Answer:
The maximum potential difference across capacitor is 20 V
Explanation:
Given;
maximum emf of the ac circuit, [tex]V_{max}[/tex] = 25 V
maximum potential difference across resistor, [tex]V_R[/tex] = 15 V
maximum potential difference across capacitor, [tex]V_C[/tex] = ?
Determine the maximum potential difference across capacitor using the equation below;
[tex]V_{max} = \sqrt{V_R^2 + V_C^2} \\\\25^2 = V_R^2 + V_C^2\\\\V_C^2 = 25^2 - V_R^2\\\\V_C^2 = 25^2 - 15^2\\\\V_C^2 = 400\\\\V_C = \sqrt{400} \\\\V_C = 20 \ V[/tex]
Therefore, the maximum potential difference across capacitor is 20 V
What is the electric potential (relative to infinity) due to these charges at the center of this square
Answer:
Zero
Explanation:
See attached file pls
Cass is walking her dog (Oreo) around the neighborhood. Upon arriving at Calina's house (a friend of Oreo's), Oreo turns part mule and refuses to continue on the walk. Cass yanks on the chain with a 67 N force at an angle of 30° above the horizontal. Determine the horizontal and vertical components of the tension force.
Answer:
Horizontal component: [tex]F_x = 58\ N[/tex]
Vertical component: [tex]F_y = 33.5\ N[/tex]
Explanation:
To find the horizontal and vertical components of the force, we just need to multiply the magnitude of the force by the cosine and sine of the angle with the horizontal, respectively.
Therefore, for the horizontal component, we have:
[tex]F_x = F * cos(angle)[/tex]
[tex]F_x = 67 * cos(30)[/tex]
[tex]F_x = 58\ N[/tex]
For the vertical component, we have:
[tex]F_y = F * sin(angle)[/tex]
[tex]F_y = 67 * sin(30)[/tex]
[tex]F_y = 33.5\ N[/tex]
So the horizontal component of the tension force is 58 N and the vertical component is 33.5 N.
The bulldog and skateboard have a combined mass of 20 kg. In case B (the middle of the three pictures of the bulldog and the well), the bulldog and skateboard have a KE of 380 J at the bottom of the well. How deep is the well in meters?
Answer:
h = 1.94 m
Explanation:
When the bull dog and skate board reach the bottom of the well, all of its potential energy is converted to the kinetic energy:
Kinetic Energy Gained by Bull Dog and Skate Board = Potential Energy Lost by Bull Dog and Skate Board
K.E = P.E
K.E = mgh
h = K.E/mg
where,
h = depth of well = ?
K.E = Kinetic Energy at bottom = 380 J
m = mass of bull dog and skate board = 20 kg
g = 9.8 m/s²
Therefore,
h = 380 J/(20 kg)(9.8 m/s²)
h = 1.94 m
1. A current of 0.001 A can be felt by the human body. 0.005 A can produce a pain response. 0.015 A can cause a loss of muscle control. In the procedures of this lesson, over 0.030 A of current traveled in the three-battery circuit. Why was this circuit safe to handle with dry hands?
Answer:
It was safe to handle the circuit with dry hands because dry skin body resistance is very high, measuring up to 500,000 ohms.
Explanation:
Given;
Current of 0.001 A to be felt
Current of 0.005 A can produce a pain response
Current of 0.015 A can cause a loss of muscle control
Total current that traveled in the three-battery circuit = 0.03 A
Thus, we can conclude that, it was safe to handle the above mentioned circuit with dry hands because dry skin body resistance is very high, measuring up to 500,000 ohms.
Ohm's Law for electrical circuits is stated VequalsRI, where V is a constant voltage, R is the resistance in ohms and I is the current in amperes. Your firm has been asked to supply the resistors for a circuit in which V will be 9 volts and I is to be 5 plus or minus 0.1 amperes. In what interval does R have to lie for I to be within 0.1 amps of the target value Upper I 0 equals 5 question mark
Answer:
The resistance interval is [tex]R = 1.8 \pm 0.037[/tex]
Explanation:
From the question we are told that
The voltage is V = 9 V
The current is [tex]I = 5 \pm 0.1[/tex]
The maximum current would be
[tex]I_{max} = 5 + 0.1 = 5.1 \ A[/tex]
The minimum current would be
[tex]I_{min} = 5 - 0.1 = 4.9 \ A[/tex]
The maximum resistance is
[tex]R_max = \frac{V}{I_{min}}[/tex]
[tex]R_max = \frac{9}{4.9}[/tex]
[tex]R_max = 1.837 \Omega[/tex]
The minimum resistance is
[tex]R_{min} = \frac{V}{I_{max}}[/tex]
[tex]R_{min} = \frac{9}{5.1}[/tex]
[tex]R_{min} = 1.765 \Omega[/tex]
and [tex]R = \frac{9}{5} = 1.8 \Omega[/tex]
The interval R lies is
[tex]R = 1.8 \pm 0.037[/tex]
A 60-watt light bulb carries a current of 0.5 ampere. The total charge passing through it in one hour is:
Answer:
Total charge = 1800C
Explanation:
Q= IT
I = currentt = timeQ = charge1. A base-ball of mass 0.3kg approaches the bat at a speed of 30 miles/hour and when the ball hits the bat for 0.5 s, it started to move away from the bat at a speed of 60 miles/hour. Find the impulse
Answer:
I = 27kg.mi/h
Explanation:
In order to calculate the impulse of the ball, you use the following formula:
[tex]I=m\Delta v[/tex] [tex]=m(v-v_o)[/tex] (1)
m: mass of the ball = 0.3kg
v: speed of the ball after the bat hit it = 60mi/h
vo: speed of the ball before the bat hit it = 30mi/h
You replace the values of all parameters in the equation (1):
[tex]I=(0.3kg)(60mi/h-(-30mi/h))=27kg\frac{mi}{h}[/tex]
where the minus sign of the initial velocity means that the motion of the ball is opposite to the final direction of such a motion.
The imulpse of the ball is 27 kg.miles/hour
The low-frequency speaker of a stereo set has a surface area of 0.07 m2 and produces 1.63 W of acoustical power. What is the intensity at the speaker (in W/m2)?
Answer:
I = 81.5 W/m^2
Explanation:
In order to calculate the intensity of the sound at the speaker, you use the following formula:
[tex]I=\frac{P}{A}[/tex] (1)
P: power of the speaker's sound = 1.63W
A: surface area of the stereo set = 0.07m^2
You assume that the intensity of the sound at the speaker depends only of the surface area of the stereo set. Furthermore, you consider that the wave front of the sound is approximately plane.
You replace the values of the parameters in the equation (1):
[tex]I=\frac{1.63W}{0.02m^2}=81.5\frac{W}{m^2}[/tex]
The intensity of the speaker's sound at the speaker is 81.5 W/m^2
The electron beam inside an old television picture tube is 0.40 mm in diameter and carries a current of 50 μA. This electron beam impinges on the inside of the picture tube screen.
Requried:
a. How many electrons strike the screen each second?
b. The electrons move with a velocity of 4.0m/s. What electric field strength is needed to accelerate electrons from rest to this velocity in a distance of 5.0 m?
c. Each electron transfers its kinetic energy to the picture tube screen upon impact. What is the power delivered to the screen by the electron beam?
Answer:
a) # _electron = 31 10¹³ electrons, b) E = 9.1 10⁻¹² N / C , c) P = 7.22 10⁻¹⁶ J
Explanation:
In this exercise we must examine the movement of electrons in an electric field.
a) they ask us for the number of electrons.
The electric current is
i = q / t
q = i t
let's calculate
q = 50 10⁻⁶ 1
q = 50 10⁻⁶ C
For this we can use a rule of direct proportions, if the charge of an electron is 1.6 10⁻¹⁹ C, the number of electrons for the charge of 50 10⁻⁶ C
#_electron = 50 10⁻⁶ C 1 / 1.6 10⁻¹⁹ C
# _electron = 31.25 10¹³ electrons
b) For this part we use kinematics to find the acceleration
v² = v₀² + 2 a x
electrons start from rest, so vo = 0
a = v² / 2x
let's calculate
a = 4²/2 5
a = 1.6 m / s²
Having the acceleration we can use Newton's second law where the force is electric
F = ma
e E = m a
E = m / e a
let's calculate
e = 9.1 10⁻³¹ / 1.6 10⁻¹⁹ 1.6
E = 9.1 10⁻¹² N / C
c) what the power on the screen
for this we must add the kinetic energy of all the electrons in the given time
P = # _electron Km / y
P = # _electron (½ m v²) / t
let's calculate
P = 31 10 13 (½ 9.1 10-31 1.6²) / 1
P = 7.22 10⁻¹⁶ J
A solid ball of radius rb has a uniform charge density rho.
A) What is the magnitude of the electric field E(r) at a distance r
B) What is the magnitude of the electric field E(r) at a distance r
C) Let E(r) represent the electric field dueto the charged ball throughout all of space. Which of the following statements about the electric field are true?
1. E(0) = 0.
2. E(rb) = 0
3. lim E(r) = 0.
4. The maximum electric field occurs when r = 0.
5. The maximum electric field occurs when r = rb.
6. The maximum electric field occurs as r to infinity.
Note: question B is incomplete.
Complete Question
A solid ball of radius rb has a uniform charge density ρ.
a. What is the magnitude of the electric field E(r) at a distance r>rb from the center of the ball? Express your answer in terms of ρ, rb, r, and ϵ0.
b. What is the magnitude of the electric field E(r) at a distance r<rb from the center of the ball? Express your answer in terms of ρ, r, rb, and ϵ0.
c. Let E(r) represent the electric field due to the charged ball throughout all of space. Which of the following statements about the electric field are true?
1. E(0) = 0.
2. E(rb) = 0
3. lim E(r) = 0.
4. The maximum electric field occurs when r = 0.
5. The maximum electric field occurs when r = rb.
6. The maximum electric field occurs as r to infinity.
Answer:
a) the magnitude of E(r)= ρr³/3 ε₀r²
b) the magnitude at distance r from the centre E(r)= ρr/3 ε ₀ ( if r<rb)
c) statements 1(E(0) = 0), 3(E(0) = 0) and 5(The maximum electric field occurs when r = rb.) are true
Explanation:
given
charge density = ρ , ε
Volume of sphere , V = (⁴/₃)πr³
a) charge density = charge/volume
ρ = q ÷ V
make charge the subject of the formula
∴q = ρ × V= ρ× (⁴/₃)πr³
where r³ = rb³(at distance rb³)
recall
E= q/4πε₀r²
E= ρ × (⁴/₃)πrb³/4πε₀r²
∴E(r)= ρrb³/3 ε ₀r²
(b) The Gaussian surface is inside the ball, therefore, surface only encloses a portion of ball's charge .
The net charge enclosed by the Gaussian surface is different to the of net charge enclosed in (a)
Recall
E= q/4πε₀r²
V= (⁴/₃)πr³
E= ρ × (⁴/₃)πr³/4πε₀r²
∴E(r)= ρr/3 ε₀
(c) E(0)= 0
limr-----∝
E(r)= 0
The maximum electric field occurs when r=rb.
What is the role of the part in the diagram labeled Y?
modulate, amplify, and send out waves
O capture, amplify, and demodulate waves
change the amplitude and frequency of waves
O change the pulse and phase of waves
Question is incomplete and image is not attached ti the question. The required image is attached below, so the complete question is:
The diagram shows a device that uses radio waves.
What is the role of the part in the diagram labeled Y?
modulate, amplify, and send out waves capture, amplify, and demodulate waves change the amplitude and frequency of waves change the pulse and phase of wavesAnswer:
2. capture, amplify, and demodulate waves
Explanation:
The part Y labeled in the diagram refers to radio receiver which capture, amplify and demodulate the radio waves.
The radio receiver seperates required radio frequency signals through antenna and consist of an amplifier that amplify or increase the power of receiving signal. At the end, demodulators present in receivers recover the information from the modulated wave.
Hence, the correct option is 2.
Answer:
B
Explanation:
edge 2020
A convex meniscus thin lens is made with the radius of curvature of the first convex surface being 25.0 cm and the second convex (towards object) surface 45.0 cm. If the glass used has index of refraction 1.500, what is the focal length of this lens?
Answer:
112.5cm
Explanation:
Using the formula
1/f = (n-1)(1/R1 + 1/R2)
1/f = (1.5 - 1)(1/25 - 1/45)
1/f = 0.5 x (4/225) = 2/225
f = 225/2
= 112.5cm
World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15m/s2. Part A How much horizontal force F must a sprinter of mass 47 kg exert on the starting blocks to produce this acceleration
Answer:
The horizontal force the sprinter exerts on the starting blocks is 705 N
Explanation:
Given;
mass of the sprinter, m = 47 kg
magnitude of the his horizontal acceleration, aₓ = 15 m/s²
The horizontal force the sprinter exerts on the starting blocks to produce this acceleration, is calculated by applying Newton's second law of motion.
∑Fₓ = maₓ
∑Fₓ = 47 x 15
∑Fₓ = 705 N
Therefore, the horizontal force the sprinter exerts on the starting blocks is 705 N
A grating having 5000 lines/cm is used with light of wavelength 633 nm. How many total maxima (count central maxima plus all those on either side of the central maxima) are produced
Answer:
The total number of maxima produced is [tex]m_T = 7[/tex] maxima
Explanation:
From the question we are told that
The number of lines per cm is [tex]n = 5000 \ lines/cm[/tex]
The wavelength of the light is [tex]\lambda = 633 nm = 633 *10^{-9} \ m[/tex]
Now the distance between the lines is mathematically evaluated as
[tex]d = \frac{1}{n}[/tex]
substituting values
[tex]d = \frac{1}{5000}[/tex]
[tex]d = \frac{1 *10^{-2}}{5000}[/tex] N/B - this statement convert it from cm to m
[tex]d = 2 *10^{ -6} \ m[/tex]
Generally the condition for diffraction i mathematically represented as
[tex]dsin(\theta ) = m \lambda[/tex]
at maximum [tex]\theta = 90 ^o[/tex]
[tex]d sin (90) = \lambda m[/tex]
here m is the number of maxima
Thus making m the subject we have
[tex]m = \frac{d sin (90)}{ \lambda }[/tex]
So [tex]m = \frac{2*10^{-6} sin (90)}{ 633 *10^{-9}}[/tex]
[tex]m = 3.2[/tex]
=> m =3
Now the total number of maxima would include the bright fringe(3) and dark fringe (3) plus the central maxima (1)
Thus
[tex]m_T = 3 + 3 +1[/tex]
[tex]m_T = 7[/tex] maxima
An electric motor drawing 10 amps at 110 V in steady state produces shaft power at 9.7 Nm and 1000 RPM. For a first Law analysis and considering the motor as the control volume How much heat will be produced from the motor (in Watts)
Answer:
The heat rate produced from the motor is 84.216 watts.
Explanation:
The electric motor receives power from electric current and releases power in the form of mechanical energy (torque) and waste heat and can be considered an stable-state system. The model based on the First Law of Thermodynamics for the electric motor is:
[tex]\dot W_{e} - \dot W_{T} -\dot Q = 0[/tex]
Where:
[tex]\dot Q[/tex] - Heat transfer from the electric motor, measured in watts.
[tex]\dot W_{e}[/tex] - Electric power, measured in watts.
[tex]\dot W_{T}[/tex] - Mechanical power, measured in watts.
The heat transfer rate can be calculated in terms of electric and mechanic powers, that is:
[tex]\dot Q = \dot W_{e} - \dot W_{T}[/tex]
The electric and mechanic powers are represented by the following expressions:
[tex]\dot W_{e} = i \cdot V[/tex]
[tex]\dot W_{T} = T \cdot \omega[/tex]
Where:
[tex]i[/tex] - Current, measured in amperes.
[tex]V[/tex] - Steady-state voltage, measured in volts.
[tex]T[/tex] - Torque, measured in newton-meters.
[tex]\omega[/tex] - Angular speed, measured in radians per second.
Now, the previous expression for heat transfer rate is expanded:
[tex]\dot Q = i \cdot V - T \cdot \omega[/tex]
The angular speed, measured in radians per second, can be obtained by using the following expression:
[tex]\omega = \frac{\pi}{30}\cdot \dot n[/tex]
Where:
[tex]\dot n[/tex] - Rotational rate of change, measured in revolutions per minute.
If [tex]\dot n = 1000\,rpm[/tex], then:
[tex]\omega = \left(\frac{\pi}{30} \right)\cdot (1000\,rpm)[/tex]
[tex]\omega \approx 104.720\,\frac{rad}{s}[/tex]
Given that [tex]i = 10\,A[/tex], [tex]V = 110\,V[/tex], [tex]T = 9.7\,N\cdot m[/tex] and [tex]\omega \approx 104.720\,\frac{rad}{s}[/tex], the heat transfer rate from the electric motor is:
[tex]\dot Q = (10\,A)\cdot (110\,V) -(9.7\,N\cdot m)\cdot \left(104.720\,\frac{rad}{s} \right)[/tex]
[tex]\dot Q = 84.216\,W[/tex]
The heat rate produced from the motor is 84.216 watts.
A simple random sample is a sample drawn in such a way that each member of the population has some chance for being included in the sample every tenth element of an arranged population is included each member of the population has equal chance for being included in the sample each member of the population has 0.10 chance for being included in the sample:__________.
A simple random sample is a sample drawn in such a way that each member of the population has equal chance for being included in the sample.
The mass percent of hydrogen in CH₄O is 12.5%.
What is the mass percent?Mass percent is the mass of the element divided by the mass of the compound or solute.
Step 1: Calculate the mass of the compound.
mCH₄O = 1 mC + 4 mH + 1 mO = 1 (12.01 amu) + 4 (1.00 amu) + 1 (16.00 amu) = 32.01 amu
Step 2: Calculate the mass of hydrogen in the compound.
mH in mCH₄O = 4 mH = 4 (1.00 amu) = 4.00 amu
Step 3: Calculate the mass percent of hydrogen in the compound.
%H = (mH in mCH₄O / mCH₄O) × 100%
%H = 4.00 amu / 32.01 amu × 100% = 12.5%
The mass percent of hydrogen in CH₄O is 12.5%.
CO2 = 1.580 grams H2O = 0.592 grams Lookup the molar mass of each element in the compound Carbon = 12.0107 Hydrogen = 1.00794 Oxygen = 15.999 Calculate the molar mass of CH4O by adding the total masses of each element used. 12.0107 + 4 * 1.00794 + 15.999 = 32.04146 Now calculate how many moles of CH4O you have by dividing by the molar mass. m = 1.15 g / 32.04146 g/mole = 0.035891 mole Now figure out how many moles of carbon and hydrogen you have. Carbon = 0.035891 moles Hydrogen = 0.035891 moles *
Therefore, The mass percent of hydrogen in CH₄O is 12.5%.
Learn more about mass percent here:
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The horse on a carousel is 3.5m from the central axis.A. If the carousel rotates at 0.13 rev/s , how long does it take the horse to go around twice?B. How fast is a child on the horse going (in m/s)?
Answer:
a. 15.4 seconds
b. 0.455 m/s
Explanation:
a. The carousel rotates at 0.13 rev/s.
This means that it takes the carousel 1 sec to make 0.13 of an entire revolution.
This means that time it will take to make a complete revolution is:
1 / 0.13 = 7.7 seconds
Therefore, the time it will take to make 2 revolutions is:
2 * 7.7 = 15.4 seconds
b. Let us calculate the linear velocity. Angular velocity is given as:
[tex]\omega = v / r[/tex]
where v = linear velocity and r = radius
The radius of the circle is 3.5 m and the angular velocity is 0.13 rev/s, therefore:
0.13 = v / 3.5
v = 3.5 * 0.13 = 0.455 m/s
Linear velocity is 0.455 m/s
A car travels 120 mi averaging a certain speed. If the car had gone 10 mph faster, the trip would have taken 1 hr less. Find the car's average speed.
Answer:
The average speed of a car will be "30 mph".
Explanation:
The given values are:
Distance = 120 mi
Let the speed be "x".
As we know,
[tex]Time = \frac{Distance}{Speed}[/tex]
[tex]=\frac{120}{x}[/tex]
According to the question,
If the speed of a car = x + 10
then the time will be:
= [tex]\frac{120}{x} -1[/tex]
Now,
⇒ [tex]Speed\times Time=(x+10)(\frac{120}{x}-1)[/tex]
⇒ [tex]120=120+\frac{1200}{x} -x-10[/tex]
On multiplying "x" both sides, we get
⇒ [tex]120x=120x+1200-x^2-10x[/tex]
⇒ [tex]x^2+10x-1200[/tex]
⇒ [tex]x^2+40x-30x-1200=0[/tex]
⇒ [tex]x(x+40)-30(x+40)=0[/tex]
⇒ [tex](x+40)(x-30)=0[/tex]
[tex]x+40=0[/tex]
[tex]x=-40[/tex]
Or,
[tex]x-30=0[/tex]
[tex]x=30[/tex]
So that the average speed will be "30 mph".
To understand the behavior of the electric field at the surface of a conductor, and its relationship to surface charge on the conductor. A conductor is placed in an external electrostatic field. The external field is uniform before the conductor is placed within it. The conductor is completely isolated from any source of current or charge.
1) Which of the following describes the electricfield inside this conductor?
A) It is in the same direction as the original external field.
B) It is in the opposite direction from that of the original external field.
C) It has a direction determined entirely by the charge on its surface.
D) It is always zero.
2) The charge density inside the conductor is:______.
a) 0.
b) non-zero;but uniform.
c) non-zero;non-uniform.
d) infinite.
3) Assume that at some point just outside the surface of the conductor, the electric field has magnitudeE and is directed toward the surface of the conductor. What is the charge density eta on the surface of the conductor at that point?
Answer:
1) Option D is correct.
The electric field inside a conductor is always zero.
2) Option A is correct.
The charge density inside the conductor is 0.
3) Charge density on the surface of the conductor at that point = η = -E ε₀
Explanation:
1) The electric field is zero inside a conductor. Any excess charge resides entirely on the surface or surfaces of a conductor.
Assuming the net electric field wasn't zero, current would flow inside the conductor and this would build up charges on the exterior of the conductor. These charges would oppose the field, ultimately (in a few nanoseconds for a metal) canceling the field to zero.
2) Since there are no charges inside a conductor (they all reside on the surface), it is logical that the charge density inside the conductor is also 0.
3) Surface Charge density = η = (q/A)
But electric field is given as
E = (-q/2πε₀r²)
q = -E (2πε₀r²)
η = (q/A) = -E (2πε₀r²)/A
For an elemental point on the surface,
A = 2πrl = 2πr²
So,
η = -E ε₀
Hope this Helps!!!
a centimeter in milimetersr difiers from each other by a fact of 100 ,1000, 1 or 10
Answer:
The correct answer is - 1.
Explanation:
The one cubic centimeter is equal to the one mililitter of a liquid. It is same de to the fact that volume of water that fits in a one cubic centimeter is equal to the liquid present in one mililiter of a liquid.
1 cm² = 1 ml
1 cm³ = 1 ml
Thus, the use of the both unit is interchangebly.
So , correct answer is - 1.
A skier traveling 9.4 m/s reaches the foot of a steady upward 16 ∘ incline and glides 12 m up along this slope before coming to rest. What was the average coefficient of friction?
Answer:
The average coefficient of friction is 0.104
Explanation:
Given;
velocity of the skier, v = 9.4 m/s
angle of inclination, θ = 16°
height of the slope, h = 12 m
Calculate the kinetic energy of the skier at the given velocity;
KE =¹/₂mv²
KE = ¹/₂m(9.4)²
KE = 44.18m
Calculate the potential energy of the skier up the slope;
PE = mghsinθ
PE = m x 9.8 x 12 x sin16
PE = 32.411 m
Calculate the friction force on the inclined plane;
Fk = μk x mgCosθ
Fk = μk x m x 9.8 x Cos16
Fk = μk x m x 9.421
Calculate the work done by friction;
Work done = Fk x d
Work done = μk x m x 9.421 x d
Work done = μk x m x 9.421 x 12
Work done = 113.052mμk
Apply the principle of conservation of energy;
KE = PE + work done by friction
44.18m = 32.411 m + 113.052mμk (factor out m and divide through by it)
44.18 = 32.411 + 113.052μk
44.18 - 32.411 = 113.052μk
11.769 = 113.052μk
μk = 11.769 / 113.052
μk = 0.104
Therefore, the average coefficient of friction is 0.104
The speed of light in a material is 1.74 x 108 m's. What is the index of refraction of this material?
Answer:
The index of refraction of this material is 1.7241
Explanation:
Recall that the index of refraction (n) of a medium is defined as the quotient between the speed of light in vacuum divided the speed of light in the medium whose index of refraction is being calculated. In mathematical terms:
[tex]n=\frac{c}{v}[/tex]
Therefore, in our case, since we know the speed of light in the medium ([tex]v=1.74\,\,10^8\,\,m/s[/tex]) and the speed of light in vacuum ([tex]c=3\,\,10^8\,\,m/s[/tex]), we can estimate the index of refraction of the medium:
[tex]n=\frac{c}{v} \\n=\frac{3\,\,10^8}{1.74\,\,10^8} \\n=1.7241[/tex]
Samantha is refinishing her rusty wheelbarrow. She moves her sandpaper back and forth 45 times over a rusty area, each time moving with a total distance of 0.12 m. Samantha pushes the sandpaper against the surface with a normal force of 2.6 N. The coefficient of friction for the metal/sandpaper interface is 0.92. How much work is done by the normal force during the sanding process
Answer:
W = 12.96 J
Explanation:
The force acting in the direction of motion of the sand paper is the frictional force. So, we first calculate the frictional force:
F = μR
where,
F = Friction Force = ?
μ = 0.92
R = Normal Force = 2.6 N
Therefore,
F = (0.92)(2.6 N)
F = 2.4 N
Now, the displacement is given as:
d = (0.12 m)(45)
d = 5.4 m
So, the work done will be:
W = F d
W = (2.4 N)(5.4 m)
W = 12.96 J
If the same quantity of heat is added to both a 2-liter and a 4-liter container of water, the temperature change for water in the 4-liter container will be A) half that of the 2-liter container. B) more than half but less than twice. C) twice that of the 2-liter container. D) no change. Please explain.
Answer:
https://brainly.com/question/10966794
Explanation:
A woman who weighs 500 N stands on an 8 m long board that weighs 100 N. The board is supported at each end. The support force at the right end is 3 times bigger than the support force at the left end. How far from the right end is the woman standing
Answer:
[tex]1.6\; \rm m[/tex].
Explanation:
Let [tex]x[/tex] denote the distance (in meters) between the person and the right end of the board.
To keep the calculations simple, consider another unknown: let [tex]y[/tex] denote the support force (in Newtons) on the left end. The support force on the right end of this board would be [tex]3 \, y[/tex] (also in Newtons.)
Now there are two unknowns. At least two equations will be required for finding the exact solutions. For that, consider this board as a lever, but with two possible fulcrums. Refer to the two diagrams attached. (Not to scale.)
In the first diagram, the support at the left end of the board is considered as the fulcrum. In the second diagram, the support at the right end of the board is considered as the fulcrum.Calculate the torque in each situation. Note that are four external forces acting on this board at the same time. (Two support forces and two weights.) Why does each of the two diagrams show only three? In particular, why is the support force at each "fulcrum" missing? The reason is that any force acting on the lever at the fulcrum will have no direct impact on the balance between torques elsewhere on the lever. Keep in mind that the torque of each force on a lever is proportional to [tex]r[/tex], the distance between the starting point and the fulcrum. Since that missing support force starts right at the fulcrum, its [tex]r[/tex] will be zero, and it will have no torque in this context.
Hence, there are three (non-zero) torques acting on the "lever" in each diagram. For example, in the first diagram:
The weight of the board acts at the center of the board, [tex](1/2) \times 8\; \rm m = 4\; \rm m[/tex] from the fulcrum. This force will exert a torque of [tex]\tau(\text{weight of board}) = 4\; \rm m \times (-100\; \rm N) = (-400\; \rm N \cdot m)[/tex] on this "lever". The negative sign indicates that this torque points downwards.The weight of the person acts at [tex]x\;\rm m[/tex] from the right end of the board, which is [tex](8 - x)\; \rm m[/tex] from the fulcrum at the other end of this board. This force will exert a torque of [tex]\tau(\text{weight of person}) = (8 - x)\; {\rm m \times (-500\; \rm N)} = (-500\, \mathnormal{(8 - x)})\; \rm N \cdot m[/tex] on this "lever". This torque also points downwards.The support on the right end of the board acts at [tex]8\; \rm m[/tex] from the fulcrum (i.e., the left end of this board.) This force will exert a torque of [tex]\tau(\text{support, right}) = 8\; {\rm m} \times (3\, \mathnormal{y})\; {\rm N} = (24\, y)\; \rm N \cdot m[/tex] on the "lever". This torque points upwards.If the value of [tex]x[/tex] and [tex]y[/tex] are correct, these three torques should add up to zero. That is:
[tex]\underbrace{(-400)}_{\text{board}} + \underbrace{(-500\, (8 - x))}_{\text{person}} + \underbrace{24\, y}_{\text{support}} = 0[/tex].
That gives the first equation of this system. Similarly, a different equation can be obtained using the second diagram:
[tex]\underbrace{(-400)}_{\text{board}} + \underbrace{(-500\,x)}_{\text{person}} + \underbrace{8\, y}_{\text{support}} = 0[/tex].
Combine these two equations into a two-by-two system. Solve the system for [tex]x[/tex] and [tex]y[/tex]:
[tex]\left\lbrace\begin{aligned}&x = 1.6\\ &y = 150\end{aligned}\right.[/tex].
In other words, the person is standing at about [tex]1.6\; \rm m[/tex] from the right end of the board. The support force at the left end of the board is [tex]150\; \rm N[/tex].
Why do some astronomers object to the new definition of a planet that was adopted in 2006?
A. New space missions show that Pluto is much larger than originally thought.
B. By this definition, Earth, Jupiter, and other planets should not be considered planets.
C. There was never a vote on whether to adopt the new definition or not.
D. It means that we now technically have over 100 planets.
Answer:
A. New space missions show that Pluto is much larger than originally thought.
Explanation:
The new definition of a planet that was adopted in 2006, defined planet as an object that orbits the sun, with sufficient mass to be round, not a satellite of another object, and has removed debris and small objects from the area around its orbit.
This new definition of a planet that was adopted in 2006, classified Pluto as "dwarf planet", because Pluto meets planetary criteria except that it has not cleared debris from its orbital neighborhood.
However, new Horizons spacecraft flew by Pluto in 2015, revealed that Pluto is much larger than originally thought
Therefore, the correct option is "A"
A. New space missions show that Pluto is much larger than originally thought.
Answer: it means that we now technically have over 100 planets
Explanation:
it’s not New space missions show that Pluto is much larger than originally thought!!!
A mass on a spring vibrates in simple harmonic motion at a frequency of 3.26 Hz and an amplitude of 5.76 cm. If the mass of the object of 0.218 kg, what is the spring constant
Answer:
91.48N/m
Explanation:
In a spring-mass system undergoing a simple harmonic motion, the inverse of the frequency f, of oscillation is proportional to the square root of the mass m, and inversely proportional to the square root of the spring constant, k. This can be expressed mathematically as follows;
[tex]\frac{1}{f}[/tex] = [tex]2\pi\sqrt{\frac{m}{k} }[/tex] -----------(i)
From the question;
f = 3.26 Hz
m = 0.218kg
Substitute these values into equation (i) as follows;
[tex]\frac{1}{3.26}[/tex] = [tex]2\pi\sqrt{\frac{0.218}{k} }[/tex] [Square both sides]
([tex]\frac{1}{3.26}[/tex])² = ([tex]2\pi[/tex])²([tex]\frac{0.218}{k}[/tex])
([tex]\frac{1}{10.6276}[/tex]) = [tex]4\pi[/tex]²([tex]\frac{0.218}{k}[/tex]) [Take [tex]\pi[/tex] to be 3.142]
([tex]\frac{1}{10.6276}[/tex]) = [tex]4(3.142)[/tex]²([tex]\frac{0.218}{k}[/tex])
([tex]\frac{1}{10.6276}[/tex]) = [tex]39.488[/tex]([tex]\frac{0.218}{k}[/tex])
([tex]\frac{1}{10.6276}[/tex]) = ([tex]\frac{8.608}{k}[/tex]) [Switch sides]
([tex]\frac{8.608}{k}[/tex]) = ([tex]\frac{1}{10.6276}[/tex]) [Re-arrange]
([tex]\frac{k}{8.608}[/tex]) = ([tex]\frac{10.6276}{1}[/tex]) [Cross-multiply]
k = 8.608 x 10.6276
k = 91.48N/m
Therefore, the spring constant of the spring is 91.48N/m
A brick is dropped with zero initial speed from the roof of a building and strikes the ground in 1.90 s. How tall is the building?
Answer:
17.69 m
Explanation:
The time it takes the brick to strike the ground is 1.90 seconds.
We can apply one of Newton's equation of linear motion to find the height of the building:
[tex]s = ut + 0.5gt^2[/tex]
where s = distance (in this case height)
u = initial velocity = 0 m/s
t = time = 1.90 s
g = acceleration due to gravity = 9.8 m/s^2
Therefore:
s = (0 * 1.9) + (0.5 * 9.8 * 1.9 * 1.9)
s = 0 + 17.68
s = 17.69 m
The height of the building is 17.69 m.
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d. Write the importance of International Bureau of
Weights and measures in the country?
As
Answer:
The effectiveness of that same country's worldwide weights, as well as Measures Bureau, has always been chosen to give elsewhere here.
Explanation:
The International association formed to unify imperial measurements in order to set and maintain basic universal requirements as well as prototypes. Check the minimum standard.It becomes an interagency body set up by that of the Metre Convention, wherein the member countries work on such measuring science including quality control concerns.what is meaning of convection
Answer:
Convection is heat transfer through the movement of liquids and gases.