Two charged particles are near each other – the charge magnitudes are +3 uC and +7 uC, respectively. The electrostatic force between these charges is 25 N. A charged particle of -1 uC is added to both of the original two charges. What is the electrostatic force between the two charged particles after the -1 uC charge is added?

Answer:

88.761

Explanation:

Answer:

54.0 x0.1=5.4 x0.03=0.162

kinetic force

If a contestant in a winter games event pushes a 54.0 kg block of ice across a frozen lake. The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03, then the minimum force F he must exert to get the block moving would be 5.2974 Nwrons.

Friction is a type of force that resists or prevents the relative motion of two physical objects when their surfaces come in contact.

As given in the problem, If a contestant in a winter games event pushes a 54.0 kg block of ice across a frozen lake. The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03,

The force required to get the block moving = μMg

                                                                          = 0.01×54×9.81

                                                                         = 5.2974 Newtons

Thus, the minimum force required to move the block would be 5.2974 Newtons.

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10¹¹ is the approximate number of atoms in a bacterium.

The atomic mass of an element is the average mass of the atoms of an element measured in atomic mass unit (amu).

Given,

Mass of a bacterium atom = 10⁻¹⁵ kg.

Mass of a hydrogen atom  = 10⁻²⁷ kg.

From the above observation ,

The average mass of an atom of the bacterium is ten times the mass of a hydrogen atom.

Atomic mass 1 bacterium atom = 10 x mass of hydrogen atom

                                                  = 10 x 10⁻²⁷ kg.

                                                  = 10⁻²⁶ kg.

Thus,

The number of atoms in a bacterium

=  [tex]\frac{Total mass}{Atomicmass of 1 bacterium}[/tex]

= [tex]\frac{10^{-15} }{10^{-26} }[/tex]

=  [tex]10^{11}[/tex]

The approximate number of atoms in a bacterium  is [tex]10^{11}[/tex].

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The velocity of the board relative to the ice is zero, since both are at rest.

Relative velocity is the velocity of an object in relation to another reference object or point.

When two objects are travelling or moving with the same velocity in the same direction, the relative velocity one relative to the other is zero.

Also, when two objects are at rest, the relative velocity one relative to the other is zero.

Therefore, the velocity of the board relative to the ice is zero, since both are at rest.

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The proper connection between kinetic energy and speed is depicted in Graph D.

The kinetic energy (KE) is defined as one-half of the mass times multiplied by the square of velocity.

[tex]\rm KE = \frac{1}{2}mv^2[/tex]

where,

KE is the kinetic energy

m is the mass of each molecule

(V) is the  velocity

As the square of the velocity is exactly proportional to kinetic energy. Consequently, the relationship between velocity and kinetic energy must be parabolic.

The proper connection between kinetic energy and speed is depicted in Graph D. Graph showing the link between kinetic energy on the y-axis and speed on the x-axis.

On the y-axis, a semi-curve line begins above the origin and ascends as it accelerates.

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The specific heat capacity of the alcohol will be 3.72  kJ/kg°C.

The amount of heat required to increase a substance's temperature by one degree Celsius is known as its "specific heat capacity."

Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.

Given data;

Mass of liquid sample of Alcohol  m₁ = 200-gram

The temperature of alcohol, T₁ =  -6°C.

Mass of liquid sample of water  m₂ = 400-gram

The temperature of the water, T₂=  20°C.

The specific heat capacity of the alcohol, S₁=?

The specific heat capacity of water is, S₂=4.19 kJ/kg.°C

As we know that;

[tex]\rm Q_{gain}= Q{loss} \\\\ Q_{alcohol} =Q_{water} \\\\\ m_1s_1\triangle T_1 = m_2S_2 \triangle S_2 \\\\ 200 \times 10^{-3} \times S_1 [ (12-(-6) ] = 40 \times 10^{-3} \times 4.19 \times 10^{-3} \times (20-12)\\\\S_1 = 2 \times 4.19 \times 10^3 \times \frac {8}{18} \\\\ S_1 = 3.72 \ kJ /kg ^0 C[/tex]

Hence the specific heat capacity of the alcohol will be 3.72  kJ/kg°C.

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Answer:

Answer is C because light travels in a sight line but when light pass through a refractor the light from the source changes direction when passes through a refractor

The total time required is 1 minute and 0.3 seconds.

Assumption: The distance between the people is 1 m and the stadium is a circle with a radius of 100m.

Here, the time taken by the person is 0.1 seconds.

Total distance covered by the wave = Circumference of the circle

So, total distance = 2 πr = 2 × 3.14 × 100 = 618 m

As the distance between each of the people is 1 m.

So, the number of personal interactions is 618.

Time taken by each person is 0.1 seconds.

So, total time, t = 0.1 × 618 = 61.8 seconds.

So, the order of the magnitude of the time required is 1 minute and 0.3 seconds for such a pulse to make one circuit around BC Place Stadium in Vancouver.

Hence, the total time required is 1 minute and 0.3 seconds.

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The voltage of the secondary will be 36 V.From the given conditions primary transformer has 3 times as many turns in the secondary coil.

Electromagnetic induction is what causes the induced voltage. Electromagnetic induction is the process of generating emf (induced voltage) by subjecting a conductor to a magnetic field.

Given data;

No turns in the primary coil,[tex]\rm N_p = 480 turns[/tex]

No turns in the secondary coil,[tex]\rm N_s = 112 turns[/tex]

The voltage of the primary coil,[tex]\rm V_p = 28 v[/tex]

The voltage of the secondary coil,[tex]\rm V_s = ?[/tex]

For a transformer,

[tex]\rm \frac{V_p}{V_s}= \frac{N_p}{N_s} \\\\ \rm \frac{28}{V_s}= \frac{480}{112} \\\\ V_s = 6.53 \ v[/tex]

Hence the voltage of the secondary will be 36 V.

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Answer: 6.53

Explanation: I had the same question and here's what I got:

Values:

Vp (Primary Voltage) = 28.0

Ns (Secondary Number of Turns) = 112

Np (Primary Number of Turns) = 480

Vs = Vp Ns/Np

= (28)(112)/(480)

Vs = 6.53 (FINAL ANSWER)

The magnitude of the reaction force of the air pushing the balloon would be equal and opposite.

A push or a pull that an object experiences as a result of interacting with another item is known as a  reaction force.

Newton's third law is officially expressed as follows: There is an equal and opposite reaction to every action. The implication of the statement is that there are always two forces acting on the two interacting objects. The force acting on the first object is equal in size to the force acting on the second. The force acting on the first object is acting in the opposite direction to the force acting on the second object. Force pairs—equal and opposing action-reaction force pairs—always exist in pairs.

Knowing that everything has an equal and opposite reaction according to Newton's second law. The preasured air that a balloon had to push out into the free air acts as the reaction force.

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Here is your answer mate,

Question,

[tex]Determine\: the\: amount\\ \: of\: power\:used\: in\\\: holding\: a\: 25\: kg\: box\:\\ , \: 1.5\: meters \: above\: the\: floor\\\: for\: 60\: seconds[/tex]

Answer,

[tex][/tex]

Solution,

[tex][/tex]

Given,

[tex]MASS \: IS\: 25\: KG\: \\ and \: HEIGHTIS\: 1.5m\: [/tex]

[tex][/tex]

WORK DONE (done against gravity) =

mass×acceleration due to gravity ×height

WORK = 25× 10× 1.5

[tex]\: \: \: \: \: \: \: \:\: \: \: \: \: \: \: [/tex]= 375 Nm

[tex][/tex]

Now

POWER =

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\frac{work}{time} [/tex]

POWER

[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:= \frac{375}{60} Watt [/tex]

[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =6.25[/tex]

[tex]Therfore\: your \: answer\: is\: 6.25[/tex]

[tex][/tex]

Check this,

[tex]Acceleration\: due\: to \: gravity\\\: can\: be\: 9.8\: m/s²\: \\As\: nothing\: mentioned\\\: in\: question\: \\I \: took \: it \: as \: 10[/tex]

[tex][/tex]

Have a good day

The term for a pair of forces described by Newton's third law is action-reaction pair. Details about Newton's law can be found below.

Newton's third law of motion states that for every action, there is an equal and opposite reaction.

This law proposed that when an object 1 acts on object 2 with a force of a particular magnitude, object 2 also acts on object 1 with an opposite force of same magnitude.

Therefore, it can be said that the term for a pair of forces described by Newton's third law is action-reaction pair.

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Disclaimer: I ended up finding what's asked for in the reverse order (e)-(a).

At time [tex]t[/tex], the horizontal position [tex]x[/tex] and vertical position [tex]y[/tex] of the ball are given respectively by

[tex]x = v_i \cos(\theta_i) t[/tex]

[tex]y = v_i \sin(\theta_i) t - \dfrac g2 t^2[/tex]

and the horizontal velocity [tex]v_x[/tex] and vertical velocity [tex]v_y[/tex] are

[tex]v_x = v_i \cos(\theta_i)[/tex]

[tex]v_y = v_i \sin(\theta_i) - gt[/tex]

The ball reaches its maximum height with [tex]v_y=0[/tex]. At this point, the ball has zero vertical velocity. This happens when

[tex]v_i \sin(\theta_i) - gt = 0 \implies t = \dfrac{v_i \sin(\theta_i)}g[/tex]

which means

[tex]y = \dfrac R6 = v_i \sin(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g - \dfrac g2 \left(\dfrac{v_i \sin(\theta_i)}g\right)^2 \\\\ \implies R = \dfrac{6{v_i}^2 \sin^2(\theta_i)}g - \dfrac{3{v_i}^2 \sin^2(\theta_i)}g \\\\ \implies R = \dfrac{3{v_i}^2 \sin^2(\theta_i)}g[/tex]

At the same time, the ball will have traveled half its horizontal range, so

[tex]x = \dfrac R2 = v_i \cos(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g \\\\ \implies R = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g[/tex]

Solve for [tex]v_i[/tex] and [tex]\theta_i[/tex] :

[tex]\dfrac{3{v_i}^2 \sin^2(\theta_i)}g = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g \\\\ \implies 3 \sin^2(\theta_i) = 2 \cos(\theta_i) \sin(\theta_i) \\\\ \sin(\theta_i) (3\sin(\theta_i) - 2 \cos(\theta_i)) = 0[/tex]

Since [tex]0^\circ<\theta_i<90^\circ[/tex], we cannot have [tex]\sin(\theta_i)=0[/tex], so we're left with (e)

[tex]3 \sin(\theta_i) - 2\cos(\theta_i) = 0 \\\\ \implies 3 \sin(\theta_i) = 2\cos(\theta_i) \\\\ \implies \tan(\theta_i) = \dfrac23 \\\\ \implies \boxed{\theta_i = \tan^{-1}\left(\dfrac23\right) \approx 33.7^\circ}[/tex]

Now,

[tex]\cos\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac3{\sqrt{13}}[/tex]

[tex]\sin\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac2{\sqrt{13}}[/tex]

so it follows that (d)

[tex]R = \dfrac{2{v_i}^2 \times\frac3{\sqrt{13}} \times \frac2{\sqrt{13}}}g \\\\ \implies {v_i}^2 = \dfrac{13Rg}{12} \\\\ \implies \boxed{v_i = \sqrt{\dfrac{13Rg}{12}}}[/tex]

Knowing the initial speed and angle, the initial vertical component of velocity is (c)

[tex]v_y = \sqrt{\dfrac{13Rg}{12}} \sin\left(\tan^{-1}\left(\dfrac23\right)\right) \\\\ \implies v_y = \sqrt{\dfrac{13Rg}{12}} \times \dfrac2{\sqrt{13}} \\\\ \implies \boxed{v_y = \sqrt{\dfrac{Rg}3}}[/tex]

We mentioned earlier that the vertical velocity is zero at maximum height, so the speed of the ball is entirely determined by the horizontal component. (b)

[tex]v_x = \sqrt{\dfrac{13Rg}{12}} \times \dfrac3{\sqrt{13}} \\\\ \implies v_x = \dfrac{\sqrt{3Rg}}{2}[/tex]

Then with [tex]v_y=0[/tex], the ball's speed [tex]v[/tex] is

[tex]v = \sqrt{{v_x}^2 + {v_y}^2} \\\\ \implies v = v_x \\\\ \implies \boxed{v = \dfrac{\sqrt{3Rg}}2}[/tex]

Finally, in the work leading up to part (e), we showed the time to maximum height is

[tex]t = \dfrac{v_i \sin(\theta_i)}g[/tex]

but this is just half the total time the ball spends in the air. The total airtime is then

[tex]2t = \dfrac{2 \times \sqrt{\frac{13Rg}{12}} \times \frac2{\sqrt{13}}}g \\\\ \implies 2t = 2\sqrt{\dfrac R{3g}}[/tex]

and the ball is in the air over the interval (a)

[tex]\boxed{0 < t < 2\sqrt{\frac R{3g}}}[/tex]

The provided question's answer is "Schwarzschild radius".

The conversion factor between mass and energy is the speed of light squared.

GM/r stands for gravitational potential energy, also known as energy per unit mass.

GM/rc² then has "mass per unit mass" units. In other words, as mass/mass splits out in a dimensional analysis, "dimensionless per unit."

The derivation yields a formula for time or space coordinate ratios requiring sqrt(1 - 2GM/rc²). This number becomes 0 when r=2GM/c2, or the formula becomes infinite if in the denominator. However, there is no justification for using c² as a conversion factor there. Consider the initial expression sqrt(1 - 2GM/rc²).

Assume that m is used as the test particle's mass instead of 1. Then you have sqrt(m - 2GMm/rc² and mass units. This expression denotes that the rest energy of the test mass m you introduced into the gravitational field is "gone" at that radius.

The 2 would be absent if the gravitational field were Newtonian. However, at the event horizon, Einstein gravity is slightly stronger than Newton gravity, resulting in the factor 2 in qualitative terms.

So, the given equation is of Schwarzschild radius.

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Answer:

The scientific method

Explanation:

The solenoid's inductance and the average emf around the solenoid if the current changes from +3.50 A to −3.50 A in 7.83 ✕ 10−3 s will be 1.01 ×10⁻⁸ H and 9.02 × 10⁻² V.

A coil of wire that carries an electric current is a solenoid. A solenoid is an electromagnet formed by a helical coil of wire.

Which generates a magnetic field when an electric current is passed through the coil.

A solenoid is a form of coil that produces a magnetic field when the electric current is passed through it. A solenoid is created when a conductive wire is used to make a loop.

Given data;

Turns ,N = 465

Length,L= 8.00 cm

Cross-sectional area,A = 3.10 ✕ 10−9 m2.

Solenoid's inductance, L=?

The average emf around the solenoid, E=?

Time of flow,t= 7.83 ✕ 10−3 s

Current changes from +3.50 A to −3.50 A

The inductance of the solenoid is found as;

[tex]\rm L = \frac{\mu_0 AN^2}{L} \\\\ \rm L = \frac{4 \times \pi \times 10^{-7}\times 3.0 \times 10^{-9} \times (465)^2 }{8.00 \times 10^{-2}} \\\\ L= 1.01 \times 10^{-8} \ H[/tex]

The average emf around the solenoid is found as;

[tex]\rm e = L \frac{I_2-I_1}{t} \\\\ \rm e = 1.01 \times 10^{-8} \times \frac{3.50-(-3.50)}{7.83 \times 10^{-3}} \\\\ e =9.02 \times 10^{-12} \ V[/tex]

Hence, the solenoid's inductance will be 1.01× 10⁻⁸ H.

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Answer:

I actually did that but I fprgot

Answer: 0.0387

Explanation: I had the same problem and here's what I got:

Values:

ε (Voltage) = 240

ω (Angular velocity) = 37.7

B (Magnitude) = 0.200

N (Number of loops) = 822

ε = NABω (Rearrange to get A)

A = ε/WBN

= (240)/(37.7)(0.200)(822)

A = 0.0387 (FINAL ANSWER)

Answer:

Let distance between two cities is x km . So total distance travelled bus x+x=2x km and time taken was=(x/40)+(x/60)=(3x+2x)/120=5x/120=x/24 hrs . So avg speed is 2x/(x/24)=48 km per hr.

or

dt1=60km1h

dt2=30km1h

2dt1+t2=r¯

d=60t1

d=30t2

60t1=30t2

2t1=t2

2dt1+(2t1)=r¯

2(60t1)t1+2t1=r¯

2(60t1)=r¯(t1+2t1)

120=r¯+2r¯

r¯=1203

r¯=40km/h

The distance x equal to 0.87 cm when the intensity of the light decreases by 9 times.

The intensity of light is given as power emitted by the light by unit area.

I = P/A

I = P/L²

I₁L₁² = I₂L₂²

L₂² = I₁L₁²/I₂

L₂² = (3 x 1.5²)/(9)

L₂² = 0.75

L₂ = 0.87 cm

Thus, the distance x equal to 0.87 cm when the intensity of the light decreases by 9 times.

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Answer:

b. B

Explanation:

Picture B has the smallest peaks among all which henceforth makes the wavelength i.e. distance between two adjacent crests or troughs, small.

Answer:b

Explanation:Wavelength can be calculated using the following formula: wavelength = wave velocity/frequency. Wavelength usually is expressed in units of meters. The symbol for wavelength is the Greek lambda λ, so λ = v/f.

Find refractive index first

[tex]\\ \rm\Rrightarrow \mu=\dfrac{c}{v}[/tex]

[tex]\\ \rm\Rrightarrow \mu=\dfrac{1.0003}{1.33}[/tex]

[tex]\\ \rm\Rrightarrow \mu =0.75[/tex]

Now

[tex]\\ \rm\Rrightarrow \dfrac{sini}{sinr}=\mu[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{sin45}{sinr}=0.75[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{sin45}{0.75}=sinr[/tex]

[tex]\\ \rm\Rrightarrow sinr=0.94[/tex]

[tex]\\ \rm\Rrightarrow r=sin^{-1}(0.94)[/tex]

[tex]\\ \rm\Rrightarrow r=70^{\circ}[/tex]

According to Piaget, that means that she has made the transition from the sensorimotor to preoperational stage

The first of Piaget's four stages of cognitive development is the sensorimotor stage.

A child's understanding that the outside world exists apart from them is what distinguishes it.

Within Piaget's stages of development, the kid will advance to the following stage after they have completely grasped this.

"Charlene has just begun to be able to form a mental representation of an object that is not visibly present.

According to Piaget, that means that she has made the transition from the sensorimotor to preoperational stage

Only recently has Charlene been able to create a mental image of something that is not physically present.

That indicates that she has moved from the sensorimotor to the preoperational level, in accordance with Piaget.

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The sled accelerates with magnitude [tex]a[/tex] such that

[tex]\left(2.3\dfrac{\rm m}{\rm s}\right)^2 = 2a(9.2\,\mathrm m) \implies a = 0.2875\dfrac{\rm m}{\mathrm s^2}[/tex]

By Newton's second law, the net force in the plane of motion (parallel to the ground) is

[tex]31\,\mathrm N - F_{\rm friction} = (17 \,\mathrm{kg}) \left(0.2875 \dfrac{\rm m}{\mathrm s^2}\right)[/tex]

so that the force of friction exerts a magnitude of

[tex]F_{\rm friction} = 26.1125 \,\mathrm N[/tex]

Perpendicular to the ground, the sled is in equilibrium, so Newton's second law says

[tex]F_{\rm normal} - (17\,\mathrm{kg})g = 0 \implies F_{\rm normal} = 166.6 \,\mathrm N[/tex]

The magnitude of friction is proportional to the magnitude of the normal force by a factor of [tex]\mu_k[/tex], the coefficient of kinetic friction. It follows that

[tex]F_{\rm friction} = \mu_k F_{\rm normal} \implies \mu_k = \dfrac{26.1125\,\rm N}{166.6\,\rm N} \approx \boxed{0.16}[/tex]

(and the coefficient is dimensionless).

The total kinetic energy of the woman when she reaches the ground is 6,198.5 J.

The total kinetic energy of the woman when she reaches the ground is calculated from the principle of conservation of energy.

P.E(top) = K.E(bottom)

P.E(top) = mgh

P.E(top) = 55 x 9.8 x 11.5

P.E(top) = 6,198.5 J

Thus, the total kinetic energy of the woman when she reaches the ground is 6,198.5 J.

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The estimated density of the golf ball is  700 kg/m³

Density is defined as Mass per unit Volume.

In displacement method,  

First , we measuring the volume of water displaced by an object which tell us the volume of the object then we will use the physical balance to determine its mass.

Then calculate the density by dividing the mass by the volume.

i.e.  D = m/V

Given,  Density of water is 1.0 g/cm³

Using displacement method  , The estimated volume of golf ball is 100 cm³  and estimated mass is 7g

Then ,

Density =  10 cm³ / 7 g= 0.07 g/ cm³ = 700 kg/m³

So the estimated density of golf ball is 700 kg/m³

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Q in = Q out

mcΔt in = mcΔt out (c = equal)

mΔt in = mΔt out

200 x  (75-25) = M x (25-5)

M = 500 g

∆T for hot water

∆T for cold water

Both are equal according law of conservation of energy

[tex]\\ \rm\Rrightarrow m_1c\delta T_0=m_2c\delta [/tex]

[tex]\\ \rm\Rrightarrow 200(50)=m_2(20)[/tex]

[tex]\\ \rm\Rrightarrow 10000=m_2(20)[/tex]

[tex]\\ \rm\Rrightarrow m_2=500g=0.5kg[/tex]

The gauge pressure at bottom of vaccine solution will be 16 kPa

Positive pressure is another name for gauge pressure. When a system's internal pressure exceeds that of its surroundings, it is said to be under positive pressure. Any leak that develops in the positively pressured system will therefore escape into the outside world. In contrast, a negative pressure chamber draws air into it.

Given As seen in the illustration, a syringe is held vertically. The container carries a 3 cm tall column of vaccine solution and has an open inner diameter of 1 cm. The needle contains a 2 cm column of vaccine solution and has an open inner diameter of 0.5 mm. At the needle's open end, the solution is exposed to the air. The vaccination solution has a density of 1200 kg/m3.

We have to find the gauge pressure at bottom of vaccine solution

Since the 5N force is applied to vaccine solution the pressure exerted will be much more

Hence the gauge pressure at bottom of vaccine solution will be 16 kPa

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Answer:

-2.5m/s^2

Explanation:

10-40/12-0=-2.5

Answer:

0.4

Explanation:

ω = km

ω = 1000 x 4

ω = 4000 divide this by 10000 and you get 0.4s

Answer:

λ = c / f     or    f = c / λ

f = 3.0E8 / 4.0E-7 = .75E15 / sec = 7.5E14 / sec = 7.5 X 10^14 /sec