Answer:
R = 0.0015Ω
Explanation:
The formula for calculating the resistivity of a material is expressed as;
ρ = RA/l
R is the resistance
ρ is the resistivity
A is the area of the wire
l is the length of the wire
Given
l = 85cm = 0.85m
A = πr²
A = 3.14*0.0018²
A = 0.0000101736m²
ρ = 1.75 × 10-8Ωm.
Substitute into the formula
1.75 × 10-8 = 0.0000101736R/0.85
1.4875× 10-8 = 0.0000101736R
R = 1.4875× 10-8/0.0000101736
R = 0.0015Ω
The distance between the two object is fixed at 5.0 m. The uncertainty distance measurement is? The percentage error in the distance is?
1. A turtle and a rabbit are to have a race. The turtle’s average speed is 0.9 m/s. The rabbit’s average speed is 9 m/s. The distance from the starting line to the finish line is 1500 m. The rabbit decides to let the turtle run before he starts running to give the turtle a head start. If the rabbit started to run 30 minutes after the turtle started, can he win the race? Explain.
Answer:no
Explanation:because 0.9*(30*60)=0.9*1800=1620
The turtle has already won the race
Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours
What will be the speed of the rabbit and the turtle?It is given
[tex]V_{t} = 0.9 \frac{m}{s}[/tex]
[tex]V_{r} = 9 \frac{m}{s}[/tex]
[tex]D=1500 m[/tex]
Time taken by turtle
[tex]T= \dfrac{D}{V_{t} }=\dfrac{1500}{0.9_{} }[/tex]
[tex]T=1666 minutes= 27 hours[/tex]
Time taken by rabbit
[tex]T= \dfrac{D}{V_{r} }=\dfrac{1500}{9_{} }[/tex]
[tex]T=166 minutes[/tex]
since rabbit started 30 minutes after turtle then
[tex]T= 136+30=196 minutes[/tex]
[tex]T= 3.2 hours[/tex]
Hence Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours
To know more about average velocity follow
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Nhiệt dung riêng của một chất là ?
Answer:
enchantment table language
Explanation:
A 1030 kg car has four 12.0 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles
Answer:
The required fraction is 0.023.
Explanation:
Given that
Mass of a car, m = 1030 kg
Mass of 4 wheels = 12 kg
We need to find the fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles.
The rotational kinetic energy due to four wheel is
[tex]=4\times \dfrac{1}{2}I\omega^2\\\\=4\times \dfrac{1}{2}\times \dfrac{1}{2}mR^2(\dfrac{v}{R})^2\\\\=mv^2[/tex]
Linear kinetic Energy of the car is:
[tex]=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times Mv^2[/tex]
Fraction,
[tex]f=\dfrac{mv^2}{\dfrac{1}{2}Mv^2}\\\\f=\dfrac{m}{\dfrac{1}{2}M}\\\\f=\dfrac{12}{\dfrac{1}{2}\times 1030}\\\\=0.023[/tex]
So, the required fraction is 0.023.
A satellite measures a spectral radiance of 8 Watts/m2/um/ster at a wavelength of 10 microns. Assuming a surface emissivity of 0.90, what would be the estimated temperature
Two identical cars, each traveling at 16 m>s, slam into a concrete wall and come to rest. In car A the air bag does not deploy and the driver hits the steering wheel; in car B the driver contacts the deployed air bag. (a) Is the impulse delivered by the steering wheel to driver A greater than, less than, or equal to the impulse delivered by the air bag to driver B
Answer:
I = - m 16 the two impulses are the same,
Explanation:
The impulse is given by the relationship
I = Δp
I = p_f - p₀
in this case the final velocity is zero therefore p_f = 0
I = -p₀
For driver A the steering wheel impulse is
I = - m v₀
I = - m 16
For driver B, the airbag gives an impulse
I = - m 16
We can see that the two impulses are the same, the difference is that in the air bag more time is used to give this impulse therefore the force on the driver is less
The pan flute is a musical instrument consisting of a number of closed-end tubes of different lengths. When the musician blows over the open ends, each tube plays a different note. The longest pipe is 0.31 m long.
What is the frequency of the note it plays? Assume room temperature of 20∘C.
Answer:
f = 276.6 Hz
Explanation:
This musical instrument can be approximated to a tube system where each tube has one end open and the other closed.
In the closed part there is a node and in the open part a belly or antinode. Therefore the wavelength is
L = λ/ 4
speed is related to wavelength and frequency
v = λ f
λ = v / f
we substitute
L = v / 4f
f = v / 4L
the speed of sound at 20ºC is
v = 343 m / s
let's calculate
f = [tex]\frac{343 }{4 \ 0.31}[/tex]
f = 276.6 Hz
Categorize each statement as true or false.
1. Electric field lines radiate away from positive charges and towards negative charges.
2. Electric field is always perpendicular to equipotential lines.
3. The electric field points in the direction of increasing electric potential.
4. The electric field inside a parallel plate capacitor decreases as it approaches the negative plate.
5. The units of electric field are either newtons per coulomb or volts per meter.
Answer:
1. True
2. True
3. False
4. True
5. True
Army is standing still on the ground; Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward. How fast does Bill see Carlos moving and in what direction?
a. 10 mis eastward
b. 5 m/s eastward
c. 15 m/s westward
d. 20 m/s westward
e. 10 m/s westward
Explanation:
Given that,
Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward.
Taking eastward as positive direction, we have:
[tex]v_B=+5\ m/s[/tex]is the velocity of Bill with respect to Amy (which is stationary)
[tex]v_c=15\ m/s[/tex] is the velocity of Carlos with respect to Amy.
Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is
[tex]v_B=+5\ m/s[/tex]
Therefore, Carlos velocity in Bill's reference frame will be
[tex]v_c'=-15\ m/s-(+5\ m/s)\\\\=-20\ m/s[/tex]
So, the magnitude is 20 m/s and the direction is westward (negative sign).
Can an electron be diffracted? Can it exhibit interference?
Answer:
Yeah, it can be diffracted. Though it depends on a diffracting medium.
It must have some magnetic fields .
Forexample; X-ray diffraction where electrons are diffracted to the target filament.
how does the use of standard units of measurement solve problems in measurement regarding validity and reliabiility? explain it
Answer:
Reliability can be estimated by comparing different versions of the same measurement. Validity is harder to assess, but it can be estimated by comparing the results to other relevant data or theory.
A 0.40-kg mass attached to the end of a string swings in a vertical circle having a radius of 1.8 m. At an instant when the string makes an angle of 40 degrees below the horizontal, the speed of the mass is 5.0 m/s. What is the magnitude of the tension in the string at this instant
Answer:
[tex]T=8.1N[/tex]
Explanation:
From the question we are told that:
Mass m=0.40
Radius r=1.8m
Angle Beneath the Horizontal \theta =40 \textdegree
Speed v=5.0m/s
The Tension Angle
[tex]\alpha=90-\theta\\\\\alpha=90-40[/tex]
[tex]\alpha=50 \textdegree[/tex]
Generally the equation for Tension is is mathematically given by
[tex]T=\frac{mv^2}{r}+mgcos \alpha[/tex]
[tex]T=\frac{0.40*5^2}{1.8}+0.40*5cos50[/tex]
[tex]T=8.1N[/tex]
Two distinct systems have the same amount of stored internal energy. 500 J are added by heat to the first system and 300 J are added by heat to the second system. What will be the change in internal energy of the first system if it does 200 J of work? How much work will the second system have to do in order to have the same internal energy?
Answer:
The change in the internal energy of the first system is 300 J
The second system will do zero work in order to have the same internal energy.
Explanation:
Given;
heat added to the first system, Q₁ = 500 J
heat added to the second system, Q₂ = 300 J
work done by the first system, W₁ = 200 J
The change in the internal energy of the system is given by the first law of thermodynamics;
ΔU = Q - W
where;
ΔU is the change in internal energy of the system
The change in the internal energy of the first system is calculated as;
ΔU₁ = Q₁ - W₁
ΔU₁ = 500 J - 200 J
ΔU₁ = = 300 J
The work done by the second system to have the same internal energy with the first.
ΔU₁ = Q₂ - W₂
W₂ = Q₂ - ΔU₁
W₂ = 300 J - 300 J
W₂ = 0
The second system will do zero work in order to have the same internal energy.
Question 4 of 5
How can the Fitness Logs help you in this class?
O A. They can't; the Fitness Logs are only useful to your teacher.
B. They show your parents how much you're learning.
C. They let you keep track of your thoughts, feelings, and progress.
D. They help you evaluate yourself for your final grade.
SUBMIT
Answer:
C is the right answer
Explanation:
fitness logs is a great way to track your progress. You can easily look back and see how you have progressed over time. In addition, it can help you plan and prepare for future workouts, as well as identify patterns of what seems to work well for you and when you have the most success
hope it was useful for you
If you double the current in a long straight wire, the magnetic field at a fixed point will... be cut in half. triple. double. quadruple.
Answer:
the magnetic field must double
Explanation:
For this exercise we use Ampere's law
∫ B . ds = μ₀ I
Where the bold indicate vectors
With this expression we can see that if we double the current, keeping the same trajectory, the magnetic field must double
A stone dropped from the top of a 80m high building strikes the ground at 40 m/s after falling for 4 seconds. The stone's potential energy with respect to the ground is equal to its kinetic energy … (use g = 10 m/s 2)
A) at the moment of impact.
B) 2 seconds after the stone is released.
C) after the stone has fallen 40 m.
D) when the stone is moving at 20 m/s.
At the moment of impact both Kinetic Energy and Potential Energy should be 0, right? So it can't be A), right? Or is this wrong? Is it indeed A)? Please show work and explain it well.
Answer:
Explanation:
The answer is C because the building is 80 meters high. Before the stone is dropped, it has ONLY potential energy since kinetic energy involves velocity and a still stone has no velocity. At impact, there is no potential energy because potential energy involves the height of the stone relative to the ground and a stone ON the ground has no height; here there is ONLY kinetic.
From the First Law of Thermodynamics, we know that energy cannot be created or destroyed, it can only change form. Therefore, that means that at the halfway point of 40 meters, half of the stone's potential energy has been lost, and it has been lost to kinetic energy. Here, at 40 meters, there is an equality between PE and KE. It only last for however long the stone is AT 40 meters, which is probably a millisecond of time, but that's where they are equal.
Two objects moving with a speed v travel in opposite directions in a straight line. The objects stick together when they collide, and move with a speed of v/2 after the collision.
Required:
a. What is the ratio of the final kinetic energy of the system to the initial kinetic energy?
b. What is the ratio of the mass of the more massive object to the mass of the less massive object?
Answer:
Explanation:
Let the mass of objects be m₁ and m₂ .
Total kinetic energy = 1/2 m₁ v² + 1/2 m₂ v²= 1/2 ( m₁ + m₂ ) v²
Total kinetic energy after collision= 1/2 ( m₁ + m₂ ) v² / 4 = 1/2 ( m₁ + m₂ ) v² x .25
final KE / initial KE = 1/2 ( m₁ + m₂ ) v² x .25 / 1/2 ( m₁ + m₂ ) v²
= 0.25
b )
Applying law of conservation of momentum to the system . Let m₁ > m₂
m₁ v - m₂ v = ( m₁ + m₂ ) v / 2
m₁ v - m₂ v = ( m₁ + m₂ ) v / 2
m₁ - m₂ = ( m₁ + m₂ ) / 2
2m₁ - 2 m₂ = m₁ + m₂
m₁ = 3m₂
m₁ / m₂ = 3 / 1
Answer:
(a) The ratio is 1 : 4.
(b) The ratio is 1 : 3.
Explanation:
Let the mass of each object is m and m'.
They initially move with velocity v opposite to each other.
Use conservation of momentum
m v - m' v = (m + m') v/2
2 (m - m') = (m + m')
2 m - 2 m' = m + m'
m = 3 m' .... (1)
(a) Let the initial kinetic energy is K and the final kinetic energy is K'.
[tex]K = 0.5 mv^2 + 0.5 m' v^2 \\\\K = 0.5 (m + m') v^2..... (1)[/tex]
[tex]K' = 0.5 (m + m') \frac{v^2}{4}.... (2)[/tex]
The ratio is
K' : K = 1 : 4
(b) m = 3 m'
So, m : m' = 3 : 1
Example 2.13 The acceleration a of a particle in a time t is given by the equation a = 2+ 5t^2. Find the instantaneous velocity after 3s. Solution
Answer:
the instantaneous velocity is 51 m/s
Explanation:
Given;
acceleration, a = 2 + 5t²
Acceleration is the change in velocity with time.
[tex]a = \frac{dv}{dt} \\\\a = 2 + 5t^2\\\\The \ acceleration \ (a) \ is \ given \ so \ we \ have \ to \ find \ the \ velocity \ (v)\\\\To \ find \ the \ velocity, \ integrate\ both \ sides \ of \ the \ equation\\\\2 + 5t^2 = \frac{dv}{dt} \\\\\int\limits^3_0 {(2 + 5t^2)} \, dt = dv\\\\v = [2t + \frac{5t^3}{3} ]^3_0\\\\v = 2(3) + \frac{5(3)^3}{3} \\\\v = 6 + 5(3)^2\\\\v = 6 + 45\\\\v = 51 \ m/s[/tex]
Therefore, the instantaneous velocity is 51 m/s
Calculate the minimum area moment of inertia for a rectangular cross-section with side lengths 6 cm and 4 cm.
52 cm4
72 cm4
32 cm4
24 cm4
2 cm4
Answer:
Minimum Area of rectangle = 24 centimeter²
Explanation:
Given:
Length of rectangle = 6 centimeter
Width of rectangle = 4 centimeter
Find:
Minimum Area of rectangle
Computation:
Minimum Area of rectangle = Length of rectangle x Width of rectangle
Minimum Area of rectangle = 6 x 4
Minimum Area of rectangle = 24 centimeter²
You throw a Frisbee of mass m and radius r so that it is spinning about a horizontal axis perpendicular to the plane of the Frisbee. Ignoring air resistance, the torque exerted about its center of mass by gravity is: __________
a. 0.
b. mgr
c. 2mgr
d. a function of the angular velocity.
e. small at first, then increasing as the Frisbee loses the torque given it by your hand.
Answer:
the correct answer is a
Explanation:
The torque is
τ = F x r
where the bold letters indicate vectors, in this case the vector of the center of mass is perpendicular to the weight of the body
τ = mg r
in body weight it is applied at the point of the center of mass, therefore as the distance of the force from the axis of rotation (center of amas) is zero, the die is zero
the correct answer is a
Need help! Need help! Need help! Need help! Need help! Need help!
Answer:
i can help you i know this answer
Answer: the side two are 50 then the other two are 140 i thank
Explanation:
Suppose that 2 J of work is needed to stretch a spring from its natural length of 34 cm to a length of 46 cm. (a) How much work is needed to stretch the spring from 36 cm to 41 cm
Answer:
0.83 J of work
Explanation:
2 J of work is required to stretch a spring from 34cm to 46cm
So that is 12cm stretched with 2 J of work
We can make that 6cm for 1 J of work
So, we need the find the work for stretching 36cm to 41cm
Which is 5cm
So, What is the work required to stretch 5cm?
1 J of work for 6cm
x work for 5cm
So, by proportion method
1 : 6 :: x : 5
6 * x = 1 * 5
6x = 5
x = 5/6
= 0.83
So to stretch 36cm to 41cm we need 0.83 J of work
The mass of a hot-air balloon and its occupants is 381 kg (excluding the hot air inside the balloon). The air outside the balloon has a pressure of 1.01 x 105 Pa and a density of 1.29 kg/m3. To lift off, the air inside the balloon is heated. The volume of the heated balloon is 480 m3. The pressure of the heated air remains the same as that of the outside air. To what temperature in kelvins must the air be heated so that the balloon just lifts off
Answer:
In order to lift off the ground, the air in the balloon must be heated to 710.26 K
Explanation:
Given the data in the question;
P = 1.01 × 10⁵ Pa
V = 480 m³
ρ = 1.29 kg/m³
M = 381 kg
we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol
let F represent the force acting upward.
Now in a condition where the hot air balloon is just about to take off;
F - Mg - m[tex]_g[/tex]g = 0
where M is the mass of the balloon and its occupants, m[tex]_g[/tex] is the mass of the hot gas inside the balloon.
the force acting upward F = Vρg
so
Vρg - Mg - m[tex]_g[/tex]g = 0
solve for m[tex]_g[/tex]
m[tex]_g[/tex] = ( Vρg - Mg ) / g
m[tex]_g[/tex] = Vρg/g - Mg/g
m[tex]_g[/tex] = ρV - M ------- let this be equation 1
Now, from the ideal gas law, PV = nRT
we know that number of moles n = m[tex]_g[/tex] / μ
where μ is the molecular mass of air
so
PV = (m[tex]_g[/tex]/μ)RT
solve for T
μPV = m[tex]_g[/tex]RT
T = μPV / m[tex]_g[/tex]R -------- let this be equation 2
from equation 1 and 2
T = μPV / (ρV - M)R
so we substitute in our values;
P = 1.01 × 10⁵ Pa
V = 480 m³
ρ = 1.29 kg/m³
M = 381 kg
we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol
T = [ (29 × 10⁻³) × (1.01 × 10⁵) × 480 ] / [ (( 1.29 × 480 ) - 381)8.31 ]
T = 1405920 / 1979.442
T = 710.26 K
Therefore, In order to lift off the ground, the air in the balloon must be heated to 710.26 K
The temperature required for the air to be heated is 710.26 K.
Given data:
The mass of a hot air-balloon is, m = 381 kg.
The pressure of air outside the balloon is, [tex]P = 1.01 \times 10^{5} \;\rm Pa[/tex].
The density of air is, [tex]\rho = 1.29 \;\rm kg/m^{3}[/tex].
The volume of heated balloon is, [tex]V = 480 \;\rm m^{3}[/tex].
The condition where the hot air balloon is just about to take off is as follows:
[tex]F-mg - m'g =0[/tex]
Here,
m' is the mass of hot gas inside the balloon and g is the gravitational acceleration and F is the force acting on the balloon in upward direction. And its value is,
[tex]F = V \times \rho \times g[/tex]
Solving as,
[tex](V \times \rho \times g)-mg - m'g =0\\\\ m'=(V \rho )-m[/tex]
Now, apply the ideal gas law as,
PV = nRT
here, R is the universal gas constant and n is the number of moles and its value is,
[tex]n=\dfrac{m'}{M}[/tex]
M is the molecular mass of gas. Solving as,
[tex]PV = \dfrac{m'}{M} \times R \times T\\\\\\T=\dfrac{P \times V\times M}{m'R}\\\\\\T=\dfrac{P \times V\times M}{(V \rho - m)R}[/tex]
Since, the standard value for the molecular mass of air is, [tex]M = 29 \times 10^{-3} \;\rm kg/mol[/tex]. Then solve for the temperature as,
[tex]T=\dfrac{(1.01 \times 10^{5}) \times 480\times 381}{(480 \times (1.29) - 381)8.31}\\\\\\T = 710.26 \;\rm K[/tex]
Thus, we can conclude that the temperature required for the air to be heated is 710.26 K.
Learn more about the ideal gas equation here:
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What would you expect to happen to the velocity of the bobber if the mass of the washers in the cylinder remained the same and the radius was doubled?
Answer:
The velocity becomes [tex]v\sqrt 2[/tex].
Explanation:
The force acting on the bobber is centripetal force.
The centripetal force is given by
[tex]F =\frac{mv^2}{r}[/tex]
when mass remains same, radius is doubled and the force is same, so the velocity is v'.
[tex]F =\frac{mv^2}{r}=\frac{mv'^2}{2r}\\\\v'=v\sqrt 2[/tex]
A building is being knocked down with a wrecking ball, which is a big metal sphere that swings on a 15-m-long cable. You are (unwisely!) standing directly beneath the point from which the wrecking ball is hung when you notice that the ball has just been released and is swinging directly toward you. How much time do you have to move out of the way? answer in seconds.
Answer:
Time to move out of the way = 1.74 s
Explanation:
Time to move out of the way is one fourth of period = 6.95/4 = 1.74 seconds.
Time to move out of the way = 1.74 s
Write a short note of the following
a) Reflection
b) Refraction
c) Diffraction
Answer:
a) Light that passes through the floor to reveal yourself (not shadow).
b) 2 rays of light that bounce between 2 transparent media.
c) I don't know what is Diffraction?
A long, current-carrying solenoid with an air core has 1550 turns per meter of length and a radius of 0.0240 m. A coil of 200 turns is wrapped tightly around the outside of the solenoid, so it has virtually the same radius as the solenoid. What is the mutual inductance of this system
Answer:
[tex]M=7.05*10^{-4}[/tex]
Explanation:
From the question we are told that:
Coil one turns N_1=1550 Turns/m
Radius [tex]r=0.0240m[/tex]
Turns 2 [tex]N_2=200N[/tex]
Generally the equation for area is mathematically given by
[tex]A=\pi*r^2[/tex]
[tex]A=\pi*0.024^2[/tex]
[tex]A=\1.81*10^{-3} m^2[/tex]
Therefore
The mutual inductance of this system is
[tex]M=\mu*N_1*N_2*A[/tex]
[tex]M=(4 \pi*10^{-7})*1550*200*1.81*10^{-3}[/tex]
[tex]M=7.05*10^{-4}[/tex]
• Explain how sound travels
Sound is a type of energy made by vibrations. These vibrations create sound waves which move through mediums such as air, water and wood. When an object vibrates, it causes movement in the particles of the medium. This movement is called sound waves, and it keeps going until the particles run out of energy.
Sound is a type of energy made by vibrations. These vibrations create sound waves which move through mediums such as air, water and wood. When an object vibrates, it causes movement in the particles of the medium. This movement is called sound waves, and it keeps going until the particles run out of energy.
At what angle torque is half of the max
Select the only true statement:
A)A beam in bending experiences tensile stresses on one side and compressive stresses on the other side.
B)A beam in bending experiences tensile stresses along the beam center and compressive stresses along the beam’s edges.
C)A beam in bending experiences only compressive stresses.
D)A beam in bending experiences only tensile stresses.
Answer:
Sorry I dont know this answer sorry
