Answer:
A₁/A₂ = 0.44
Explanation:
The emissive power of the bulb is given by the formula:
P = σεAT⁴
where,
P = Emissive Power
σ = Stefan-Boltzman constant
ε = Emissivity
A = Surface Area
T = Absolute Temperature of Surface
FOR BULB 1:
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₁T₁⁴ ----------- equation 1
where,
A₁ = Surface Area of Bulb 1
T₁ = Temperature of Bulb 1 = 3000 k
FOR BULB 2:
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₂T₂⁴ ----------- equation 2
where,
A₂ = Surface Area of Bulb 2
T₂ = Temperature of Bulb 1 = 2000 k
Dividing equation 1 by equation 2, we get:
P/P = σεA₁T₁⁴/σεA₂T₂⁴
1 = A₁(3000)²/A₂(2000)²
A₁/A₂ = (2000)²/(3000)²
A₁/A₂ = 0.44
A low C (f = 65Hz) is sounded on a piano. If the length of the piano wire is 2.0 m and its mass density is 5.0 g/m2, determine the tension of the wire.
Answer:
Tension of the wire(T) = 169 N
Explanation:
Given:
f = 65Hz
Length of the piano wire (L) = 2 m
Mass density = 5.0 g/m² = 0.005 kg/m²
Find:
Tension of the wire(T)
Computation:
f = v / λ
65 = v / 2L
65 = v /(2)(2)
v = 260 m/s
T = v² (m/l)
T = (260)²(0.005/2)
T = 169 N
Tension of the wire(T) = 169 N
An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by parallel lines drawn with a density N lines per m2 that are perpendicular to and away from the sheet. The charge per unit area on the sheet is doubled. How should the density of the electric field lines be changed
Complete Question
An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by parallel lines drawn with a density N lines per m2 that are perpendicular to and away from the sheet. The charge per unit area on the sheet is doubled. How should the density of the electric field lines be changed?
A It should stay the same
B It should be quadrupled.
C It should be quintupled
D It should be doubled.
E It should be tripled
Answer:
Option D is the correct option
Explanation:
Generally electric field is mathematically represented as
[tex]E = \frac{\sigma}{\epsilon_o}[/tex]
Where [tex]\sigma[/tex] is the charge per unit area (Charge density )
From the question we are told that [tex]\sigma[/tex] is doubled hence the
[tex]E = \frac{2 \sigma }{\epsilon_o}[/tex]
Looking the equation above we see that the value of the electric field will also double given that it is directly proportional to the charge density
Assuming 100% efficient energy conversion, how much water stored behind a 50 centimeter high hydroelectric dam would be required to charge the battery?
Complete question is;
Assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charge the battery with power rating, 12 V, 50 Ampere-minutes.
Answer:
Amount of water required to charge the battery = 7.35 m³
Explanation:
The formula for Potential energy of the water at that height = mgh
Where;
m = mass of the water
g = acceleration due to gravity = 9.8 m/s²
h = height of water = 50 cm = 0.5 m
We know that in density, m = ρV
Where;
ρ = density of water = 1000 kg/m³
V = volume of water
So, potential energy is now given as;
Potential energy = ρVgH = 1000 × V × 9.8 × 0.5 = (4900V) J
Now, formula for energy of the battery is given as;
E = qV
We are given;
q = 50 A.min = 50 × 60 = 3,000 C
V = 12 V
Thus;
qV = 3,000 × 12 = 36,000 J
E = 36,000 J
At a 100% conversion rate, the energy of the water totally powers the battery.
Thus;
(4900V) = (36,000)
4900V = 36,000
V = 36,000/4900
V = 7.35 m³
Question 10
Air with a density of 1.20 kg/m3 flows through a 75.0 cm diameter pipe with a velocity of 2.00 m/s. What is the mass flow rate?
Answer:
75.0 cm
Explanation:
becouse i don,t no the right answer
EASY HELP
As a space shuttle climbs, _____.
its mass increases
its mass decreases
its weight increases
its weight decreases
Answer: it's weight decreases
Explanation:
ASK YOUR TEACHER A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 51.5-gram mass is attached at the 16.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick
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
what is the speed of light in quartz
Answer:
1.95 x 10^8 m/s.
Explanation:
Answer:
the answer is 1.95 x 10^8 m/s
Explanation:
Use Hooke's Law to determine the work done by the variable force in the spring problem. Nine joules of work is required to stretch a spring 0.5 meter from its natural length. Find the work required to stretch the spring an additional 0.40 meter.
Answer:
29.16 J
Explanation:
From Hook's law,
W = 1/2(ke²)..................... Equation 1
Where W = work done, k = Spring constant, e = extension.
Given: W = 9 J, e = 0.5 m.
Substitute into equation 1
9 = 1/2(k×0.5²)
Solve for k
k = 18/0.5²
k = 72 N/m.
The work done required to stretch the spring by additional 0.4 m is
W = 1/2(72)(0.4+0.5)²
W = 36(0.9²)
W = 29.16 J.
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?
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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A type of friction that occurs when air pushes against a moving object causing it to negatively accelerate
a) surface area
b) air resistance
c) descent velocity
d) gravity
Answer:
Air resistance
Answer B is correct
Explanation:
The friction that occurs when air pushes against a moving object causing it to negatively accelerate is called as air resistance.
hope this helps
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Two plates with area 7.00×10−3 m27.00×10−3 m2 are separated by a distance of 4.80×10−4 m4.80×10−4 m . If a charge of 5.40×10−8 C5.40×10−8 C is moved from one plate to the other, calculate the potential difference (voltage) between the two plates. Assume that the separation distance is small in comparison to the diameter of the plates.
Answer:
The voltage is [tex]V = 418.60 \ Volts[/tex]
Explanation:
From the question we are told that
The area of the both plate is [tex]A = 7.00 *10^{-3} \ m^2[/tex]
The distance between the plate is [tex]d = 4.80*10^{-4}\ m[/tex]
The magnitude of the charge is [tex]q = 5.40 *10^{-8} \ C[/tex]
The capacitance of the capacitor that consist of the two plates is mathematically represented as
[tex]C = \frac{\epsilon _o A}{d}[/tex]
Where [tex]\epsilon_o[/tex] is the permitivity of free space with a value [tex]e = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
So
[tex]C = \frac{8.85*10^{-12} * (7* 10^{-3})}{ 4.8*10^{-4}}[/tex]
[tex]C = 1.29 *10^{-10} \ F[/tex]
The potential difference between the plate is mathematically represented as
[tex]V = \frac{ Q}{C }[/tex]
[tex]V = \frac{ 5.4*10^{-8}}{1.29 *10^{-10}}[/tex]
[tex]V = 418.60 \ Volts[/tex]
Volume of an block is 5 cm3. If the density of the block is 250 g/cm3, what is the mass of the block ?
Answer:
The mass of the block is 1250g.
Explanation:
Given that the formula for density is ρ = mass/volume. Then you have to substitute the values into the formula :
[tex]ρ = \frac{mass}{volume} [/tex]
Let density = 250,
Let volume = 5,
[tex]250 = \frac{m}{5} [/tex]
[tex]m = 250 \times 5[/tex]
[tex]m = 1250g[/tex]
We say that the displacement of a particle is a vector quantity. Our best justification for this assertion is: A. a displacement is obviously not a scalar. B. displacement can be specified by a magnitude and a direction. C. operating with displacements according to the rules for manipulating vectors leads to results in agreement with experiments. D. displacement can be specified by three numbers. E. displacement is associated by motion.
Answer:
Option B - displacement can be specified by a magnitude and a direction.
Explanation:
A Vector quantity is defined as a physical quantity characterized by the presence of both magnitude as well as direction. Examples include displacement, force, torque, momentum, acceleration, velocity e.t.c
Whereas a scalar quantity is defined as a physical quantity which is specified with the magnitude or size alone. Examples include length, speed, work, mass, density, etc.
Displacement is the difference between the initial position and the final position of a body. Displacement is a vector quantity and not a scalar quantity because it can be described by using both magnitude as well as direction.
Looking at the options, the only one that truly justifies this definition is option B.
Although these quantities vary from one type of cell to another, a cell can be 2.2 micrometers in diameter with a cell wall 40 nm thick. If the density (mass divided by volume) of the wall material is the same as that of pure water, what is the mass (in mg) of the cell wall, assuming the cell to be spherical and the wall to be a very thin spherical shell?
Answer:
m = 6.082 x 10⁻¹⁶ kg = 6.082 x 10⁻¹⁰ mg
Explanation:
First, we find the the surface area of the cell wall. Since, the cell is spherical in shape. Therefore, surface area of cell wall will be:
A = 4πr²
where,
A = Surface Area = ?
r = Radius of Cell = Diameter/2 = 2.2 μm/2 = 1.1 μm = 1.1 x 10⁻⁶ m
Therefore,
A = 4π(1.1 x 10⁻⁶ m)²
A = 15.2 x 10⁻¹² m²
Now, we find the volume of the cell wall. For that purpose, we use formula:
V = At
where,
V = Volume of the Cell Wall = ?
t = Thickness of Wall = 40 nm = 4 x 10⁻⁸ m
Therefore,
V = (15.2 x 10⁻¹² m²)(4 x 10⁻⁸ m)
V = 60.82 x 10⁻²⁰ m³
Now, to find mass of cell wall, we use formula:
ρ = m/V
m = ρV
where,
ρ = density of water = 1000 kg/m³
m = Mass of Wall = ?
Therefore,
m = (1000 kg/m³)(60.82 x 10⁻²⁰ m³)
m = 6.082 x 10⁻¹⁶ kg = 6.082 x 10⁻¹⁰ mg
The mass of the cell wall in mg is 6.082 × 10⁻¹⁰ mg
Since we assume the cell to be spherical and the wall to be a thin spherical shell, the volume of the cell wall V = At where
A = surface area of cell = 4πR² where R = radius of cell = 2.2 μm/2 = 1.1 × 10⁻⁶ m and t = thickness of cell wall = 40 nm = 40 × 10⁻⁹ m.Volume of cell wallSo, V = 4πR²t
Substituting the values of the variables into the equation, we have
V = 4πR²t
V = 4π(1.1 × 10⁻⁶ m)² × 40 × 10⁻⁹ m.
V = 4π(1.21 × 10⁻¹² m²) × 40 × 10⁻⁹ m.
V = 193.6π × 10⁻²¹ m³
V = 608.21 × 10⁻²¹ m³
V = 6.0821 × 10⁻¹⁹ m³
V ≅ 6.082 × 10⁻¹⁹ m³
Mass of the cell wallWe know that density of cell wall, ρ = m/v where m = mass of cell wall and V = volume of cell wall.
Making m subject of the formula, we have
m = ρV
Since we assume the density of the cell wall to be equal to that of pure water, ρ = 1000 kg/m³
So, m = ρV
m = 1000 kg/m³ × 6.082 × 10⁻¹⁹ m³
m = 6.082 × 10⁻¹⁶ kg
Converting to mg, we have
m = 6.082 × 10⁻¹⁶ kg × 10⁶ mg/kg
m = 6.082 × 10⁻¹⁰ mg
So, the mass of the cell wall in mg is 6.082 × 10⁻¹⁰ mg
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Compare the energy consumption of two commonly used items in the household. Calculate the energy used by a 1.40 kW toaster oven, Wtoaster , which is used for 5.40 minutes , and then calculate the amount of energy that an 11.0 W compact fluorescent light (CFL) bulb, Wlight , uses when left on for 10.50 hours .
Energy = (power) x (time)
-- For the toaster:
Power = 1.4 kW = 1,400 watts
Time = 5.4 minutes = 324 seconds
Energy = (1,400 W) x (324 s) = 453,600 Joules
-- For the CFL bulb:
Power = 11 watts
Time = 10.5 hours = 37,800 seconds
Energy = (11 W) x (37,800 s) = 415,800 Joules
-- The toaster uses energy at 127 times the rate of the CFL bulb.
-- The CFL bulb uses energy at 0.0079 times the rate of the toaster.
-- The toaster is used for 0.0086 times as long as the CFL bulb.
-- The CFL bulb is used for 116.7 times as long as the toaster.
-- The toaster uses 9.1% more energy than the CFL bulb.
-- The CFL bulb uses 8.3% less energy than the toaster.
A projectile is defined as
Answer:
By definition, a projectile has a single force that acts upon it - the force of gravity.
Explanation:
A projectile is any object that once projected or dropped continues in motion by its own inertia and is influenced only by the downward force of gravity.
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A worker pushes on a crate that experiences a net force of 45.0 N. If it accelerates at 0.500 m/s2 what is the weight?
Answer:
882 N
Explanation:
F = ma
45.0 N = m (0.500 m/s²)
m = 90.0 kg
mg = 882 N
determine the smallest mass of lead that when tied using a string to a wooden boat on a pond will be enough to sink the toy boat. assuming specific gravity of wood is 0.5 and density of water is 1000kg per cubic metre?
Which person will most likely hear the loudest sound?
A
B
C
D
Answer:
The youngest person
Explanation:
Hearing worsens with age
Please mark brainliest
Answer:
A
Explanation:
The person closest to the origin of the sound will most likely hear the loudest sound. ^^
EASY! WILL REWARD BRAINLIEST!
Electrical current is defined as _____.
the capacity to store charge
the flow of electric charge per unit time
the amount of stored electric energy
the voltage of the battery
Electrical current is defined as the flow of electric charge per unit time.
A uniform thin spherical shell of mass M=2kg and radius R=0.23m is given an initial angular speed w=18.3rad/s when it is at the bottom of an inclined plane of height h=3.5m, as shown in the figure. The spherical shell rolls without slipping. Find wif the shell comes to rest at the top of the inclined plane. (Take g-9.81 m/s2, Ispherical shell = 2/3 MR2 ).Express your answer using one decimal place.
Answer:
47.8rad/s
Explanation:
For energy to be conserved.
The potential energy sustain by the object would be equal to K.E
P.E = m× g× h = 2 × 9.81× 3.5= 68.67J
Now K.E = 1/2 × I × (w1^2 - w0^2)
I = 2/3 × M × R2
= 2/3 × 2 × (0.23)^2= 0.0705
Hence
W1 = final angular velocity
Wo = initial angular velocity
From P.E = K.E we have;
68.67J = 1/2 × 0.0705 × (w1^2 - w0^2)
(w1^2 - w0^2) = 1948.09
W1^2 = 1948.09 + (18.3^2)
W1^2=2282.98
W1 = √2282.98
=47.78rad/s
= 47.8rad/s to 1 decimal place.
A certain type of laser emits light that has a frequency of 4.6 x 1014 Hz. The light, however, occurs as a series of short pulses, each lasting for a time of 3.1 x 10-11s. The light enters a pool of water. The frequency of the light remains the same, but the speed of light slows down to 2.3 x 108 m/s. In the water, how many wavelengths are in one pulse
Answer:
14,260
Explanation:
Relevant data provided for computing the wavelengths are in one pulse is here below:-
The number of wavelengths in Ls = [tex]4.6\times 10_1_4[/tex]
Therefore the Number of in time = Δt = [tex]3.1\times 10_-_1_1[/tex]
The number of wavelengths are in one pulse is shown below:-
[tex]Number\ of\ wavelengths = \triangle t\times f[/tex]
[tex]= 3.1\times 10_-_1_1\times 4.6\times 10_1_4[/tex]
= 14,260
Therefore for computing the number of wavelengths are in one pulse we simply applied the above formula.
student conducted an experiment and find the density of an ICEBERGE. A students than recorded the following readings. Mass 425 25 g Volume 405 15 mL What experimental value should be quoted for the density of the ICEBERG? Compare your answer with the density of water, which is 3 1.00 10 kg . Show any calculations necessary to justify your answer
Complete Question
The complete question is shown on the first uploaded image
Answer:
The experimental value of density is [tex]\rho = 1.05*10^{3} \ kg/m^3 \pm 101 \ kg/m^3[/tex]
Comparing it with the value of density of water ([tex]1.0*10^{3} \ kg/m^3[/tex]) we can see that the density of ice is greater
Explanation:
From the question we are told
The mass is [tex]M = (425 \pm 25) \ g =(0.425 \pm 0.025) \ kg[/tex]
The volume is [tex]V = (405 \pm 15 ) \ mL = (0.000405 \pm 1.5*10^{-5}) \ m^3[/tex]
The experimental value of density is mathematically evaluated as
[tex]\rho = \frac{M}{V}[/tex]
[tex]\rho = \frac{0.425}{0.000405}[/tex]
[tex]\rho = 1.05 *10^{3} \ kg/m^3[/tex]
The possible error in this experimental value of density is mathematically evaluated as
[tex]\frac{\Delta \rho}{\rho} = \frac{\Delta M}{M} +\frac{\Delta V}{V}[/tex]
substituting value
[tex]\frac{\Delta \rho}{1.05*10^{3}} = \frac{0.025}{0.425} +\frac{1.5*10^{-5}}{0.000405}[/tex]
[tex]\Delta \rho = 101 \ kgm^{-3}[/tex]
Thus the experimental value of density is
[tex]\rho = 1.05*10^{3} \ kg/m^3 \pm 101 \ kg/m^3[/tex]
A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 45 ∘ with the vertical. Air resistance is negligible. Part A What is the speed of the rock when the string passes through the vertical position
Answer:
v = 3.33 m/s
Explanation:
In the position of 45 degrees, all the energy of the rock is gravitational, then we have:
E = m*g*L*cos(angle)
and in the vertical position of the string, all the energy is kinetic, so we have:
E = m*v^2/2
If there is no dissipation, both energies are equal, so we have:
m*g*L*cos(45) = m*v^2/2
9.81 * 0.8 * 0.7071 * 2 = v^2
v^2 = 11.0986
v = 3.33 m/s
g 95 N force exerted at the end of a 0.50 m long torque wrench gives rise to a torque of 15 N • m. What is the angle (assumed to be less than 90°) between the wrench handle and the direction of the applied force?
Answer:
Angle = 18.41°
Explanation:
Torque = F•r•sin θ
where;
F = force
r = distance from the rotation point
θ = the angle between the force and the radius vector.
We are given;
Torque = 15 N.m
F = 95 N
r = 0.5 m
Thus, plugging in the relevant values ;
15 = 95 × 0.5 × sin θ
sin θ = 15/(95 × 0.5)
sin θ = 0.3158
θ = sin^(-1)0.3158
θ = 18.41°
state Ohm`s law as applied in electricity
Answer:
Ohm's Law (E = IR) is as fundamentally important as Einstein's Relativity equation (E = mc²) is to physicists. When spelled out, it means voltage = current x resistance, or volts = amps x ohms, or V = A x Ω.
how does the statement " silence is golden " relate to ethics in communicating at the workplace.?
Answer:
Being silent most of the time is a good virtue under certain circumstances and environment. It is always advisable to remain quite silent and not be too quick to respond to situations or issues so as to avoid making and saying wrong words.
The ethics in a workplace involves communicating with others with less amount of talking as possible and more of body languages and signs. This is because the workplace is meant to be a serene place.
The amount of friction divided by the weight of an object forms a unit less number called the
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]
Observe: Air pressure is equal to the weight of a column of air on a particular location. Airpressure is measured in millibars (mb). Note how the air pressure changes as you move StationB towards the center of the high-pressure system.
A. What do you notice?
B. Why do you think this is called a high-pressure system?
Answer:
a) When moving towards a high pressure center the pressure values increase in the equipment
b) This area is called high prison since the weight of the atmosphere on top is maximum
Explanation:
A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values 85.5 kPa.
When moving towards a high pressure center the pressure values increase in the equipment
B) This area is called high prison since the weight of the atmosphere on top is maximum
in general they are areas of good weather
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?
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
