⁹)(51×10⁻⁹) [ 18.5 / ((-1.085 + 1.715)²) + (30.5 / (-1.085)² ] × 10⁻⁹
(Fnet3)x = -3.3287 × 10⁻⁵ N
After crossing the finish line, a race car slows down from 47 m/s to 32m/s in 3seconds. What is the car’s acceleration?
Answer: -5 m/s^2
Explanation: a = v - u/t
= 32 - 47/3
= -15/3
= -5 m/s^2
A rough value of deceleration of a skidding automobile is about 7.0m\s^2.using this how long does it take for a car going at 30m\s to stop after the skid starts.How far dose the car go in this time??
Explanation:
Given:
v₀ = 30 m/s
v = 0 m/s
a = -7.0 m/s²
Find: t and Δx
v = at + v₀
0 m/s = (-7.0 m/s²) t + 30 m/s
t = 4.3 seconds
v² = v₀² + 2aΔx
(0 m/s)² = (30 m/s)² + 2 (-7.0 m/s²) Δx
Δx = 64 meters
Jennifer has taught her pet rat to run a maze. She thinks that the rat will
go faster if she puts its favorite treat at the end. She has the rat run the
maze ten times with the favorite treat and ten times with a regular food
pellet. She uses a stopwatch to measure how long it takes for the rat to get
to the reward.
O Replication
O Repetition
for an ideal gas, a plot of In P versus 1 / T (in Kelvin) yields a straight line with a slope equal to
Which of the following does a microscope use to produce an image?
a) reflection
b) refraction
c) total internal reflection
d) two of A, B, and C
e) all of A, B, and C
Answer:
All of a,b, and c hope this helps
g The velocity (V) of a particle is given by 5t2 meters, where t is in s. Find the acceleration of the particle at t=3s.
Answer:
Acceleration, [tex]a=45\ m/s^2[/tex]
Explanation:
The velocity of a particle is given by :
[tex]V=5t^2[/tex]
t is time in seconds
The acceleration in terms of velocity is given by :
[tex]a=\dfrac{dv}{dt}\\\\a=\dfrac{d(5t^2)}{dt}\\\\a=5\times \dfrac{t^3}{3}[/tex]
We need to find the acceleration of the particle at t = 3 s. Put t = 3 s in the expression of a.
So,
[tex]a=\dfrac{5\times 3^3}{3}\\\\a=45\ m/s^2[/tex]
So, the acceleration of the particle at t = 3 s is [tex]45\ m/s^2[/tex].
Which of the following statements are true according to the uncertainty principle? A. It is impossible to simultaneously determine both the energy and the momentum of a particle with arbitrary accuracy. B. It is impossible to determine a particle's energy with arbitrary accuracy in a finite amount of time. C. It is impossible to simultaneously determine both the position and the momentum of a particle with arbitrary accuracy. D. It is impossible to measure the position of a particle with arbitrary accuracy in a finite amount of time. E. It is impossible to simultaneously measure both the energy and position of a particle with arbitrary accuracy.
Answer and Explanation: Heisenberg's Uncertainty Principle states that if the position of the particle is known, its momentum is unknown and vice-versa.
So, it is impossible to determine both the position and the momentum of a particle with arbitrary accuracy.
The statement, " It is impossible to simultaneously determine both the position and the momentum of a particle with arbitrary accuracy" is correct. Hence, option (c) is correct.
The given problem is based on the Heisenberg's uncertainty principle. As per the Heisenberg's uncertainty principle, "It is not possible to obtain both the momentum and position of particles at same time, if one is obtained with full certainty then other comes uncertain". And the expression for the Heisenberg's principle is,
[tex]\Delta x \times \Delta p \geq \dfrac{h}{4\pi}[/tex]
Here,
[tex]\Delta x[/tex] is the uncertainty in position.
[tex]\Delta p[/tex] is the uncertainty in momentum.
h is the Planck's constant.
So, as per the above definition the statement, " It is impossible to simultaneously determine both the position and the momentum of a particle with arbitrary accuracy" is justified.
Thus, we can conclude that the statement, " It is impossible to simultaneously determine both the position and the momentum of a particle with arbitrary accuracy" is correct. Hence, option (c) is correct.
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light of wavelength 587.5 nm illuminates a slit of width 0.75 mm. at what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be 0.85 mm from the central maximum?
1.085m
Explanation:
Using
a= lambda/sinစ
Sinစ= (587.5*10^-9) x 0.75*10^-3
= 0.000783
Sinစ=0.875*10^-3/d
0.000783= 0.875/d
d= 1.085m
A laser with a power of 1.0 mW has a beam radius of 1.0 mm. What is the peak value of the electric field in that beam
Answer:
The peak value of the electric field is 489.64 V/m
Explanation:
Given;
power of the laser, P = 1.0 mW = 1 x 10⁻³ W
Radius of the beam, R = 1.0 mm = 1 x 10⁻³ m
Area of the beam = πr² = π(1 x 10⁻³ )² = 3.142 x 10⁻⁶ m²
The average intensity of the light = P / A
The average intensity of the light = ( 1 x 10⁻³) / (3.142 x 10⁻⁶)
The average intensity of the light = 318.27 W/m²
The peak value of the electric field is given by;
[tex]E_o = \sqrt{\frac{2I_{avg}}{c\epsilon_o}}\\\\E_o = \sqrt{\frac{2(318.27)}{(3*10^8)(8.85*10^{-12})}}\\\\E_o = 489.64 \ V/m[/tex]
Therefore, the peak value of the electric field is 489.64 V/m.
Explain in your own words the interaction between the electric and magnetic fields that make up a light wave.
Answer:
They oscillates perpendicularly to one another, the oscillation of one field generates the other field.
Explanation:
In a light wave, an oscillating electric field of a light wave produces a magnetic field, and the magnetic field also oscillates to produce an electric field. The magnetic field and the electric field of a light wave both oscillates perpendicularly to one another. The resultant energy and direction of the wave generated as a result of these oscillating fields is propagated perpendicularly to both fields.
g A ball is thrown against the wall and bounces back with the same velocity. What type of collision is this
Answer:
Perfectly elastic collision
Explanation:
In a closed system, an elastic collision is a type of collision between two bodies, where the total momentum and kinetic energy are conserved.
We are told from the problem that the ball bounces back with its original velocity. For the ball to bounce back to the thrower in the first place, this is our first hint that the collision is elastic. If the collision was inelastic, the ball would most likely have stuck to the wall.
In addition to that, the velocity of the ball remains fairly unchanged even after the collision. This confirms that the kinetic energy it had before the collision is the same as the kinetic energy it has after the collision.
As a result of this, the collision is perfectly elastic