Answer:
g' = g/9 = 1.09 m/s²
Explanation:
The magnitude of free fall acceleration at the surface of earth is given by the following formula:
g = GM/R² ----- equation 1
where,
g = free fall acceleration
G = Universal Gravitational Constant
M = Mass of Earth
R = Distance between the center of earth and the object
So, in our case,
R = R + 2 R = 3 R
Therefore,
g' = GM/(3R)²
g' = (1/9) GM/R²
using equation 1:
g' = g/9
g' = (9.8 m/s)/9
g' = 1.09 m/s²
Answer:
The magnitude of the free-fall acceleration [tex]g_h = 1.09m/s^2[/tex]Explanation:
Surface of earth,
[tex]g = \frac{GM}{R^2}\\\\g = 9.8m/s^2[/tex]
free fall acceleration at height h,
[tex]g_h = \frac{GM}{(R+h)^2}[/tex]
where
G = gravitational constant
R = Radius of earth
M = mass of earth
therefore,
[tex]\frac{g_h}{g} = \frac{\frac{GM}{(R+h)^2}}{\frac{GM}{R^2}}\\\\ \frac{g_h}{g} = \frac{R^2}{(R+h)^2}\\\\g_h = g\frac{R^2}{(R+h)^2}[/tex]
Where height h = 2R
[tex]g_h = 9.8\frac{R^2}{(R+2R)^2}\\\\g_h = 9.8\frac{R^2}{(3R)^2}\\\\g_h = 9.8\frac{R^2}{(9R^2}\\\\g_h = 1.09m/s^2[/tex]
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What is the separation in meters between two slits for which 594 nm orange light has its first maximum at an angle of 32.8°?
Answer:
1.1micro meter
Explanation:
Given that
Constructive interference is
ma = alpha x sin theta
Alpha = 1 x 594 x10^ -9/ sin 32.8°
= 1.1 x 10^ -6m
Explanation:
A particle moves along line segments from the origin to the points (1, 0, 0), (1, 5, 1), (0, 5, 1), and back to the origin under the influence of the force field. F(x, y, z)= z^2i + 4xyj + 5y^2kFind the work done.
Answer:
0 J
Explanation:
Since work done W = ∫F.dr and F(x, y, z)= z²i + 4xyj + 5y²k and dr = dxi + dyj + dzk
F.dr = (z²i + 4xyj + 5y²k).(dxi + dyj + dzk) = z²dx + 4xydy + 5y²dz
W = ∫F.dr = ∫z²dx + 4xydy + 5y²dz = z²x + 2xy² + 5y²z
We now evaluate the work done for the different regions
W₁ = work done from (0,0,0) to (1,0,0)
W₁ = {z²x + 2xy² + 5y²z}₀₀₀¹⁰⁰ = 0²(1) + 2(1)(0)² + 5(0)²(0) - [(0)²(0) + 2(0)(0)² + 5(0)²(0)] = 0 - 0 = 0 J
W₂ = work done from (1,0,0) to (1,5,1)
W₂ = {z²x + 2xy² + 5y²z}₁₀₀¹⁵¹ = (1)²(1) + 2(1)(5)² + 5(5)²(1) - [0²(1) + 2(1)(0)² + 5(0)²(0)] = 1 + 50 + 125 - 0 = 176 J
W₃ = work done from (1,5,1) to (0,5,1)
W₃ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁵¹ = 1²(0) + 2(0)(5)² + 5(5)²(1) - [(1)²(1) + 2(1)(5)² + 5(5)²(1)] = 125 - (1 + 50 + 125) = 125 - 176 = -51 J
W₄ = work done from (0,5,1) to (0,0,0)
W₄ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁰⁰ = (0)²(0) + 2(0)(0)² + 5(0)²(0) - [1²(0) + 2(0)(5)² + 5(5)²(1)] = 0 - 125 = -125 J
The total work done W is thus
W = W₁ + W₂ + W₃ + W₄
W = 0 J + 176 J - 51 J - 125 J
W = 176 J - 176 J
W = 0 J
The total work done equals 0 J
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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Suppose you want a telescope that would allow you to see distinguishing features as small as 3.5 km on the Moon some 384,000 km away. Assume an average wavelength of 550 nm for the light received.Required:What is the minimum diameter mirror on a telescope?
Explanation:
[tex]\theta=1.22 \frac{\lambda}{D}[/tex]
And, from equation ( 2 ), we get
[tex]\theta=\frac{x}{d}[/tex]
Thus,
[tex]\frac{x}{d} &=1.22 \frac{\lambda}{D}[/tex]
[tex]D &=1.22 \frac{\lambda d}{x}[/tex]
[tex]=1.22 \frac{550 \times 10^{-9} 3.84 \times 10^{8}}{5 \times 10^{3}}[/tex]
[tex]=0.0515 \mathrm{m}[/tex]
Thus, the diameter of the telescope's mirror that would allow us to see details as small as is
g In the atmosphere, the shortest wavelength electromagnetic waves are called A. infrared waves. B. ultraviolet waves. C. X-rays. D. gamma rays. E.
Answer:gamma ray
Explanation:
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
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be observed when the detector is 3 meters from the sample?
Using the sample in above question how many counts per second would be observed when the detector is 10 meters away from the sample?
Answer:
At 3 meter distance, the per-second count is 222.22 and at a 10 meter distance, the per-second count is 20.
Explanation:
The number of particles (N) counts are inversely proportional to the distance between the source and the detector.
By using the below formula we can find the number of counts.
[tex]N2 = \frac{(D1)^2}{(D2)^2} \times N1 \\N1 = 2000 \\D 1 = 1 \ meter \\D2 = 3 \\[/tex]
The number of count per second, when the distance is 3 meters.
[tex]= \frac{1}{3^2} \times 2000 \\= 222.22[/tex]
Number of count per second when the distance is 10 meters.
[tex]= \frac{1}{10^2} \times 2000 \\= 20[/tex]
Scouts at a camp shake the rope bridge they have just crossed and observe the wave crests to be 9.70 m apart. If they shake the bridge twice per second, what is the propagation speed of the waves (in m/s)?
Answer:
The speed of the wave is 19.4 m/s
Explanation:
The wave's crest to crest distance (the wavelength of this rope's wave) λ= 9.70 m
The bridge is shaken twice, meaning that two wavelengths passed a given point on the rope per sec. The frequency of a wave is the amount of that wave that passes a given point in a second.
this means that the frequency f = 2 Hz
The speed of a wave = fλ = 9.70 x 2 = 19.4 m/s
Polarized light passes through a polarizer. If the electric vector of the polarized light is horizontal what, in terms of the initial intensity I0, is the intensity of the light that passes through a polarizer if the polarizer is tilted 22.5° from the horizontal?
Answer: I0*0.853
Explanation:
Ok, the Malus's law says that:
If you have light polarized along a given line with an intensity I0, and it passes through a polaroid which axis of polarization forms an angle θ with respect to the polarization of the light, then the intensity of the resulting beam is:
I(θ) = I0*cos^2(θ)
For example, if the axis of the polaroid is exactly the same as the axis of polarization of the light beam that will impact it, then we have θ = 0°, and the equation above says that the intensity of the beam will not change.
In this particular case, we have that the intensity of the light is I0, and the angle is θ = 22.5°
Then:
I(22.5°) = I0*cos^2(22.5°) = I0*0.853
The physics of wind instruments is based on the concept of standing waves. When the player blows into the mouthpiece, the column of air inside the instrument vibrates, and standing waves are produced. Although the acoustics of wind instruments is complicated, a simple description in terms of open and closed tubes can help in understanding the physical phenomena related to these instruments. For example, a flute can be described as an open-open pipe because a flutist covers the mouthpiece of the flute only partially. Meanwhile, a clarinet can be described as an open-closed pipe because the mouthpiece of the clarinet is almost completely closed by the reed.
1. Consider a pipe of length 80.0 cm open at both ends. What is the lowest frequency f of the sound wave produced when you blow into the pipe?
2. A hole is now drilled through the side of the pipe and air is blown again into the pipe through the same opening. The fundamental frequency of the sound wave generated in the pipe is now:______.
a. the same as before.
b. lower than before.
c. higher than before.
3. If you take the original pipe in Part A and drill a hole at a position half the length of the pipe, what is the fundamental frequency of the sound that can be produced in the pipe?
4. What frequencies, in terms of the fundamental frequency of the original pipe in Part A, can you create when blowing air into the pipe that has a hole halfway down its length?
4-1. Recall from the discussion in Part B that the standing wave produced in the pipe must have an antinode near the hole. Thus only the harmonics that have an antinode halfway down the pipe will still be present.
A. Only the odd multiples of the fundamental frequency.
B. Only the even multiples of the fundamental frequency.
C. All integer multiples of the fundamental frequency.
E. What length of open-closed pipe would you need to achieve the same fundamental frequency as the open pipe discussed in Part A?
A. Half the length of the open-open pipe.
B. Twice the length of the open-open pipe.
C. One-fourth the length of the open-open pipe.
D. Four times the length of the open-open pipe.
E. The same as the length of the open-open pipe.
F. What is the frequency of the first possible harmonic after the fundamental frequency in the open-closed pipe described in Part E?
F-1. Recall that possible frequencies of standing waves that can be generated in an open-closed pipe include only odd harmonics. Then the first possible harmonic after the fundamental frequency is the third
harmonic.
Answer:
1) f = 214 Hz , 2) answer is c , 3) f = 428 Hz , 4) f₂ = 428 Hz , f₃ = 643Hz
Explanation:
1) A tube with both ends open, the standing wave has a maximum amplitude and a node in its center, therefore
L = λ / 2
λ = 2L
λ = 2 0.8
λ = 1.6 m
wavelength and frequency are related to the speed of sound (v = 343 m / s)
v =λ f
f = v / λ
f = 343 / 1.6
f = 214 Hz
2) In this case the air comes out through the open hole, so we can assume that the length of the tube is reduced
λ' = 2 L ’
as L ’<L₀
λ' <λ₀
f = v / λ'
f' > fo
the correct answer is c
3) in this case the length is L = 0.40 m
λ = 2 0.4 = 0.8 m
f = 343 / 0.8
f = 428 Hz
4) the different harmonics are described by the expression
λ = 2L / n n = 1, 2, 3
λ₂ = L
f₂ = 343 / 0.8
f₂ = 428 Hz
λ₃ = 2 0.8 / 3
λ₃ = 0.533 m
f₃ = 343 / 0.533
f₃ = 643 Hz
4,1) as we have two maximums at the ends, all integer multiples are present
the answer is C
E) the length of an open pipe created that has a wavelength of lam = 1.6 m is requested
in this pipe there is a maximum in the open part and a node in the closed part, so the expression
L = λ / 4
L = 1.6 / 4
L = 0.4 m
the answer is C
F) in this type of pipe the general expression is
λ = 4L / n n = 1, 3, 5 (2n + 1)
therefore only odd values can produce standing waves
λ₃ = 4L / 3
λ₃ = 4 0.4 / 3
λ₃ = 0.533
f₃ = 343 / 0.533
f₃ = 643 Hz
Monochromatic light of wavelength, λ is traveling in air. The light then strikes a thin film having an index of refraction n1 that is coating a material having an index of refraction n2. If n1 is larger than n2, what minimum film thickness will result in minimum reflection of this light?
Answer:
tmin= lambda/2
Explanation:
See attached file pls
"If a beam of monochromatic light is passed though a slit of width 15 μm and the second order dark fringe of the diffraction pattern is at an angle of 5.2o from the central axis, what is the wavelength of the light?"
Answer:
λ= 5.4379 10⁻⁷ m = 543.79 nm
Explanation:
The phenomenon of diffraction is described by the expression for destructive diffraction is
a sin θ = (m + 1/2) λ
λ = a sin θ / (m + 1/2)
let's reduce the magnitudes to the SI system
a = 15 um = 15 10⁻⁶ m
m = 2
θ = 5.2º
Let's calculate
λ = 15 10⁻⁶ sin 5.2 / (2 +1/2)
λ = 5.4379 10⁻⁷ m
Let's reduce to nm
λ= 5.4379 10⁻⁷ m = 543.79 nm
Improved balance is a primary benefit of regular cardiovascular exercise .
Answer:
Cardiovascular exercise is the activity use that aerobic metabolism and cellular reaction.
Explanation:
Cardiovascular exercise is activity increase heart rate and raises oxygen large muscle group of the body.Cardiovascular exercise is that contain cardio improve to the health mental health, heart health.Cardiovascular exercise such as walking, swimming, running is that exercise is benefit to the health.Cardiovascular exercise to the internal body organs that the healthy heart for the function and performance of the heart.Cardiovascular exercise that having involve feet of the ground this type of activity is called high impact of cardio.Cardio is a good and maintaining exercise for the lungs and heart or healthy bones.Cardio exercise is performed that to a water in reduce to the gravity of that pull on the body weight.Cardiovascular daily to build the stronger muscle and that control the blood pressure.Four charges each of magnitude 15 µC are arranged on the corners of a square of side 5 cm. What is the total potential energy of the system?
Answer:
-105J
Explanation:
See attached file
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 sinusoidal sound wave moves through a medium and is described by the displacement wave function s(x, t) = 1.99 cos(15.2x − 869t) where s is in micrometers, x is in meters, and t is in seconds. (a) Find the amplitude of this wave. µm (b) Find the wavelength of this wave. cm (c) Find the speed of this wave. m/s (d) Determine the instantaneous displacement from equilibrium of the elements of the medium at the position x = 0.050 9 m at t = 2.94 ms. µm (e) Determine the maximum speed of a element's oscillatory motion. mm/s
Answer:
a) A = 1.99 μm , b) λ = 0.4134 m , c) v = 57.2 m / s , d) s = - 1,946 nm ,
e) v_max = 1,739 mm / s
Explanation:
A sound wave has the general expression
s = s₀ sin (kx - wt)
where s is the displacement, s₀ the amplitude of the wave, k the wave vector and w the angular velocity, in this exercise the expression given is
s = 1.99 sin (15.2 x - 869 t)
a) the amplitude of the wave is
A = s₀
A = 1.99 μm
b) wave spectrum is
k = 2π /λ
in the equation k = 15.2 m⁻¹
λ = 2π / k
λ = 2π / 15.2
λ = 0.4134 m
c) the speed of the wave is given by the relation
v = λ f
angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 869 / 2π
f = 138.3 Hz
v = 0.4134 138.3
v = 57.2 m / s
d) To find the instantaneous velocity, we substitute the given distance and time into the equation
s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)
s = 1.99 sin (0.77368 - 2.55486)
remember that trigonometry functions must be in radians
s = 1.99 (-0.98895)
s = - 1,946 nm
The negative sign indicates that it shifts to the left
e) the speed of the oscillating part is
v = ds / dt)
v = - s₀(-w) cos (kx -wt)
the maximum speed occurs when the cosines is 1
v_maximo = s₀w
v_maximum = 1.99 869
v_maximo = 1739.31 μm / s
let's reduce to mm / s
v_maxio = 1739.31 miuy / s (1 mm / 103 mu)
v_max = 1,739 mm / s
a) A is = 1.99 μm , b) λ is = 0.4134 m , c) v is = 57.2 m / s , d) s is = - 1,946 nm, e) v_max is = 1,739 mm / s
Calculation of Wavelength
When A sound wave has the general expression is:
Then, s = s₀ sin (kx - wt)
Now, where s is the displacement, Then, s₀ is the amplitude of the wave, k the wave vector, and w the angular velocity, Now, in this exercise the expression given is
s is = 1.99 sin (15.2 x - 869 t)
a) When the amplitude of the wave is
A is = s₀
Thus, A = 1.99 μm
b) When the wave spectrum is
k is = 2π /λ
Now, in the equation k = 15.2 m⁻¹
Then, λ = 2π / k
After that, λ = 2π / 15.2
Thus, λ = 0.4134 m
c) When the speed of the wave is given by the relation is:
Then, v = λ f
Now, the angular velocity and frequency are related is:
w is = 2π f
Then, f = w / 2π
After that, f = 869 / 2π
Now, f = 138.3 Hz
Then, v = 0.4134 138.3
Thus, v = 57.2 m / s
d) Now, To find the instantaneous velocity, When we substitute the given distance and time into the equation
Then, s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)
After that, s = 1.99 sin (0.77368 - 2.55486)
Then remember that trigonometry functions must be in radians
After that, s = 1.99 (-0.98895)
Thus, s = - 1,946 nm
When The negative sign indicates that it shifts to the left
e) When the speed of the oscillating part is
Then, v = ds / dt)
Now, v = - s₀(-w) cos (kx -wt)
When the maximum speed occurs when the cosines is 1
Then, v_maximo = s₀w
After that, v_maximum = 1.99 869
v_maximo = 1739.31 μm / s
Now, let's reduce to mm / s
Then, v_maxio = 1739.31 miuy / s (1 mm / 103 mu)
Therefore, v_max = 1,739 mm / s
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It takes 85N of force to accelerate a ball at a rate of 15 m/s². What is the mass of the ball?
Explanation:
F = ma
85 N = m (15 m/s²)
m ≈ 5.7 kg
Inside the wall of a house, an L-shaped section of hot-water pipe consists of three parts: a straight, horizontal piece 28.0 cm long, an elbow, and a straight vertical piece ℓ = 159 cm long. A stud and a second-story floorboard hold the ends of this section of copper pipe stationary. Find the magnitude and direction of the displacement of the pipe elbow when the water flow is turned on, raising the temperature of the pipe from 18.0°C to 40.2°C. (The coefficient of linear expansion of copper is
Answer:
The magnitude and direction are
7.638×10-4m
80.01°
Explanation:
We know that the coefficient of linear expansion for copper = 16.6×10^-6 m/m-C
ΔT = 40.2 - 18.0 = 28.5 C°
The expansion of horizontal pipe length can be calculated as
= (0.28)(16.6×10^-6)(28.5) = 13247×10^-8
= 0.0001325 m
The expansion of vertical pipe length = (1.28)(16.6×10^-6)(28.5) = 60557×10^-8 = 0.000752229 m
horizontal displacement = 0.1325 mm
= 1.356×10^-4m
vertical displacement = 0.75223mm
=7.5223×10-4m
size of total displacement can be calculated as
√(x²+y²)
Where x and y are vertical and horizontal displacement respectively
= √(0.1325)²+(0.75223)² =
= 0.7638 mm
= 7.638×10-4m
Angle below horizontal = arctan Θ
= 0.75223/0.1325
=5.6772
= arctan (5.6772)
= 80.01°
Therefore, the the magnitude and direction of the displacement of the pipe elbow when the water flow is turned at (7.638×10-4m) 0.7638 mm and 80.01°
1.
(a)
P
center
Figure 1
A ball is released at point P with a tangential velocity of 5 ms to move in a circular track in a
vertical plane as shown in the Figure 1. Can the ball reach the highest point of the circular track
of radius 1.0 m? Give reasons. (4 marks]
Answer:
No.
Explanation:
Given the following :
Velocity (V) of ball = 5m/s
Radius = 1m
Can the ball reach the highest point of the circular track
of radius 1.0 m?
The highest point in the track could be considered as the diameter of the circle :
Radius = diameter / 2;
Diameter = (2 * Radius) = (2*1) = 2
Maximum height which the ball can reach :
Using the relation :
Kinetic Energy = Potential Energy
0.5mv^2 = mgh
0.5v^2 = gh
0.5(5^2) = 9.8h
0.5 * 25 = 9.8h
12.5 = 9.8h
h = 12.5 / 9.8
h = 1.2755
h = 1.26m
Therefore maximum height which can be reached is 1.26m.
Since h < Diameter
What is the observed wavelength of the 656.3 nm (first Balmer) line of hydrogen emitted by a galaxy at a distance of 2.40 x 108 ly
Answer:
λ = 667.85 nm
Explanation:
Let f be the frequency detected by the observer
Let v be the speed at which the observer is moving.
Now, when the direction at which the observer is moving is away from the source, we have the frequency as;
f = f_o√((1 - β)/(1 + β))
From wave equations, we know that the wavelength is inversely proportional to the frequency. Thus, wavelength is now;
λ = λ_o√((1 + β)/(1 - β))
Where, β = Hr/c
H is hubbles constant which has a value of 0.0218 m/s • ly
c is speed of light = 3 × 10^(8) m/s
r is given as 2.40 x 10^(8) ly
Thus,
β = (0.0218 × 2.4 x 10^(8))/(3 × 10^(8))
β = 0.01744
Since we are given λ_o = 656.3 nm
Then;
λ = 656.3√((1 + 0.01744)/(1 - 0.01744))
λ = 667.85 nm
6. What is the bulk modulus of oxygen if 32.0 g of oxygen occupies 22.4 L and the speed of sound in the oxygen is 317 m/s?
Answer:
[tex] \boxed{\sf Bulk \ modulus \ of \ oxygen \approx 143.5 \ kPa} [/tex]
Given:
Mass of oxygen (m) = 32.0 g = 0.032 kg
Volume occupied by oxygen (V) = 22.4 L = 0.0224 m³
Speed of sound in oxygen (v) = 317 m/s
To Find:
Bulk modulus of oxygen
Explanation:
[tex]\sf Density \ of \ oxygen \ (\rho) = \frac{m}{V}[/tex]
[tex]\sf \implies Bulk \ modulus \ of \ oxygen \ (B) = v^{2} \rho[/tex]
[tex]\sf \implies B = v^{2} \times\frac{m}{V}[/tex]
[tex]\sf \implies B = {(317)}^{2} \times \frac{0.032}{0.0224} [/tex]
[tex]\sf \implies B = {(317)}^{2} \times 1.428[/tex]
[tex]\sf \implies B = 100489 \times 1.428[/tex]
[tex]\sf \implies B = 143498.292 \: Pa[/tex]
[tex]\sf \implies B \approx 143.5 \: kPa[/tex]
A loud sound is produced in the downtown section of a city. Which of the following is least likely to occur with the sound waves?
A. The sound wave will reflect off Buildings and automobiles.
B. The air will transmit the sound in longitudinal waves of energy.
C. All those sound waves will be absorbed by the surroundings.
D. The sound will bend spread between buildings by the fraction.
Answer:
A. The sound wave will reflect off Buildings and automobiles.
Explanation:
This is because the sound waves would more likely propagate through diffraction through buildings and transmission through the air. It is also more likely to be absorbed by buildings than for multiple reflections to occur off buildings and automobiles. In the process of reflection, these materials would absorb the sound energy thereby reducing its ability to reflect.
Ocean waves with a wavelength of 120 m are coming in at a rate of 8 per minute. What is their speed?
Explanation:
We know that,
[tex]v(wave \: speed) = f(frequency) \times \alpha (wavelength)[/tex]
frequency (f) = 1 / t (sec) = 8/60 = 0.13 Hz
V ( wave speed) = 0.13 * 120 = 16 m/sec
The speed of the given wave is equal to 15.96 m/s.
What are frequency and wavelength?The frequency of the wave can be defined as the number of oscillations of a wave in one second. The frequency has S.I. units which can be expressed as per second or hertz (Hz).
The wavelength can be described as the distance between the two adjacent points in phase. Two crests or two troughs of a wave are separated by a distance is called wavelength.
The relationship between wavelength (λ), frequency (ν), and wave speed (V):
V = νλ
Given, the frequency of the wave, ν = 8 min⁻¹ = 0.133 s⁻¹
The wavelength of the wave, λ = 120 m
The speed of the waves can calculate from the above-mentioned relationship:
V = νλ = 120 × 0.133 = 15.96 m/s
Therefore, the speed of the wave is equal to 15.96 m/s.
Learn more about wavelength and frequency, here:
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The concentration of sodium and potassium ions in the blood and body fluid is regulated by :
Answer:
Kidney
Explanation:
One of the main function of the kidney is to maintain the homeostasis of sodium and potassium ions in the blood and body.
Aldosterone is a key steroid hormone that balances sodium and potassium ions in the blood and body fluid. Potassium and sodium ions generate electric impulse in the body which helps to perform different activities such as muscles flexing.
Kidney function for reabsorption and secretion, in which reabsorption of Na is done nd balances the sodium and potassium in the blood and body.
A generator consists of 1000 turns of wire around a coil of area 10cm2 (0.0010m 2 ). Calculate the peak output voltage of this generator when operating at 100 cycles/second in a 0.50 tesla field.
Answer:
The peak output voltage of this generator is 314.2 V.
Explanation:
Given;
the number of turns of the coil, N = 1000 turns
area of the coil, A = 0.001 m²
angular frequency of the coil, f = 100 cycles/seconds
magnitude of the magnetic field, B = 0.5 T
The peak output voltage of this generator is given by;
E = NBAω
Where;
ω is the angular velocity = 2πf
E = NBA(2πf)
E = 1000 x 0.5 x 0.001(2 x π x 100)
E = 314.2 V
Therefore, the peak output voltage of this generator is 314.2 V.
Two sources of light of wavelength 700 nm are 9 m away from a pinhole of diameter 1.2 mm. How far apart must the sources be for their diffraction patterns to be resolved by Rayleigh's criterion
Answer:
The distance is [tex]D = 0.000712 \ m[/tex]
Explanation:
From the question we are told that
The wavelength of the light source is [tex]\lambda = 700 \ nm = 700 *10^{-9} \ m[/tex]
The distance from a pin hole is [tex]x = 9\ m[/tex]
The diameter of the pin hole is [tex]d = 1.2 \ mm = 0.0012 \ m[/tex]
Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is
mathematically represented as
[tex]D = \frac{1.22 \lambda }{d }[/tex]
substituting values
[tex]D = \frac{1.22 * 700 *10^{-9} }{ 0.0012 }[/tex]
[tex]D = 0.000712 \ m[/tex]
We observe that a moving charged particle experiences no magnetic force. From this we can definitely conclude that:_______
a. no magnetic field exists in that region of space.
b. the particle must be moving parallel to the magnetic field.
c. the particle is moving at right angles to the magnetic field.
d. either no magnetic field exists or the particle is moving parallel to the magnetic field.
e. either no magnetic field exists or the particle is moving perpendicular to the magnetic field.
Answer:
b. the particle must be moving parallel to the magnetic field.
Explanation:
The magnetic force on a moving charged particle is given by;
F = qvBsinθ
where;
q is the charge of the particle
v is the velocity of the particle
B is the magnetic field
θ is the angle between the magnetic field and velocity of the moving particle.
When is the charge is stationary the magnetic force on the charge is zero.
Also when the charge is moving parallel to the magnetic field, the magnetic force is zero.
Therefore, when a moving charged particle experiences no magnetic force, we can definitely conclude that the particle must be moving parallel to the magnetic field.
b. the particle must be moving parallel to the magnetic field.
What is the maximum speed (in units of m/s) with which a car can round a
flat horizontal curve of radius (r=60 m), if the coefficient of static friction between tires and
road is (0.4)?
A) 5
B) 15.5
€) 240
D) 25.1
Answer:
B) 15.5 m/s
Explanation:
r = 60m
μs = 0.4
using the formula max V = √r*g*μs (flat roadway)
v = sqrt(60 * 10 * 0.4)
v = 15.5 m/s
A scuba diver fills her lungs to capacity (6.0 L) when 10.0 m below the surface of the water and begins to ascend to the surface. Assume the density of the water in which she is swimming is 1000 kg/m3 and use g = 10 m/s2A. Assuming the temperature of the air in her lungs is constant, to what volume must her lungs expand when she reaches the surface of the water?B. What effect would the warming of the air in her lungs have on the volume needed when she surfaces?C. Assuming the temperature of the air in her lungs is constant, what effect does her ascent have on the vrms of the air molecules in her lungs?
Answer:
Explanation:
As temperature is constant , we shall apply Boyle's law
P₁V₁ = P₂V₂
P₁ = pressure at depth of 10 m
= P + hdg , h = 10 , d = 10³ , g = 10
P is atmospheric pressure which is 10⁵ Pa
P₁ = 10⁵ + 10 x 10³ x 10
= 2 x 10⁵
applying the formula
2 x 10⁵ x 6 = 10⁵ x v
v = 2 x 6 = 12 L
volume will be doubled at the surface .
B )
warming of air at the surface will increase the volume of air in her lungs so so she will need more lung capacity .
C )
The rms value of a gas depends upon the temperature of the gas . As temperature of the gas is constant , the rms value of the gas particles will remain constant when she goes to the surface .
The lungs will expand 12 L when she reaches the surface of the water, and the warming of the air results in more lung capacity, and [tex]\rm V_{rms}[/tex] the value remains the same.
What is Boyle's law?According to the law, the pressure of the gas is inversely proportional to the volume of the gas. In other words when the pressure of the gas increases the volume of the gas decreases.
We know the pressure at the 10 meters depth:
[tex]\rm P_1 = P+h\times \rho\times g[/tex]
Where P = Atmospheric pressure
h = Depth
ρ =Density of the water
We have: [tex]\rm P = 10^5 \ Pa[/tex], h = 10 meters, and [tex]\rm \rho = 1000 \ kg/m^3[/tex], and [tex]\rm g = 10 \ m/s^2[/tex]
Putting the values in the above equation, we get:
[tex]\rm P_1 = 10^5+ 10\times 1000\times 10[/tex]
[tex]\rm P_1 = 2\times 10^5[/tex]
From the Boyle's law:
[tex]\rm P_1\times V_1 = P_2\times V_2[/tex]
[tex]\rm 2\times10^5\times 6 = 10^5\times V_2[/tex]
[tex]\rm V_2 = 12 \ L[/tex]
We know that as the air at the surface warms, the volume of air in her lungs expands, requiring more lung capacity.
The temperature of the gas is constant and [tex]\rm V_{rms}[/tex] values for gas depend on the temperature of the gas, but here the temperature of the gas is constant thus, the [tex]\rm V_{rms}[/tex] will remains constant.
Thus, the lungs will expand 12 L when she reaches the surface of the water, and the warming of the air results in more lung capacity, and [tex]\rm V_{rms}[/tex] the value remains the same.
Learn more about the Boyle's law here:
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The resistor used in the procedures has a manufacturer's stated tolerance (percent error) of 5%. Did you results from Data Table agree with the manufacturer's statement? Explain.
Resistor Measured Resistance
100 99.1
Answer:
e% = 0.99% this value is within the 5% tolerance given by the manufacturer
Explanation:
Modern manufacturing methods establish a tolerance in order to guarantee homogeneous characteristics in their products, in the case of resistors the tolerance or error is given by
e% = | R_nominal - R_measured | / R_nominal 100
where R_nominal is the one written in the resistance in your barcode, R_measured is the real value read with a multimeter and e% is the tolerance also written in the resistors
let's apply this formula to our case
R_nominal = 10 kΩ = 10000 Ω
R_measured = 100 99 Ω
e% = | 10000 - 10099.1 | / 10000 100
e% = 0.99%
this value is within the 5% tolerance given by the manufacturer
