zero field spot for opposite unequal charges

What is the magnitude of force required to accelerate a car of mass 1.7 × 10³ kg by 4.75 m/s²

Answer:

F = 8.075 N

Explanation:

Formula for force is;

F = ma

Where;

m is mass

a is acceleration

F = 1.7 × 10³ × 4.75

F = 8.075 N

Option d (T₁ > T₂ > T₃) correctly describes the tension in the three rope system.    

Let's evaluate each tension.

Case T₃.

[tex] T_{3} - W_{3} = 0 [/tex]

For the system to be in equilibrium, the algebraic sum of the tension force (T) and the weight (W) must be equal to zero. The minus sign of W is because it is in the opposite direction of T.          

[tex] T_{3} = W_{3} [/tex]          

Since W₃ = mg, where m is for mass and g is for the acceleration due to gravity, we have:                

[tex] T_{3} = W_{3} = mg [/tex]  (1)                                                                                                     Case T₂.

[tex] T_{2} - (T_{3} + W_{2}) = 0 [/tex]    

[tex] T_{2} = T_{3} + W_{2} [/tex]   (2)

By entering W₂ = 2mg and equation (1) into eq (2) we have:

[tex] T_{2} = T_{3} + W_{2} = mg + 2mg = 3mg [/tex]

Case T₁.

[tex] T_{1} - (T_{2} + W_{1}) = 0 [/tex]  

[tex] T_{1} = T_{2} + W_{1} [/tex]    (3)

Knowing that W₁ = 3mg and T₂ = 3mg, eq (3) is:

[tex] T_{1} = 3mg + 3mg = 6mg [/tex]        

Therefore, the correct option is d: T₁ > T₂ > T₃.

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I hope it helps you!

Correct answer: D. [tex]T_{1} > T_{2} > T_{3}[/tex]

First, we must construct the Equations of Equilibrium for each mass based on Newton's Laws of Motion, then we solve the resulting system for every Tension force:

Mass m:

[tex]\Sigma F = T_{3}-m\cdot g = 0[/tex] (1)

Mass 2m:

[tex]\Sigma F = T_{2}-2\cdot m \cdot g -T_{3} = 0[/tex] (2)

Mass 3m:

[tex]\Sigma F = T_{1}-3\cdot m\cdot g - T_{2} = 0[/tex] (3)

The solution of this system is: [tex]T_{3} = m\cdot g[/tex], [tex]T_{2} = 3\cdot m\cdot g[/tex] and [tex]T_{1} = 6\cdot m\cdot g[/tex], which means that [tex]T_{1} > T_{2} > T_{3}[/tex]. (Correct answer: D.)

Answer:

(a) the tangential speed of a point at the edge is 3.14 m/s

(b) At a point halfway to the center of the disc, tangential speed is 1.571 m/s

Explanation:

Given;

angular speed of the disc, ω = 500 rev/min

diameter of the disc, 120 mm

radius of the disc, r = 60 mm = 0.06 m

(a) the tangential speed of a point at the edge is calculated as follows;

[tex]\omega = 500 \ \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1 \min}{60 \ s} = 52.37 \ rad/s[/tex]

Tangential speed, v = ωr

                               v = 52.37 rad/s  x 0.06 m

                              v = 3.14 m/s

(b) at the edge of the disc, the distance of the point = radius of the disc

   at half-way to the center, the distance of the point = half the radius.

r₁ = ¹/₂r = 0.5 x 0.06 m = 0.03 m

The tangential velocity, v = ωr₁

                                       v = 52.37 rad/s x 0.03 m

                                       v = 1.571 m/s

Answer:

Minimum Area of rectangle = 24 centimeter²

Explanation:

Given:

Length of rectangle = 6 centimeter

Width of rectangle = 4 centimeter

Find:

Minimum Area of rectangle

Computation:

Minimum Area of rectangle = Length of rectangle x Width of rectangle

Minimum Area of rectangle = 6 x 4

Minimum Area of rectangle = 24 centimeter²

Answer:

height is 78.4m

Explanation:

h=u.t + 0.5.g.t^2

= 0 + 0.5x9.8x4^2

= 78.4m

Hooke's law gives the relationship between the force applied and observed compression and expansion

(a) Hooke's law states that force applied is proportional to the deformation of an object

(b) The tube becomes warm from the excess of the energy absorbed but not given off back as the straightening of the tyre

The reason for the above explanation are as follows;

(a) Hooke's law states that for little deformations, the change in the dimension, Δx, of an extended or compressed object is directly proportional to the applied force, F

Mathematically, we get;

F = k × ΔL

[tex]k = \mathbf{\dfrac{F}{\Delta L}}[/tex]

Given that the material cross sectional area = A, and the original length of the material = L, we get;

F/A = Stress = σ, ΔL/L = strain = ε

[tex]\mathbf{Young's \ Modulus, \ E} = \mathbf{\dfrac{\sigma}{\varepsilon}} = \dfrac{\left(\dfrac{F}{A} \right) }{\left(\dfrac{\Delta L}{L} \right) } = \mathbf{\dfrac{F}{\Delta L } \times \dfrac{L}{A}}[/tex]

Therefore, Hooke's law can be expressed as a form of Young's Modulus for a given length to area ratio of a material

(b) The given graph of stress to strain curve, is attached, from which area under the curve gives the energy absorbed by the the material during deformation

Therefore during bending, the stress is increasing as shown in the top of the two curve, and the energy absorbed is given by the area under the curve

As the tyre straightens, the path of the stress change curve is given by the lower curve

The area under the top (bending) curve is larger than the area under the lower (straightening) curve, therefore, the energy absorbed during bending is larger than the energy given off during straightening

Energy absorbed < Energy released

The balance energy is transformed into other forms of energy, including  heat energy, which is observed by the raising in temperature, warming, of the tyre

Energy absorbed = Energy released + Heat energy

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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

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Answer:

44 N

Explanation:

Given that,

If you stand next to a wall on a frictionless skateboard and push the wall with a force of 44 N, then we need to find the force the wall push on you.

It is based on Newton's third law of motion which states that for an action there is an equal and opposite reaction. If the you push the wall with a force of 44 N, the wall push on you is 44 N also as it is based on Newton's third law of motion.

Explanation:

Given that,

Five ideal-gas molecules chosen at random are found to have speeds of 500, 600, 700, 800, and 900 m/s.

The rms speed for this collection is as follows :

[tex]v_{rms}=\sqrt{\dfrac{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2}{5}} \\\\v_{rms}=\sqrt{\dfrac{500^2+600^2+700^2+800^2+900^2}{5}} \\\\v_{rms}=714.14[/tex]

The average speed of these molecules is :

[tex]v_{a}=\dfrac{v_1+v_2+v_3+v_4+v_5}{5}} \\\\v_{a}={\dfrac{500+600+700+800+900}{5}} \\\\v_{a}=700\ m/s[/tex]

So, the rms speed is 714.14 m/s abd the average speed is 700 m/s.

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Answer:

B = 0.088 T

Explanation:

Given that,

The length of a solenoid, l = 0.4 m

No. of turns of wire, N = 356

Current, I = 79 A

We need to find the magnitude of the magnetic field at the center of the solenoid. It is given by the formula.

[tex]B=\mu_onI\\\\B=\mu_o \dfrac{N}{l}\times I\\\\B=4\pi \times 10^{-7}\times \dfrac{356}{0.4}\times 79\\\\B=0.088\ T[/tex]

So, the magnitude of the magnetic field at the center of the solenoid is 0.088 T.

Answer:

Change in energy Δu = 40 J

Explanation:

Given:

System heat ΔQ = 150 J

Missing value of Lost heat ΔW = 110 J

Find:

Change in energy Δu

Computation:

Change in energy Δu = System heat ΔQ - Missing value of Lost heat ΔW

Change in energy Δu = 150 - 110

Change in energy Δu = 40 J

Answer:

[tex]q=41.62\ \mu C[/tex]

Explanation:

Given that,

Force between two objects, F = 11.2 N

Distance between objects, d = 1.18 m

We need to find the charge on each objects. The force between charges is as follows :

[tex]F=\dfrac{kq^2}{r^2}\\\\q=\sqrt{\dfrac{Fr^2}{k}} \\\\q=\sqrt{\dfrac{11.2\times (1.18)^2}{9\times 10^9}} \\\\q=41.62\ \mu C[/tex]

So, the charge on each sphere is [tex]41.62\ \mu C[/tex].

Answer:

i. -4m

ii. 20m

Explanation:

The car travels 8m to the east, then travels 12m to the west which is the opposite of the east. Going west, the car travels 8m back to the origin point and then another 4m due west to make 12m. The displacement from the origin point is -4 (the negative sign shows the direction because displacement is a vector quantity)

Total distance = 8m going east + 8m back to origin + 4m west = 20m

Explanation:

Given that,

Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward.

Taking eastward as positive direction, we have:

[tex]v_B=+5\ m/s[/tex]is the velocity of Bill with respect to Amy (which is stationary)

[tex]v_c=15\ m/s[/tex] is the velocity of Carlos with respect to Amy.

Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is

[tex]v_B=+5\ m/s[/tex]

Therefore, Carlos velocity in Bill's reference frame will be

[tex]v_c'=-15\ m/s-(+5\ m/s)\\\\=-20\ m/s[/tex]

So, the magnitude is 20 m/s and the direction is westward (negative sign).

Answer with explanation:

*Pre-condition: The mass of all blocks are evenly distributed.

As you add more and more blocks, the center of mass of the system changes.

Let the first block be the support. The maximum distance that can be unsupported (hung over) by the second block is equal to half the length of second block.

Why?

Let's take a look at our formula for torque. Torque, [tex]\tau[/tex], is given by:

[tex]\tau=rF\sin \theta[/tex], where [tex]r[/tex] is radius, [tex]F[/tex] is force, and [tex]\theta[/tex] is the angle between the radius and level arm.

[tex]\sin \theta[/tex] is used to calculate the relevant component of a force producing torque. In this case, the only force acting on the object is the force of gravity. Therefore, [tex]\theta =90^{\circ}[/tex] and recall [tex]\sin 90^{\circ}=1[/tex].

Conceptually, it's more important to look at the [tex]r[/tex] term here. From our formula, we can see that if [tex]r=0[/tex], there is no torque.

The point of pivot will be at the center of mass. Here's the important part:

As long as the point of pivot is supported, [tex]r[/tex] will remain zero and no torque will be created from the force of gravity. As you keep stacking blocks, as long as the center of mass of the entire system remains supported from your first support, the tower will not fall.

In the given picture shown, that first block will be your support.

The 6 blocks on top of that first block form a center of mass that is still on that first block, thus allowing the tower to remain standing.

Answer:

Explanation:

The answer is C because the building is 80 meters high. Before the stone is dropped, it has ONLY potential energy since kinetic energy involves velocity and a still stone has no velocity. At impact, there is no potential energy because potential energy involves the height of the stone relative to the ground and a stone ON the ground has no height; here there is ONLY kinetic.

From the First Law of Thermodynamics, we know that energy cannot be created or destroyed, it can only change form. Therefore, that means that at the halfway point of 40 meters, half of the stone's potential energy has been lost, and it has been lost to kinetic energy. Here, at 40 meters, there is an equality between PE and KE. It only last for however long the stone is AT 40 meters, which is probably a millisecond of time, but that's where they are equal.

Answer:

Fq = k q Q / R^2

Fp = k q Q / (6 R)^2

The force exerted between S and p is 1 / 36 of that between S and q

or the force between S and q is 36 times that between S and p.

Answer:

t = 2.26 x 10⁵ s

Explanation:

The energy supplied to the water will be equal to the heat required for the boiling of water:

E = ΔQ

Pt = mL

where,

P = Power = 10 W

t = time = ?

m = mass of water = 1 kg

L = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg

Therefore,

[tex](10\ W)t = (1\ kg)(2.26\ x\ 10^6\ J/kg)\\\\t = \frac{2.26\ x\ 10^6\ J}{10\ W}\\\\[/tex]

t = 2.26 x 10⁵ s

This time will be less than the actual time taken due to some heat loss during the transmission of this heat energy to the container in which water is held.

Answer:

a) Light that passes through the floor to reveal yourself (not shadow).

b) 2 rays of light that bounce between 2 transparent media.

c) I don't know what is Diffraction?

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:

Explanation:

a)

Final velocity v = 0 ; initial velocity u = 30 m/s , time t = .0015 s

v = u + a t

0 = 30 m/s + a x .0015 s

a = - 30 / .0015

= - 20000 m / s²

b )

Magnitude of force = m x a

= .140 kg x 20,000 m / s²

= 2800 N = 2.8 kN.

c )

The force applied by baseball = 2.8 kN .

d )

Since ball can withstand a force of 1.3 kN so it will break if 2.8 kN force acts on it . SO, head will fracture.

Answer:

Option (D) is correct.

Explanation:

Let the speed is v.

[tex]\Delta t = \gamma \Delta t'\\\\\Delta t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\times \frac{\Delta t}{2}\\\\\sqrt{1-\frac{v^2}{c^2}} =\frac{1}{2}\\\\1-\frac{v^2}{c^2}=\frac{1}{4}\\\\\frac{3}{4}c^2 = v^2\\\\v = 0.87 c[/tex]

Option (D) is correct.

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:

The answer is "Option d".

Explanation:

In the given scenario, when a surface is drawn in a magnetic field, the number of magnetic field lines that enter a surface must be equal to the number left because we know that magnetic field lines are always a closed curve hence the number of magnetic field lines that join the surface should match the total left.

The stress in Magapascals is 80 Magapascals

This question can be solved by applying Young's modulus.

Young's modulus : Young's modulus states that, stress  is directly proportional to strain.

Where Stress is pressure acting the body. The s.i unit of stress is N/m² or Pascal. And it can be expressed mathematically as,

From the question,

[tex]P = 10000/(125*10^{-6} )[/tex]

[tex]P =[/tex] [tex]8*10^{7}[/tex] Pascals

[tex]P =[/tex] 80 Magapascals

Hence the stress in Magapascals is 80 Magapascals

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The stress experimented by the tie-bar is 80 megapascals.

The tie-bar is under axial load. Under the assumption that force is distributed uniformly in the cross-section area, we can use the following definition of normal stress ([tex]\sigma[/tex]), in megapascals:

[tex]\sigma = \frac{F}{A}[/tex] (1)

Where:

[tex]F[/tex] - Axial force, in meganewtons.

[tex]A[/tex] - Cross-section area, in square meters.

If we know that [tex]F = 10\times 10^{-3}\,MN[/tex] and [tex]A = 125\times 10^{-6}\,m^{2}[/tex], then the stress experimented by the tie-bar is:

[tex]\sigma = \frac{10\times 10^{-3}\,MN}{125\times 10^{-6}\,m^{2}}[/tex]

[tex]\sigma = 80\,MPa[/tex]

The stress experimented by the tie-bar is 80 megapascals.

Answer:

a: true.

Explanation:

We can define an equipotential surface as a surface where the potential at any point of the surface is constant.

For example, for a punctual charge, the equipotential surfaces are spheres centered at the punctual charge.

Or in the case of an infinite plane of charge, the equipotential surfaces will be planes parallel to our plane of charge.

Now we want to see if the electric field is always perpendicular to these equipotential surfaces.

You can see that in the two previous examples this is true, but let's see for a general case.

Now suppose that you have a given field, and you have a test charge in one equipotential surface.

So, now we can move the charge along the equipotential surface because the potential in the surface is constant, then the potential energy of the charge does not change. And because there is no potential change, then there is no work done by the electric field as the charge moves along the equipotential surface.

But the particle is moving and the electric field is acting on the particle, so the only way that the work can be zero is if the force (the one generated by the electric field, which is parallel to the electric field) and the direction of motion are perpendiculars.

Then we can conclude that the electric field will be always perpendicular to the equipotential surfaces.

The correct option is a.

Answer: [tex]247.12\ N[/tex]

Explanation:

Given

Magnitude of the charges

[tex]q_1=4.47\times 10^{-6}\ C[/tex]

[tex]q_2=1.86\times 10^{-6}\ C[/tex]

Distance between them [tex]d=17.4\ mm[/tex]

As both charges are of same sign, they must repel each other

Force experienced by second charge is

[tex]\Rightarrow F_{21}=\dfrac{kq_1q_2}{d^2}\\\\\Rightarrow F_{21}=\dfrac{9\times 10^9\times 4.47\times 10^{-6}\times 1.86\times 10^{-6}}{(17.4\times 10^{-3})^2}\\\\\Rightarrow F_{21}=\dfrac{74.82\times 10^{-3}}{302.76\times 10^{-6}}\\\\\Rightarrow F_{21}=0.2471\times 10^3\\\Rightarrow F_{21}=247.12\ N[/tex]

Thus, charge 2 experience a force of [tex]247.12\ N[/tex]

Answer:

The force between the two charges is 247.15 N.

Explanation:

Charge, q = 4.47 x 10^-6 C

charge, q' = 1.86 x 10^-6 C

distance, d = 17.4 mm

Let the force is F.

The force is given by the Coulomb's law:

[tex]F = \frac{K q q'}{r^2}\\\\F =\frac{9\times 10^9\times 4.47\times 10^{-6}\times1.86\times 10^{-6}}{(17.4\times 10^{-3})^2}\\\\F = 247.15 N[/tex]