Answer:
correct answer is C
Explanation:
The time constant of an RC circuit is
τ = RC
so to find the capacitance
C = τ/ R
C = 2.150 / 5.20 10³
C = 4.13 10⁻⁴ F
to find the error we use the worst case
ΔC = | [tex]|\frac{dC}{d \tau }| \ \Delta \tau + | \frac{dC}{dR} | \ \Delta R[/tex]
the absolute value guarantees that we find the worst case, we evaluate the derivatives
ΔC = 1 /R Δτ + τ/R² ΔR
the absolute values of the errors are
Δτ = 0.002 s
ΔR = 0.3 kΩ
we substitute
ΔC = 0.002 /5.20 10³ + 2.150/(5.20 10³)² 0.3 10³
ΔC = 3.8 10⁻⁷ + 1.74 10⁻⁵
ΔC = 1.77 10⁻⁵ F
the uncertainty or error must be expressed with a significant figure
ΔC = 2 10⁻⁵ F
the percentage error is
Er% =[tex]\frac{\Delta C}{C} \ 100[/tex]
Er% = [tex]\frac{2 \ 10^{-5} }{ 4.13 \ 10^{-4} } \ 100[/tex]
Er% = 4.8%
the correct answer is C
A rock is thrown by applying a force of 2.50N while doing 6.00J of work. Over what distance was the force applied?
O 0.417m
8.50m
2.40m
15.0m
Answer:
[tex]\boxed {\boxed {\sf 2.40 \ meters}}[/tex]
Explanation:
Work is the product of force and distance, so the formula is:
[tex]W= F \times d[/tex]
The work is 6.00 Joules, but
1 Joule (J) is equal to 1 Newton meter (N*m). Therefore, the work of 6.00 J is equal to 6.00 N*mThe force is 2.50 Newtons.
The known values are:
W= 6.00 N*m F= 2.50 NSubstitute the values into the formula.
[tex]6.00 \ N*m= 2.50 \ N *d[/tex]
We are solving for distance, so we must isolate the variable, d. It is being multiplied by 2.50 Newtons and the inverse of multiplication is division. Divide both sides by 2.50 N.
[tex]\frac {6.00 \ N*m}{2.50 \ N}=\frac{2.50 \ N * d}{2.50 \ N}[/tex]
[tex]\frac {6.00 \ N*m}{2.50 \ N}=d[/tex]
The units of Newtons cancel out.
[tex]\frac {6.00 m}{2.50 }=d[/tex]
[tex]2.40 \ m =d[/tex]
The force was applied to the rock over a distance of 2.40 meters and choice C is correct.
Coherent light with wavelength 591 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringe. For what wavelength of light will the first-order dark fringe be observed at this same point on the screen?
Answer:
The wavelength will be "1.182 μm".
Explanation:
The given values are:
Wavelength
[tex]\lambda=591 \ nm[/tex]
or,
[tex]=591\times 10^-9 \ m[/tex]
Distance,
[tex]d = 3.00 m[/tex]
[tex]n = 1[/tex]
Distance of fringe from center,
[tex]y = 4.84 \ mm[/tex]
We have to find the wavelength of first order dark fringe,
[tex]\lambda = ?[/tex]
As we know,
⇒ [tex]y_{bright} =\frac{1\times \lambda\times L}{d}[/tex]
On putting the given values in the formula, we get
[tex]0.00484=\frac{1\times (591\times 10^{-9})\times 3}{d}[/tex]
On applying the cross multiplication, we get
[tex]\lambda = \frac{0.00484\times 000036632}{0.5\times 3}[/tex]
[tex]=1182\times 10^{-9}[/tex]
or,
[tex]=1.182 \ \mu m[/tex]
Three liquids that do not mix are poured into a cylindrical container with a diameter of 10.0 cm. The densities and volumes of the liquids are as follows.
Liquid 1: ????1 = 2.80 ✕ 103 kg/m3 and V1 = 2.00 ✕ 10−3 m3
Liquid 2: ????2 = 1.00 ✕ 103 kg/m3 and V2 = 1.50 ✕ 10−3 m3
Liquid 3: ????3 = 0.600 ✕ 103 kg/m3 and V3 = 1.00 ✕ 10−3 m3
Determine the pressure on the bottom of the container.
Answer:
P = 9622.9 Pa = 9.62 KPa
Explanation:
First, we will calculate the mass of all three liquids:
m = ρV
where,
m = mass of liquid
ρ = density of liquid
V = Volume of liquid
FOR LIQUID 1:
m₁ = (2.8 x 10³ kg/m³)(2 x 10⁻³ m³) = 5.6 kg
m₂ = (1 x 10³ kg/m³)(1.5 x 10⁻³ m³) = 1.5 kg
m₃ = (0.6 x 10³ kg/m³)(1 x 10⁻³ m³) = 0.6 kg
The total mass will be:
m = m₁ + m₂+ m₃ = 5.6 kg + 1.5 kg + 0.6 kg
m = 7.7 kg
Hence, the weight of the liquids will be:
W = mg = (7.7 kg)(9.81 m/s²) = 75.54 N
Now, we calculate the base area:
A = πr² = π(0.05 m)²
A = 7.85 x 10⁻³ m²
Now the pressure will be given as:
[tex]P = \frac{F}{A}\\\\P = \frac{75.54\ N}{7.85\ x\ 10^{-3}\ m^2}[/tex]
P = 9622.9 Pa = 9.62 KPa
1. Determine the potential energy of a 2kg rock at the top of a hill that is 20m high.
2. Determine the kinetic energy of a 2000 kg roller coaster car that is moving with a speed of 25 m/s.
3. A 80kg freezer is located in an office on the 59 floor of an office building 200 meters above the ground. What is the potential energy of the freezer?
Answer:
1) 392 joules
2) 625,000 joules
3) 156,800 joules
Explanation:
Gravitational potential energy = mgh. m = mass in kg, g = acceleration due to gravity and h = height in meters.
Kinetic energy = 1/2(m)(v)^2, m = mass in kg, v = velocity in meters per second
1) P.E = 2 x 9.8 x 20 = 392 joules
2) K.E = 1/2 x 2000 x (25)^2 = 625,000 joules
3) P.E = 80 x 9.8 x 200 = 156,800 joules
Answer:
[tex]1)\:392\:\text{J}\:(400\:\text{J with one significant figure)},\\2)\:625,000\:\text{J}\:(600,000\:\text{J with one significant figure)},\\3)\:156,800\:\text{J}\:(200,000\:\text{J with one significant figure)},[/tex]
Explanation:
1. The potential energy of an object is given by [tex]PE=mgh[/tex]. Substituting given values, we have:
[tex]PE=2\cdot 9.8\cdot 20=\boxed{392\:\text{J}}[/tex]
2. The kinetic energy of an object is given by [tex]KE=\frac{1}{2}mv^2[/tex]. Substituting given values, we have:
[tex]KE=\frac{1}{2}\cdot 2000\cdot 25^2 =\boxed{625,000\: \text{J}}[/tex]
3. 1. The potential energy of an object is given by [tex]PE=mgh[/tex]. Substituting given values, we have:
[tex]PE=80\cdot 9.8\cdot 200=\boxed{156,800\:\text{J}}[/tex]
In which mode of heat transfer is the convection heat transfer coefficient usually higher, natural convection or forced convection?why?
use a trigonometric equation to determine the leg of this triangle
C=90°
A=30°
c=10m
What is a?
Answer: 5
Explanation: B is for sure 60°, c* cosB = 10*1/2 =5
The school is 3, 000m meters from the mall. What is the distance in kilometers?
Answer:
3,000 meters = 3 kilometers
Explanation:
At what speed is the kinetic energy of a particle twice its Newtonian value?
Which of the following statements correctly describes the index of refraction of a material? Select all that apply.
The index of refraction is the ratio of the speed of light in a material to the speed of light in a vacuum.
The index of refraction of a material must be greater than 1.
The index of refraction is the ratio of the speed of light in a vacuum to the speed of light in the material.
The index of refraction of a material must be less than 1.
The index of refraction of water is less than the index of refraction of air.
Answer:
Option C and D only
Explanation:
Option A is incorrect because refractive index of a material is the ratio of speed of light in vacuum to the speed of light in a any given medium
Option B is correct as the speed of light in vacuum is always greater than the speed of light in any given medium.
Option C is correct
Option D is incorrect
Option E is incorrect because the denser the medium the more is the refractive index. Water is denser than air, hence it should have more refractive index as compared to that of air.
The radius of curvature of both sides of a converging lens is 18 cm. One side of the lens is coated withsilver so that the inner surface is reflective. When light is incident on the uncoated side it passes throughthe lens, reflects off the silver coating, and passes back through the lens. The overall effect is that of amirror with focal length 5.0 cm. What is the index of refraction of the lens material
Answer:
n = 1.4
Explanation:
Given,
R1 = 18 cm, R2 = -18 cm
From lens makers formula
1/f = (n - 1)(1/18 + 1/18) = (n-1)/9
f = 9/(n-1)
Power, P = 1/f ( in m) = (n-1)/0.09
Now, this lens is in with conjunction with a concave mirror which then can be thought of as to be in conjunction with another thin lens
Power of concave mirror = P' = 1/f ( in m) = 2/R = 2/0.18 = 1/0.09
Net power of the combination = 2P + P' = 2(n-1)/0.09 + 1/0.09 = 1/0.05
n = 1.4
Can anybody help meh :(
A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing through the cross-sectional area in 10s is
Answer:
[tex]n=1.56\times 10^{17}[/tex] electrons
Explanation:
Given that,
Total charge = 9 mC = 0.009 C
0.009 C of charge passes through a wire in 3.6 s.
Let q' is the charge that passes through it in 10 s.
So,
[tex]\dfrac{0.009 }{3.6}=\dfrac{q'}{10}\\\\q'=\dfrac{0.009 \times 10}{3.6}\\\\q'=0.025\ C[/tex]
We know that,
q = ne
Where
n is the number of electrons
So,
[tex]n=\dfrac{q}{e}\\\\n=\dfrac{0.025}{1.6\times 10^{-19}}\\\\n=1.56\times 10^{17}[/tex]
So, [tex]1.56\times 10^{17}[/tex]electrons must pass through the cross-sectional area.
What is the answer to this problem?
Answer:
The answer is A
Explanation:
An insulator doesn't have to be a metal
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Answer:
a substance which does not readily allow the passage of heat or sound.
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Intensity: Radiation of a single frequency reaches the upper atmosphere of the earth with an intensity of 1350 W/m2. What is the maximum value of the electric field associated with this radiation? (c = 3.00 × 108 m/s, μ0 = 4π × 10-7 T ∙ m/A, ε0 = 8.85 × 10-12 C2/N ∙ m2)
Answer:
The right answer is "1010 V/m".
Explanation:
The given values are:
Intensity,
[tex]I=1350 \ W/m^2[/tex]
[tex]c = 3.00\times 108 \ m/s[/tex]
[tex]\mu_0= 4\pi\times 10^{-7} \ T.m/A[/tex]
Now,
The electric field's maximum value will be:
= [tex]\sqrt{2\times u\times c\times I}[/tex]
On substituting the values in the above formula, we get
= [tex]\sqrt{2\times 4\times \pi\times 10^{-7}\times 3\times 10^8\times 1350}[/tex]
= [tex]\sqrt{32400\times 3.14\times 10^{-7}\times 10^8}[/tex]
= [tex]1010 \ V/m[/tex]
The maximum value of the electric field associated with this radiation is :
1010 V/mGiven data :
Intensity ( I ) = 1350 W/m²
C = 3.00 * 10⁸ m/s
The maximum value of the electric field can be calculated using the equation below
[tex]E_{max} = \sqrt{2*u*c*I}[/tex] ----- ( 1 )
where : I = 1350 W/m², μ = 4π * 10⁻⁷, c = 3.00 * 10⁸
Insert values into equation ( 1 )
∴ [tex]E_{max}[/tex] = 1010 V/m.
Hence we can conclude that the maximum value of the electric field is 1010 V/m .
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For a damped oscillator with a mass of 360 g, a spring constant 130 N/m and a damping coefficient of 86 g/s, what is the ratio of the amplitude of the damped oscillations to the initial amplitude at the end of 11 cycles
Answer:
A/A₀ = 0.5106
Explanation:
To do this, we need to use several formulas and expressions. First, we need to know the period of time of the oscillator. This can be calculated using the following expression:
ω = 2π/T (1)
But angular frequency (ω) can be also be calculated using:
ω = √(k/m) (2)
Using (1) and (2), we can solve for the period T:
2π/T = √(k/m) (mass in kg)
2π/T = √(130/0.360)
2π/T = √361.11
2π/T = 19.003
T = 2π/19.003 = 0.331 s
Now, the amplitude A at a time t, is:
A = x exp(-bt/2m) (3)
At time 0, A = x. so A₀ = x
The problem states that we have 11 cycles respect to the initial amplitude, so expression (3) can be rewritten as:
A = x exp(-b(17t/2m)) using b as kg/s = 0.086 kg/s
Replacing the data we have:
A = x exp(-0.086(17*0.331)/2*0.36)
A = x exp(-0.086 * 7.815)
A = x exp(-0.6721)
A = 0.5106x (4)
Now, doing the ratio with the innitial we have:
A / A₀ = 0.5106x / x
The ratio is:
A/A₀ = 0.5106Hope this helps
How are satellites used in GPS?
to keep the International Space Station in orbit
to communicate with the International Space Station
to communicate with possible extraterrestrial life
to pinpoint specific locations on Earth
Answer:
To pinpoint specific locations on Earth
Explanation:
The satellite has a full specific view of a specific area just like other satellites do. Satellites communicate together which also communicates with your GPS to give you an accurate location where you are or where you want to be. That is how satellites are used in GPS.
Hope this helps you out :)
11. A candle is placed in front of a plane mirror. The calculated value of m,
the lateral magnification, is positive. What does the positive sign indicate
about the image?
O The image is enlarged.
The image is on the same side of the mirror as the object.
The image distance is greater than the object distance.
The image is upright.
Answer: If the magnification is positive, the image is upright compared to the object; if m is negative, the image is inverted compared to the object.
(I think, I was also stuck on this question for a bit)
The positive sign indicate about the image that
b) The image is on the same side of the mirror as the object.
d) The image is upright.
What is magnification ?
Magnification is a quantification of comparing the size of the image with respect to the size of the object. It gives us information about the image in terms of how large or small is the image formed.
magnification = height of image / height of object
since , height of object is always positive as it is always kept upright hence , in order to make m positive , height of image need to be positive
also magnification = -(v/u )
v = image distance
u = object distance
u is always positive hence , in order to make m positive , v needs to be negative , which implies on the same side of the mirror as the object
correct option
b) The image is on the same side of the mirror as the object.
d) The image is upright.
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1.) Rn-222 decays from 400 grams to 6.25 grams in 240 minutes. How long is one “half-life.
Answer:
40
Explanation:
Can someone help me with this please
Answer:
yea im sorry
Explanation:
PLEASE HELP WITH THIS ONE QUESTION
Which of the following must be true for a current to be induced in a wire passing through a magnetic field?
A) the magnetic field and the direction of motion must be perpendicular
B) the magnetic field and direction of motion must be parallel
Answer:
A) the magnetic field and the direction of motion must be perpendicular
For a current to be induced in a wire passing through a magnetic field, the magnetic field and the direction of motion must be perpendicular. The correct option is A.
What is magnetic field?The magnetic field is the region of space where another object experiences magnetic force and the current is induced in it.
According to the Fleming's right hand rule, the direction of motion, magnetic force and magnetic field are mutually perpendicular.
Thus, the correct option is A.
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50 POINTS!!!!!!!
Two charged objects are positioned 5 cm away from each other. Describe the change in the force between these two objects when:
(a) the charge on one of the objects is increased
(b) the distance between the objects is increased to 10 cm
(c) the charge on both of the objects is decreased
Please answer all 3
Answer:
A. The shorter the distance the greater the force, so now the force would have increased because they haven't moved away from each other.
B. The force would decrease because they are now further apart.
C. the force wouldn't be as great as before but the force would still be high because they are not far apart from each other.
I hope this helps you( ◜‿◝ )♡
A car travels 8km in 7 minutes. Find the speed of the car.
Answer:
42.6083 mi/h
Explanation:
Given: A car travels 8km in 7 minutes.
To find: Find the speed of the car.
Formula: [tex]Speed = \frac{Distance}{Time}[/tex]
Solution: Since the formula for the speed of an object (which is the car) is speed = distance ÷ time, divide the distance (8km) by the time (7min)
Speed = 42.6083 miles per hour
Which term refers to how often a person works out?
Answer: Frequency
Explanation: Frequency is term which best describes how often a person exercises.
Answer:
frequency
Explanation:
I took the. test
Which of the following describes an electric insulator?
O A. A material that has a low resistance and prevents charges from
moving freely
B. A material that has a low resistance and allows charges to move
freely
C. A material that has a high resistance and prevents charges from
moving freely
D. A material that has a high resistance and allows charges to move
freely
The following that describes an electric insulator are :
C. A material that has a high resistance and prevents charges from moving freely.
The following that describes an electric insulator is a material that has a high resistance and prevents charges from moving freely. The conductors exhibit low resistance, which allows for the flow of current through it freely.Thus, the correct answer is C.
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A microscope has an objective lens with diameter 1.04 cm. You wish to resolve an object 9.09 micrometers in size. You are using visible light of wavelength 562 nm. Using the Rayleigh criterion, what is the distance that the objective lens must be from the object to resolve it
Answer:
L = 0.1379 m = 13.79 cm
Explanation:
The Rayleigh criterion establishes that two objects are separated when the maximum of diffraction for slits coincides with the minimum of the other point, therefore the expression for the diffraction
a sin θ = m λ
the first zero occurs when m = 1
let's use trigonometry to find the angle
tan θ = y / L
as in these experiments the angles are very small
tan θ = sin θ /cos θ = sin θ
sin θ = y / L
we substitute
a y /L = λ
In the case of circular aperture the system must be solved in polar coordinates, for which a numerical constant is introduced
a y / L = 1.22 λ
L = a y / 1.22 λ
We search the magnitudes to the SI system
a = 1.04 cm = 1.04 10⁻² m
y = 9.09 10⁻⁶ m
λ = 562 10⁻⁹ m
let's calculate
L = [tex]\frac{1.04 \ 10^{-2} \ 9.09 \ 10^{-6} }{1.22 \ 562 \ 10^{-9} }[/tex]
L = 1.379 10⁻¹ m
L = 0.1379 m = 13.79 cm
How does the intensity of a sound wave change if the distance from the
source is increased by a factor of 3?
O A. The intensity decreases by a factor of 3.
O B. The intensity decreases by a factor of 9.
O c. The intensity increases by a factor of 9.
O D. The intensity increases by a factor of 3.
Answer: C- The intensity increases by a factor 9
Explanation: The intensity of a sound wave follows an inverse square law, that means that it is inversely proportional to the square of the distance: so the new distance is the intensity will increase by a factor 9.
My sentence- I hope that helped!
Intensity of a sound wave decreases by a factor of 9, if the distance from the source is increased by a factor of 3. Hence option B is correct.
Intensity of Sound is inversely proportional to the square of the distance from the sound source. Since sound waves carry its energy though a two-dimensional or three-dimensional medium, the intensity of the sound wave decreases with increasing as second power of distance form the source.
Mathematically,
Intensity I ∝ 1/D²
If the distance from the source is increased by a factor of 3, Then
I ∝ 1/3² ∝ 1/9
Intensity ∝ 1/9
Hence option B is correct.
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Consider the system consisting of the box and the spring, but not Earth. How does the energy of the system when the spring is fully compressed compare to the energy of the system at the moment immediately before the box hits the ground? Justify your answer.
Answer:
the energy when it reaches the ground is equal to the energy when the spring is compressed.
Explanation:
For this comparison let's use the conservation of energy theorem.
Starting point. Compressed spring
Em₀ = K_e = ½ k x²
Final point. When the box hits the ground
Em_f = K = ½ m v²
since friction is zero, energy is conserved
Em₀ = Em_f
1 / 2k x² = ½ m v²
v = [tex]\sqrt{ \frac{k}{m} }[/tex] x
Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.
Based on the law of conservation of energy, the elastic potential energy of the system when the spring is fully compressed is equal to the kinetic energy of the system at the moment immediately before the box hits the ground.
What is the energy in a compressed spring?The energy in a compressed spring is elastic potential energy given by the formula:
Ek = 1/2 Kx^2where
K is spring constant x is displacement of the springWhat is the kinetic energy of a body?The kinetic energy of a body is the energy the body the has due to it's motion.
Kinetic energy, KE, is givenby the formula below:
KE = 1/2mv^2How does the energy of the system when the spring is fully compressed compare to the energy of the system at the moment immediately before the box hits the ground?From the law of conservation of energy, the total energy in a closed system is conserved.
Based on this law, all the energy in the compressed spring is converted to the kinetic energy of the box just before it reaches the ground.
Therefore, the elastic potential energy of the system when the spring is fully compressed is equal to the kinetic energy of the system at the moment immediately before the box hits the ground.
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A force of 15N is action on an area with pressure of 10pa. What is the area covered?
Answer:
150
Explanation:
P=F/A
10=15/A
10*15=15/A*15
150=A
I HOPE I HELPED
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18. Un avión de rescate de animales que vuela hacia el este a 36.0 m/s deja caer una paca de
heno desde una altitud de 60.0 m. Si la paca de heno pesa 175 N, ¿cuál es el momentum
de la paca antes de que golpee el suelo?
Answer:
Definimos momento como el producto entre la masa y la velocidad
P = m*v
(tener en cuenta que la velocidad es un vector, por lo que el momento también será un vector)
Sabemos que el peso de la paca de heno es 175N, y el peso es masa por aceleración gravitatoria, entonces.
Peso = m*9.8m/s^2 = 175N
m = (175N)/(9.8m/s^2) = 17.9 kg
Ahora debemos calcular la velocidad de la paca justo antes de tocar el suelo.
Sabemos que la velocidad horizontal será la misma que tenía el avión, que es:
Vx = 36m/s
Mientras que para la velocidad vertical, usamos la conservación de la energía:
E = U + K
Apenas se suelta la caja, esta tiene velocidad cero, entonces su energía cinética será cero y la caja solo tendrá energía potencial (Si bien la caja tiene velocidad horizontal en este punto, por la superposición lineal podemos separar el problema en un caso horizontal y en un caso vertical, y en el caso vertical no hay velocidad inicial)
Entonces al principio solo hay energía potencial:
U = m*g*h
donde:
m = masa
g = aceleración gravitatoria
h = altura
Sabemos que la altura inicial es 60m, entonces la energía potencial es:
U = 175N*60m = 10,500 N
Cuando la paca esta próxima a golpear el suelo, la altura h tiende a cero, por lo que la energía potencial se hace cero, y en este punto solo tendremos energía cinética, entonces:
10,500N = (m/2)*v^2
De acá podemos despejar la velocidad vertical justo antes de golpear el suelo.
√(10,500N*(2/ 17.9 kg)) = 34.25 m/s
La velocidad vertical es 34.25 m/s
Entonces el vector velocidad se podrá escribir como:
V = (36 m/s, -34.25 m/s)
Donde el signo menos en la velocidad vertical es porque la velocidad vertical es hacia abajo.
Reemplazando esto en la ecuación del momento obtenemos:
P = 17.9kg*(36 m/s, -34.25 m/s)
P = (644.4 N, -613.075 N)
Help please ( with out links ).
Young's double slit experiment is one of the quintessential experiments in physics. The availability of low cost lasers in recent years allows us to perform the double slit experiment rather easily in class. Your professor shines a green laser (568 nm) on a double slit with a separation of 0.112 mm. The diffraction pattern shines on the classroom wall 3.5 m away. Calculate the fringe separation between the second order and central fringe.
Answer:
Y = 17.75 x 10⁻³ m = 17.75 mm
Explanation:
Using Young's Double Slit formula below:
[tex]Y = \frac{\lambda L}{d}[/tex]
where,
Y = fringe separation = ?
λ = wavelength = 568 nm = 5.68 x 10⁻⁷ m
L = distance between slits and screen = 3.5 m
d = slit separation = 0.112 mm = 1.12 x 10⁻⁴ m
Therefore,
[tex]Y = \frac{(5.68\ x\ 10^{-7}\ m)(3.5\ m)}{1.12\ x\ 10^{-4}\ m}[/tex]
Y = 17.75 x 10⁻³ m = 17.75 mm