Answer:
ΔD = 2.29 10⁻⁵ m
Explanation:
This is a problem of thermal expansion, if the temperature changes are not very large we can use the relation
ΔA = 2α A ΔT
the area is
A = π r² = π D² / 4
we substitute
ΔA = 2α π D² ΔT/4
as they do not indicate the initial temperature, we assume that ΔT = 75ºC
α = 1.7 10⁻⁵ ºC⁻¹
we calculate
ΔA = 2 1.7 10⁻⁵ pi (1.8 10⁻²) ² 75/4
ΔA = 6.49 10⁻⁷ m²
by definition
ΔA = A_f- A₀
A_f = ΔA + A₀
A_f = 6.49 10⁻⁷ + π (1.8 10⁻²)² / 4
A_f = 6.49 10⁻⁷ + 2.544 10⁻⁴
A_f = 2,551 10⁻⁴ m²
the area is
A_f = π D_f² / 4
A_f = [tex]\sqrt{4 A_f /\pi }[/tex]
D_f = [tex]\sqrt{4 \ 2.551 10^{-4} /\pi }[/tex]
D_f = 1.80229 10⁻² m
the change in diameter is
ΔD = D_f - D₀
ΔD = (1.80229 - 1.8) 10⁻² m
ΔD = 0.00229 10⁻² m
ΔD = 2.29 10⁻⁵ m
A Child stands on the bus Remains Still When The bus is at rest. When the bus moves forward AndeaThe bus is at rest. When the bus moves forward And then slows down, the children the Contnues moving forward at the original speed. This is an example of
Answer:
inertia
Explanation:
Need an answer in hurry u can make the pic big
find the resistance of wire of
0.65m Radius 0.25
and
resistivity 3x10-6 OHM
Complete Question:
Find the resistance of a wire of length 0.65 m, radius 0.25 mm and resistivity 3 * 10^{-6} ohm-metre.
Answer:
Resistance = 9.95 Ohms
Explanation:
Given the following data;
Length = 0.65 m
Radius = 0.25 mm to meters = 0.00025 m
Resistivity = 3 * 10^{-6} ohm-metre.
To find the resistance of the wire;
Mathematically, resistance is given by the formula;
[tex] Resistance = P \frac {L}{A} [/tex]
Where;
P is the resistivity of the material. L is the length of the material.A is the cross-sectional area of the material.First of all, we would find the cross-sectional area of the wire.
Area of circle = πr²
Substituting into the equation, we have;
Area = 3.142 * (0.00025)²
Area = 3.142 * 6.25 * 10^{-8}
Area = 1.96 * 10^{-7} m²
Now, to find the resistance of the wire;
[tex] Resistance = 3 * 10^{-6} * \frac {0.65}{1.96 * 10^{-7}} [/tex]
[tex] Resistance = 3 * 10^{-6} * 3316326.531 [/tex]
Resistance = 9.95 Ohms
Một vô lăng sau khi bắt đầu quay được một phút thì thu được vận tốc 700
vòng/phút. Tính gia tốc góc của vô lăng
According to ____________ , the randomness of the universe is constantly increasing.
a. The first law of thermodynamics
b. The zeroth law of thermodynamics
c. The second law of thermodynamics
Answer:
According to " The second law of thermodynamics", the randomness of the universe is constantly increasing?
Explanation:
So answer option C. Have a great summer.
A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if the length of the runway is 2.00 km.ii) At this acceleration, how much time would the plane need from starting to takeoff. iii) What force must the engines exert to attain this acceleration
Answer:
i) the minimum acceleration to take off is 22500 km/h²
ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs
iii) required force that the engine must exert to attain acceleration is 625 kN
Explanation:
Given the data in the question;
mass of plane m = 360,000 kg
take of speed v = 300 km/hr = 83.33 m/s
i)
What should be the minimum acceleration to take off if the length of the runway is 2.00 km
from Newton's equation of motion;
v² = u² + 2as
we know that a plane starts from rest, so; u = 0
given that distance S = 2 km
we substitute
(300)² = 0² + ( 2 × a × 2 )
90000 = 4 × a
a = 90000 / 4
a = 22500 km/h²
Therefore, the minimum acceleration to take off is 22500 km/h²
ii) At this acceleration, how much time would the plane need from starting to takeoff.
from Newton's equation of motion;
v = u + at
we substitute
300 = 0 + 22500 × t
t = 300 / 22500
t = 0.0133 hrs
Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs
iii) What force must the engines exert to attain this acceleration
we know that;
F = ma
acceleration a = 22500 km/hr² = 1.736 m/s²
so we substitute
F = 360,000 kg × 1.736 m/s²
F = 624960 N
F = 625 kN
Therefore, required force that the engine must exert to attain acceleration is 625 kN
45. Pressure in air undergoes a decrease when the air
a) rises to higher altitudes.
b) accelerates to higher speed.
c) fills a greater space.
d) All of these.
A typical ceiling fan running at high speed has an airflow of about 2.00 ✕ 103 ft3/min, meaning that about 2.00 ✕ 103 cubic feet of air move over the fan blades each minute.
Determine the fan's airflow in m3/s.
Answer:
0.94 m³/s
Explanation:
From the question given above, the following data were obtained:
Air flow (in ft³/min) = 2×10³ ft³/min
Air flow (in m³/s) =.?
Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:
35.315 ft³/min = 1 m³/min
Therefore,
2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min
2×10³ ft³/min = 56.63 m³/min
Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:
1 m³/min = 1/60 m³/s
Therefore,
56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min
56.63 m³/min = 0.94 m³/s
Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.
The peak value of the electric field component of an electromagnetic wave is E. At a particular instant, the intensity of the wave is of 0.020 W/m2. If the electric field were increased to 5E, what would be the intensity of the wave?
Answer:
[tex]I_2=0.50 w/m^2[/tex]
Explanation:
From the question we are told that:
initial Intensity [tex]I_1=0.020 w/m^2[/tex]
Final Electric field [tex]E_2=5E[/tex]
Generally the equation for Relation ship between intensity and Electric field is mathematically given by
[tex]\frac{I_1}{I_2}= \frac{E_1^2}{E_2^2}[/tex]
Therefore
[tex]I_2=\frac{I_1}{ \frac{E_1^2}{E_2^2}}[/tex]
[tex]I_2=\frac{0.020}{ \frac{E^2}{5E^2}}[/tex]
[tex]I_2=0.50 w/m^2[/tex]
20 pts.
A man forgets that he set his coffee cup on top of his car. He starts to drive and the coffee CUP rolls off the car onto the road. How does this scenario demonstrate the first law of motion? Be specific and use the words from the law in your answer.
Answer:
The cup is acted upon by an unbalanced force which is the acceleration of the car, but before it was an object at rest that stayed at rest.
Explanation:
Newton's first law of motion states, "if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force."
Since the cup is at rest while sitting on top of the car, it stays at rest as the car begins to move. Since the car is accelerating and the cup is not, the cup falls off of the car.
Help! plz!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Now!!!!
Answer:
Find answers below.
Explanation:
A graph can be defined as the graphical representation of data (informations) on horizontal and vertical lines i.e x-axis and y-axis respectively.
On the x-coordinates, we have the following points or values;
1. Smiley face = 2
2. Diamond = 2
3. Sun = 0
4. Heart = 3.75
On the y-coordinates, we have the following points or values;
1. Smiley face = 1
2. Diamond = 3.75
3. Sun = 3.75
4. Heart = 0
Note: to read the points on a graph, you would look at the exact points marked on the x-coordinates (x-axis) and y-coordinates (y-axis) respectively.
The scale used for this graph is 0.75 for each unit on the x-coordinates and y-coordinates respectively.
How can a wire become magnetic?
add a resistor
point it north
heat it up
run a current through it
Answer:
Moving electrons always create a magnetic field. Electrons moving along a wire make a magnetic field that goes in circles around the wire. When you bend the wire into a coil, the magnetic fields around each loop of the coil add up to make a long , thin magnet with north at one end and south at the other.
Explanation:
Which phenomenon occurs when one wave is superimposed on another?
A. Interference
B. Refraction
C. Diffraction
D. Polarization
Answer:Alternativa A. Damos o nome de interferência a superposição de efeitos que ocorre ao ser produzido dois pulsos de onda, que serão propagados e acabarão inevitavelmente por se encontrar. No instante em que os pulsos se cruzarem, há então, uma superposição de efeitos individuais de cada um deles. Se durante o cruzamento, houver um reforço das ondas, estará ocorrendo a este fenômeno.
Which of the following best describes the upper respiratory tract?
O It takes air in from outside the body.
O It is where oxygen and carbon dioxide are exchanged.
O It is located inside the thorax.
O It is not directly involved in respiration.
The velocity of an object traveling in a circle is quadrupled and its radius is tripled The acceleration of this object will change by factor of?
Answer:
The process of solving a circular motion problem is much like any other problem in physics class. The process involves a careful reading of the problem, the identification of the known and required information in variable form, the selection of the relevant equation(s), substitution of known values into the equation, and finally algebraic manipulation of the equation to determine the answer. Consider the application of this process to the following two circular motion problems.
Sample Problem #1
A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. Determine the acceleration and the net force acting upon the car.
The solution of this problem begins with the identification of the known and requested information.
Known Information:
m = 900 kg
v = 10.0 m/s
R = 25.0 m
Requested Information:
a = ????
Fnet = ????
To determine the acceleration of the car, use the equation a = v2 / R. The solution is as follows:
a = v2 / R
a = (10.0 m/s)2 / (25.0 m)
a = (100 m2/s2) / (25.0 m)
a = 4 m/s2
To determine the net force acting upon the car, use the equation Fnet = m•a. The solution is as follows.
Fnet = m • a
Fnet = (900 kg) • (4 m/s2)
Fnet = 3600 N
Sample Problem #2
A 95-kg halfback makes a turn on the football field. The halfback sweeps out a path that is a portion of a circle with a radius of 12-meters. The halfback makes a quarter of a turn around the circle in 2.1 seconds. Determine the speed, acceleration and net force acting upon the halfback.
The solution of this problem begins with the identification of the known and requested information.
Known Information:
m = 95.0 kg
R = 12.0 m
Traveled 1/4-th of the circumference in 2.1 s
Requested Information:
v = ????
a = ????
Fnet = ????
To determine the speed of the halfback, use the equation v = d / t where the d is one-fourth of the circumference and the time is 2.1 s. The solution is as follows:
v = d / t
v = (0.25 • 2 • pi • R) / t
v = (0.25 • 2 • 3.14 • 12.0 m) / (2.1 s)
v = 8.97 m/s
To determine the acceleration of the halfback, use the equation a = v2 / R. The solution is as follows:
a = v2 / R
a = (8.97 m/s)2 / (12.0 m)
a = (80.5 m2/s2) / (12.0 m)
a = 6.71 m/s2
To determine the net force acting upon the halfback, use the equation Fnet = m•a. The solution is as follows.
Fnet = m*a
Fnet = (95.0 kg)*(6.71 m/s2)
Fnet = 637 N
In Lesson 2 of this unit, circular motion principles and the above mathematical equations will be combined to explain and analyze a variety of real-world motion scenarios including amusement park rides and circular-type motions in athletics.
Your hand and wrist curl in toward the center of your body (chest and stomach) to prepare to throw the frisbee.
O True
O False
True
Hope this helps! :)
Answer:
true because when trow the frisbee gives u level
A 0.500 m length of wire with a cross-sectional area of 3.14 * 10 ^ - 6 meters squared is found to have a resistance of 2.53 * 10 ^ - 3 ohms according to the resisting chart, from what material is the wire made?
PLEASE HELP MEEEEEEE
Answer:
The wire is made of silver (ρ = 1.59×10⁻⁸ ohms/m)
Explanation:
Applying,
R = ρL/A................. Equation 1
Where R = Resistance length of the wire, ρ = Resistivity of the wire, L = Length of the wire, A = crosssectional area of the wire
make ρ the subject of the equation
ρ = RA/L............. Equation 2
From the question,
Given: R = 2.53×10⁻³ ohms, A = 3.14×10⁻⁶ m², L = 0.5 m
Substitute these values into equation 2
ρ = (2.53×10⁻³)(3.14×10⁻⁶)/0.5
ρ = 1.59×10⁻⁸ ohms/m
Hence from the resistivity chart, the wire is made of silver
all
What is Velocity
Answer:
noun
the speed of something in a given direction.
"the velocities of the emitted particles"
(in general use) speed.
"the tank shot backwards at an incredible velocity"
Similar:
speed
pace
rate
tempo
momentum
impetus
swiftness
swift/fast pace
fastness
quickness
speediness
rapidity
briskness
expeditiousness
expedition
dispatch
acceleration
clip
fair old rate
fair lick
steam
nippiness
fleetness
celerity
ECONOMICS
the rate at which money changes hands within an economy.
noun: velocity of circulation; plural noun: velocities of circulation
Answer:
i hope this helps you
Explanation:
hii
How does gravity affect your ability to live on a planet?
What does it mean when work is positive?
O Velocity is greater than kinetic energy
O Kinetic energy is greater than velocity
The environment did work on an object.
O An object did work on the environment.
d. An object did work on the environment.
Explanation:Work is defined in many contexts. Some of these are;
i. Work is the product of force and displacement. In this case, work done is positive if the force applied on an object or body and the displacement caused by the force are in the same direction. If instead the force and displacement are in opposite direction, then the work done will be negative. If it is the case the force and the displacement are perpendicular to each other, the work done is zero.
ii. In the first law of thermodynamics, the internal energy of a system is the sum of the work done and the heat exchanged between the system and the environment. Therefore, work done is the difference between the internal energy of a system and the heat exchanged between the system and the environment.
In this case, work is said to be positive if work is done by the system (object) on the environment. It is negative if work is done by the environment on the system (object).
Answer:
its c
Explanation:
Consider the following possibilities and select the correct choice.
1. Tx Ty > Tz
2. Tx Ty < Tz
3. Tx Ty = Tz
Answer:
Tx not but mybe
Explanation:
for that reason its just trying to help
The laboratory exercise for this chapter addresses kinematic motion. You have experienced these motions in everyday life. Instead of a discussion board requiring posts in a forum, this assignment has been modified to accept your response to the following questions in this assignment. Be sure to clearly address each of the points below and show all of your work • What is the difference between a vector and a scalar quantity? • List two examples each of vector and scalar quantities • Write the due date of this assignment: Month and Day (For example, July 15 would be Month - 7. Day = 15) • For a building having a height equal to the quantity you have recorded for Day in meters in our example 15 meters), compute the time required for the ball to fall to the ground while experiencing acceleration due to gravity (g=9.8m/s) • How fast was the ball traveling when it hit the ground? Submit the kinetics Assignment by 11:59 p.m. (ET) on Monday,
Answer:
A) vectors: veloicty, force
scalar: speed, work
B) t = 1.75 s, C) v = - 17 2 m / s
Explanation:
We answer each part separately
A) A vector magnitude has magnitude and direction instead a scalar magnitude has only magnitude
vector quantities: the speed of a car number is the magnitude and direction is where it goes
Force, the number is the magnitude and above that applies gives direction
Scalar magnitude: how quickly the number of the speedometer of the car
Temperature, work
B) I = 15 m height to the soil and get to calculate time = 0
y = y₀ + v₀ t - ½ g t²
as the ball is loose its initial velocity is zero
0 = 0 +0 - ½ g t²
t = [tex]\sqrt{2y_o/g}[/tex]
t = [tex]\sqrt{2 \ 15/ 9.8}[/tex]
t = 1.75 s
C) the velocity to the reach the floor
v = vo - g t
v = 0 - g t
v = - 9.8 1.74
v = - 17 2 m / s
The negative signt iindicates that the speed goes down
In the Biomedical and Physical Sciences building at MSU there are 135 steps from the ground floor to the sixth floor. Each step is 16.6 cm tall. It takes 5 minutes and 30 seconds for a person with a mass of 73.5 kg to walk all the way up. How much work did the person do?
Answer:
W = 16.4 kJ
Explanation:
Given that,
There are 135 steps from the ground floor to the sixth floor.
Each step is 16.6 cm tall.
The mass of a person, m = 73.5 kg
We need to find the work done by the person. We know that,
Work done = Fd
Where
d is the displacement, d = 135 × 0.166 = 22.41
So,
W = 73.5 × 10 × 22.41
= 16471.35 J
or
W = 16.4 kJ
So, 16.4 kJ is the work done by the person.
Which statements describe using genetic factors to influence the growth of organisms? Select the three (3) that apply.
-increasing use of hybrid crops
-altering genes in DNA to create new plants
-increasing human population
-increasing climate change
-developing disease or pest resistant crops
Answer:
- increasing use of hybrid crops
- altering genes in DNA to create new plants
- developing disease or pest resistant crops
Explanation:
The use of genetic factors to influence the growth of a plant encompasses manipulating the genetic constituent (gene) of such plant.
For example,
- Increasing use of hybrid crops entails mating two pure bred plants based on a gene of interest responsible for a particular trait, to form a hybrid.
- Altering genes in DNA to create new plants is also a genetic factor as it has to with gene modification.
- developing disease or pest resistant crops means that the genetic make up of such plant has been modified to be resistant to pest/disease.
A television of mass 8 kg sits on a table. The coefficient of static friction
between the table and the television is 0.48. What is the minimum applied
force that will cause the television to slide?
O A. 38 N
O B. 62 N
O C. 78 N
D. 55 N
The television has weight (8 kg) g = 78.4 N, and the magnitude of the normal force between the table and television would be the same, 78.4 N. This mean the maximum magnitude of static friction between the table and television is
0.48 (78.4 N) ≈ 37.6 N ≈ 38 N
and this is the minimum required force needed to get the television to slide.
Two identical loudspeakers 2.30 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standing 5.00 m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. Part A What is the lowest possible frequency of sound for which this is possible
Answer:
By the Pythagorean Theorem the distances from the speakers os
5 and 5.5 (rounding) meters - let y be the wavelength in the solution
n y = 5 n is number of wavelengths from speaker
(n + m) y = 5.5 m must be integral for constructive interference
m y = .5 subtracting equations
m = 2 and y = ,25 for the above conditions
(n + 2) y = 5.5 substituting for m
n = 5.5 / .25 - 2 = 20
f = v / y using frequency of sound
f = 340 / .25 = 1360 / sec for lowest frequency
Check: D1 = y n = ,25 * 20 = 5
and D2 = .25 * 22 = 5.5 for the distances traveled
what is simple definition of democracy
it's a form of government where people elect their representatives
Answer:
The word democracy itself means rule by the people.
PLEASE HELP MEE THIS IS DUE IN 45 MINS
Answer:
The distance travelled does not depend on the mass of the vehicle. Therefore, [tex]s = d[/tex]
Explanation:
This deceleration situation can be analyzed by means of Work-Energy Theorem, where change in translational kinetic energy is equal to the work done by friction:
[tex]\frac{1}{2}\cdot m\cdot v^{2}-\mu\cdot m\cdot g \cdot s = 0[/tex] (1)
Where:
[tex]m[/tex] - Mass of the car, in kilogram.
[tex]v[/tex] - Initial velocity, in meters per second.
[tex]\mu[/tex] - Coefficient of friction, no unit.
[tex]s[/tex] - Travelled distance, in meters.
Then we derive an expression for the distance travelled by the vehicle:
[tex]\frac{1}{2}\cdot v^{2} = \mu \cdot g \cdot s[/tex]
[tex]s = \frac{v^{2}}{\mu\cdot g}[/tex]
As we notice, the distance travelled does not depend on the mass of the vehicle. Therefore, [tex]s = d[/tex]
When a mass of 3.0-kg is hung on a vertical spring, it stretches by 0.085 m. Determine
the period of oscillation of a 4.0-kg object suspended from this spring.
Answer:
the period of oscillation of the given object is 0.14 s
Explanation:
Given;
mass of the object, m = 3 kg
extension of the spring, x = 0.085 m
The spring constant is calculated as follows;
[tex]F = mg = \frac{1}{2} ke^2\\\\2mg = ke^2\\\\k = \frac{2mg}{e^2} \\\\k = \frac{2\times 3 \times 9.8}{(0.085)^2} \\\\k = 8,138.41 \ N/m[/tex]
The angular speed of a 4 kg object is calculated as follows;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\frac{2\pi }{T} = \sqrt{\frac{k}{m} } \\\\T= 2\pi \sqrt{\frac{m}{k} } \\\\T = 2\pi \sqrt{\frac{4}{8138.41} }\\\\T = 0.14 \ s[/tex]
Therefore, the period of oscillation of the given object is 0.14 s
Unpolarized light of intensity 0.0288 W/m2 is incident on a single polarizing sheet. What is the rms value of the electric field component transmitted
Answer:
the rms value of the electric field component transmitted is 3.295 V/m
Explanation:
Given;
intensity of the unpolarized light, I = 0.0288 W/m²
For unpolarized light, the relationship between the amplitude electric field and intensity is given as;
[tex]E_{max} = \sqrt{2\mu_0cI} \\\\E_{max} = \sqrt{2(4\pi \times 10^{-7})(3\times 10^8)(0.0288)} \\\\E_{max} = 4.66 \ V/m[/tex]
The relationship between the rms value of the electric field and the amplitude electric field is given as;
[tex]E_{rms} = \frac{E_0}{\sqrt{2} } =\frac{E_{max}}{\sqrt{2} } \\\\E_{rms} = \frac{4.66}{\sqrt{2} }\\\\E_{rms} = 3.295 \ V/m[/tex]
Therefore, the rms value of the electric field component transmitted is 3.295 V/m
