Answer:
To cover a larger distance we use km(kilometre) , hm (hectometre) , and dac(decametre). These are called the multiples of Metre. For the distance is smaller , we use Dm (decimetre , cm(centimetre) and mm (millimetre) . These are called the submultiples of
1. A positive electric charge in a region of changing electric potential will: A. move in the direction of decreasing potential B. move in the direction of increasing potential C. move in an undetermined way; we need more information D. remain at rest
Answer:
The correct option is (B).
Explanation:
The electric potential is the amount of work energy needed to move a unit of electric charge from a reference point to the specific point in an electric field. The electric potential due to a point charge is given by :
[tex]V=\dfrac{kQ}{r}[/tex]
Where
k is the electrostatic constant
Q is the electric charge
r is the distance from the charge
So, a positive electric charge in a region of changing electric potential will move in the direction of increasing potential.
Two distinct systems have the same amount of stored internal energy. 500 J are added by heat to the first system and 300 J are added by heat to the second system. What will be the change in internal energy of the first system if it does 200 J of work? How much work will the second system have to do in order to have the same internal energy?
Answer:
The change in the internal energy of the first system is 300 J
The second system will do zero work in order to have the same internal energy.
Explanation:
Given;
heat added to the first system, Q₁ = 500 J
heat added to the second system, Q₂ = 300 J
work done by the first system, W₁ = 200 J
The change in the internal energy of the system is given by the first law of thermodynamics;
ΔU = Q - W
where;
ΔU is the change in internal energy of the system
The change in the internal energy of the first system is calculated as;
ΔU₁ = Q₁ - W₁
ΔU₁ = 500 J - 200 J
ΔU₁ = = 300 J
The work done by the second system to have the same internal energy with the first.
ΔU₁ = Q₂ - W₂
W₂ = Q₂ - ΔU₁
W₂ = 300 J - 300 J
W₂ = 0
The second system will do zero work in order to have the same internal energy.
Question 4 of 5
How can the Fitness Logs help you in this class?
O A. They can't; the Fitness Logs are only useful to your teacher.
B. They show your parents how much you're learning.
C. They let you keep track of your thoughts, feelings, and progress.
D. They help you evaluate yourself for your final grade.
SUBMIT
Answer:
C is the right answer
Explanation:
fitness logs is a great way to track your progress. You can easily look back and see how you have progressed over time. In addition, it can help you plan and prepare for future workouts, as well as identify patterns of what seems to work well for you and when you have the most success
hope it was useful for you
Consider an airplane with a total wing surface of 50 m^2. At a certain speed the difference in air pressure below and above the wings is 4.0 % of atmospheric pressure.
Required:
Find the lift on the airplane.
Answer:
[tex]F=202650N[/tex]
Explanation:
From the question we are told that:
Area [tex]a=50m^2[/tex]
Difference in air Pressure [tex]dP=4.0\% atm=>0.04*101325=>4035Pa[/tex]
Generally the equation for Force is mathematically given by
[tex]F=dP*A[/tex]
[tex]F=4053*50[/tex]
[tex]F=202650N[/tex]
In which type of mixture do the physically distinct component parts each have distinct properties?
Answer:
In heterogeneous mixture do the physically distinct component parts each have distinct properties.
• Explain how sound travels
Sound is a type of energy made by vibrations. These vibrations create sound waves which move through mediums such as air, water and wood. When an object vibrates, it causes movement in the particles of the medium. This movement is called sound waves, and it keeps going until the particles run out of energy.
Sound is a type of energy made by vibrations. These vibrations create sound waves which move through mediums such as air, water and wood. When an object vibrates, it causes movement in the particles of the medium. This movement is called sound waves, and it keeps going until the particles run out of energy.
The distance between the two object is fixed at 5.0 m. The uncertainty distance measurement is? The percentage error in the distance is?
A long, current-carrying solenoid with an air core has 1550 turns per meter of length and a radius of 0.0240 m. A coil of 200 turns is wrapped tightly around the outside of the solenoid, so it has virtually the same radius as the solenoid. What is the mutual inductance of this system
Answer:
[tex]M=7.05*10^{-4}[/tex]
Explanation:
From the question we are told that:
Coil one turns N_1=1550 Turns/m
Radius [tex]r=0.0240m[/tex]
Turns 2 [tex]N_2=200N[/tex]
Generally the equation for area is mathematically given by
[tex]A=\pi*r^2[/tex]
[tex]A=\pi*0.024^2[/tex]
[tex]A=\1.81*10^{-3} m^2[/tex]
Therefore
The mutual inductance of this system is
[tex]M=\mu*N_1*N_2*A[/tex]
[tex]M=(4 \pi*10^{-7})*1550*200*1.81*10^{-3}[/tex]
[tex]M=7.05*10^{-4}[/tex]
When an external magnetic flux through a conducting loop decreases in magnitude, a current is induced in the loop that creates its own magnetic flux through the loop. How does that induced magnetic flux affect the total magnetic flux through the loop
Answer:
Len's law
Explanation:
We can explain this exercise using Len's law
when the magnetic flux decreases, a matic flux appears that opposes the decrease, thus maintaining the value of the initial luxury.
An artificial satellite circling the Earth completes each orbit in 125 minutes. (a) Find the altitude of the satellite.(b) What is the value of g at the location of this satellite?
Answer:
(a) Altitude = 1.95 x 10⁶ m = 1950 km
(b) g = 5.9 m/s²
Explanation:
(a)
The time period of the satellite is given by the following formula:
[tex]T^2 = \frac{4\pi^2r^3}{GM_E}[/tex]
where,
T = Time period = (125 min)([tex]\frac{60\ s}{1\ min}[/tex]) = 7500 s
r = distance of satellite from the center of earth = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
[tex]M_E[/tex] = Mass of Earth = 6 x 10²⁴ kg
Therefore,
[tex](7500\ s)^2 = \frac{4\pi^2r^3}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(6\ x\ 10^{24}\ kg)}\\\\r^3 = \frac{(7500\ s)^2(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(6\ x\ 10^{24}\ kg)}{4\pi^2}\\\\r = \sqrt[3]{5.7\ x\ 10^{20}\ m^3} \\[/tex]
r = 8.29 x 10⁶ m
Hence, the altitude of the satellite will be:
[tex]Altitude = r - radius\ of\ Earth\\Altitude = 8.29\ x\ 10^6\ m - 6.34\ x\ 10^6\ m[/tex]
Altitude = 1.95 x 10⁶ m = 1950 km
(b)
The weight of the satellite will be equal to the gravitational force between satellite and Earth:
[tex]Weight = Gravitational\ Force\\\\M_sg = \frac{GM_EM_s}{r^2}\\\\g = \frac{GM_E}{r^2}\\\\g = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(6\ x\ 10^{24}\ kg)}{(8.23\ x\ 10^6\ m)^2}[/tex]
g = 5.9 m/s²
The mass of a hot-air balloon and its occupants is 381 kg (excluding the hot air inside the balloon). The air outside the balloon has a pressure of 1.01 x 105 Pa and a density of 1.29 kg/m3. To lift off, the air inside the balloon is heated. The volume of the heated balloon is 480 m3. The pressure of the heated air remains the same as that of the outside air. To what temperature in kelvins must the air be heated so that the balloon just lifts off
Answer:
In order to lift off the ground, the air in the balloon must be heated to 710.26 K
Explanation:
Given the data in the question;
P = 1.01 × 10⁵ Pa
V = 480 m³
ρ = 1.29 kg/m³
M = 381 kg
we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol
let F represent the force acting upward.
Now in a condition where the hot air balloon is just about to take off;
F - Mg - m[tex]_g[/tex]g = 0
where M is the mass of the balloon and its occupants, m[tex]_g[/tex] is the mass of the hot gas inside the balloon.
the force acting upward F = Vρg
so
Vρg - Mg - m[tex]_g[/tex]g = 0
solve for m[tex]_g[/tex]
m[tex]_g[/tex] = ( Vρg - Mg ) / g
m[tex]_g[/tex] = Vρg/g - Mg/g
m[tex]_g[/tex] = ρV - M ------- let this be equation 1
Now, from the ideal gas law, PV = nRT
we know that number of moles n = m[tex]_g[/tex] / μ
where μ is the molecular mass of air
so
PV = (m[tex]_g[/tex]/μ)RT
solve for T
μPV = m[tex]_g[/tex]RT
T = μPV / m[tex]_g[/tex]R -------- let this be equation 2
from equation 1 and 2
T = μPV / (ρV - M)R
so we substitute in our values;
P = 1.01 × 10⁵ Pa
V = 480 m³
ρ = 1.29 kg/m³
M = 381 kg
we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol
T = [ (29 × 10⁻³) × (1.01 × 10⁵) × 480 ] / [ (( 1.29 × 480 ) - 381)8.31 ]
T = 1405920 / 1979.442
T = 710.26 K
Therefore, In order to lift off the ground, the air in the balloon must be heated to 710.26 K
The temperature required for the air to be heated is 710.26 K.
Given data:
The mass of a hot air-balloon is, m = 381 kg.
The pressure of air outside the balloon is, [tex]P = 1.01 \times 10^{5} \;\rm Pa[/tex].
The density of air is, [tex]\rho = 1.29 \;\rm kg/m^{3}[/tex].
The volume of heated balloon is, [tex]V = 480 \;\rm m^{3}[/tex].
The condition where the hot air balloon is just about to take off is as follows:
[tex]F-mg - m'g =0[/tex]
Here,
m' is the mass of hot gas inside the balloon and g is the gravitational acceleration and F is the force acting on the balloon in upward direction. And its value is,
[tex]F = V \times \rho \times g[/tex]
Solving as,
[tex](V \times \rho \times g)-mg - m'g =0\\\\ m'=(V \rho )-m[/tex]
Now, apply the ideal gas law as,
PV = nRT
here, R is the universal gas constant and n is the number of moles and its value is,
[tex]n=\dfrac{m'}{M}[/tex]
M is the molecular mass of gas. Solving as,
[tex]PV = \dfrac{m'}{M} \times R \times T\\\\\\T=\dfrac{P \times V\times M}{m'R}\\\\\\T=\dfrac{P \times V\times M}{(V \rho - m)R}[/tex]
Since, the standard value for the molecular mass of air is, [tex]M = 29 \times 10^{-3} \;\rm kg/mol[/tex]. Then solve for the temperature as,
[tex]T=\dfrac{(1.01 \times 10^{5}) \times 480\times 381}{(480 \times (1.29) - 381)8.31}\\\\\\T = 710.26 \;\rm K[/tex]
Thus, we can conclude that the temperature required for the air to be heated is 710.26 K.
Learn more about the ideal gas equation here:
https://brainly.com/question/18518493
A wheel rotates at an angular velocity of 30rad/s. If an acceleration of 26rad/s2 is applied to it, what will its angular velocity be after 5.0s
An eagle flying at 35 m/s emits a cry whose frequency is 440 Hz. A blackbird is moving in the same direction as the eagle at 10 m/s. (Assume the speed of sound is 343 m/s.)
(a) What frequency does the blackbird hear (in Hz) as the eagle approaches the blackbird?
Hz
(b) What frequency does the blackbird hear (in Hz) after the eagle passes the blackbird?
Hz
Answer:
a) [tex]F=475.7Hz[/tex]
b) [tex]F'=410.899Hz[/tex]
Explanation:
From the question we are told that:
Velocity of eagle [tex]V_1=35m/s[/tex]
Frequency of eagle [tex]F_1=440Hz[/tex]
Velocity of Black bird [tex]V_2=10m/s[/tex]
Speed of sound [tex]s=343m/s[/tex]
a)
Generally the equation for Frequency is mathematically given by
[tex]F=f_0(\frac{v-v_2}{v-v_1})[/tex]
[tex]F=440(\frac{343-10}{343-35})[/tex]
[tex]F=475.7Hz[/tex]
b)
Generally the equation for Frequency is mathematically given by
[tex]F'=f_0(\frac{v+v_2}{v+v_1})[/tex]
[tex]F'=440(\frac{343+10}{343+35})[/tex]
[tex]F'=410.899Hz[/tex]
If you tethered a space station to the earth by a long cable, you could get to space in an elevator that rides up the cable much simpler and cheaper than riding to space on a rocket. There's one big problem, however: There is no way to create a cable that is long enough. The cable would need to reach 36,000 km upward, to the height where a satellite orbits at the same speed as the earth rotates; a cable this long made of ordinary materials couldn't even support its own weight. Consider a steel cable suspended from a point high above the earth. The stress in the cable is highest at the top; it must support the weight of cable below it.
What is the greatest length the cable could have without failing?
Answer:
[tex]l=12916.5m[/tex]
Explanation:
Distance [tex]d=3600km[/tex]
Since
Density of steel [tex]\rho=7900kg/m^3[/tex]
Stress of steel [tex]\mu= 1*10^9[/tex]
Generally the equation for Stress on Cable is mathematically given by
[tex]S=\frac{F}{A}[/tex]
[tex]S=\frac{\rho Alg}{A}[/tex]
Therefore
[tex]l=\frac{s}{\rhog}[/tex]
[tex]l=\frac{ 1*10^9}{7900kg/m^3*9.8}[/tex]
[tex]l=12916.5m[/tex]
Can an electron be diffracted? Can it exhibit interference?
Answer:
Yeah, it can be diffracted. Though it depends on a diffracting medium.
It must have some magnetic fields .
Forexample; X-ray diffraction where electrons are diffracted to the target filament.
g A CD is spinning on a CD player. You open the CD player to change out the disk and notice that the CD comes to rest after 15 revolutions with a constant deceleration of 120 r a d s 2 . What was the initial angular speed of the CD
Answer:
[tex]\omega_1=150rads/sec[/tex]
Explanation:
From the question we are told that:
Number of Revolution [tex]N=15=30\pi[/tex]
Deceleration [tex]d= -120 rads/2[/tex]
Generally the equation for initial angular speed [tex]\omega_1[/tex] is mathematically given by
[tex]\omega_2^2=\omega_1^2 +2(d)(N)[/tex]
[tex]0=\omega_1^2 +2(-120)(20 \pi)[/tex]
[tex]\omega_1^2=7200 \pi[/tex]
[tex]\omega_1=150rads/sec[/tex]
how does the use of standard units of measurement solve problems in measurement regarding validity and reliabiility? explain it
Answer:
Reliability can be estimated by comparing different versions of the same measurement. Validity is harder to assess, but it can be estimated by comparing the results to other relevant data or theory.
When a car's starter is in use, it draws a large current. The car's lights draw much less current. As a certain car is starting, the current through the battery is 54.0 A and the potential difference across the battery terminals is 9.18 V. When only the car's lights are used, the current through the battery is 2.10 A and the terminal potential difference is 12.6 V. Find the battery's emf.
Answer:
12.74 V
Explanation:
We are given that
Current, I1=54 A
Potential difference, V1=9.18V
I2=2.10 A
V2=12.6 V
We have to find the battery's emf.
[tex]E=V+Ir[/tex]
Using the formula
[tex]E=9.18+54r[/tex] ....(1)
[tex]E=12.6+2.10r[/tex] .....(2)
Subtract equation (1) from (2)
[tex]0=3.42-51.9r[/tex]
[tex]3.42=51.9r[/tex]
[tex]r=\frac{3.42}{51.9}=0.0659ohm[/tex]
Using the value of r in equation (1)
[tex]E=9.18+54(0.0659)[/tex]
[tex]E=12.74 V[/tex]
Two objects moving with a speed v travel in opposite directions in a straight line. The objects stick together when they collide, and move with a speed of v/2 after the collision.
Required:
a. What is the ratio of the final kinetic energy of the system to the initial kinetic energy?
b. What is the ratio of the mass of the more massive object to the mass of the less massive object?
Answer:
Explanation:
Let the mass of objects be m₁ and m₂ .
Total kinetic energy = 1/2 m₁ v² + 1/2 m₂ v²= 1/2 ( m₁ + m₂ ) v²
Total kinetic energy after collision= 1/2 ( m₁ + m₂ ) v² / 4 = 1/2 ( m₁ + m₂ ) v² x .25
final KE / initial KE = 1/2 ( m₁ + m₂ ) v² x .25 / 1/2 ( m₁ + m₂ ) v²
= 0.25
b )
Applying law of conservation of momentum to the system . Let m₁ > m₂
m₁ v - m₂ v = ( m₁ + m₂ ) v / 2
m₁ v - m₂ v = ( m₁ + m₂ ) v / 2
m₁ - m₂ = ( m₁ + m₂ ) / 2
2m₁ - 2 m₂ = m₁ + m₂
m₁ = 3m₂
m₁ / m₂ = 3 / 1
Answer:
(a) The ratio is 1 : 4.
(b) The ratio is 1 : 3.
Explanation:
Let the mass of each object is m and m'.
They initially move with velocity v opposite to each other.
Use conservation of momentum
m v - m' v = (m + m') v/2
2 (m - m') = (m + m')
2 m - 2 m' = m + m'
m = 3 m' .... (1)
(a) Let the initial kinetic energy is K and the final kinetic energy is K'.
[tex]K = 0.5 mv^2 + 0.5 m' v^2 \\\\K = 0.5 (m + m') v^2..... (1)[/tex]
[tex]K' = 0.5 (m + m') \frac{v^2}{4}.... (2)[/tex]
The ratio is
K' : K = 1 : 4
(b) m = 3 m'
So, m : m' = 3 : 1
Consider a uniform electric field of 50 N/C directed toward the east. If the voltage measured relative to ground at a given point in the field is 80 V, what is the voltage at a point 1.0 m directly east of the point
Answer:
30 V
Explanation:
Given that:
The uniform electric field = 50 N/C
Voltage = 80 V
distance = 1.0 m
The potential difference of the electric field = Δ V
E_d = V₁ - V₂
50 × 1 = 80V - V₂
50 - 80 V = - V₂
-30 V = - V₂
V₂ = 30 V
abrief history of hand writing
A building is being knocked down with a wrecking ball, which is a big metal sphere that swings on a 15-m-long cable. You are (unwisely!) standing directly beneath the point from which the wrecking ball is hung when you notice that the ball has just been released and is swinging directly toward you. How much time do you have to move out of the way? answer in seconds.
Answer:
Time to move out of the way = 1.74 s
Explanation:
Time to move out of the way is one fourth of period = 6.95/4 = 1.74 seconds.
Time to move out of the way = 1.74 s
A monochromatic light falls on two narrow slits that are 4.50 um apart. The third destructive fringes which are 35° apart were formed at a screen 2m from the light source. The light source is 0.30 m from the slits. () Calculate Ym. (4 marks) Compute the wavelength of the light. (4 marks)
Answer:
y = 1.19 m and λ = 8.6036 10⁻⁷ m
Explanation:
This is a slit interference problem, the expression for destructive interference is
d sin θ = m λ
indicate that for the angle of θ = 35º it is in the third order m = 3 and the separation of the slits is d = 4.50 10⁻⁶ m
λ = d sin θ / m
let's calculate
λ = 4.50 10⁻⁶ sin 35 /3
λ = 8.6036 10⁻⁷ m
for the separation distance from the central stripe, we use trigonometry
tan θ= y / L
y = L tan θ
the distance L is measured from the slits, it indicates that the light source is at x = 0.30 m from the slits
L = 2 -0.30
L = 1.70 m
let's calculate
y = 1.70 tan 35
y = 1.19 m
Two positive charges ( 8.0 mC and 2.0 mC) are separated by 300 m. A third charge is placed at distance r from the 8.0 mC charge in such a way that the resultant electric force on the third charge due to the other two charges is zero. The distance r is
Answer:
[tex]r=200m[/tex]
Explanation:
From the question we are told that:
Charges:
[tex]Q_1=8.0mC[/tex]
[tex]Q_2=2.0mC[/tex]
[tex]Q_3=8.mC[/tex]
Distance [tex]d=300m[/tex]
Generally the equation for Force is mathematically given by
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
Therefore
[tex]F_{32}=F_{31}[/tex]
[tex]\frac{q_2}{(300-r)^2}=\frac{q_1}{r^2}[/tex]
[tex]\frac{2*10^{-3}}{(300-r)^2}=\frac{8*10^{-3}}{r^2}[/tex]
[tex]r=2(300-r)[/tex]
[tex]r=200m[/tex]
You throw a Frisbee of mass m and radius r so that it is spinning about a horizontal axis perpendicular to the plane of the Frisbee. Ignoring air resistance, the torque exerted about its center of mass by gravity is: __________
a. 0.
b. mgr
c. 2mgr
d. a function of the angular velocity.
e. small at first, then increasing as the Frisbee loses the torque given it by your hand.
Answer:
the correct answer is a
Explanation:
The torque is
τ = F x r
where the bold letters indicate vectors, in this case the vector of the center of mass is perpendicular to the weight of the body
τ = mg r
in body weight it is applied at the point of the center of mass, therefore as the distance of the force from the axis of rotation (center of amas) is zero, the die is zero
the correct answer is a
Nhiệt dung riêng của một chất là ?
Answer:
enchantment table language
Explanation:
Hi,A body changes its velocity from 60 km/hr to 72 km/hr in 2 sec.Find the acceleration and distance travelled.
Answer:
Initial velocity, u = 60 km/h = 16.7 m/s
Final velocity, v = 72 km/h = 20 m/s
time, t = 2 sec
From first equation of motion:
[tex]{ \bf{v = u + at}}[/tex]
Substitute the variables:
[tex]{ \tt{20 = 16.7 + (a \times 2)}} \\ { \tt{2a = 3.3}} \\ { \tt{acceleration = 1.65 \: {ms}^{ - 2} }}[/tex]
A 0.20 kg mass on a horizontal spring is pulled back a certain distance and released. The maximum speed of the mass is measured to be 0.30 m/s. If, instead, a 0.40 kg mass were used in this same experiment, choose the correct value for the maximum speed.
a. 0.40 m/s.
b. 0.20 m/s.
c. 0.28 m/s.
d. 0.14 m/s.
e. 0.10 m/s.
Answer:
b. 0.20 m/s.
Explanation:
Given;
initial mass, m = 0.2 kg
maximum speed, v = 0.3 m/s
The total energy of the spring at the given maximum speed is calculated as;
K.E = ¹/₂mv²
K.E = 0.5 x 0.2 x 0.3²
K.E = 0.009 J
If the mass is changed to 0.4 kg
¹/₂mv² = K.E
mv² = 2K.E
[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2\times 0.009}{0.4} } \\\\v = 0.21 \ m/s\\\\v \approx 0.20 \ m/s[/tex]
Therefore, the maximum speed is 0.20 m/s
In a calorimetry experiment, three samples A, B, and C with TA> TB> Tc are placed in thermal contact. When the samples have reached thermal equilibrium at a common temperature T, which one of the following must be true?
a. QA > QB >QC
b. QA< 0, QB <0, and Qc > 0
c. T> TB
d. T
e. TA > T> Tc
Answer:
e. TA>T>Tc
Explanation:
a) In this case, we cannot say for sure QA>QB>QC. This is because the magnitude of the heat flow will depend on the specific heat and the mass of each sample. Due to the equation:
[tex]Q=mC_{p}(T_{f}-T_{0})[/tex]
if we did an energy balance of the system, we would get that>
QA+QB+QC=0
For this equation to be true, at least one of the heats must be negative. And one of the heats must be positive.
We don't know either of them, so we cannot determine if this statement is true.
b) We can say for sure that QA<0, because when the two samples get to equilibrum, the temperatrue of A must be smaller than its original temperature. Therefore, it must have lost heat. But we cannot say for sure if QB<0 because sample B could have gained or lost heat during the process, this will depend on the equilibrium temperature, which we don't know. So we cannot say for sure this option is correct.
c) In this case we don't know for sure if the equilibrium temperature will be greater or smaller than TB. This will depend on the mass and specific heat of the samples, just line in part a.
d) is not complete
e) We know for sure that A must have lost heat, so its equilibrium temperature must be smaller than it's original temperature. We know that C must have gained heat, therefore it's equilibrium temperature must be greater than it's original temperature, so TA>T>Tc must be true.
Put the balloon near (BUT NOT TOUCHING) the wall. Leave about as much space as the width of your pinky finger between the balloon and wall. Does the balloon move, if so which way
Answer:
Move towards the wall.
Explanation:
When the balloon is kept near to the wall not touching the wall, there is a force of electrostatic attraction so that the balloon moves towards the wall and stick to it.
As there is some charge on the balloon and the wall is uncharged so the force is there due to which the balloon moves towards the wall.
