Answer:
Part A : 336 N/m
Part B: 19.75 cm
Explanation:
Part A: Using hooks law,
F = ke................... Equation 1
Where F = Force applied to the spring, k = spring constant, e = extension.
Note: From the question, the force applied to the spring is the weight of the mass hung on it.
Hence,
F = mg............... Equation 2
Substitute equation 2 into equation 1.
mg = ke
make k the subject of the equation
k = mg/e.............. Equation 3
Given: m = 2.4 kg, e = 18-11 = 7 cm = 0.07 m
Constant: g = 9.8 m/s²
Substitute these value into equation 3
k = 2.4(9.8)/(0.07)
k = 336 N/m.
Part B,
Similarly,
mg = ke.
make e the subject of the equation
e = mg/k................ Equation 4
Given: m = 3.0 kg, k = 336 N/m, g = 9.8 m/s²
Substitute into equation 4
e = 3.0(9.8)/336
e = 0.0875 m
e = 8.75 cm
Length of the spring = original length+ extension
Length of the spring = 11+8.75
Length of the spring = 19.75 cm.
?
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.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.
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
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!!!
The image formed by a convex lens is 5.0 cm from the midline. The image is 10.0 cm in height. If the object is 2.0 cm tall, where is the object located? 1.0 cm from the lens 2.5 cm from the lens 10.0 cm from the lens 5.0 cm from the lens
Answer:
1.0 cm from the lens
Explanation:
Answer:
1.0cm from the lens
Explanation:
I had this in a lesson and got it right with this answer.
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 body weighing 8 N is supported by two cables whose voltages T1 and T2 form angles of 60 ° and 30 ° with the ceiling. How much are these tensions worth? Choose the correct option. Select one: a. 4N and 7N b. 5N and 7N c. 4N and 10N
Answer:
a. 4N and 7N
Explanation:
Draw a free body diagram.
Sum of the forces in the x direction:
∑F = ma
T₂ cos 30° − T₁ cos 60° = 0
T₂ cos 30° = T₁ cos 60°
T₂ (½√3) = T₁ (½)
T₁ = T₂ √3
Sum of the forces in the y direction:
∑F = ma
T₂ sin 30° + T₁ sin 60° − mg = 0
T₂ sin 30° + (T₂ √3) sin 60° − mg = 0
½ T₂ + 1.5 T₂ − mg = 0
2 T₂ = mg
T₂ = 4 N
T₁ = 4√3 N
T₁ ≈ 7 N
1. 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
Consider a steel tape measure with cross-sectional area, A = 0.0625 inches squared, and length L = 3, 600 inches at room temperature. How much error will occur if this tape measure is used on a hot day? Assume it is 130F and the coefficient of thermal expansion is α = 5× 10 −6 1/F. Does the error depend on the distance being measured?
Answer:
e = -0.00031 ( the -ve sign is due to the increase in length)
The error depends on the distance measured
Explanation:
Cross Sectional Area of the tape, A = 0.0625 in²
Length of the steel tape, L = 3600 in
Normal room temperature, T₁ = 68°F
Temperature of the hot day, T₂ = 130°F
ΔT = T₂ - T₁ = 130 - 68
ΔT = 62°F
Coefficient of Linear expansion [tex]\alpha = 5 * 10^{-6} F^{-1}[/tex]
The coefficient of linear expansion is given by the formula:
[tex]\alpha = \frac{\triangle L}{L* \triangle T} \\\triangle L = \alpha L \triangle T\\\triangle L = 5* 10^{-6} * 3.6* 10^3 * 62\\\triangle L = 1.11 6 in[/tex]
Since the length is increased, the error will be given by the formula:
[tex]e = \frac{-\triangle L}{L} \\\\e = \frac{-1.116}{3600}[/tex]
e = -0.00031 ( the -ve sign is due to the increase in length)
Since the error is a function of length and change in length, it depends on the distance measured
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]
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.
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:
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Explanation:
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
A ski starts from rest and slides down a 28∘ incline 100 m long. If the coefficient of friction is 0.090, what is the ski's speed at the base of the incline? If the snow is level at the foot of the incline and has the same coefficient of friction, how far will the ski travel along with the level? Use energy methods.
Answer:
Explanation:
Frictional force acting on incline = μ mg cosθ
μ is coefficient of friction , m is mass of object , θ is incline
= .09 x m x 9.8 x cos 28
= .78 m
work done by friction
= frictional force x displacement
= - .78m x 100
= - 78m
Potential energy of sli at height
= mgh
= m x 9.8 x 100 sin 28
= 460.08 m
net energy at the base
= 460.08m - 78 m
= 382.08 m
This will be in the form of kinetic energy .
1/2 m v² = 382.08 m
.5 x v² = 382.08
v = 27.64 m/s
After that it travels on plane surface .
Let the distance travelled be d
work done by frictional force
= μ mg x d
= .09 x m x 9.8 x d
This will be equal to kinetic energy at the base
.09 x m x 9.8 x d = 382.08 m
d = 433.2 meter .
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
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
Find the buoyant force on a brass block 10.5 cm long by 12.3 cm wide by 15.0 cm high when placed in (a) water, (b) glycerine, and (c) mercury
Answer:
(a) F = 18.99 N
(b) F = 23.96 N
(c) F = 258.56 N
Explanation:
First we need to find the volume of the block:
Volume of Block = V = (Length)(Width)(Height)
V = (10.5 cm)(12.3 cm)(15 cm)
V = (0.105 m)(0.123 m)(0.15 m)
V = 0.00194 m³
Now, the buoyant force i given by the formula:
F = (Density of Fluid)(V)(g)
(a)
F = (Density of Water)(V)(g)
F = (1000 kg/m³)(0.00194 m³)(9.8 m/s²)
F = 18.99 N
(b)
F = (Density of Glycerine)(V)(g)
F = (1260 kg/m³)(0.00194 m³)(9.8 m/s²)
F = 23.96 N
(c)
F = (Density of Mercury)(V)(g)
F = (13600 kg/m³)(0.00194 m³)(9.8 m/s²)
F = 258.56 N
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.
what is meaning of convection
Answer:
Convection is heat transfer through the movement of liquids and gases.
A solenoid passes through the center of a wire loop, as shown in (Figure 1). The solenoid has 1200 turns, a diameter of 2.0 cm, and is 7.5 cm long. The resistance of the loop is 0.032 Ω.
The induced current in the loop at the given dimension is 10.25 A.
The given parameters:
number of turns of the solenoid, n = 1200 turnsdiameter of the solenoid, d = 2.0 cmlength of the solenoid, L = 7.5 cmresistance of the loop, R = 0.032 Ω.current in solenoid, I = 1.3 Atime, t = 30 msThe area are of the solenoid is calculated as;
[tex]A = \frac{\pi d^2}{4} \\\\A = \frac{\pi \times 0.02^2}{4} = 0.000314 \ m^2[/tex]
The emf induced in the solenoid is calculated as;
[tex]emf = N\frac{d\phi}{dt} \\\\emf = N (\frac{BA}{t} )\\\\emf = N(\frac{\mu_0 NI \times A} {L\times t} )\\\\emf = \frac{N^2 \mu_0 I A}{Lt} \\\\emf = \frac{(1200)^2 \times (4\pi \times 10^{-7}) \times (1.3) \times (0.000314)}{0.075 \times 30\times 10^{-3}} \\\\emf = 0.328 \ V[/tex]
The induced current in the loop is calculated as follows;
[tex]I = \frac{emf}{R} \\\\I = \frac{0.328}{0.032} \\\\I = 10.25 \ A[/tex]
Thus, the induced current in the loop is 10.25 A.
"Your question is not complete, it seems to be missing the following information;"
A solenoid passes through the center of a wire loop, as shown in (Figure 1). The solenoid has 1200 turns, a diameter of 2.0 cm, and is 7.5 cm long. The resistance of the loop is 0.032 Ω.If the current in the solenoid is increased by 1.3 A in 30 ms, what is the induced current in the loop?
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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%.
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The electric field just above the surface of the earth has been measured to be 139 N/C downward. What total charge (sign and magnitude) on the Earth is implied by this measurement
Answer:
156KC
Explanation:
Pls see attached file
156 C is the total charge on the Earth is implied by measurement.
What is Gauss law ?It states that the electric flux Φ across any closed surface is proportional to the net electric charge q enclosed by the surface;
i.e. Φ = q/ε0
According to the question,
Electric field of Earth is 139 N/C
ε[tex]_{o}[/tex][tex]= 8.85 *10^{-12} C^{2} /Nm^{2}[/tex]
[tex]R_{e} = 6.37 * 10^{6}[/tex]
Using the formula by Gauss's Law
Φ = q/ε[tex]_{o}[/tex] = EA
q = ε[tex]_{o}[/tex]EA = ε[tex]_{o}[/tex]E*r²
= [tex]= 8.85*10^{12} *139*(6.37*10^{6} )^2\\\\= 156,905.9*10^0 = 156C[/tex]
Therefore ,
The total charge on the Earth is implied by this measurement is 156 C.
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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.
metal sphere A has 4 units of negative charge and metal sphere B has 2 units of positive charge. The two spheres are brought into contact. What is the final charge state of each sphere
In final state, every metal sphere carries 1 unit of negative charge.
When the spheres brought into contact, the charges try to get an equilibrium distribution on both spheres.It means that same amount of positive charges neutralise the same amount of negative charges.Given that, metal sphere A has 4 units of negative charge and metal sphere B has 2 units of positive charge. So that, 2 units of negative charge will neutralize the 2 unit of positive charge of the second sphere. The others 2 units of negative charge of the first sphere will distribute equally on both spheres.Hence, In final state, every metal sphere carries 1 unit of negative charge.
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A 4-meter plank weighing 8 kg is lying on a dock with 3.58 meters of its length on the dock, and the rest of its length extending off the dock. If a child weighing 57.2 kg walks out onto the part of the plank extending off the dock, what is the maximum distance in meters that the child can walk on the plank from the edge of the dock before he falls into the water
Answer:
From the edge of the dock this distance is 0.221 m
Explanation:
Length of the plank = 4 m
mass of the plank = 8 kg
mass of the child = 57.2 kg
We will assume that:
the mass of the plank acts at the center of the plank.plank is balanced at 3.58 m from the the end of the plank on the dock.This means that the moment of the mass of the plank acts at (3.58 - 2 = 1.58 m) from the balance point.
For the maximum point the boy can walk while still maintaining stability, we balance the moment due to the mass of the plank against the moment that will be generated due to the mass of the boy, at the maximum distance at which stability is possible.
moment of the mass of the plank about the 3.58 m mark is
==> 8 x 1.58 = 12.64 kg-m
moment of the boy about the 3.58 m mark is
==> 57.2 x d = 57.2d
where d is the maximum point at which stability is still possible
equating the two moments,
12.64 = 57.2d
d = 12.64/57.2 = 0.221 m away from the 3.58 m mark
the maximum distance at which stability is still possible is the maximum distance that the boy can walks before he falls into the water.
From the edge of the dock this distance is 0.221 m
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.
After fixing a flat tire on a bicycle you give the wheel a spin. Its initial angular speed was 5.45 rad/s and it rotated 14.4 revolutions before coming to rest.
1. What was its average angular acceleration?
2. For what length of time did the wheel rotate?
Answer:
(a) α = -0.16 rad/s²
(b) t = 33.2 s
Explanation:
(a)
Applying 3rd equation of motion on the circular motion of the tire:
2αθ = ωf² - ωi²
where,
α = angular acceleration = ?
ωf = final angular velocity = 0 rad/s (tire finally stops)
ωi = initial angular velocity = 5.45 rad/s
θ = Angular Displacement = (14.4 rev)(2π rad/1 rev) = 28.8π rad
Therefore,
2(α)(28.8π rad) = (0 rad/s)² - (5.45 rad/s)²
α = -(29.7 rad²/s²)/(57.6π rad)
α = -0.16 rad/s²
Negative sign shows deceleration
(b)
Now, we apply 1st equation of motion:
ωf = ωi + αt
0 rad/s = 5.45 rad/s + (-0.16 rad/s²)t
t = (5.45 rad/s)/(0.16 rad/s²)
t = 33.2 s
when electricity is transmitted through the air the resulting spark is called ?
Answer:
Lightning
Explanation:
Air is a non-conductor of electricity because it does not have free electrons to carry the current. But when there is a high voltage, as in the case of lightning, the molecules of air get ionized and electrons are available to flow to make the current and conduct electricity through the air.
Answer:
corona discharge
Explanation:
A corona discharge is an electrical discharge caused by the ionization of a fluid such as air, surrounding a conductor that is electrically charged.
The linear charge density on the inner conductor is and the linear charge density on the outer conductor is
Complete Question
The complete question is shown on the first uploaded image (reference for Photobucket )
Answer:
The electric field is [tex]E = -1.3 *10^{-4} \ N/C[/tex]
Explanation:
From the question we are told that
The linear charge density on the inner conductor is [tex]\lambda _i = -26.8 nC/m = -26.8 *10^{-9} C/m[/tex]
The linear charge density on the outer conductor is
[tex]\lambda_o = -60.0 nC/m = -60.0 *10^{-9} \ C/m[/tex]
The position of interest is r = 37.3 mm =0.0373 m
Now this position we are considering is within the outer conductor so the electric field at this point is due to the inner conductor (This is because the charges on the conductor a taken to be on the surface of the conductor according to Gauss Law )
Generally according to Gauss Law
[tex]E (2 \pi r l) = \frac{ \lambda_i }{\epsilon_o}[/tex]
=> [tex]E = \frac{\lambda _i }{2 \pi * \epsilon_o * r}[/tex]
substituting values
[tex]E = \frac{ -26 *10^{-9} }{2 * 3.142 * 8.85 *10^{-12} * 0.0373}[/tex]
[tex]E = -1.3 *10^{-4} \ N/C[/tex]
The negative sign tell us that the direction of the electric field is radially inwards
=> [tex]|E| = 1.3 *10^{-4} \ N/C[/tex]