In the figure, if Q = 52 µC q =10 µC and d = 55 cm, what is the magnitude of the electrostatic force on q?​

Answer:

Analysis

Explanation:

Because you must Analysis each and every cold too find out which virus caused this.

It’s weird because Interpretation and Analysis have the meaning of examination

Answer:

Noah caught 35 Noah on Satuday.

Explanation:

Given that,

Kieran caught 53 Pokemon

Kieran caught 18 more Pokémon on Saturday than  Noah did.

Let Noah caught x Pokémon on Saturday.

ATQ,

18 + x = 53

Thie is the equation that can be used to find the value of x.

Subtract 18 from both sides.

18 + x -18 = 53 - 18

x = 35

Hence, Noah caught 35 Noah on Satuday.

Answer:

See explanation

Explanation:

Now we have;

vi = (-30, 0) m/s

vf = (0, 20) m/s

Δvav= vf - vi= (30, 20) m/s

magnitude of Δvav= √30^2 + 20^2 = 36.0 m/s

Direction = tan-1(20/30) = 33.69°

For aav

aav= Δvav/t = (30/5, 20/5) = (6,4) m/s^2

magnitude of  aav = √6^2 + 4^2 = 7.2 m/s^2

direction of aav = tan-1(4/6) = 33.69°

Explanation:

Given that,

A girl walks 20.0 m east then 70.0 meters north.

Displacement is the shortest path covered by an object. Let it is d. It is calculated as :

[tex]d=\sqrt{20^2+70^2} \\\\=72.80\ m[/tex]

For direction,

[tex]\theta=\tan^{-1}(\dfrac{70}{20})\\\\\theta=74.05^{\circ}[/tex]

Girl's distance = 20 m + 70 m

= 90 m

Hence, this is the required solution.

if we hold a very hot glass tumbler under cold water, it cracks. This is because the outer surface of the glass comes in direct contact with cold water and contracts more as compared to the inner surface. We observed that water expanded on heating.

Railway tracks consist of two parallel metal rails joined together. Small gaps, called expansion gaps, are deliberately left between the rails as there is an expansion of the rails in hot weather. Water expands on heating.

Answer:

The energy increase, and the lines of force are denser

Explanation:

The rest of the answers

1.) The field energy will increase.

3.) It points toward the field of earths magnetic poles.

4.) l, ll, and lll only

5.) ll, lV, l, lll

Based on the information given, the energy and the lines of force change as the energy increases, and the lines of force are dense.

In physics, it should be noted that energy is the quantitative property which must be transferred to a body or physical system to perform work on the body.

From the information, the bar magnet was placed underneath a sheet of paper where a pile of iron filings sits and in the presence of the energy stored in the magnetic field, the iron filings arranged themselves, creating lines of force. Therefore, when a stronger bar magnet is used, the energy increases and the lines of force are denser.

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Answer:

d

Explanation:

The quantity that must be squared in the equation of gravitational force is distance d.

According to the universal gravitational law, the square of the distance between two objects is inversely proportional to the force of gravity.

The formula is given as:

  Fg  = [tex]\frac{G m_{1} m_{2} }{d^{2} }[/tex]  

So d is the quantity that must be squared

Answer:

conduction I believe if not its convection

Answer: Conduction transfers energy from the spoon to the hot chocolate.

Explanation: Heated water molecules and steam rise in the beaker, carrying heat by convection.

Answer:

Take a bar magnet and place a steel pin at some distance. ... Now, bring the steel pin near the pole of the bar magnet. We notice that pin sticks to the magnet. This experiment shows that maximum magnetic force acts at the poles of the magnet.

Explanation:

hope that is help

Answer:

Explanation:

Force between two charges can be expressed as follows

F = k q₁ q₂ / d²

q₁ and q₂ are two charges , d is distance between them , k is a constant whose value is 9 x 10⁹

distance between charges is fixed which is 4 -2 = 2 cm = 2 x 10⁻² m

force between 1μC and  4μC

= 9 x 10⁹ x 1 x 4 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 4μC and  1μC

= 9 x 10⁹ x 4 x 1 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 2μC and  2μC

= 9 x 10⁹ x 2 x 2 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 1μC and  2μC

= 9 x 10⁹ x 1 x 2 x 10⁻¹² / ( 2 x 10⁻² )²

= 4.5 x 10 = 45 N

force between 1μC and  8μC

= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²

= 18 x 10 = 180 N

force between 2μC and  8μC

= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²

= 36 x 10 = 360 N

Left Charge   Right Charge Resulting force(N)

1μC                     4μC                  90 N

4μC                   1μC                    90 N

2μC                  2μC                    90 N

1μC                    2μC                   45 N

1μC                  8μC                    180 N

2μC                  8μC                  360 N

Answer:

[tex] \underline{ \boxed{ yes}}\\[/tex]

Explanation:

[tex]given : initial \: velocity \: (u )= 40 {ms}^{ - 1} \\ given : final \: velocity \: (u )= 0 {ms}^{ - 1} \\ given : - (acceleration) \: (a_r) = 2 {ms}^{ - 2} \\ given : distance \: (s) \: = \: ? : \\ but \: {v}^{2} = {u}^{2} + 2( a)s\\ {0}^{2} = {40}^{2} + 2( - 2)s \\ - {40}^{2} = - 4s \\ s = \frac{ - {40}^{2} }{ - 4} \\ s = \frac{1600}{4} \\s = 400 \: m[/tex]

Answer:

the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

Explanation:

Given that;

speed of car V  = 120 km/h = 33.3333 m/s

Reaction time of an alert driver = 0.8 sec

Reaction time of an alert driver = 3 sec

extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec

now, extra distance that car will travel in case of sleepy driver  will be'

S_d = V × 2.2 sec

S_d = 33.3333 m/s × 2.2 sec

S_d = 73.3333 m

hence, number of car of additional car length  n will be;

n = S_n / car length

n = 73.3333 m / 5m

n = 14.666 ≈ 15

Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

Answer:

Explanation:

Total displacement for entire trip = final position - initial position

= 50 m - 30 m = 20 m

Total time = 27 + 47 = 74 s

Average velocity = Total displacement / total time

= 20 / 74 = .27 m /s

Total distance covered in entire trip = 50 + 30 = 80 m

Total time = 74 s

Average speed = Total distance covered / total time

= 80 / 74 = 1.08 m /s .

Answer:

Density of Air = ([tex]P_{1}[/tex]  - [tex]P_{2}[/tex] )/(g x h)

Density of Air =  1.27 Kg/[tex]m^{3}[/tex]

Explanation:

Note: This question is not complete, and lacks its first part in which it contains important data to solve for the density of air. But, I have found the similar question and its data. So. I will be solving the question for the sack of understanding and concept.

Missing part: A weather balloon has an absolute-pressure sensor attached. On the ground the sensor reads [tex]P_{1}[/tex]  = 1.01x [tex]10^{5}[/tex] Pa. At a height of h = 950 m, the sensor reads [tex]P_{2}[/tex]=8.92x[tex]10^{4}[/tex] Pa.

Solution:

Let

[tex]P_{1}[/tex] be the pressure of the balloon at ground.

[tex]P_{2}[/tex] be the pressure of the balloon at height h = 950 m

g = acceleration due to gravity,

In order to derive the expression, we need to find the pressure difference:

Pressure difference = ΔP

ΔP = [tex]P_{1}[/tex]  - [tex]P_{2}[/tex]

As we know that,

Pressure difference = density x acceleration due to gravity x height.

So,

ΔP = [tex]P_{1}[/tex]  - [tex]P_{2}[/tex]  = (Density of Air) x (g) x (h)

We need expression for the density of air, so,

Density of Air = ΔP / (g x h)

Hence, the expression is:

Density of Air = ([tex]P_{1}[/tex]  - [tex]P_{2}[/tex] )/(g x h)

Now, we can calculate the density of air as well, by putting the values given above in the data.

[tex]P_{1}[/tex]  =  1.01 x [tex]10^{5}[/tex]

[tex]P_{2}[/tex] = 8.92 x [tex]10^{4}[/tex]

g = 9.8 m/s

h = 950 m

So,

Density of Air = ((1.01 x [tex]10^{5}[/tex]) - (8.92 x [tex]10^{4}[/tex]) )/ (9.8 x 950)

Density of Air =  1.27 Kg/[tex]m^{3}[/tex]

An expression for an estimate of the density of air is; ρ = ( P₁ - P₂)/gh

According to pascal's principle;

ΔP = ρgh

Where;

ΔP is change in pressure = P₁ - P₂

g is acceleration due to gravity

h is height of pressure change

Now, we want to find an expression for an estimate of the density of air. This means we want to make density the subject of the formula. Thus;

ΔP = ρgh

⇒ divide both sides by gh to get;

ρ = ΔP/(gh)

Recall that ΔP = P₁ - P₂

Thus; ρ = ( P₁ - P₂)/gh

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Answer:

[tex]\theta \approx 59.036^{\circ}[/tex], [tex]T_{2} \approx 23.324\,N[/tex]

Explanation:

First we build the Free Body Diagram (please see first image for further details) associated with the mass, we notice that system consist of a three forces that form a right triangle (please see second image for further details): (i) The weight of the mass, (ii) two tensions.

The requested tension and angle can be found by the following trigonometrical and geometrical expressions:

[tex]\theta = \tan^{-1} \frac{W}{T_{2}}[/tex] (1)

[tex]T_{1} = \sqrt{W^{2}+T_{2}^{2}}[/tex] (2)

Where:

[tex]W[/tex] - Weight of the mass, measured in newtons.

[tex]T_{1}[/tex], [tex]T_{2}[/tex] - Tensions from the mass, measured in newtons.

If we know that [tex]W = 20\,N[/tex] and [tex]T_{2} = 12\,N[/tex], then the requested values are, respectively:

[tex]\theta = \tan^{-1} \frac{20\,N}{12\,N}[/tex]

[tex]\theta \approx 59.036^{\circ}[/tex]

[tex]T_{2} = \sqrt{(20\,N)^{2}+(12\,N)^{2}}[/tex]

[tex]T_{2} \approx 23.324\,N[/tex]

Answer:

Explanation:

Expression for fundamental  frequency of tone produced in a wire under tension of T and length L is given as follows

[tex]f=\frac{1}{2L} \times \sqrt{\frac{T}{ m} }[/tex]

m is mass per unit length .

We shall apply this formula for given wires .

For shorter wire

[tex]60 =\frac{1}{2L} \times \sqrt{\frac{T}{ m} }[/tex]

For longer wire for second harmonic

length of wire is 2L , tension is 4T ,

[tex]f =\frac{2}{4L} \times \sqrt{\frac{4T}{ m} }[/tex]

[tex]f =\frac{2\times 2}{4L} \times \sqrt{\frac{T}{ m} }[/tex]

f = 2 x 60 = 120 Hz .

Answer:

bipolar disorder

Explanation:

bipolar disorder, is previously known as manic depression, is a mental disorder characterized by periods of depression and periods of abnormally elevated mood that last from days to weeks.

Answer:

1.5 Amp is rated for 5 W so it would not be possible

Answer:

[tex]\mathbf{ C_d = 0.86}[/tex]

Explanation:

Given that:

diameter of the orifice D = 5 cm = 5 × 10⁻² m

discharge rate Q = 0015 m³/s

Actual velocity V = 9 m/s

By using the formula

[tex]Q = C_d \times A \times \sqrt{2gh}[/tex]

where;

[tex]v = \sqrt{2gh}[/tex] = 9  m/s

[tex]Q = C_d \times A \times v}[/tex]

where;

[tex]A = \dfrac{\pi D^2}{4}[/tex]

[tex]A = \dfrac{\pi (5 \times 10^{-2} ) ^2}{4}[/tex]

[tex]A = 0.00196 \ m^2[/tex]

Again,

[tex]Q = C_d \times A \times v}[/tex]

[tex]0.015 = C_d \times 0.00196 \times 9[/tex]

[tex]C_d = \dfrac{0.015 }{ 0.00196 \times 9}[/tex]

[tex]\mathbf{ C_d = 0.86}[/tex]

Hence, the coefficient of discharge [tex]\mathbf{ C_d = 0.86}[/tex]

Answer:

t=  16.75 s

Explanation:

We will solve this exercise using the kinematic expressions

corridor that goes at constant speed, suppose that its speed is v₁ = 20 m/s, it does not appear in the statement, we start counting the time when it passes the policeman.

           x₁ = v₁ t

The policeman starts from rest, so his initial velocity is zero and he has an acceleration a = 2.70 m /s², to use the same time counter we take into account that the policeman left at = 1.00 s after passing the corridor

           x₂ = v₀ (t-t₀) + ½ a (t-t₀)²

           x₂ = ½ a (t-1)²

at the point where the two meet, the position must be the same

           x₁ = x₂

          v₁ t = ½ a (t-1)²

          (t-1)² = [tex]\frac{2 v_1 t}{a}[/tex]

           t² - 2t + 1 - \frac{2 v_1 t}{a} +1 = 0

           t² - 2(1 + [tex]\frac{v_1}{a}[/tex]) t  +1

let's we solve the second degree equation

          t² - 2 ( 1 + [tex]\frac{20}{2.7}[/tex]) t + 1=0

          t² - 16.81 t +1=0

          t = [ 16.81 ± [tex]\sqrt{ 16.81^2 - 4 )}[/tex] ] /2

          t = [16.81 ± 16.695]/2

          t₁=  16.75 s

          t2= 0.06 s

Time t₂ is less than the reaction time of humans, so the correct answer is the first time

            t=  16.75 s

Answer:

The water cycle describes how water evaporates from the surface of the earth, rises into the atmosphere, cools and condenses into rain or snow in clouds, and falls again to the surface as precipitation.The water cycle involves the exchange of energy, which leads to temperature changes.

Answer:

so be more specific please

Answer:

Explanation:

The force occurring on a spring F = mg

If x should be the expansion in the spring.

Then F = mg = -kx

where;

[tex]x = \dfrac{mg}{k}[/tex]

[tex]x = \dfrac{2.5 \ kg*9.81 \ m/s^2 }{58 \ N/m}[/tex]

x = 0.423 m

x = 42.3 cm

The final reading of the spring when the end of the spring lines up with the ruler mark is = 42.3 cm + 17 cm

= 59.3 cm

≅ 60 cm

Answer:

1)FALSE

2)TRUE

3)FALSE

4) FALSE

Explanation:

1) Retrieving an entry at a given position is not slow ( it is done using index number ) hence the answer is FALSE

2) Adding an entry at the end of the list is Fast this is because in using array implementation a new entry is made immediately after the last position and it is done very fast as well. hence the answer is TRUE

3) Retrieving an entry at a given position ( from an ADT list ) that is been implemented using Vector is very fast this is simply because Indexing in vector is done at a fixed time  hence the answer is FALSE

4) Adding an entry at the end of the list is slow using vector  hence answer is FALSE

Answer:

Acceleration = 4.8 m/s²

Explanation:

Given:

Change in velocity = 19 m/s

Change in time = 4 s

Find:

Acceleration

Computation:

Acceleration = Change in velocity / Change in time

Acceleration = 19/4

Acceleration = 4.8 m/s²

Positive acceleration

Answer:

D is the answer  I think (0 w 0 )

Explanation:

The  glaciers move gradually. Hence, option (C) is correct.

A glacier is a long-lasting mass of heavy ice that is perpetually moving. When the ablation of snow is greater than the accumulation over a long period of time, frequently centuries, a glacier forms.

As it slowly flows and deforms under forces brought on by its weight, it gains distinctive features like crevasses and seracs. Cirques, moraines, and fjords are the result of the erosion of rock and debris from its substrate as it travels.

The considerably thinner sea ice and lake ice that form on the surface of bodies of water are not the same as glaciers, which form only on land and may flow into water bodies.

The huge ice sheets, commonly referred to as "continental glaciers," in the polar areas of the planet contain 99 percent of the planet's glacial ice.

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#SPJ2

Answer:

1/4πε₀[Hx/ (√x² + b²)^3/2]i.

Explanation:

So, without mincing words let's dive straight into the solution to the question above. There is need to determine the electric field on-axis of a uniformly-charged disk sitting.

The electric field in the x-component, dεₓ = 1/4πε₀[H/ x² + b²] cos .

Thus, the total electric field in the x-component, εₓ = 1/4πε₀ [ xdH/ (x^2 + b^2)^3/2.

Therefore, the electric field = 1/4πε₀ [ xdH/ (x^2 + b^2)^3/2i.

Where x=0

Answer:

A The mass of the object

Explanation:

i hope it's helpful