Answer:
the value of the final pressure is 0.168 atm
Explanation:
Given the data in the question;
Let p₁ be initial pressure, v₁ be initial volume.
After expansion, p₂ is final pressure and v₂ is final volume.
So using the following equations;
p₁v₁ = nRT
p₂v₂ = nRT
hence, p₁v₁ = p₂v₂
we find p₂
p₂ = p₁v₁ / v₂
given that; initial volume v₁ = 0.175 m³, Initial pressure p₁ = 0.350 atm,
final volume v₂ = 0.365 m³
we substitute
p₂ = ( 0.350 atm × 0.175 m³ ) / 0.365 m³
p₂ = 0.06125 atm-m³ / 0.365 m³
p₂ = 0.168 atm
Therefore, the value of the final pressure is 0.168 atm
what is the light synthesis ?
Answer:
Photosynthesis, the process by which green plants and certain other organisms transform light energy into chemical energy. ... During photosynthesis in green plants, light energy is captured and used to convert water, carbon dioxide, and minerals into oxygen and energy-rich organic compounds.
Explanation:
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the earth is on the month of June
An electron travels 1.49 m in 7.4 µs (microsecWhat is its speed if 1 inch = 0.0254 m? Answer in units of in/min.
Explanation:
Write what you know
Speed = Distance / Time
micro- = 10^-6
write your conversions as fractions
1 in / 0.0254 m
1 min / 60 sec
First convert time to regular seconds
7.4 x 10^-6 seconds
Use Velocity
1.49m / (7.4 x 10^-6) s
We've written our conversions in fractions because units cancel out just like numbers
[tex] \frac{1.49m}{7.4 \times {10}^{ - 6} } \times \frac{1in}{0.0254m} \times \frac{60sec}{1min} [/tex]
Multiply all the fractions accross and youll have your answer
Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.390 mm wide. The diffraction pattern is observed on a screen 3.10 m away. Define the width of a bright fringe as the distance between the minima on either side.
Answer:
Y = 5.03 x 10⁻³ m = 5.03 mm
Explanation:
Using Young's Double-slit formula:
[tex]Y = \frac{\lambda L}{d}[/tex]
where,
Y = Fringe Spacing = Width of bright fringe = ?
λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m
L = Screen distance = 3.1 m
d = slit width = 0.39 mm = 3.9 x 10⁻⁴ m
Therefore,
[tex]Y = \frac{(6.33\ x\ 10^{-7}\ m)(3.1\ m)}{3.9\ x\ 10^{-4}\ m}[/tex]
Y = 5.03 x 10⁻³ m = 5.03 mm
comparison between copper properties and aluminium properties
According to Newton's second law, how are mass and acceleration related?
A. They are directly proportional to each other
B. They are inversely proportional to each other
Answer:
B. They are inversely proportional to each other
[tex] \frac{momentum}{time} = force \\ \\ \frac{mass \times velocity}{time} = force \\ \\ \frac{mass \times velocity}{time} = mass \times acceleration[/tex]
One of the earliest vertebrate animal groups that evolved in the early Paleozoic Era
are
Physics part 2
These the other questions 14 - 17
Answer:
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What is the advantage of a capacitor as it stores charge?
If a spider can travel 3.5 meters in 25 minutes, how fast can they go?
If we convert a circuit into a current source with parallel load it is called?
Answer:
If we convert a circuit into a current source with parallel load it is called source transformation
Consider a solid sphere and a solid disk with the same radius and the same mass. Explain why the solid disk has a greater moment of inertia than the solid sphere, even though it has the same overall mass and radius.
Answer:
Explanation:
In a Solid sphere; the moment of inertia around its geometrical axis can be expressed by using the formula:
[tex]\mathtt{I_s = \dfrac{2}{5} M_s R^2_s}[/tex]
For the solid disk; the moment of inertia around the central axis is:
[tex]\mathtt{I_D= \dfrac{1}{2}M_DR_D^2}[/tex]
Suppose [tex]M_D = M_S[/tex]; then we can say both to be equal to M
As well as [tex]R_D = R_S[/tex]; then that too can be equal to R
Now;
[tex]\mathtt{I_s = \dfrac{2}{5} M R^2} --- (1)[/tex]
[tex]\mathtt{I_D= \dfrac{1}{2}MR^2}---(2)[/tex]
Multiplying equation (1) by 2, followed by dividing it by 2; we have:
[tex]\mathtt{I_s= \dfrac{2}{5}MR^2} \times \dfrac{2}{2}[/tex]
[tex]I_s = \dfrac{4}{5} \times \dfrac{1}{2}MR^2 \\ \\ I_s = \dfrac{4}{5}\times I_D \\ \\ I_s > I_D[/tex]
The linear magnification produced by a spherical mirror is 1/4.Analysing this value state the (i) type of mirror and (i) position of the object with respect to the pole of the mirror. Draw
ray diagram to justify your answer
“The magnification produced by a spherical mirror is -3”. List four information you obtain from this statement about the mirror/image.
An elevator filled with passengers has a mass of 1603 kg. (a) The elevator accelerates upward from rest at a rate of 1.20 m/s2 for 2.50 s. Calculate the tension in the cable (in N) supporting the elevator.
Answer:
T = 17649.03 N = 17.65 KN
Explanation:
The tension in the cable must be equal to the apparent weight of the passenger. For upward acceleration:
[tex]T = W_A = m(g+a)\\[/tex]
where,
T = Tension in cable = ?
[tex]W_A[/tex] = Apparent weight
m = mass = 1603 kg
g = acceleration due to gravity = 9.81 m/s²
a = acceleration of elevator = 1.2 m/s²
Therefore,
[tex]T = (1603\ kg)(9.81\ m/s^2+1.2\ m/s^2)\\\\[/tex]
T = 17649.03 N = 17.65 KN
16. Olympic ice skaters are able to spin at about 5 rev/s.
(a) What is their angular velocity in radians per second?
(b) What is the centripetal acceleration of the skater's nose it
it is 0.120 m from the axis of rotation?
Answer:
a) w = 31.4 rad / s, b) a = 118.4 m / s²
Explanation:
a) let's reduce to the SI system
w = 5 rev / s (2pi rad / 1 rev)
w = 31.4 rad / s
b) the expression for the centripetal acceleration is
a = v² / r
linear and angular variables are related
v = w r
we substitute
a = w² r
a = 31.4² 0.120
a = 118.4 m / s²
What is measured by the change in velocity of a moving object?
Answer:
acceleration is measured
The decibel level of the sound of a subway train was measured at 92 dB. Find the intensity in watts per square meter (W/m2). (Give your answer in scientific notation, correct to one decimal place.)
Answer:
I = 1.58 x 10⁻³ watt/m²
Explanation:
Here, we will use the following formula:
[tex]\beta = 10\ log_{10}(\frac{I}{I_o})[/tex]
where,
β = decibel level = 92 dB
I = Intenisty of sound in watt/m² = ?
I₀ = reference intensity = 10⁻¹² watt/m²
Therefore,
[tex]92\ dB =10\ log_{10}(\frac{I}{10^{-12}\ watt/m^2} )\\\\[/tex]
[tex]10^{9.2} = \frac{I}{10^{-12}}\ watt/m^2\\\\I = (1.58\ x\ 10^9)(10^{-12}\ watt/m^2)[/tex]
I = 1.58 x 10⁻³ watt/m²
A mass m, which is connected to a spring of spring constant k, is released from x = A to perform
a simple harmonic motion. Another mass 2m, which is connected to another spring of the same
spring constant k, is also released from x = A to perform a simple harmonic motion. Compare the
values of total mechanical energy stored in these two spring-mass systems.
A Carnot engine with an efficiency of 30% operates with a high-temperature reservoir at 188oC and exhausts 2000 J of heat each cycle. What are (a) the heat input per cycle and (b) the Celcius temperature of the low-temperature reservoir
Answer:
a) The heat input per cycle is 2857.143 joules.
b) The temperature of the low-temperature reservoir is 49.655 °C.
Explanation:
a) The efficiency of the Carnot engine is defined by the following formula:
[tex]\eta_{th} = 1-\frac{T_{L}}{T_{H}} = 1 - \frac{Q_{L}}{Q_{H}}[/tex] (1)
Where:
[tex]T_{L}[/tex] - Low temperature reservoir, in Kelvin.
[tex]T_{H}[/tex] - High temperature reservoir, in Kelvin.
[tex]Q_{L}[/tex] - Heat output, in joules.
[tex]Q_{H}[/tex] - Heat input, in joules.
[tex]\eta_{th }[/tex] - Engine efficiency, no unit.
If we know that [tex]\eta_{th} = 0.3[/tex] and [tex]Q_{L} = 2000\,J[/tex], the heat input of the Carnot engine is:
[tex]\eta_{th} = 1 - \frac{Q_{L}}{Q_{H}}[/tex]
[tex]\frac{Q_{L}}{Q_{H}} = 1 - \eta_{th}[/tex]
[tex]Q_{H} = \frac{Q_{L}}{1-\eta_{th}}[/tex]
[tex]Q_{H} = \frac{2000\,J}{1-0.3}[/tex]
[tex]Q_{H} = 2857.143\,J[/tex]
The heat input per cycle is 2857.143 joules.
b) If we know that [tex]T_{H} = 461.15\,K[/tex] and [tex]\eta_{th} = 0.3[/tex], then the temperature of the low-temperature reservoir:
[tex]\eta_{th} = 1 - \frac{T_{L}}{T_{H}}[/tex]
[tex]\frac{T_{L}}{T_{H}} = 1 - \eta_{th}[/tex]
[tex]T_{L} = T_{H}\cdot (1-\eta_{th})[/tex]
[tex]T_{L} = (461.15\,K)\cdot (1-0.3)[/tex]
[tex]T_{L} = 322.805\,K[/tex]
[tex]T_{L} = 49.655\,^{\circ}C[/tex]
The temperature of the low-temperature reservoir is 49.655 °C.
A laser of wavelength 720 nm illuminates a double slit where the separation between the slits is 0.22 mm. Fringes are seen on a screen 0.85 m away from the slits. How far apart are the second and third bright fringes
Answer:
The appropriate solution is "2.78 mm".
Explanation:
Given:
[tex]\lambda = 720 \ nm[/tex]
or,
[tex]= 720\times 10^{-9} \ m[/tex]
[tex]D=0.85 \ m[/tex]
[tex]d = 0.22 \ mm[/tex]
or,
[tex]=0.22 \times 10^{-3} \ m[/tex]
As we know,
Fringe width is:
⇒ [tex]\beta=\frac{\lambda D}{d}[/tex]
hence,
Separation between second and third bright fringes will be:
⇒ [tex]\theta=\beta=\frac{\lambda D}{d}[/tex]
[tex]=\frac{720\times 10^{-9}\times 0.85}{0.22\times 10^{-3}}[/tex]
[tex]=2.78\times 10^{-3} \ m[/tex]
or,
[tex]=2.78 \ mm[/tex]
The loudness of a sound is the wave's _______
Answer:
amplitude
Explanation:
The loudness of a musical sound is a measure of the sound wave's ?
is amplitude explanation:- The loudness of a sound depends upon the amplitude.Loudness of a sound depends on the amplitude of the vibration producing that sound. Greater is the amplitude of vibration, louder is the sound produced by it. if you find this answer helpful please rate positive thank you so much.
What kind of model is shown below?
O A. A mathematical model
O B. An experimental model
O C. A computer model
D. A physical model
The type of model shown here is an experimental model. The correct option is B.
What is an experimental model?Animals are used in experimental modeling to model the development and progression of diseases and to test new treatments before they are administered to humans.
This result stems from the distinction that, whereas experiments are versions of the real world captured within an artificial laboratory environment, models are artificial worlds constructed to represent the real world.
Theories are plausible explanatory propositions developed to connect potential causes to their effects.
Models are schematic representations of reality or one's view of a possible world that are built to improve one's understanding of the world and/or to make predictions.
Thus, the correct option is B.
For more details regarding model, visit:
https://brainly.com/question/28381011
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Your question seems incomplete, the probable complete question is:
II. Magnetic fields Magnets and magnetic fields EM 115 We have observed that magnets interact even when they are not in direct contact. In electrostatics we used the idea of an electric field to account for the interaction between charges that were separated from one another. For magnetic interactions, we similarly define a magnetic field. A. Obtain a compass from a tutorial instructor. I. Use the compass to explore the region around a bar magnet. Describe the behavior of the compass needle both near the poles of the magnet and in the region between the poles.
Solution :
We all know that a bar magnet have two poles, the north pole and the south pole. These poles interacts with each other. The ends of the magnets having similar poles will push each other away while the poles with like charges will pull each others towards it.
The compass needle is also a magnet having south polarity as well as north polarity. When the compass needle is close to the bar magnet, it is opposite to the poles or along the poles. The compass needle shows the direction or is pointed towards the north. So when the compass needle is placed near the north pole of the bar magnet, the pointer of the compass needle points towards the north, i.e. it gets deflected because of he like charges. And when it is placed near the south pole of the magnet, it gets attracted towards it and is pointed towards the pole.
Now as we move the compass needle from the poles to the region that is between the poles, the compass needle pointer points towards the north direction every time. It show a deflection always. If we place the magnetic lines, we will see that the magnetic lines will exit from the north poles and enters the south pole of the bar magnet.
Two long, straight wires are fixed parallel to one another a distance do apart. The wires carry equal constant currents 1, in the same direction. The attractive magnetic force per unit length between them if f = F/L. What is the force per unit length between the wires if their separation is 2d, and each carries current 2I0?
A. f/4
B. f/2
C. 3f/2
D.) 2f
Answer:
Option D
Explanation:
From the question we are told that:
The attractive magnetic force per unit length as
[tex]f = F/L[/tex]
Separation Distance [tex]x=2d[/tex]
Generally the equation for Magnetic force between two current carrying wire is mathematically given by
[tex]\frac{F}{\triangle l}=\frac{\mu_0I_1I_2}{\mu \pi x}[/tex]
[tex]\frac{F}{\triangle l }=\frac{I_1I_2}{ x}[/tex]
Where
[tex]x=2r[/tex]
And
[tex]I_1=I_2=>2I[/tex]
Then
[tex]\frac{F}{\triangle l}=>\frac{2*2}{2}*f[/tex]
[tex]\frac{F}{\triangle l}=>2f[/tex]
Therefore s the force per unit length between the wires if their separation is 2d
[tex]\frac{F}{\triangle l}=>2f[/tex]
Option D
Explain what resonance is and give one example of a destructive outcome of resonance. Also give one example of a non-destructive outcome of resonance. Use details to support your answer.
Answer:
resonance is when a body is made to vibrate with the frequency of another body without touching it.
Explanation:
example of destructive: soldiers matching on a bridge unanimously can make the bridge collapse.
Answer:hi
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and lemon
and poetry
Explanation:
A 0.50-m long solenoid consists of 500 turns of copper wire wound with a 4.0 cm radius. When the current in the solenoid is 22 A, the magnetic field at a point 1.0 cm from the central axis of the solenoid is
Answer: The magnetic field at a point 1.0 cm from the central axis of the solenoid is 0.0276 T.
Explanation:
Given: Length = 0.50 m
No. of turns = 500
Current = 22 A
Formula used to calculate magnetic field is as follows.
[tex]B = \mu_{o}(\frac{N}{L})I[/tex]
where,
B = magnetic field
[tex]\mu_{o}[/tex] = permeability constant = [tex]4\pi \times 10^{-7} Tm/A[/tex]
N = no. of turns
L = length
I = current
Substitute the values into above formula as follows.
[tex]B = \mu_{o}(\frac{N}{L})I\\= 4 \pi \times 10^{-7} Tm/A \times (\frac{500}{0.5 m}) \times 22\\= 0.0276 T[/tex]
Thus, we can conclude that magnetic field at a point 1.0 cm from the central axis of the solenoid is 0.0276 T.
Suppose that the position of a particle is given by s=f(t)=5t3+6t+9. (a) Find the velocity at time t.
This question is incomplete, the complete question is;
Suppose that the position of a particle is given by s=f(t)=5t³ + 6 t+ 9.
(a) Find the velocity at time t.
(b) Find the velocity at time t=3 seconds
Answer:
a) the velocity at time t is ( 15t² + 6 ) m/s
b) Velocity at time t=3 seconds is 141 m/s
Explanation:
Give the data in the question;
position of a particle is given by;
s = f(t) = 5t³ + 6t + 9
Velocity at t;
we differentiate with respect to t
so
V(t) = f'(t) = d/dt ( 5t³ + 6t + 9 )
V(t) = f(t) = 5(3t²) + 6(1) + 0 )
V(t) = f(t) = ( 15t²+6 ) m/s
Therefore, the velocity at time t is 15t²+6 m/s
b) Velocity at t = 3 seconds
V(t) = f(t) = ( 15t²+6 ) m/s
we substitute
V(3) = ( 15(3)² + 6 ) m/s
V(3) = ( (15 × 9) + 6 ) m/s
V(3) = ( 135 + 6 ) m/s
V(3) = 141 m/s
Therefore, Velocity at time t=3 is 141 m/s
compare the time period of two pendulums of length 4m and 9m
area= length × length
area = 4m × 9m
ans 36
List five instruments of mechanical fluid
[tex]\sf{The~ different~ types~ of~ measuring~ instruments~ are:-}[/tex]
Calipers.Micrometer.Laser Measure.Ruler.Compass.A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor stops. The rocket altitude 20 s after launch is 6600 m. You can ignore any effects of air resistance.
Required:
a. What was the rocket's acceleration during the first 16s?
b. What is the rocket's speed as it passes through acloud 5100 m above the ground?
Answer:
a) a = 34.375 m / s², b) v_f = 550 m / s
Explanation:
This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.
a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed
v_f = [tex]\frac{x-x_1}{t}[/tex]
we substitute the values
v_f = [tex]\frac{ 6600 -x_1}{4}[/tex]
The initial part of the movement is carried out with acceleration
v_f = v₀ + a t
x₁ = x₀ + v₀ t + ½ a t²
the rocket starts from rest v₀ = 0 with an initial height x₀ = 0
x₁ = ½ a t²
v_f = a t
we substitute the values
x₁ = 1/2 a 16²
x₁ = 128 a
v_f = 16 a
let's write our system of equations
v_f = [tex]\frac{6600 - x_1}{4}[/tex]
x₁ = 128 a
v_f = 16 a
we substitute in the first equation
16 a = [tex]\frac{6600 -128 a}{4}[/tex]
16 4 a = 6600 - 128 a
a (64 + 128) = 6600
a = 6600/192
a = 34.375 m / s²
b) let's find the time to reach this height
x = ½ to t²
t² = 2y / a
t² = 2 5100 / 34.375
t² = 296.72
t = 17.2 s
We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part
v_f = 16 a
v_f = 16 34.375
v_f = 550 m / s