Answers
Answer:
0.114 kg or 114 g
Explanation:
From the diagram attaches,
Taking the moment about the fulcrum,
sum of clockwise moment = sum of anticlockwise moment.
Wd = W'd'
Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.
make W' the subject of the equation
W' = Wd/d'................ Equation 1
Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm
Substitute these values into equation 1
W' = 0.5047(23.2)/10.5
W' = 1.115 N.
But,
m' = W'/g
m' = 1.115/9.8
m' = 0.114 kg
m' = 114 g
Related Questions
an object's resistance to any change in motion is the_________ of the object.
Answers
An object's resistance to any change in motion is the Inertia of the object.
This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating strings. Assume that the rod is initially electrically neutral. For convenience we will refer to the left end of the rod as end A, and the right end of the rod as end B. In the answer options for this problem, "strongly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two charged balls.A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?
Answers
Answer:
rod end A is strongly attracted towards the balls
rod end B is weakly repelled by the ball as it is at a greater distance
Explanation:
When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.
Therefore rod end A is strongly attracted towards the balls and
rod end B is weakly repelled by the ball as it is at a greater distance
A student writes down several steps of scientific method. Put the steps in the best order
Answers
Answer:
Make a hypothesis, conduct an experiment, Analyze the experimental data..
In an RC-circuit, a resistance of R=1.0 "Giga Ohms" is connected to an air-filled circular-parallel-plate capacitor of diameter 12.0 mm with a separation distance of 1.0 mm. What is the time constant of the system?
Answers
Answer:
[tex]\tau = 1\ ms[/tex]
Explanation:
First we need to find the capacitance of the capacitor.
The capacitance is given by:
[tex]C = \epsilon_0 * area / distance[/tex]
Where [tex]\epsilon_0[/tex] is the air permittivity, which is approximately 8.85 * 10^(-12)
The radius is 12/2 = 6 mm = 0.006 m, so the area of the capacitor is:
[tex]Area = \pi * radius^{2}\\Area = \pi * 0.006^2\\Area = 113.1 * 10^{-6}\ m^2[/tex]
So the capacitance is:
[tex]C = \frac{8.85 * 10^{-12} * 113.1 * 10^{-6}}{0.001}[/tex]
[tex]C = 10^{-12}\ F = 1\ pF[/tex]
The time constant of a rc-circuit is given by:
[tex]\tau = RC[/tex]
So we have that:
[tex]\tau = 10^{9} * 10^{-12} = 10^{-3}\ s = 1\ ms[/tex]
In each pair, select a substance that is a better heat conductor.
1. copper wire / wood 3. water / iron
2. water / air 4. iron / glass
Answers
Answer:
1)copper wire
Explanation:
it is the best electric conductor
Cooper is a good conductor of heat and electricity. Cooper is used in making wires
The amount of friction divided by the weight of an object forms a unit less number called the
Answers
Answer:
Coefficient of friction.
Explanation:
The amount of friction divided by the weight of an object is equal to the coefficient of friction. It is a dimensional less number. It can be given by :
[tex]F=\mu N[/tex]
N is normal force.
[tex]\mu[/tex] = coefficient of friction
[tex]\mu=\dfrac{F}{N}[/tex]
A block is supported on a compressed spring, which projects the block straight up in the air at velocity . The spring and ledge it sits on then retract. You can win a prize by hitting the block with a ball. When should you throw the ball and in what direction to be sure the ball hits the block? (Assume the ball can reach the block before the blochk reaches the ground and that the ball is thrown from a height equal to the release position of the block.)
A. At the instant when the block is at the highest point, directed at the spring.
B. At the instant when the block is at the highest point, directed at the block.
C. At the instant when the block leaves the spring, directed at the spring.
D. At the instant when the block leaves the spring, directed at the block.
E. When the block is back at the spring's original position, directed at that position.
Answers
Answer:
B. At the instant when the block is at the highest point, directed at the block.
Explanation:
Motion of an object is the change in the position of the object with respect to time. On the earth, gravity has a great influence on the motion of an object (especially in a vertical direction).
When the block is projected up in the air, it moves with a varying velocity until the velocity becomes zero due to gravity. Which make the object to rest a little in the air (when velocity = gravity) and starts to fall freely.
To ensure hitting the block by the ball, it is thrown at the block when the block is at its highest point in the air. Since the block would be at rest at this instant before it start to fall at a constant acceleration under gravity.
A beam of light is incident upon a flat piece of glass (n = 1.50) at an angle of incidence of 30.00. Part of the beam is transmitted and part is reflected. Determine the angle between the reflected and transmitted rays
Answers
Answer:
130.528779365 degrees
Explanation:
The angle of incidence is 30 degrees. From this, we can use Snell's Law to calculate the angle of refraction.
n1/n2 = sin(theta2)/sin(theta1)
let theta1 be 30 degrees, and n1 be the refractive index of air = 1
1/1.5 = sin(theta2)/sin(30deg)
solve:
sin(theta2) = 2/3 sin(30deg) = 1/3
theta2 = arcsin (1/3) = 19.4712206345 degrees
The angle of reflection will always be equal to the angle of incidence, in this case, 30 degrees.
Because these angles are measured relative to the normal, the angle formed between the two rays is the difference between the normal line (180 degrees) and the sum of the two angle measures.
Angle between = 180-30-19.4712206345 = 130.528779365 degrees
The angle between the reflected and transmitted rays 130.5287 degrees
What is the refraction of light?The angle of incidence is 30 degrees. From this, we can use Snell's Law to calculate the angle of refraction.
[tex]\dfrac{n_1}{n_2} = \dfrac{sin(\theta_2)}{sin(\theta_1)}[/tex]
let [tex]\theta_1[/tex] be 30 degrees, and n1 be the refractive index of air = 1
[tex]\dfrac{1}{1.5} = \dfrac{sin(\theta_2)}{sin(30)}[/tex]
solve:
[tex]sin(\theta_2) = \dfrac{2}{3} sin(30) = \dfrac{1}{3}[/tex]
[tex]\theta_2 = sin ^{-1}\dfrac{1}{3} = 19.4712 \ degrees[/tex]
The angle of reflection will always be equal to the angle of incidence, in this case, 30 degrees.
Because these angles are measured relative to the normal, the angle formed between the two rays is the difference between the normal line (180 degrees) and the sum of the two angle measures.
Angle between = 180-30-19.4712206345 = 130.528779365 degrees
Hence the angle between the reflected and transmitted rays 130.5287 degrees
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please help
Complete the first and second sentences, choosing the correct answer from the given ones.
1. A temperature of 100 K corresponds on a Celsius scale to 100 ° C / 0 ° C / 173 ° C / –173 ° C.
2. At 50 ° C, it corresponds to a Kelvin scale of 150 K / 323 K / 273 K / 223 K.
Answers
1) 100 ° C
2) 323 K
hope it helps youuuuuu
N capacitors are connected in parallel to form a "capacitor circuit". The capacitance of first capacitor is C, second one is C/2 and third one is C/4, forth one is C/8 and so on. Namely, capacitance of a capacitor is one-half of the previous one. What is the equivalent capacitance of this parallel combination when N goes to inifinity?
Answers
Answer:
2C
Explanation:
The equivalent capacitance of a parallel combination of capacitors is the sum of their capacitance.
So, if the capacitance of each capacitor is half the previous one, we have a geometric series with first term = C and rate = 0.5.
Using the formula for the sum of the infinite terms of a geometric series, we have:
Sum = First term / (1 - rate)
Sum = C / (1 - 0.5)
Sum = C / 0.5 = 2C
So the equivalent capacitance of this parallel connection is 2C.
Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all automobiles are brought to rest.
Car A: 500 kg, 10 m/s,
Car B: 2000 kg, 5 m/s,
Car C: 500 kg, 20 m/s,
Car D: 1000 kg, 20 m/s,
Car E: 4000 kg, 5 m/s, and
Car F: 1000 kg, 10 m/s.
(a) Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest.
(b) Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest.
Answers
Answer:
a)Car E = Car D > (Car F = Car B = Car C) > Car A
b)Car E = Car D > (Car F = Car B = Car C) > Car A
Explanation:
Car A: mass = 500 kg; speed = 10 m/s
Car B: mass = 2000 kg;speed = 5 m/s
Car C:mass = 500 kg; speed = 20 m/s
Car D: mass = 1000 kg; speed = 20 m/s
Car E:mass = 4000 kg; speed = 5 m/s
Car F: mass = 1000 kg; speed = 10 m/s
Part a) Now we know that momentum of each car is product of mass and velocity , so we will have
CarA:
[tex]P_1 = m \times v\\P_1 = (500)(10)\\P_1 = 5 \times 10^3 kg m/s[/tex]
Car B:
[tex]P_2 = m v\\P_2 = (2000)(5)\\P_2 = 10^4 kg m/s[/tex]
Car C:
[tex]P_3 = m v\\P_3 = (500)(20)\\P_3 = 10^4 kg m/s[/tex]
Car D:
[tex]P_4 = m v\\P_4 = (1000)(20)\\P_4 = 2\times 10^4 kg m/s[/tex]
Car E:
[tex]P_5 = m v\\P_5 = (4000)(5)\\P_5 = 2\times 10^4 kg m/s[/tex]
Car F:
[tex]P_6 = m v\\P_6 = (1000)(10)\\P_6 = 10^4 kg m/s[/tex]
So the momentum is given as ,
Car E = Car D > (Car F = Car B = Car C) > Car A
Part b)Impulse is given as change in momentum so here we can say that final momentum of all the cars will be zero as they all stops and hence the impulse is same as initial momentum of the car
so the order of impulse from largest to least is given as
Car E = Car D > (Car F = Car B = Car C) > Car A
What is the highest point at which weather will generally occur?
Answers
Answer:
At thestratosphere: it 20- 25km
Sara walks part way around a swimming pool. She walks 50 yards north, then
20 yards east, then 50 yards south. The magnitude of her total displacement
during this walk is
yards.
Answers
Answer:
20 Yards
Explanation:
|---20----|
| |
| 50 |50
|---D--->|
Start End
Total displacement(D) 20 yards (East).
Of one of the planets becomes a black hole , what would the escape speed be?
Answers
Answer:
If, instead, that rocket was on a planet with the same mass as Earth but half the diameter, the escape velocity would be 15.8 km/s Any object that is smaller than its Schwarzschild radius is a black hole – in other words, anything with an escape velocity greater than the speed of light is a black hole.
Explanation:
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A tank with a constant volume of 3.72 m3 contains 22.1 moles of a monatomic ideal gas. The gas is initially at a temperature of 300 K. An electric heater is used to transfer 4.5 × 104 J of energy into the gas. It may help you to recall that CV = 12.47 J/K/mole for a monatomic ideal gas, and that the number of gas molecules is equal to Avagadros number (6.022 × 1023) times the number of moles of the gas.
a) What is the temperature of the gas after the energy is added?___K
b) What is the change in pressure of the gas?____Pa
c) How much work was done by the gas during this process?____J
Answers
Answer:
a) 463.29 K
b) 8065.65 Pa
c) 0 J
Explanation:
The parameters given are;
Volume of the tank, V = 3.72 m³
Number of moles of gas present in the tank, n = 22.1 moles
Temperature of the gas before heating, T₁ = 300 k
Heat added to the gas, ΔQ = 4.5 × 10⁴ J
Specific heat capacity at constant volume, [tex]c_v[/tex], for monatomic gas = 12.47 J/K/mole
Avogadro's number = 6.022 × 10²³ particles per mole
a) ΔQ = n × [tex]c_v[/tex] × ΔT
Where:
ΔT = T₂ - T₁
T₂ = Final temperature of the gas
Hence, by plugging in the values, we have;
4.5 × 10⁴ = 22.1 × 12.47 × (T₂ - 300)
[tex]T_{2} - 300 = \frac{4.5\times 10^{4}}{22.1\times 12.47}[/tex]
T₂ = 300 + 163.29 = 463.29 K
b) The pressure of the gas is found from the relation;
P×V = n×R×T
[tex]P = \dfrac{n \times R \times T}{V}[/tex]
Where:
P = Pressure of the gas
R = Universal gas constant = 8.3145 J/(mol·K)
T = Temperature of the gas
V = Volume of the gas = 3.72 ³ (constant)
n = Number of moles of gas present = 22.1 moles (constant)
Hence the change in pressure is given by the relation;
[tex]\Delta P = \dfrac{n \times R \times (T_2 - T_1)}{V} = \dfrac{n \times R \times \Delta T}{V}[/tex]
Plugging in the values, we have;
[tex]\Delta P = \dfrac{22.1 \times 8.3145 \times 163.29}{3.72} = 8065.65 \, Pa[/tex]
c) Work done, W, by the gas is given by the area under the pressure to volume graph which gives;
W = f(P) × ΔV
The volume given in the question is constant
∴ ΔV = 0
Hence, W = f(P) × 0 = 0 J
No work done by the gas during the process.
Now consider a different electromagnetic wave, also described by: Ex(z,t) = Eocos(kz - ω t + φ) In this equation, k = 2π/λ is the wavenumber and ω = 2π f is the angular frequency. In this case, though, assume φ = +30o and Eo = 1 kV/m. What is the value of Ex(z,t) when z/λ = 0.25 and ft = 0.125?
Answers
Answer:
Explanation:
Ex(z,t) = Eocos(kz - ω t + φ)
k = 2π/λ , ω = 2π f
φ = +30° , E₀ = 10³ V .
z/λ = 0.25 , ft = 0.125
Ex(z,t) = Eocos(2πz/λ - 2πf t + φ)
Putting the values given above
Ex(z,t) = 10³ cos ( 2π / 4 - 2π x .125 + 30⁰ )
= 1000cos (90⁰ - 45+30)
= 1000 cos 75
=258.8 V .
Which of the following statements is true of a gas?
It has a fixed volume, but not a fixed shape
It has closely packed molecules
It can change into a liquid by adding heat
It takes the shape and size of a container
Answers
Answer:
it takes the shape and size of the container that it is in
Explanation:
Answer:
it takes the shape and size of a container
What is the period of a wave if the frequency is? 5 Hz
Answers
Answer: If the woodpecker drums upon a tree 5 times in one second, then the frequency is 5 Hz; each drum must endure for one-fifth a second, so the period is 0.2 s.
The International Space Station is about 90 meters across and about 380 kilometers away. One night t appears to be the same angular size as Jupiter. Jupiter is 143,000 km in size. Use serxa to figure out how far away Jupiter is in AU Note: 1 AU= 1.5 x 10-km
a) 6.0 x 10 Au
b) 4.0 AU
c) 9.1 x 1010 AU
d) 4.0 x 10 AU
Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
The distance is [tex]r_2 = 4 \ AU[/tex]
Explanation:
From the question we are told that
The size of Jupiter is [tex]s_2 = 143,000 \ km[/tex]
The length of the International Space Station is [tex]r_1 = 380\ km[/tex]
The size of the International Space Station is [tex]s_1 = 90 \ m =0.09 \ km[/tex]
The angular size where the same one night and this angular size is mathematically represented as
[tex]\theta = \frac{s}{r}[/tex]
Since [tex]\theta[/tex] is constant
[tex]\frac{s_1}{r_1} = \frac{s_2}{r_2}[/tex]
substituting values
[tex]\frac{0.09}{380} = \frac{143000}{r_2}[/tex]
=> [tex]r_2 = 6.04 * 10^{9} \ km[/tex]
Now we are told to convert to AU and 1 AU [tex]= 1.5 * 10^8 \ km[/tex]
So
[tex]r_2 = \frac{6.04 * 10^8}{1.5*10^{8}}[/tex]
[tex]r_2 = 4 \ AU[/tex]
6. When a positive charge is released and moves along an electric field line, it moves to a position of A) lower potential and lower potential energy. B) lower potential and higher potential energy. C) higher potential and lower potential energy. D) higher potential and higher potential energy.
Answers
Answer:
Since you would have to do work on the charge to bring it back to its original position, the charge moves to a position of lower potential and lower potential energy.
The positive charge is released from a point such that it will move along an uniform electric field to the position of lower potential and lower potential energy. Therefore, option (A) is correct,
When a positive charge (say +Q) is released from a point (say A) and moves in an uniform electric field to reach the point (say B), then some work is done on the charge. This work done is given as,
[tex]W=+Q(V_{A}-V_{B})[/tex]
Here, [tex]V_{A}[/tex] and [tex]V_{B}[/tex] are the potential differences between the points A and B respectively..
This means the charge is moving from higher potential to lower potential. And since it is moving along the uniform electric field, therefore the electric potential energy of charged system is decreased.
Thus, we conclude that on releasing the positive charge from a point, it starts moving along the electric field towards the direction of lower electric potential and lower electric potential energy. Hence, option (A) is correct.
Learn more about electric potential and electric potential energy here:
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A skateboarder, starting from rest, rolls down a 12.8-m ramp. When she arrives at the bottom of the ramp her speed is 8.89 m/s. (a) Determine the magnitude of her acceleration, assumed to be constant. (b) If the ramp is inclined at 32.6 ° with respect to the ground, what is the component of her acceleration that is parallel to the ground?
Answers
Answer:
a) a = 3.09 m/s²
b) aₓ = 2.60 m/s²
Explanation:
a) The magnitude of her acceleration can be calculated using the following equation:
[tex] V_{f}^{2} = V_{0}^{2} + 2ad [/tex]
Where:
[tex]V_{f}[/tex]: is the final speed = 8.89 m/s
[tex]V_{0}[/tex]: is the initial speed = 0 (since she starts from rest)
a: is the acceleration
d: is the distance = 12.8 m
[tex] a = \frac{V_{f}^{2}}{2d} = \frac{(8.89 m/s)^{2}}{2*12.8 m} = 3.09 m/s^{2} [/tex]
Therefore, the magnitude of her acceleration is 3.09 m/s².
b) The component of her acceleration that is parallel to the ground is given by:
[tex] a_{x} = a*cos(\theta) [/tex]
Where:
θ: is the angle respect to the ground = 32.6 °
[tex] a_{x} = 3.09 m/s^{2}*cos(32.6) = 2.60 m/s^{2} [/tex]
Hence, the component of her acceleration that is parallel to the ground is 2.60 m/s².
I hope it helps you!
A skateboarder, starting from rest, rolls down a 12.8-m ramp the magnitude of the skateboarder's acceleration is approximately 3.07 [tex]m/s^2[/tex], the component of her acceleration that is parallel to the ground is approximately 1.66 [tex]m/s^2[/tex].
(a) The following kinematic equation can be used to calculate the skateboarder's acceleration:
[tex]v^2 = u^2 + 2as[/tex]
[tex](8.89)^2 = (0)^2 + 2a(12.8)[/tex]
78.72 = 25.6a
a = 78.72 / 25.6
a = 3.07 [tex]m/s^2[/tex]
(b) Trigonometry can be used to calculate the part of her acceleration that is parallel to the ground. We are aware that the ramp's angle with the ground is 32.6°.
[tex]a_{parallel }= a * sin(\theta)[/tex]
Plugging in the values:
[tex]a_{parallel[/tex] = 3.07 [tex]m/s^2[/tex]* sin(32.6°)
[tex]a_{parallel[/tex]≈ 1.66 [tex]m/s^2[/tex]
Therefore, the component of her acceleration that is parallel to the ground is approximately 1.66 [tex]m/s^2[/tex].
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You are comparing a reaction that produces a chemical change and one that produces a physical change. What evidence could you use to determine which type of change is occurring?
Answers
Answer: A chemical change results from a chemical reaction, while a physical change is when matter changes forms but not chemical identity. Examples of chemical changes are burning, cooking, rusting, and rotting. Examples of physical changes are boiling, melting, freezing, and shredding. Often, physical changes can be undone, if energy is input.
Explanation: hope this helps have a good day
Answer:
If the reaction is a chemical change, new substances with different properties and identities are formed. This may be indicated by the production of an odor, a change in color or energy, or the formation of a solid.
3. A ray of light incident on one face of an equilateral glass prism is refracted in such a way that it emerges from the opposite surface at an angle of 900 to the normal. Calculate the i. angle of incidence. ii. minimum deviation of the ray of light passing through the prism [n_glass=1.52]
Answers
Answer:
i) angle of incidence;i = 29.43°
ii) δm = 38.92°
Explanation:
Prism is equilateral so angle of prism (A) = 60°
Refractive index of glass; n_glass = 1.52
A) Let's assume the incident angle = i and Critical angle = θc
We know that, sin θc = 1/n
Thus;
sin θc = 1/n_glass
θc = sin^(-1) (1/n_glass)
θc = sin^(-1) (1/1.52)
θc = 41.14°
Now, the angle of prism will be the sum of external angle that is critical angle and reflected angle.
Thus;
A = r + θc
r = A - θc
So;
r = 60° - 41. 14°
r = 18.86°
From, Snell's law. If we apply it to this question, we will have;
(sin i)/(sin r) = n_glass
Where;
i is angle of incidence and r is angle of reflection.
Let's make i the subject;
i = sin^(-1) (n_glass × sin r)
i = sin^(-1) (1.52 × sin 18.86)
i = sin^(-1) 0.4914
i = 29.43°
B) The formula to calculate minimum deviation would be from;
μ = [sin ((A + δm)/2)]/(sin A/2)
Where;
μ is Refractive index
δm is minimum angle of deviation
A is angle of prism
Now Refractive index is given by a formula; μ = (sin i)/(sin r)
So; μ = (sin 29.43)/(sin 18.86)
μ = 1.52
Thus;
1.52 = [sin ((60 + δm)/2)]/(sin 60/2)
1.52 * sin 30 = sin ((60 + δm)/2)
0.76 = sin ((60 + δm)/2)
sin^(-1) 0.76 = ((60 + δm)/2)
49.46 × 2 = (60 + δm)
98.92 - 60 = δm
δm = 38.92°
Some types of spiders build webs that consist of threads made of dry silk coated with a solution of a variety of compounds. This coating leaves the threads, which are used to capture prey, hygroscopic - that is, they attract water from the atmosphere. It has been hypothesized that this aqueous coating makes the threads good electrical conductors. To test the electrical properties of coated thread, researchers placed a 5-mm length of thread between two electrical contacts. The researchers stretched the thread in 1-mm increments to more than twice its original length, and then allowed it to return to its original length, again in 1-mm increments. Some of the resistance measurements are shown.If the conductivity of the thread results from the aqueous coating only, how does the cross-sectional area A of the coating compare when the thread is 13 mm long versus the starting length of 5 mm? Assume that the resistivity of the coating remains constant and the coating is uniform along the thread.If the conductivity of the thread results from the aqueous coating only, how does the cross-sectional area of the coating compare when the thread is 13 long versus the starting length of 5 ? Assume that the resistivity of the coating remains constant and the coating is uniform along the thread.A13mm is about 1/10 A5mm.A13mm is about 1/4 A5mm. === correct answer... I figured it out. R = pL/A. L is 2.5 times. Therefore, A must be 1/4 times.A13mm is about 2/5 A5mm.A13mm is the same as A5mm.
Answers
Answer:
A13 mm is about 1/4 A5 mm
Explanation:
Find the attachment
a vector has components x=6 m and y=8 m. what is its magnitude and direction?
Answers
Answer: 10m
Explanation:
The magnitude of the vector would be 10
[tex]\sqrt{6^{2}+8^{2} } =10[/tex]
I need help physics
Answers
Because the actual was greater than the estimated, therefore they underestimated the population in 2010
Water flows at 0.850 m/s from a hot water heater, through a 450-kPa pressure regulator. The pressure in the pipe supplying an upstairs bathtub 3.70m above the heater is 414-kPa. What's the flow speed in this pipe?
Answers
Answer:
The velocity is [tex]v_2= 0.45 \ m/s[/tex]
Explanation:
From the question we are told that
The initial speed of the hot water is [tex]v_1 = 0.85 \ m/s[/tex]
The pressure from the heater [tex]P_1 = 450 \ KPa = 450 *10^{3} \ Pa[/tex]
The height of the hot water before flowing is [tex]h_1 = 0 \ m[/tex]
The height of bathtub above the heater is [tex]h_2 = 3.70 \ m[/tex]
The pressure in the pipe is [tex]P_2 = 414 KPa = 414 *10^{3} \ Pa[/tex]
The density of water is [tex]\rho = 1000 \ kg/m^3[/tex]
Apply Bernoulli equation
[tex]P_1 + \rho gh_1 +\frac{1}{2} \rho v_1^2 = \rho g h_2 + \frac{1}{2}\rho v_2 ^2[/tex]
Substituting values
[tex](450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) = (1000 * *9.8*3.70) + (0.5*1000*v_2^2 )[/tex]
=> [tex]v_2^2 = \frac{ (450 *10^{3}) + (1000 * 9.8 *0 ) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}[/tex]
=> [tex]v_2= \sqrt{ \frac{ (450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}}[/tex]
=> [tex]v_2= 0.45 \ m/s[/tex]
A point charge is located at the center of a thin spherical conducting shell of inner and outer radii r1 and r2, respectively. A second thin spherical conducting shell of inner and outer radii R1 and R2, respectively, is concentric with the first shell. The flux is as follows for the different regions of this arrangement. Ф -10.3 103 N-m2/C for
0 for r<2 4:
-36.8 x 10נ N-m2/c
0 for r > R2
36.8 x 10נ N-m2/c
Determine the magnitude ond sign of the point chorge ond the charge on the surface of the two shels point charge inner shell outer shel.
Answers
Answer:
the magnitude is 7 and sign of the point charge on the surface shell is -13
Explanation:
someone please help me with this thanks
Answers
Two large insulating parallel plates carry charge of equal magnitude, one positive and the other negative, that is distributed uniformly over their inner surfaces. Rank the points 1 through 5 according to the magnitude of the electric field at the points, least to greatest.
A. 1, 2, 3, 4, 5
B. 2, then 1, 3, and 4 tied, then 5
C. 1, 4, and 5 tie, then 2 and 3 tie
D. 2 and 3 tie, then 1 and 4 tie, then 5
E. 2 and 3 tie, then 1, 4, and 5 tie
Answers
Answer:
The correct answer is C 1, 4, and 5 tie, then 2 and 3 tie
Explanation:
Solution
The electric field due to sheets E₁ positive =б/2E₀
E₂ is negative = б/2E₀
Now,
At the point 1, 4, 5 the electric field due to the sheets are in the opposite direction
At the point 1, the net field = -E₁ + E₂ =0
At the point A, the net field = -E₁ - E₂ = 0
Now,
At nay point inside between them, the electric field is seen to be at the same direction.
At the 2, 3 points the field is seen at the right
Thus,
E net = E₁ + E₂
= б/2E₀ + σ/2E₀
=б/E₀
Note: Kindly find an attached copy of the complete question to the solution
The correct answer is option C
The rank of the points according to the magnitude of the electric field is 1, 4, and 5 tie, then 2 and 3 tie
The magnitude of the electric field:
Let sheet 1 has positive surface charge density and sheet 2 has a negative surface charge density
The electric field (without direction) due to sheets will be
E₁ =σ/2E₀
E₂= σ/2E₀
Now,
At the point 1, 4, 5 the electric field due to the sheets is given by:
E = E₁ - E₂
E = σ/2E₀ - σ/2E₀
since the positive charge plate will have electric field lines away from the sheet and the negative charge plate will have electric field lines towards the sheet
E = 0
Now,
At points 2, 3 which are between the plates,
The net electric field is:
E = E₁ + E₂
since the electric field due to both the plates will be from positive to negative ( towards the negatively charged plate)
E = σ/2E₀ + σ/2E₀
E = σ/E₀
Learn more about surface charge density:
https://brainly.com/question/8966223?referrer=searchResults