Answer
OPTION A is correct
a)increases by a factor of more than 2.
Explanation:
adiabatic process is a process that takes place without the transfer of heat or mass between a thermodynamic system and its surroundings.
It is a process in which no heat transfer takes place. It does not denote that the temperature is constant.
Then from an adiabatic system,
we know that
PV= constant
PV^γ= constant
γ=ratio of specific heat for the gas
Where P= pressure of the system
V= volume of the system
We can say P= 1/(V^γ)
Where γ >1
Which means the pressure is inversely proportional to the volume raise to power of gamma sign and whenever the pressure increases the volume decreases with V^γ.
The question says the volume is decreased by factor of two then V^γ will also decreased by factor of 2.
That means (2^γ) > 2
Since our γ > 1 Therefore, the pressure increases by a factor of more than 2.
Answer:
a) increases by a factor of more than 2.
Explanation:
In an adiabatic process,
[tex]PV^{Y} = constant[/tex]
we can say that the volume is inverse to the pressure as
[tex]V^{Y}[/tex]∝ [tex]\frac{1}{P}[/tex]
where
[tex]P[/tex] is the pressure
[tex]V[/tex] is the volume
[tex]Y[/tex] is the adiabatic index, and it is always greater than 1
If volume V is decreased by a factor of 2, then it is actually decreased by a factor of [tex]2^{Y}[/tex] in proportion to the pressure, and since [tex]Y[/tex] is always greater than 1, it is actually decreased by a factor greater than 2 in proportion to the pressure.
This means that the pressure will also increase by a factor greater than 2.
Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of the system by a factor of
Answer:
The frequency changes by a factor of 0.27.
Explanation:
The frequency of an object with mass m attached to a spring is given as
[tex]f[/tex] = [tex]\frac{1}{2\pi } \sqrt{\frac{k}{m} }[/tex]
where [tex]f[/tex] is the frequency
k is the spring constant of the spring
m is the mass of the substance on the spring.
If the mass of the system is increased by 14 means the new frequency becomes
[tex]f_{n}[/tex] = [tex]\frac{1}{2\pi } \sqrt{\frac{k}{14m} }[/tex]
simplifying, we have
[tex]f_{n}[/tex] = [tex]\frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }[/tex]
[tex]f_{n}[/tex] = [tex]\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }[/tex]
if we divide this final frequency by the original frequency, we'll have
==> [tex]\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }[/tex] ÷ [tex]\frac{1}{2\pi } \sqrt{\frac{k}{m} }[/tex]
==> [tex]\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }[/tex] x [tex]2\pi \sqrt{\frac{m}{k} }[/tex]
==> 1/3.742 = 0.27
A large number of very industrious people make a very long pole. It is 10.0 light years long! ( As they measure it. ) Soon a spaceship flies along the length of the pole at 90% the speed of light. How much time passes on the spaceship from the moment the ship passes the first end of the pole to the moment the ship passes the second end of the pole
Answer:
L = L0 ( 1 - v^2/c^2))1/2 where L0 is the proper length
L = 10 L-y (1 - .9^2)^1/2 = 4.36 L-y length of pole measured by ship
t = 4.36 L-y / .9 c = 4.84 y since the ship travels at .9 c
Find the sum of 46 and -46
Answer:
0
Explanation:
A 22-g bullet traveling 240 m/s penetrates a 2.0-kg block of wood and emerges going 150 m/s. If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?
Answer:
After the bullet emerges the block moves at 0.99 m/s
Explanation:
Given;
mass of bullet, m₁ = 22 g = 0.022 kg
initial speed of the bullet, u₁ = 240 m/s
final speed of the bullet, v₁ = 150 m/s
mass of block, m₂ = 2.0 kg
initial speed of the block, u₂ = 0
Let the final speed of the block = v₂
Apply principles of conservation of linear momentum;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.022 x 240 + 2 x 0 = 0.022 x 150 + 2v₂
5.28 = 3.3 + 2v₂
5.28 - 3.3 = 2v₂
1.98 = 2v₂
v₂ = 1.98 / 2
v₂ = 0.99 m/s
Therefore, after the bullet emerges the block moves at 0.99 m/s
The body moves at a speed of 2.61m/s after the bullet emerges.
According to the law of collision which states that the momentum of the body before the collision is equal to the momentum of the body after the collision.
The formula for calculating the collision of a body is expressed as:
p = mv
m is the mass of the body
v is the velocity of the body
Based on the law above;
[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]
v is the final velocity of the body after the collision
Substitute the given parameters into the formula as shown:
[tex]0.022(240) + 2(0) = (0.022+2)v\\ 5.28 = 2.022v\\v=\frac{5.28}{2.022}\\v= 2.611m/s[/tex]
This shows that the body moves at a speed of 2.61m/s after the bullet emerges.
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In the previous part, you determined the maximum angle that still allows the crate to remain at rest. If the coefficient of friction is less than 0.7, what happens to this angle? A. The maximum angle increases.B. The maximum angle decreases.C. The maximum angle remains the same.D. Simulation q not sure if needed.
Answer:
B. The maximum angle decreases
Explanation:
If θ be the maximum angle of a slope that allows a crate placed on it to remain at rest , following condition exists .
tanθ = μ , θ is called angle of repose . μ is coefficient of static friction .
So the tan of angle of repose θ is proportional to coefficient of static friction.
If coefficient of static friction is less than .7 , naturally angle of repose will also become less ,ie, it at lower angle of inclination , the object will start slipping .
A uniform meter rule with a mass of 200g is suspended at zero mark pivotes at 22.0cm mark. calculate the mass of the rule.
pls answer quickly. Thanks
Answer:
The mass of the rule is 56.41 g
Explanation:
Given;
mass of the object suspended at zero mark, m₁ = 200 g
pivot of the uniform meter rule = 22 cm
Total length of meter rule = 100 cm
0 22cm 100cm
-------------------------Δ------------------------------------
↓ ↓
200g m₂
Apply principle of moment
(200 g)(22 cm - 0) = m₂(100 cm - 22 cm)
(200 g)(22 cm) = m₂(78 cm)
m₂ = (200 g)(22 cm) / (78 cm)
m₂ = 56.41 g
Therefore, the mass of the rule is 56.41 g
If you converted 0.000013 to scientific notation, what would the prefix be to the correct number of significant digits?
Answer:
1.3
Explanation:
it will taken to as in from of standard form
Answer:
1.3 * 10^-5
Explanation:
We are learning about scientific notation.
When a number becomes a decimal followed before with zeroes, we know that the value of that number is decreasing. So instead of usually doing a positive exponent, we will do a negative exponent indicating we are going back.
So let's not only count the amount of zeroes followed before 13, but the decimal.
0.000013
The original number "1.3" went back 5 spaces, therefore making our exponent 5.
1.3 * 10^-5
What is the shortest possible time in which a bacterium to travel distance of 8.4cm across a Petri dish at a constant velocity of 1.2 cm/s
Answer:
[tex] \boxed{\sf Shortest \ possible \ time = 7 \ seconds} [/tex]
Given:
Distance travelled (s) = 8.4 cm
velocity (v) = 1.2 cm/s
To Find:
Shortest possible time (t) in which a bacterium travel a distance 8.4 cm across a Petri Dish
Explanation:
[tex] \boxed{ \bold{\sf Time \ (t) = \frac{Distance \ travelled \ (s)}{Velocity \ (v)}}}[/tex]
Substituting values of Distance travelled (s) & Velocity (v) in the equation:
[tex] \sf \implies t = \frac{8.4}{1.2} [/tex]
[tex] \sf \implies t = \frac{7 \times \cancel{1.2}}{ \cancel{1.2}} [/tex]
[tex] \sf \implies t = 7 \: s[/tex]
Assume that helium behaves as an ideal monatomic gas. If 2 moles of helium undergo a temperature increase of 100 K at constant pressure, how much energy has been transferred to the helium as heat
Answer:
6235.5J
Explanation:
Using ( nစ)p= ncp x change in temp
But cp= ( 1+ f/2)R
So cp= ( 1+ 3/2R
Cp= 5R/2
So = n x 5R/2x 150k
= 2 x 5/2x 8.314 x150
= 6235.5J
Answer:
2500 J
Explanation:
Q=(3/2)nRΔT
Q=(3/2)*2 mol*(8.314 J/mol*k)*100 k
Q=2494 J
A uniform crate with a mass of 22 kg must be moved up along the 15° incline without tipping. The force P is horizontal. Determine the corresponding magnitude of force P.
Answer:
[tex]F_x=208.25\ N[/tex]
Explanation:
Given that,
Mass of a crate is 22 kg
It moved up along the 15 degrees incline without tipping.
We need to find the corresponding magnitude of force P. The force P is acting in horizontal direction.
It means that the horizontal component of force is given by :
[tex]F_x=F\cos\theta\\\\F_x=mg\cos\theta\\\\F_x=22\times 9.8\times \cos(15)\\\\F_x=208.25\ N[/tex]
So, the horizontal component of force is 208.25 N.
Monochromatic light of wavelength, lambda, 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 minumim film thickness will result in minimum reflection of this light?A. lambda/(4*n2)B. lambda/n2C. lambda/4D. lambda(2*n1)E. lambdaF. lambda/(2*n2)G. lambda/n1H. lambda/(4n1)I. lambda/2
Answer:
The correct option is H
Explanation:
From the question we are told that
The index of refraction of coating is [tex]n_1[/tex]
The index of refraction of material is [tex]n_2[/tex]
Generally the condition for constructive for a thin film interference is mathematically represented
[tex]2 * t = [ m + \frac{1}{2}] \frac{\lambda}{n_1 }[/tex]
Here t represents the thickness
For minimum thickness m = 0
So
[tex]2 * t =0 + \frac{1}{2}\frac{\lambda}{n_1 }[/tex]
=> [tex]t =\frac{\lambda}{4n_1 }[/tex]
Air in a 124 km/h wind strikes head-on the face of a building 42 m wide by 73 m high and is brought to rest. If air has a mass of 1.3 kg per cubic centimeter, determine the average force of wind on the building.
Answer:
The average force of wind on the building is 4.728 x 10¹² N
Explanation:
Given;
speed of the air wind, v = 124 km/h
dimension of the building, 42 m wide by 73 m high
density of the air, ρ = 1.3 kg/cm³ =
speed of the air in m/s = 124/3.6 = 34.44 m/s
Area of the building, A = 42 m x 73 m = 3066 m²
density of the air in (k.g/m³);
[tex]\rho = \frac{1.3 \ kg}{cm^3} *(\frac{100\ cm}{1 \ m} )^3\\\\\rho = \frac{1.3 \ kg}{cm^3} *\frac{10^6\ cm^3}{1 \ m^3} = \frac{1.3*10^6 \ kg}{m^3}[/tex]
The average force of wind on the building;
F = mass flow rate x velocity
F = (ρvA) x V
F = ρAv²
F = 1.3 x 10⁶ x 3066 x (34.44)²
F = 4.728 x 10¹² N
Therefore, the average force of wind on the building is 4.728 x 10¹² N
Select the correct answer. Physics is explicitly involved in studying which of these activities? A. the mixing of metals to form an alloy B. the metabolic functions of a living organism C. the motion of a spacecraft under gravitational influence D. the depletion of the atmospheric ozone layer due to pollutants E. the killing of cancerous cells by radiation therapy
Answer:
C. the motion of a spacecraft under gravitational influence
#1 A boy pushes forward a cart of groceries with a total mass of 40 kg. What is
the acceleration of the cart if the net force on the cart is 60 N?
Explanation:
∑F = ma
60 N = (40 kg) a
a = 1.5 m/s²
What type of information is available to scientists through a Global Positioning System (GPS) device?
Answer:
GPS receivers provide location in latitude, longitude, and altitude. They also provide the accurate time. GPS includes 24 satellites that circle Earth in precise orbits.
A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm light from a He-Ne laser. Because the lecture hall is very large, the interference pattern will be projected on a wall that is 5.0 m from the slits. For easy viewing by all students in the class, the professor wants the distance between the m=0 and m=1 maxima to be 35 cm. What slit separation is required in order to produce the desired interference pattern?
Answer:
The distance of separation is [tex]d = 9.04 *10^{-6 } \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 633\ nm = 633 *10^{-9} \ m[/tex]
The distance of the screen is [tex]D = 5.0 \ m[/tex]
The distance between the fringes is [tex]y = 35 \ cm = 0.35 \ m[/tex]
Generally the distance between the fringes is mathematically represented as
[tex]y = \frac{ \lambda * D }{d }[/tex]
Here d is the distance of separation between the slit
=> [tex]d = \frac{ \lambda * D }{y }[/tex]
=> [tex]d = \frac{ 633 *10^{-9} * 5 }{ 0.35 }[/tex]
=> [tex]d = 9.04 *10^{-6 } \ m[/tex]
Water enters a student's house 10.0 m above the ground through a pipe with a cross section area of 1.00 x 10-4m2 at ground. Inside the house the pipe's cross section area is 0.50 x 10-4m2. The student in the house want to know the water pressure inside the pipe at the ground level. He first measured the volume of the bath tank that equals to 45.0 L. Then he fill the tank (the tank is 10 meters above the ground) inside the house with 90.0 seconds. The pipe inside the house is open with the sea level pressure The density of water is 1000 kgm3.
(a) Calculate the water speed at the ground pipe with larger cross section area and the water speed inside the house with smaller cross section area.
(b) Calculate the water pressure in the pipeline at the ground level.
Answer:
(a). V₁ = 10m/s (velocity inside the house), V₂ = 5m/s (velocity at ground level)
(b). P₂ = 236500 Pa
Explanation:
This is quite straight-forward so let us begin by defining the terms given.
Given that;
The cross-section area inside the student's house A₁ = 0.50 0.50 x 10-4m2.
Let us make the velocity of water inside the house be V₁
such that the Volume of water entering the per second is = A₁V₁
Therefore, in 90sec:
45 L = 90 A₁V₁
V₁ = 45 * 10⁻³m³ / 90*0.5*10⁻⁴
V₁ = 10m/s (velocity of water inside the house)
From the continuity equation we have that;
A₁V₁ = A₂V₂
0.5*10⁻⁴ * 10 = 1*10⁻⁴ V₂
V₂ = 5m/s (velocity at ground level)
(b). We are told to calculate the water pressure in the pipeline at the ground level.
Using Bernoulli's equation;
P₁ + pgh₁ + 1/2PV₁² (inside) = P₂ + pgh₂ + 1/2PV₂² (ground level)
1.01*10⁵ + 1000*9.8*10 + 1/2*1000*(10)² = P₂ + 0 + 1/2*1000*(5)²
P₂ (pressure) = 1.01*10⁵Pa
Therefore we have;
101000 + 98000 + 50000 = P₂ + 12500
P₂ = 236500 Pa
cheers I hope this helped !!
Please help and if you have the answer if you can please explain how you got it :)!
Answer:
The mass of ball C is greater than the mass of ball A but less than the mass of ball B.
Explanation:
From Newton's second law, net force = mass × acceleration.
Using the data for ball B, the acceleration of gravity near the surface of the moon is:
∑F = ma
9.6 N = (6 kg) a
a = 1.6 m/s²
Therefore, the mass of ball C is:
∑F = ma
6.6 N = m (1.6 m/s²)
m = 4.1 kg
You've recently read about a chemical laser that generates a 20.0-cm-diameter, 26.0 MW laser beam. One day, after physics class, you start to wonder if you could use the radiation pressure from this laser beam to launch small payloads into orbit. To see if this might be feasible, you do a quick calculation of the acceleration of a 20.0-cm-diameter, 110 kg, perfectly absorbing block.
Required:
a. What speed would such a block have if pushed horizontally 100 m along a frictionless track by such a laser?
b. Does this seem like a promising method for launching satellites?
Answer :
(a). The speed of the block is 0.395 m/s.
(b). No
Explanation :
Given that,
Diameter = 20.0 cm
Power = 26.0 MW
Mass = 110 kg
diameter = 20.0 cm
Distance = 100 m
We need to calculate the pressure due to laser
Using formula of pressure
[tex]P_{r}=\dfrac{I}{c}[/tex]
[tex]P_{r}=\dfrac{P}{Ac}
Put the value into the formula
[tex]P_{r}=\dfrac{26.0\times10^{6}}{\pi\times(10\times10^{-2})^2\times3\times10^{8}}[/tex]
[tex]P_{r}=2.75\ N/m^2[/tex]
We need to calculate the force
Using formula of force
[tex]F=P\times A[/tex]
[tex]F=P\times \pi r^2[/tex]
Put the value into the formula
[tex]F=2.75\times\pi (0.01)^2[/tex]
[tex]F=0.086\ N[/tex]
We need to calculate the acceleration
Using formula of force
[tex]F=ma[/tex]
Put the value into the formula
[tex]0.086=110\times a[/tex]
[tex]a=\dfrac{0.086}{110}[/tex]
[tex]a=0.000781\ m/s^2[/tex]
[tex]a=7.81\times10^{-4}\ m/s^2[/tex]
(a). We need to calculate speed of the block
Using equation of motion
[tex]v^2=u^2+2ad[/tex]
Put the value into the formula
[tex]v=\sqrt{2\times7.81\times10^{-4}\times100}[/tex]
[tex]v=0.395\ m/s[/tex]
(b). No because the velocity is very less.
Hence, (a). The speed of the block is 0.395 m/s.
(b). No
The tires of a car make 77 revolutions as the car reduces its speed uniformly from 92.0 km/h to 60.0 km/h. The tires have a diameter of 0.84 m.
1. What was the angular acceleration of the tires?
2. If the car continues to decelerate at this rate, how much more time is required for it to stop?
3. If the car continues to decelerate at this rate, how far does it go? Find the total distance.
The angular acceleration of the tires is -2.2 rad/s².
If the car continues to decelerate at this rate, the time required to stop is 27.66 s.
The total distance traveled by the car before stopping is 210.96 revolutions.
The given parameters;
number of revolutions of the tire, N = 77 revinitial linear speed of the car, u = 92 km/h = 25.56 m/sfinal linear speed of the tire, v = 60 km/h = 16.67 m/sdiameter of the tire, d = 0.84 mradius of the tire, r = 0.42 mThe angular acceleration of the tire is calculated as follows;
[tex]\omega _f^2 = \omega _i ^2 + 2\alpha \theta\\\\(\frac{16.67}{0.42} )^2 = (\frac{25.56}{0.42} )^2 + 2( 77 \ rev \times \frac{2 \pi \ rad}{1 \ rev} ) \alpha \\\\1575.33 = 3703.59 \ + \ 967.736 \alpha \\\\-2128.26 = 967.736 \alpha\\\\\alpha = \frac{-2128.26}{967.736} \\\\\alpha = - 2.2 \ rad/s^2[/tex]
When the car stops, the final angular speed = 0. The time for the motion is calculated as;
[tex]\omega _f = \omega _i + \alpha t\\\\0 = \omega _i + \alpha t\\\\0 = 60.86 + (-2.2)t\\\\0 = 60.86 - 2.2t\\\\2.2t = 60.86\\\\t = \frac{60.86}{2.2} \\\\t = 27.66 \ s[/tex]
The total distance traveled by the car before stopping;
[tex]\theta = \omega_i t + \frac{1}{2} \alpha t^2\\\\\theta = (60.86 \times 27.66) \ + \ (0.5 \times -2.2\times 27.66^2)\\\\\theta = 841.8 \ rad\\\\\theta = 841.8 \ rad \times\frac{1 \ rev}{2\pi \ rad} = 133.96 \ rev[/tex]
total distance = 133.96 + 77 = 210.96 revolutions.
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1. The most likely injuries in an Anatomy class are (circle all that apply)
a. chemical spill
b. cut from scalpel
c, burn from
open
flame
d. foreign object or splash in eye
e, animal bite
What are the density, specific gravity and mass of the air in a room whose dimensions are 4 m * 6 m * 8 m at 100 kPa and 25 C.
Answer:
Density = 1.1839 kg/m³
Mass = 227.3088 kg
Specific Gravity = 0.00118746 kg/m³
Explanation:
Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³
Now, from tables, density of air at 25°C is 1.1839 kg/m³
Now formula for density is;
ρ = mass(m)/volume(v)
Plugging in the relevant values to give;
1.1839 = m/192
m = 227.3088 kg
Formula for specific gravity of air is;
S.G_air = density of air/density of water
From tables, density of water at 25°C is 997 kg/m³
S.G_air = 1.1839/997 = 0.00118746 kg/m³
Two parallel very long straight wires carrying current of 5A each are kept at a separation of 1m. If the currents are in the same direction, the force per unit length between them is __________
Answer:
The force per unit between the two parallel wires with same current flowing in the same direction is 5 x 10⁻⁶ N/m repulsive force.
Explanation:
Given;
current though the two parallel wires, I₁ and I₂ = 5A
distance between the two wires, R = 1 m
The force per unit of the wires is calculated as;
[tex]\frac{F}{L} = \frac{\mu_o I_1I_2}{2\pi R}[/tex]
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
Substitute in the given values into the equation and determine the force per unit length (F/L).
[tex]\frac{F}{L} = \frac{\mu_o I_1I_2}{2\pi R} \\\\ \frac{F}{L} = \frac{4\pi *10^{-7}*5*5}{2\pi *1}\\\\ \frac{F}{L} = 5*10^{-6} \ N/m \ (repulsive)[/tex]
Therefore , the force per unit between the two parallel wires with same current flowing in the same direction is 5 x 10⁻⁶ N/m repulsive force.
Zoning laws establish _______. a. what types of buildings can be built in an area b. the uses an area of land can be put to c. who can live in an area d. the types of business that can occupy a building Please select the best answer from the choices provided A B C D
Answer:
its B
Explanation:
Answer:
It's B
Explanation: hope it helps ^w^
What is the heat-loss rate through the slab if the ground temperature is 5 ∘C while the interior of the house is 25 ∘C?
Complete question :
A 12 m x 15 m house is built on a 12-cm-thick concrete slab.
What is the heat-loss rate through the slab if the ground temperature is 5°C while the interior of the house is 25°C
Answer:
3kW
Explanation:
Given the following :
Dimension of house :
Length = 12m
Width = 15m
Thickness of concrete slab (t) = 12cm
t in metres :
100cm = 1m
12cm = (12/100)m
= 0.12m
Ground temperature (Tg) = 5°C
Interior temperature = (Th) = 25°C
Thermal conductivity of concrete (K) is approximately 1 Wm/k
Using the relation:
Q = KA * [ (Th - Tg) / d]
A = Length * width = (12 *15) = 180
Q = (1 * 180) * [(25°C - 5°C) / 0.12]
Q = 180 * (20/0.12)
Q = 180 * 16.6666
Q = 3,000W = 3kW
The heat-loss rate is 3kW
Given that,
Dimension of house :
Length = 12m
Width = 15m
Thickness of concrete slab (t) = 12cm
We know that
100cm = 1m
so,
12cm = (12/100)m
= 0.12m
And,
Ground temperature (Tg) = 5°C
Interior temperature = (Th) = 25°C
calculation of heat loss rate:Q = KA * [ (Th - Tg) / d]
A = Length * width = (12 *15) = 180
Q = (1 * 180) * [(25°C - 5°C) / 0.12]
Q = 180 * (20/0.12)
Q = 180 * 16.6666
Q = 3,000W
= 3kW
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Consider the waveform expression. y(x,t)=ymsin(801t+3.38+0.503x) The transverse displacement ( y ) of a wave is given as a function of position ( x in meters) and time ( t in seconds) by the expression. Determine the wavelength, frequency, period, and phase constant of this waveform.
Answer:
f = 127.48 Hz , T = 7.844 1⁻³ s , Ф = 3.38 , λ = 12.49 m
Explanation:
The general equation for the motion of a wave in a string is
y = A sin (kx -wt + fi)
the expression they give is
y = ym sin (0.503x + 801 t + 3.38 )
the veloicda that accompanies time is
w = 801 rad / s
angular velocity is related to frequency
w = 2π f
f = w / 2π
f = 801 / 2π
f = 127.48 Hz
The period is the inverse of the frequency
T = 1 / f
T = 1 / 127.48
T = 7.844 10⁻³ s
the csntnate of phase fi is the independent term
Ф = 3.38
the wave vector accompanies the position k = 0.503 cm
ka = 2pi /λ
λ = 2 π / k
λ = 2 π / 0.503
λ = 12.49 m
sound source of frequency f moves with constant velocity (less than the speed of sound) through a medium that is at rest. A stationary observer hears a sound whose frequency is appreciably different from f because
Answer:
A static observer hears a sound whose frequency is appreciably different from actual frequency because here the change in frequency of the sound due to doppler effect.
Explanation:
Given that,
Frequency = f
We know that,
The sound source of frequency f moves with constant velocity through a medium that is at rest.
A static observer hears a sound whose frequency is significantly different from actual frequency due to doppler effect.
We know that,
The doppler effect is defined as
[tex]f=f_{0}(\dfrac{v+v_{0}}{v-v_{s}})[/tex]
Where, f₀ = actual frequency
f = observe frequency
v = speed of sound
[tex]v_{o}[/tex] = speed of observer
[tex]v_{s}[/tex] = speed of source
Hence, A static observer hears a sound whose frequency is significantly different from actual frequency because here the change in frequency of the sound due to doppler effect.
If you are driving 95 km????h along a straight road and you look to the side for 2.0 s, how far do you travel during this inattentive period?
Answer:
52.7 m
Explanation:
Given that
speed of the vehicle, v = 95 km/h
time of inattentiveness, t = 2 s
distance travelled, s = ?
Since we have the speed in km/h and the time in s, it would be best if we converted one of them to make sure we have all units in the same rank.
95 km/h = 95 * 1000/3600 m/s
95 km/h = 95000/3600 m/s
95 km/h = 26.38 m/s
Now, we use our derived speed in m/s
Speed of a moving vehicle is given by,
v = s/t, where
v = speed in m/s
s = distance travelled, in m
t = time spent, in s
if we make d the subject of formula by rearranging the equation, we have
s = v * t
distance travelled, s = 26.38 * 2
distance travelled, s = 52.7 m
therefore, during this inattentive period, 52.7 m was travelled.
Turning the barrel of a 50-mm-focal-length lens on a manual-focus camera moves the lens closer to or farther from the sensor to focus on objects at different distances. The lens has a stated range of focus from 0.70 m infinity.
How far does the lens move between these two extremes?
Answer:
Explanation:
To focus object at .7m , the image distance can be measured as follows
object distance u = .7m
focal length f = .05 m
image distance v = ?
from lens formula
[tex]\frac{1}{v} -\frac{1}{u} = \frac{1}{f}[/tex]
[tex]\frac{1}{v} +\frac{1}{.7} = \frac{1}{.05}[/tex]
[tex]\frac{1}{v} =\frac{1}{.05} -\frac{1}{.7}[/tex]
v = .054 m
= 54 mm
when the object is at infinity , image is formed at focus ie at distance of
50 mm .
So lens position from sensor where image is formed , varies from 54 mm to 50 mm .
J. Henry Alston was the first African American to publish his research findings on the perception of heat and cold in a major US psychology journal. Please select the best answer from the choices provided T F
Answer:
True
Explanation:
J. Henry Alston was known as a famous African American psychologist. He was known through his thorough study of the sensations of heat and cold.
He thereby became the first African American to publish his research findings on the perception of heat and cold in a major US psychology journal and was an important figure in the field.