Many cultures around the world still use a simple weapon called a blowgun, a tube with a dart that fits tightly inside. A sharp breath into the end of the tube launches the dart. When exhaling forcefully, a healthy person can supply air at a gauge pressure of 6.0 kPa.

Required:
What force does this pressure exert on a dart in a 1.5-cm-diameter tube?

Answer:

a) The distance from the cornea vertex to the retina is 2.37 cm

b) This distance is shorter than for the normal eye.

Explanation:

a) Let refractive index of air,

n(air) = x = 1

Let refractive index of lens,

n(lens) = y = 1.4

Object distance, s = 36 cm

Radius of curvature, R = 0.65 cm

The distance from the cornea vertex to the retina is the image distance because image is formed in the retina.

Image distance, s' = ?

(x/s) + (y/s') = (y-x)/R

(1/36) + (1.4/s') = (1.4 - 1)/0.65

1.4/s' = 0.62 - 0.028

1.4/s' = 0.592

s' = 1.4/0.592

s' = 2.37 cm

Distance from the cornea vertex to the retina is 2.37 cm

(b) For a normal eye, the distance between the cornea vertex and the retina is 2.60 cm. Since 2.37 < 2.60, this distance is shorter than for normal eye.

Sun's energy comes from the nuclear fusion taking place inside. In nuclear fusion two light nuclei fuses together to form a heavy nuclei with the release of greater amount of energy.

Nuclear fusion is the process of combining two light nuclei to form a heavy nuclei. In this nuclear process, tremendous energy is released. This is the source of heat and light in stars.

On the other hand, nuclear fission is the process of breaking of a heavy nuclei into two lighter nuclei. Fission also produces massive energy. But in comparison, more energy is produced by nuclear fusion.

Nuclear fission is used in nuclear power generators. The light energy and  heat energy comes form the nuclear fusion of hydrogens to form helium nuclei. Hence, option D is correct.

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Answer:

a. 94.54 N

b. 0.356 m/s^2

Explanation:

Given:-

- The mass of the inclined block, M = 100 kg

- The mass of the vertically hanging block, m = 10 kg

- The angle of inclination, θ = 20°

- The coefficient of friction of inclined surface, u = 0.3

Find:-

a) The magnitude of tension in the cable

b) The acceleration of the system

Solution:-

- We will first draw a free body diagram for both the blocks. The vertically hanging block of mass m = 10 kg tends to move "upward" when the system is released.

- The block experiences a tension force ( T ) in the upward direction due the attached cable. The tension in the cable is combated with the weight of the vertically hanging block.

- We will employ the use of Newton's second law of motion to express the dynamics of the vertically hanging block as follows:

                        [tex]T - m*g = m*a\\\\[/tex]  ... Eq 1

Where,

              a: The acceleration of the system

- Similarly, we will construct a free body diagram for the inclined block of mass M = 100 kg. The Tension ( T ) pulls onto the block; however, the weight of the block is greater and tends down the slope.

- As the block moves down the slope it experiences frictional force ( F ) that acts up the slope due to the contact force ( N ) between the block and the plane.

- We will employ the static equilibrium of the inclined block in the normal direction and we have:

                        [tex]N - M*g*cos ( Q )= 0\\\\N = M*g*cos ( Q )[/tex]

- The frictional force ( F ) is proportional to the contact force ( N ) as follows:

                        [tex]F = u*N\\\\F = u*M*g *cos ( Q )[/tex]

- Now we will apply the Newton's second law of motion parallel to the plane as follows:

                       [tex]M*g*sin(Q) - T - F = M*a\\\\M*g*sin(Q) - T -u*M*g*cos(Q) = M*a\\[/tex] .. Eq2

- Add the two equation, Eq 1 and Eq 2:

                      [tex]M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g = a* ( M + m )\\\\a = \frac{M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g}{M + m} \\\\a = \frac{100*9.81*sin ( 20 ) - 0.3*100*9.81*cos ( 20 ) - 10*9.81}{100 + 10}\\\\a = \frac{-39.12977}{110} = -0.35572 \frac{m}{s^2}[/tex]

- The inclined block moves up ( the acceleration is in the opposite direction than assumed ).

- Using equation 1, we determine the tension ( T ) in the cable as follows:

                     [tex]T = m* ( a + g )\\\\T = 10*( -0.35572 + 9.81 )\\\\T = 94.54 N[/tex]

Answer:

1.424 μC

Explanation:

I'm assuming here, that the charged ball is suspended by the string. If the string also is deflected by the angle α, then the forces acting on it would be: mg (acting downwards),

tension T (acting along the string - to the pivot point), and

F (electric force – acting along the line connecting the charges).

We then have something like this

x: T•sin α = F,

y: T•cosα = mg.

Dividing the first one by the second one we have

T•sin α/ T•cosα = F/mg, ultimately,

tan α = F/mg.

Since we already know that

q1=q2=q, and

r=2•L•sinα,

k=9•10^9 N•m²/C²

Remember,

F =k•q1•q2/r², if we substitute for r, we have

F = k•q²/(2•L•sinα)².

tan α = F/mg =

= k•q²/(2•L•sinα)² •mg.

q = (2•L•sinα) • √(m•g•tanα/k)=

=(2•0.5•0.486) • √(0.0142•9.8•0.557/9•10^9) =

q = 0.486 • √(8.61•10^-12)

q = 0.486 • 2.93•10^-6

q = 1.424•10^-6 C

q = 1.424 μC.

Answer:

Explanation:

this is the answer to your question

Answer:

a. 1000N/C

Explanation:

Data mentioned in the question

Electrical field magnitude = 1000 NC

Perpendicular distance = 0.1 m

Perpendicular distance = 0.2 m

Based on the above information, the electric field is

As we know that

[tex]E = \frac{\sigma}{2\times E_o}[/tex]

where,

[tex]\sigma[/tex] = surface charge density

E = distance from nearby point to sheet i.e be independent

The distance at 0.1 and 0.2, the electric field would remain the same

So,

Based on the above explanation, the first option is correct

Answer:

the final temperature is [tex]\mathbf{T_f = -377.2^0 C}[/tex]

Explanation:

The change in length of a bar can be expressed with the relation;

[tex]\Delta L = L_f - L_i[/tex]   ---- (1)

Also ; the relative or fractional increase in length  is proportional to the change in temperature.

Mathematically;

ΔL/L_i ∝ kΔT

where;

k is replaced with ∝ (the proportionality constant )

[tex]\dfrac{ \Delta L}{L_i}=\alpha \Delta T[/tex]    ---- (2)

From (1) ;

[tex]L_f = \Delta L + L_i[/tex]  ---  (3)

from (2)

[tex]{ \Delta L}=\alpha \Delta T*{L_i}[/tex]  ---- (4)

replacing (4) into (3);we have;

[tex]L_f =(\alpha \Delta T*{L_i} ) + L_i[/tex]

On re-arrangement; we have

[tex]L_f = L_i + \alpha L_i (\Delta T )[/tex]

from the given question; we can say that :

[tex](L_f)_{brass}}} = (L_f)_{Al}[/tex]

So;

[tex]L_{brass} + \alpha _{brass} L_{brass}(\Delta T) = L_{Al} + \alpha _{Al} L_{Al}(\Delta T)[/tex]

Making the change in temperature the subject of the formula; we have:

[tex]\Delta T = \dfrac{L_{Al}-L_{brass}}{\alpha _ {brass} L_{brass}-\alpha _{Al}L_{Al}}[/tex]

where;

[tex]L_{Al}[/tex] = 10.02 cm

[tex]L_{brass}[/tex] = 10.00 cm

[tex]\alpha _{brass}[/tex] = 19 × 10⁻⁶ °C ⁻¹

[tex]\alpha_{Al}[/tex] = 24  × 10⁻⁶ °C ⁻¹

[tex]\Delta T = \dfrac{10.02-10.00}{19*10^{-6} \ \ {^0}C^{-1} *10.00 -24*10^{-6} \ \ {^0}C^{-1} *10.02}[/tex]

[tex]\Delta T[/tex] = −396.1965135 ° C

[tex]\Delta T[/tex] ≅ −396.20  °C

Given that the initial temperature [tex]T_i = 19^0 C[/tex]

Then ;

[tex]\Delta T = T_f - T_i[/tex]

[tex]T_f = \Delta T + T_I[/tex]

Thus;

[tex]T_f =(-396.20 + 19.0)^0 C[/tex]

[tex]\mathbf{T_f = -377.2^0 C}[/tex]

Thus; the final temperature is [tex]\mathbf{T_f = -377.2^0 C}[/tex]

Answer:

v = 4.08m/s₂

Explanation:

Answer:

563.64 m

Explanation:

Given that as per the question

x = 5 cm = 0.05 m

D = 4.2 × 107 m

d = smallest aperture size

As per the situation the solution of the smallest aperture telescope that she can get away with is below :-

We will use Rayleigh's diffraction limit which is

[tex]d\frac{x}{D} = 1.22\lambda[/tex]

The equation will be

[tex]d\frac{0.05}{4.2\times 10^7} = 1.22[550\times 10^{-9}][/tex]

d = 563.64 m

So, the answer is d = 563.64 m

Answer:

τ = 1679.68Nm

Explanation:

In order to calculate the required torque you first take into account the following formula:

[tex]\tau=I\alpha[/tex]           (1)

τ: torque

I: moment of inertia of the merry-go-round

α: angular acceleration

Next, you use the following formulas for the calculation of the angular acceleration and the moment of inertia:

[tex]\omega=\omega_o+\alpha t[/tex]         (2)

[tex]I=\frac{1}{2}MR^2[/tex]           (3)       (it is considered that the merry-go-round is a disk)

w: final angular speed = 3.1 rad/s

wo: initial angular speed = 0 rad/s

M: mass of the merry-go-round = 432 kg

R: radius of the merry-go-round = 2.3m

You solve the equation (2) for α. Furthermore you calculate the moment of inertia:

[tex]\alpha=\frac{\omega}{t}=\frac{3.1rad/s}{2.1s}=1.47\frac{rad}{s^2}\\\\I=\frac{1}{2}(432kg)(2.3)^2=1142.64kg\frac{m}{s}[/tex]

Finally, you replace the values of the moment of inertia and angular acceleration in the equation (1):

[tex]\tau=(1142.64kgm/s)(1.47rad/s^2)=1679.68Nm[/tex]

The required torque is 1679.68Nm

Answer:

the two elements are resistor R and inductor L

answers in two significant figures

R = 9.6Ω

L = 54mH

Explanation:

Answer:

46.67 N  Upwards (with a clockwise moment)

Explanation:

length of board = 2.5 m

weight of board = 120 N

the board has two supports,  say support A and support B

support A is at one end,

support B is at 100 m from the other end.

weight of diver = 100 N

diver stands on the other end of the board.

Magnitude of support A at the end of the board

To get the magnitude and force exerted by the support at the end of the board (support A, we take moment of the forces about support B

Moment of a force is the product of force and perpendicular distance of the force about a center.

The weight of the board acts at the center of the board (1.25 m from each end of the board). That is 2.5 m from the support B.

moment of board's weight about support B is  120 x 0.25 = 30 N-m

The moment due to the weight of the board acts anticlockwise.

Weight of the diver acts at the opposite side of the board, and it acts 1 m from support B.

Moment of diver about support B is 100 x 1 = 100 N-m

Th moment due to the diver acts clockwise.

The moment due to the reaction at support A acts at a distance 1.5 m from support B

If the reaction force on support A is Fa, then the reaction about support B is Fa x 1.5 = 1.5Fa.

The moment due to support A acts clockwise.

According to moment laws, the total clockwise movement must be equal to the total anticlockwise movement.

Total clockwise movements = 100 N-m + 1.5Fa

Total anticlockwise moment = 30 N-m

according to moment laws,

100 + 1.5Fa = 30

1.5 Fa = 30 - 100 = -70

Fa = -70/1.5 = -46.67 N

The magnitude of the force exerted at support A is equal but opposite to the reaction at support A and is equal to 46.67 N

Answer:

E = V/5 x10⁻³

Explanation:

if the potential difference is V

then electric field E is given by

E = V/d

d = 0.5cm = 5 x 10⁻³m

E = V/5 x10⁻³

Answer:

Option B:

The normal force

Explanation:

The normal force does no work as the box slides down the ramp.

Work can only be done when the force succeeds in moving the object in the direction of the force.

All the other forces involved have a component that is moving the box in their direction.

However, the normal force does not, as it points downwards into the ramp. Since the normal force is pointing into the ramp, and the box is sliding down the ramp, we can say that no work is being done by the normal force because the box is not moving in its direction (which would have been the box moving into the ramp)

Answer:

-18896.49 V/m

Explanation:

Distance between the two plates = 10 cm = 10 x [tex]10^{-2}[/tex] m = 0.1 m

Also, one of the plates is taken as zero volt.

a. The potential strength between the zero volt plate, and 7.05 cm (0.0705 m) away is 393 V

b. The potential strength between the other plate, and 2.95 cm (0.0295 m) away is 393 V

Potential field strength = -dV/dx

where dV is voltage difference between these points,

dx is the difference in distance between these points

For the first case above,

potential field strength = -393/0.0705 = -5574.46 V/m

For the second case ,

potential field strength = -393/0.0295 = -13322.03 V/m

Magnitude of the field strength across the plates will be

-5574.46 + (-13322.03) = -5574.46 + 13322.03 = -18896.49 V/m

Answer:

The two value of the wavelength for the out of tune guitar is  

[tex]\lambda _2 = (6.48,6.52) \ cm[/tex]

Explanation:

From the question we are told that

     The wavelength of the note is [tex]\lambda = 6.50 \ cm = 0.065 \ m[/tex]

     The difference in beat frequency is [tex]\Delta f = 17.0 \ Hz[/tex]

Generally the frequency of the note played by the guitar that is in tune is  

        [tex]f_1 = \frac{v_s}{\lambda}[/tex]

Where [tex]v_s[/tex] is the speed of sound with a constant value [tex]v_s = 343 \ m/s[/tex]

       [tex]f_1 = \frac{343}{0.0065}[/tex]

      [tex]f_1 = 5276.9 \ Hz[/tex]

The difference in beat is mathematically represented as

       [tex]\Delta f = |f_1 - f_2|[/tex]

Where [tex]f_2[/tex] is the frequency of the sound from the out of tune guitar

     [tex]f_2 =f_1 \pm \Delta f[/tex]

substituting values

      [tex]f_2 =f_1 + \Delta f[/tex]

      [tex]f_2 = 5276.9 + 17.0[/tex]  

     [tex]f_2 = 5293.9 \ Hz[/tex]

The wavelength for this frequency is

      [tex]\lambda_2 = \frac{343 }{5293.9}[/tex]

     [tex]\lambda_2 = 0.0648 \ m[/tex]

    [tex]\lambda_2 = 6.48 \ cm[/tex]

For the second value of the second frequency

     [tex]f_2 = f_1 - \Delta f[/tex]

     [tex]f_2 = 5276.9 -17[/tex]

      [tex]f_2 = 5259.9 Hz[/tex]

The wavelength for this frequency is

   [tex]\lambda _2 = \frac{343}{5259.9}[/tex]

   [tex]\lambda _2 = 0.0652 \ m[/tex]

   [tex]\lambda _2 = 6.52 \ cm[/tex]

This question involves the concepts of beat frequency and wavelength.

The possible wavelengths of the out-of-tune guitar are "6.48 cm" and "6.52 cm".

The beat frequency is given by the following formula:

[tex]f_b=|f_1-f_2|\\\\[/tex]

f₂ = [tex]f_b[/tex] ± f₁

where,

f₂ = frequency of the out-of-tune guitar = ?

[tex]f_b[/tex] = beat frequency = 17 Hz

f₁ = frequency of in-tune guitar = [tex]\frac{speed\ of\ sound\ in\ air}{\lambda_1}=\frac{343\ m/s}{0.065\ m}=5276.9\ Hz[/tex]

Therefore,

f₂ = 5276.9 Hz ± 17 HZ

f₂ = 5293.9 Hz (OR) 5259.9 Hz

Now, calculating the possible wavelengths:

[tex]\lambda_2=\frac{speed\ of\ sound}{f_2}\\\\\lambda_2 = \frac{343\ m/s}{5293.9\ Hz}\ (OR)\ \frac{343\ m/s}{5259.9\ Hz}\\\\[/tex]

λ₂ = 6.48 cm (OR) 6.52 cm

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Answer:

0.15 m

Explanation:

First calculating the center of mass from the saw horse

[tex]\frac{3.15}{2} -1=0.575 m[/tex]

from the free body diagram we can write

Taking moment about the saw horse

55.9×9.81×y=14.5×0.575×9.81

y= 0.15 m

So, the painter walk from the saw horse on the right until the board begins to tip is 0.15 m far.

Answer:

a) correct answer is b higher , b) correct answer is b higher , c) correct answer is b faster , d)  traveling wave , e)

Explanation:

A traveling wave is described by the expression

            y = A sin (kx - wt)

where k is the wave vector and w is the angular velocity

let's examine every situation presented

a) a new faster pulse is generated

A faster pulse should have a higher angular velocity

equal speed is related to the period and frequency

            w = 2π f = 2π / T

therefore in this case the period must decrease so that the angular velocity increases

the correct answer is c narrower

b) Generate a pulse, but move your hand more.

Moving the hand increases the amplitude (A) of the pulse

the correct answer is b higher

c) generates a pulse but the force is tightened

Set means that more tension force is applied to the string, so the velicate changes

       v = √ (T /μ)

the correct answer is b faster

d) move your hand back and forth

in this case you would see a pulse series whose sum corresponds to a traveling wave

e) Advance a frame the movie

in this case the wave will be displaced a whole period to the right

the correct answer is b

f) move your hand faster

the waves will have a maximum fast, so they are closer

answer C

g) decrease wave frequency

Since the speed of the wave is a constant m ak, decreasing the frequency must increase the wavelength to keep the velocity constant.

the correct answer is b increases its wavelength

Answer:

0.48 cm

Explanation:

given data

wavelength = 480 nm

wavelength = 560 nm

slit spacing = 0.040 mm

distance between double slits and the screen  = 1.2 m

solution

we know that  (1 nm= [tex]10^{-9}[/tex] m)

we wil take here equation of equations of interference that is

ym = R × (m λ)/d    ..........................1

here m = 2 R  i.e distance of screen and slit

 so put here value and we get

separation between the second-order bright fringes = 0.48 cm

Answer:

the coefficient of volume expansion of the glass is [tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]

Explanation:

Given that:

Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³

temperature of the glass flask and mercury= 1.00° C

After heat is applied ; the final temperature = 52.00° C

Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C

Volume of the mercury overflow = 8.50 cm^3 = 8.50 ×  10⁻⁶ m³

the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K

The increase in the volume of the mercury =  10⁻³ m³ ×  51.00 × 1.80 × 10⁻⁴

The increase in the volume of the mercury = [tex]9.18*10^{-6} \ m^3[/tex]

Increase in volume of the glass =  10⁻³ × 51.00 × [tex]\beta _{glass}[/tex]

Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask

the mercury overflow = [tex](9.18*10^{-6} - 51.00* \beta_{glass}*10^{-3})\ m^3[/tex]

[tex]8.50*10^{-6} = (9.18*10^{-6} -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]

[tex]8.50*10^{-6} - 9.18*10^{-6} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]

[tex]-6.8*10^{-7} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]

[tex]6.8*10^{-7} = ( 51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]

[tex]\dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}= ( \beta_{glass} )[/tex]

[tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]

Thus; the coefficient of volume expansion of the glass is [tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]

Answer: [tex]\frac{P_B}{P_A}[/tex] = 8.5264

Explanation: Power is the rate of energy transferred per unit of time: P = [tex]\frac{E}{t}[/tex]

The energy from the engine is converted into kinetic energy, which is calculated as: [tex]KE = \frac{1}{2}.m.v^{2}[/tex]

To compare the power of the two cars, first find the Kinetic Energy each one has:

K.E. for Model A

[tex]KE_A = \frac{1}{2}.m.v^{2}[/tex]

K.E. for model B

[tex]KE_B = \frac{1}{2}.m.(2.92v)^{2}[/tex]

[tex]KE_B = \frac{1}{2}.m.8.5264v^{2}[/tex]

Now, determine Power for each model:

Power for model A

[tex]P_{A}[/tex] = [tex]\frac{m.v^{2} }{2.t}[/tex]

Power for model B

[tex]P_B = \frac{m.8.5264.v^{2} }{2.t}[/tex]

Comparing power of model B to power of model A:

[tex]\frac{P_B}{P_A} = \frac{m.8.5264.v^{2} }{2.t}.\frac{2.t}{m.v^{2} }[/tex]

[tex]\frac{P_B}{P_A} =[/tex] 8.5264

Comparing power for each model, power for model B is 8.5264 better than model A.

Answer:

More current will be loss through the metal wire strands if the force on them was repulsive, and more stress will be induced on the wire strands due to internal and external flexing.

Explanation:

A wire bundle is made up of wire strands bunched together to increase flexibility that is not always possible in a single solid metal wire conductor. In the strands of wire carrying a high voltage power, each strand carries a certain amount of current, and the current through the strands all travel in the same direction. It is know that for two conductors or wire, separated by a certain distance, that carries current flowing through them in the same direction, an attractive force is produced on these wires, one on the other. This effect is due to the magnetic induction of a current carrying conductor. The forces between these strands of the high voltage wire bundle, pulls the wire strands closer, creating more bond between these wire strands and reducing internal flex induced stresses.

If the case was the opposite, and the wires opposed themselves, the effect would be that a lot of cost will be expended in holding these wire strands together. Also, stress within the strands due to the repulsion, will couple with external stress from the flexing of the wire, resulting in the weakening of the material.

The biggest problem will be that more current will be lost in the wire due to increased surface area caused by the repulsive forces opening spaces between the strand. This loss is a s a result of the 'skin effect' in wire transmission, in which current tends to flow close to the surface of the metal wire. The skin effect generates power loss as heat through the exposed surface area.

Answer:

f = 2200 Hz

Explanation:

It is given that,

Speed of a wave is 242 m/s

The distance between crests is 0.11 m

We need to find the frequency of the wave. The distance between crests is called wavelength of a wave. So,

[tex]v=f\lambda\\\\f=\dfrac{v}{\lambda}\\\\f=\dfrac{242}{0.11}\\\\f=2200\ Hz[/tex]

So, the frequency of the wave is 2200 Hz.

Answer:2200 hz

Explanation:

Answer:

[tex]F = -1.107*10^{-8} N\\|F| = 1.107*10^{-8} N[/tex]

Explanation:

[tex]q_1 = 4.5 * 10^{-9} C[/tex]

[tex]q_2 = -2.8 * 10^{-9} C[/tex]

The distance separating the two charges, r = 3.2 m

According to Coulomb's law of electrostatic attraction, the electrostatic force between the two charges can be given by the formula:

[tex]F = \frac{kq_{1} q_{2} }{r^2}[/tex]

Where [tex]k = 9.0 * 10^9 Nm^2/C^2[/tex]

[tex]F = \frac{9*10^9 * 4.5*10^{-9} * (-2.8*10^{-9}}{3.2^2} \\F = \frac{-113.4*10^{-9}}{10.24}\\F = -11.07 *10^{-9}\\F = -1.107*10^{-8}N[/tex]

Answer:

F = 1.1074 × [tex]10^{-8}[/tex] N

Explanation:

An electrostatic force is either a force or attraction or repulsion between two charges. It can be determined by:

F = [tex]\frac{kq_{1}q_{2} }{r^{2} }[/tex]

where: F is the force, k is a constant, [tex]q_{1}[/tex] is the first charge, [tex]q_{2}[/tex] is the second charge and r the distance between the charges.

Given that: k = 9 × [tex]10^{9}[/tex] N[tex]m^{2}[/tex][tex]C^{-2}[/tex], [tex]q_{1}[/tex] = 4.5 × [tex]10^{-9}[/tex]C, [tex]q_{2}[/tex] = -2.8 × [tex]10^{-9}[/tex]C and r = 3.2 m.

Then,

F = [tex]\frac{9*10^{9}*4.5*10^{-9}*2.8*10^{-9} }{3.2^{2} }[/tex]

 = [tex]\frac{1.134*10^{-7} }{10.24}[/tex]

 = 1.1074 × [tex]10^{-8}[/tex]

The electrostatic force exerted is 1.1074 × [tex]10^{-8}[/tex] N, and it is a force of attraction.

Answer:

the acceleration [tex]a^{\to} = (0.0159 \ \ m/s^2 )i[/tex]

Explanation:

Given that:

the initial speed v₁ = 0 m/s i.e starting from rest ; since the car accelerates at a distance Δx = 6 miles in order to teach that final speed v₂ of 63.15 km/h.

So;  the acceleration for the first 6 miles can be calculated by using the formula:

v₂² = v₁² + 2a (Δx)

Making acceleration  a the subject of the formula in the above expression ; we have:

v₂² - v₁² = 2a (Δx)

[tex]a = \dfrac{v_2^2 - v_1^2 }{2 \Delta x}[/tex]

[tex]a = \dfrac{(63.15 \ km/s)^2 - (0 \ m/s)^2 }{2 (6 \ miles)}[/tex]

[tex]a = \dfrac{(17.54 \ m/s)^2 - (0 \ m/s)^2 }{2 (9.65*10^3 \ m)}[/tex]

[tex]a =0.0159 \ m/s^2[/tex]

Thus;

Assume the car moves in the +x direction;

the acceleration [tex]a^{\to} = (0.0159 \ \ m/s^2 )i[/tex]

Answer:

By convention a negative torque leads to clockwise rotation and a positive torque leads to counterclockwise rotation.

here weight of the child =21kgx9.8m/s2 = 205.8N

the torque exerted by the child Tc = - (1.8)(205.8) = -370.44N-m ,negative sign is inserted because this torque is clockwise and is therefore negative by convention.

torque exerted by adult Ta = 3(151) = 453N , counterclockwise torque.

net torque Tnet = -370.44+453 =82.56N , which is positive means counterclockwise rotation.

b) Ta = 2.5x151 = 377.5N-m

Tnet = -370.44+377.5 = 7.06N-m , positive ,counterclockwise rotation.

c)Ta = 2x151 = 302N-m

Tnet = -370.44+302 = -68.44N-m, negative,clockwise rotation.

Answer:

D Remains constant

Explanation:

Answer:

(a) k = 1684.38 N/m = 1.684 KN/m

(b) Vi = 0.105 m/s

(c) F = 1010.62 N = 1.01 KN

Explanation:

(a)

First, we find the deceleration of the car. For that purpose we use 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = ?

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = 0.35 m/s

s = distance covered by train before stopping = 2 m

Therefore,

a = [(0 m/s)² - (0.35 m/s)²]/(2)(2 m)

a = 0.0306 m/s²

Now, we calculate the force applied on spring by train:

F = ma

F = (1.1 x 10⁵ kg)(0.0306 m/s²)

F = 3368.75 N

Now, for force constant, we use Hooke's Law:

F = kΔx

where,

k = Force Constant = ?

Δx = Compression = 2 m

Therefore.

3368.75 N = k(2 m)

k = (3368.75 N)/(2 m)

k = 1684.38 N/m = 1.684 KN/m

(c)

Applying Hooke's Law with:

Δx  = 0.6 m

F = (1684.38 N/m)(0.6 m)

F = 1010.62 N = 1.01 KN

(b)

Now, the acceleration required for this force is:

F = ma

1010.62 N = (1.1 kg)a

a = 1010.62 N/1.1 x 10⁵ kg

a = 0.0092 m/s²

Now, we find initial velocity of train by using 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = -0.0092 m/s² (negative sign due to deceleration)

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = ?

s = distance covered by train before stopping = 0.6 m

Therefore,

-0.0092 m/s² = [(0 m/s)² - Vi²]/(2)(0.6 m)

Vi = √(0.0092 m/s²)(1.2 m)

Vi = 0.105 m/s

Answer:

[tex]\mathbf{1.51\times10^{-15}N}[/tex]

Explanation:

The computation of the weight of the paper in newtons is shown below:

On the paper, the induced charge is of the same magnitude as on the initial charges and in sign opposite.

Therefore the paper charge is

[tex]q_{paper}=-4.1\times10^{-15}C[/tex]

Now the distance from the charge is

[tex]r=1cm=0.01m[/tex]

Now, to raise the paper, the weight of the paper acting downwards needs to be managed by the electrostatic force of attraction between both the paper and the charge, i.e.

[tex]mg=\frac{k_{e}q_{1}q_{2}}{r^{2}}[/tex]

[tex]\Rightarrow W=mg[/tex]

[tex]=\frac{9\times10^{9}\times(4.1\times10^{-15})^{2}}{0.01^{2}}[/tex]

[tex]=\mathbf{1.51\times10^{-15}N}[/tex]

Answer:

If instead a charge 2q were placed at that same point the force will be 2F.

Explanation:

The electric force is equal to:

[tex] F = q*E [/tex]    (1)

Where:

F: is the electric force

q: is the charge

E: is the electric field

We can see that in equation (1) the electric force (F) is proportional to the charge q, thus, if now the charge it's the double (2q) then the force will be the double too:    

Initially:

[tex] F_{1} = q_{1}*E [/tex]

Now,

[tex](2q_{1}})*E = 2(q_{1}*E) = 2F_{1}[/tex]  

Therefore, if instead a charge 2q were placed at that same point the force will be 2F.

I hope it helps you!