Urgent please help me

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.

Answer:

24 meters

Explanation:

Find the final velocity. 12m/s

d=[final-initial]/2×time

D=(6m/s)×4=24 m/s

Answer:

The correct answer is "[tex]21.344^{\circ}[/tex]" and "[tex]39.56^{\circ}[/tex]".

Explanation:

According to the question,

Slit width,

[tex]d=\frac{1}{910 \ lines/mm}[/tex]

  [tex]=\frac{1}{910\times 10^3}[/tex]

  [tex]=1.099\times 10^{-6} \ m[/tex]

The condition far first order maxima will be:

⇒ [tex]d Sin \theta = 1 \lambda[/tex]

Now,

⇒ [tex]\Theta_{min} = Sin^{-1} (\frac{\lambda}{d} )[/tex]

             [tex]=Sin^{-1} (\frac{400\times 10^{-9}}{1.099\times 10^{-6}} )[/tex]

             [tex]=21.344^{\circ}[/tex]

⇒ [tex]\Theta_{max} = Sin^{-1} (\frac{\lambda}{d} )[/tex]

             [tex]=Sin^{-1} (\frac{700\times 10^{-9}}{1.099\times 10^{-6}} )[/tex]

             [tex]=39.56^{\circ}[/tex]

Answer:

1) Law of Universal Gravitation     Gravity is an attractive force

5) General relativity               Gravity is due to the curvature of spacetime

Explanation:

In this exercise you are asked to relate the correct theory and its explanation

Theory Explanation

1) Law of Universal Gravitation              Gravity is an attractive force

2) Law of universal gravitation              Mass and distance affect force

3) Classical mechanics                           time and space are absolute

4) Special relativity                                 Time and space are relative

5) General relativity                                Gravity is due to the curvature of

                                                               spacetime

6) General relativity                                 Mass affects the curvature of space - time

Answer:

Explanation:

edge2022

Answer:

enchantment table language

Explanation:

Answer:

[tex]T=8.1N[/tex]

Explanation:

From the question we are told that:

Mass m=0.40

Radius r=1.8m

Angle Beneath the Horizontal \theta =40 \textdegree

Speed v=5.0m/s

The Tension Angle

 [tex]\alpha=90-\theta\\\\\alpha=90-40[/tex]

 [tex]\alpha=50 \textdegree[/tex]

Generally the equation for Tension is is mathematically given by

 [tex]T=\frac{mv^2}{r}+mgcos \alpha[/tex]

 [tex]T=\frac{0.40*5^2}{1.8}+0.40*5cos50[/tex]

 [tex]T=8.1N[/tex]

Answer:

28 j

Explanation:

because when you add you get 28

Answer:no

Explanation:because 0.9*(30*60)=0.9*1800=1620

The turtle has already won the race

Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours

It is given

[tex]V_{t} = 0.9 \frac{m}{s}[/tex]

[tex]V_{r} = 9 \frac{m}{s}[/tex]

[tex]D=1500 m[/tex]

Time taken by turtle  

 [tex]T= \dfrac{D}{V_{t} }=\dfrac{1500}{0.9_{} }[/tex]

[tex]T=1666 minutes= 27 hours[/tex]

Time taken by  rabbit

[tex]T= \dfrac{D}{V_{r} }=\dfrac{1500}{9_{} }[/tex]

[tex]T=166 minutes[/tex]

since rabbit started 30 minutes after turtle then

[tex]T= 136+30=196 minutes[/tex]

[tex]T= 3.2 hours[/tex]

Hence Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours

To know more about average velocity follow

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Answer:  

t = 1.27 x 10⁹ s  

Explanation:  

First, we will find the volume of the wire:

Volume = V = AL  

where,  

A = Cross-sectional area of wire = πr² = π(1 cm)² = π(0.01 m)² = 3.14 x 10⁻⁴ m²  

L = Length of wire = 150 km = 150000 m  

Therefore,    

V = 47.12 m³

Now, we will find the number of electrons in the wire:  

No. of electrons = n = (Electrons per unit Volume)(V)  

n = (8.43 x 10²⁸ electrons/m³)(47.12 m³)  

n = 3.97 x 10³⁰ electrons  

Now, we will use the formula of current to find out the time taken by each electron to cross the wire:

[tex]I =\frac{q}{t}[/tex]  

where,  

t = time = ?  

I = current = 500 A  

q = total charge = (n)(chareg on one electron)  

q = (3.97 x 10³⁰ electrons)(1.6 x 10⁻¹⁹ C/electron)  

q = 6.36 x 10¹¹ C  

[tex]500\ A = \frac{6.36\ x\ 10^{11}\ C}{t}\\\\t = \frac{6.36\ x\ 10^{11}\ C}{500\ A}[/tex]

Therefore,

t = 1.27 x 10⁹ s

Answer:

[tex]F_b= 0.720 N[/tex]

Explanation:

From the question we are told that:

Weight [tex]W=600N[/tex]

Average density [tex]\rho=1.20kg/m^3[/tex]

Mass

[tex]m=\frac{W}{g}[/tex]

[tex]m=\frac{600}{9.81}[/tex]

[tex]m=61.22kg[/tex]

Generally the equation for Volume is mathematically given by

[tex]V =\frac{ mass}{density}[/tex]

[tex]V= \frac{61.22}{1000}[/tex]

[tex]V=0.06122 m^3[/tex]

Therefore

Buoyant force [tex]F_b[/tex]

[tex]F_b=\rho*V*g[/tex]

[tex]F_b= rho_air*V*g[/tex]

[tex]F_b= 0.720 N[/tex]

Answer:

So the correct answer is letter e)

Explanation:

The electric field of an infinite yz-plane with a uniform surface charge density  (σ) is given by:

[tex]E=\frac{\sigma }{2\epsilon_{0}}[/tex]

Where ε₀ is the electric permitivity.

As we see, this electric field does not depend on distance, so the correct answer is letter e)

I hope it helps you!

Answer:

0.28 J

Explanation:

Let the mass of the object is 0.5 kg

The maximum velocity of the object is 1.06 m/s.

We need to find the kinetic energy at that time. It is given by :

[tex]K=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times 0.5\times (1.06)^2\\\\K=0.28\ J[/tex]

So, the required kinetic energy is equal to 0.28 J.

Answer:

The velocity becomes [tex]v\sqrt 2[/tex].

Explanation:

The force acting on the bobber is centripetal  force.

The centripetal force is given by

[tex]F =\frac{mv^2}{r}[/tex]

when mass remains same, radius is doubled and the force is same, so the velocity is v'.

[tex]F =\frac{mv^2}{r}=\frac{mv'^2}{2r}\\\\v'=v\sqrt 2[/tex]

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

Answer:

The required fraction is 0.023.

Explanation:

Given that

Mass of a car, m = 1030 kg

Mass of 4 wheels = 12 kg

We need to find the fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles.

The rotational kinetic energy due to four wheel is

[tex]=4\times \dfrac{1}{2}I\omega^2\\\\=4\times \dfrac{1}{2}\times \dfrac{1}{2}mR^2(\dfrac{v}{R})^2\\\\=mv^2[/tex]

Linear kinetic Energy of the car is:

[tex]=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times Mv^2[/tex]

Fraction,

[tex]f=\dfrac{mv^2}{\dfrac{1}{2}Mv^2}\\\\f=\dfrac{m}{\dfrac{1}{2}M}\\\\f=\dfrac{12}{\dfrac{1}{2}\times 1030}\\\\=0.023[/tex]

So, the required fraction is 0.023.

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

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]

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

Answer:

solid cylinder

Explanation:

the object that arrives first is the object that has more speed, let's use the concepts of energy

starting point. Highest point

         Em₀ = U = m g h

final point. Lowest point

         Em_f = K = ½ mv² + ½ I w²

since the body has rotational and translational movement

how energy is conserved

         m g h = ½ mv² + ½ I w²

linear and angular velocity are related

          v = w r

          w = v / r

we substitute

          m g h = ½ mv² + ½ I (v/r) ²

          mg h = ½ v² (m + I /r²)

          v = [tex]\sqrt{2gh \ \frac{m}{m + \frac{I}{r^2} } }[/tex]

the tabulated moments of inertia for the bodies are

solid cylinder     I = ½ m r²

hollow cylinder  I = m r²

we look for the speed for each body

solid cylinder

          v₁ = [tex]\sqrt {2gh} \ \sqrt{\frac{m}{m + m/2} }[/tex]

          v₁ = [tex]\sqrt{2gh} \ \sqrt{2/3}[/tex]

let's call    v₀ = [tex]\sqrt{2gh}[/tex]

         v₁ = 0.816 v₀

hollow cylinder

          v₂ = [tex]\sqrt{2gh } \ \sqrt{\frac{m}{m+ m} }[/tex]

          v₂ = v₀ √½

          v₂ = 0.707 v₀

Therefore, the body that has the highest speed is the solid cylinder and since time is the inverse of speed, this is the body that spends less time to reach the bottom, that is, it is the first to arrive

Answer:

[tex]t=0.50cm[/tex]

Explanation:

From the question we are told that:

Wavelength [tex]\lamda=3c[/tex]m

Refraction Index [tex]n=1.50[/tex]

Generally the equation for Destructive interference for Normal incidence is mathematically given by

[tex]2nt=m(\frac{1}{2})\lambda[/tex]

Since  Minimum Thickness occurs at

At [tex]m=0[/tex]

Therefore

[tex]t=\frac{\lambda}{2}[/tex]

[tex]t=\frac{3}{4(1.50)}[/tex]

[tex]t=0.50cm[/tex]

Answer:

Momentum of object A = Momentum of object C < momentum of B.

Explanation:

The momentum of an object is equal to the product of mass and velocity.

Object A has a mass m and a speed v. Its momentum is :

p = mv

Object B has a mass m/2 and a speed 4v. Its momentum is :

p = (m/2)×4v = 2mv

Object C has a mass 3m and a speed v/3. Its momentum is :

p = (3m)×(v/3) = mv

So,

Momentum of object A = Momentum of object C < momentum of B.

Answer:

 I = - m 16  the two impulses are the same,

Explanation:

The impulse is given by the relationship

         I = Δp

         I = p_f - p₀

in this case the final velocity is zero therefore p_f = 0

        I = -p₀

For driver A the steering wheel impulse is

        I = - m v₀

        I = - m 16

For driver B, the airbag gives an impulse

        I = - m 16

We can see that the two impulses are the same, the difference is that in the air bag more time is used to give this impulse therefore the force on the driver is less

Answer:

η = 0.5 = 50%

Explanation:

The efficiency of the power cycle is given by the following formula:

[tex]\eta = \frac{W}{Q_1}\\\\\eta = \frac{Q_1-Q_2}{Q_1}[/tex]

where,

where,

η = efficiency = ?

Q₁ = heat received from hot reservoir = 1000 KJ

Q₂ = heat discharged to cold reservoir = 500 KJ

Therefore,

[tex]\eta = \frac{1000\ KJ-500\ KJ}{1000\ KJ}[/tex]

η = 0.5 = 50%

Answer:

3.55 T

Explanation:

Applying,

F = BILsin∅.............. Equation 1

Where F = Force, B = magnetic Field, I = current, L = Length of the wire, ∅ = Angle between the wire and the magnetic field

make B the subject of the equation

B = F/ILsin∅.................. Equation 2

From the question,

Given: F = 2.45 N, L = 3.00 cm = 0.03 m, I = 23.0 A, ∅ = 90° (Perpendicular)

Substitute these values into equation 2

B = 2.45/(0.03×23×sin90)

B = 2.45/0.69

B = 3.55 T

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

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.

Answer:

1.621 kN

Explanation:

Since each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal, the horizontal component of the force due to the first horse along the canal is F= 839cos15° N and its vertical component is F' = 839sin15° N(it is positive since it is perpendicular to the centerline of the canal and points upwards).

The horizontal component of the force due to the second horse along the canal is f = 839cos15° N and its vertical component is f' = -839sin15° N (it is negative since it is perpendicular to the centerline of the canal and points downwards).

So, the resultant horizontal component of force R = F + f = 839cos15° N + 839cos15° N = 2(839cos15°) N = 2(839 × 0.9659) = 2 × 810.412 = 1620.82 N

So, the resultant vertical component of force R' = F' + f' = 839sin15° N + (-839sin15° N) = 839sin15° N - 839sin15° N = 0 N

The magnitude of the resultant force which is the sum of the two forces is R" = √(R² + R'²)

= √(R² + 0²)  (since R' = 0)

= √R²

= R  

= 1620.82 N

= 1.62082 kN

≅ 1.621 kN

So, the sum of these  two forces on the barge is 1.621 kN

Answer: [tex]283.2\times 10^{-9}\ nC[/tex]

Explanation:

Given

Cross-sectional area [tex]A=0.4\ cm^2[/tex]

Dielectric constant [tex]k=4[/tex]

Dielectric strength [tex]E=2\times 10^8\ V/m[/tex]

Distance between capacitors [tex]d=5\ mm[/tex]

Maximum charge that can be stored before dielectric breakdown is given by

[tex]\Rightarrow Q=CV\\\\\Rightarrow Q=\dfrac{k\epsilon_oA}{d}\cdot (Ed)\quad\quad [V=E\cdot d]\\\\\Rightarrow Q=k\epsilon_oAE\\\\\Rightarrow Q=4\times 8.85\times 10^{-12}\times 0.4\times 10^{-4}\times 2\times 10^8\\\\\Rightarrow Q=28.32\times 10^{-8}\\\\\Rightarrow Q=283.2\times 10^{-9}\ nC[/tex]

Answer:

The maximum charge is 7.08 x 10^-8 C.

Explanation:

Area, A = 0.4 cm^2

K = 4

Electric field, E = 2 x 10^8 V/m

separation, d = 5 mm = 0.005 m

Let the capacitance is C and the charge is q.

[tex]q = CV\\\\q=\frac{\varepsilon o A}{d}\times E d\\\\q = \varepsilon o A E\\\\q = 8.85\times 10^{-12}\times0.4\times 10^{-4}\times 2\times 10^8\\\\q = 7.08\times 10^{-8}C[/tex]