A magnifying glass produces a maximum angular magnification of 5.4 when used by a young person with a near point of 20 cm. What is the maximum angular magnification obtained by an older person with a near point of 65 cm

Answer:

49 N

Explanation:

Given,

Mass ( m ) = 50 kg

To find : Weight ( W ) = ?

Take the value of acceleration due to gravity as 9.8 m/s^2

Formula : -

W = mg

W = 50 x 9.8

W = 49 N

I guess that we want to find how much money you get each week.

We know that the job pays $8.60 per hour.

We know that you work 20 hours per week.

Then the gross pay (the total money that you earn) in a week is 20 times $8.60, or:

20*$8.60 = $172.

Now we know that your employer witholds:

10% + 7.65% + 5% = 22.65%

Then your employer withholds 22.65% of your gross pay.

if the 100% of your gross pay is $172

Then the 22.65% will be:

(22.65%/100%)*$172 = 0.2265*$172 = $38.96

This means that your employer withholds $38.96 of your weekly gross pay.

Then each week you get:

$172 - $38.96 = $133.04

If you want to learn more, you can read:

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Answer:

Force A newton Law first law

F = M.A which Force in 0 N as you Questions Above

Force B

Newton Law 3

Action = -Reaction

Hope you can explain this formula as you want to scribe to explaining

Answer:

the required solution is; x(t) = 0.675sin( 2.222t )

Explanation:

Given the data in the question;

Using both Newton's and Hooke's law;

m[tex]x^{ff[/tex] + k[tex]x[/tex] = 0, [tex]x[/tex](0) = 0, [tex]x^f[/tex](0) = 1.5

given that mass m = 9 kg

[tex]x[/tex] = 1.8 m

k is F / x

hence

k = F / x

given that, F = 80 N

we substitute

k = 80 / 1.8

k = 44.44

so

m[tex]x^{ff[/tex] + k[tex]x[/tex] = 0,

we input

9[tex]x^{ff[/tex] + 44.44[tex]x[/tex] = 0,

[tex]x^{ff[/tex] + 4.9377[tex]x[/tex] = 0

so auxiliary equation is,

r² + 4.9377 = 0

r² = -4.9377

r = √-4.9377

r = ±2.222i

hence, the solution will  be;

x(t) = A×cos( 2.222t ) + B×sin( 2.222t )

⇒ [tex]x^t[/tex](t) = -2.222Asin( 2.222t ) + 2.222Bcos( 2.222t )

using initial conditions

x(0) = 0

⇒ 0 = A

[tex]x^t[/tex](t) = 1.5

1.5 = 2.222B

so

B = 1.5 / 2.222 = 0.675

Hence, the required solution is; x(t) = 0.675sin( 2.222t )

Answer:

The ball moves from lowest to highest point:

W = M g h = 3 * 9.8 * 4 = 118 J

This is work done "against" gravity so work done by gravity is -118 J

The tension of the string does no work because the tension does not

move thru any distance   W = T * x = 0 because the length of the string is fixed.

Observer Frequency = sound frequency x ( speed of sound / speed of sound - speed of car)

= 250 x (340/( 340-15))

= 261.54 Hz

Answer:

[tex]T=3*10^-3 N/m[/tex]

Explanation:

From the question we are told that:

Radius :

[tex]R_1=4=>0.04\\\\R_2=6=>0.06[/tex]

Work [tex]W=1.5 * 10^{-4}[/tex]

Generally the equation for Work done  is mathematically given by

[tex]W=TdA[/tex]

Where

[tex]dA=A_2-A_1\\\\dA=(2 \pi r_2^2)(2 \pi r_1^2)[/tex]

[tex]dA=8 \pi*(r_2^2-r_1^2)\\\\dA=8*3.142*(0.06^2-0.04^2)[/tex]

[tex]dA=0.050m^2[/tex]

Therefore

[tex]W=TdA[/tex]

[tex]T=\frac{1.5 * 10^{-4}}{0.05m^2}[/tex]

[tex]T=3*10^-3 N/m[/tex]

Answer:

d = 68.5 x 10⁻⁶ m = 68.5 μm

Explanation:

The complete question is as follows:

An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is  1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?

The answer can be given by using the formula derived from Young's Double Slit Experiment:

[tex]y = \frac{\lambda L}{d}\\\\d =\frac{\lambda L}{y}\\\\[/tex]

where,

d = slit separation = ?

λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m

L = distance from screen (detector) = 1.7 m

y = distance between bright fringes = 15.7 mm = 0.0157 m

Therefore,

[tex]d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\[/tex]

d = 68.5 x 10⁻⁶ m = 68.5 μm

Answer:

answer : option (b) 0.016 amp

explanation : resistance of resistor , R = 10 Ω

inductance of inductor , X_LX

L

= 20H

voltage of AC circuit , V = 120volts

frequency, ff =60Hz

so, angular frequency, \omega=2\pi fω=2πf = 2 × π × 60 = 120π rad/s

now, current , i=\frac{V}{\sqrt{R^2+\omega^2L^2}}i=

R

2

+ω

2

L

2

V

= 120/√{10² + (120π)² × 20²}

= 120/√{100 + 14400π² × 400}

after solving this we get, i = 0.016 amp

Answer:

kinetic and static

Explanation:

hope it helps! ^w^

Answer:

Mass, M = 1000 kg

Speed, v = 90 km/h = 25 m/s

time, t = 6 sec.

Distance:

[tex]{ \tt{distance = speed \times time }} \\ { \tt{distance = 25 \times 6}} \\ { \tt{distance = 150 \: m}}[/tex]

Force:

[tex]{ \tt{force = mass \times acceleration}} \\ { \bf{but \: for \: acceleration : }} \\ from \: second \: equation \: of \: motion : \\ { \bf{s = ut + \frac{1}{2} {at}^{2} }} \\ \\ { \tt{150 = (0 \times 6) + ( \frac{1}{2} \times a \times {6}^{2} ) }} \\ \\ { \tt{acceleration = 8.33 \: {ms}^{ - 2} }} \\ \\ { \tt{force = 1000 \times 8.33}} \\ { \tt{force = 8333.3 \: newtons}}[/tex]

Answer:

thus the coulomb force is F – 8.19x10-8N. this is also an attractive force, although it is traditionally shown as positive since gravitational force is always attractive. the ratio of the magnitude of the electrostatic force to gravitational force in this case is,thus,FFG – 2.27x1039 F F G – 2.27x 10 39.

Answer:

C

Explanation:

Similarity

Answer:

ωf = 8.8 rad/s

v = 2.2 m/s

Explanation:

We will use the third equation of motion to find the maximum angular velocity of the wheel:

[tex]2\alpha \theta = \omega_f^2 -\omega_I^2[/tex]

where,

α = angular acceleration = 6 rad/s²

θ = angular displacemnt = 1 rev = 2π rad

ωf = max. final angular velocity = ?

ωi = initial angular velocity = 1.5 rad/s

Therefore,

[tex]2(6\ rad/s^2)(2\pi\ rad)=\omega_f^2-(1.5\ rad/s)^2\\\omega_f^2=75.4\ rad/s^2+2.25\ rad/s^2\\\omega_f = \sqrt{77.65\ rad/s^2}[/tex]

ωf = 8.8 rad/s

Now, for linear velocity:

v = rω = (0.25 m)(8.8 rad/s)

v = 2.2 m/s

Answer:

If the person is to remain the floor the reaction force will be equal to the normal force exerted by the floor.

F(normal) - F(reaction) = 0

That means the person is not moving with respect to the elevator.

Expanding the applied forces we have:

Fw - Fn = 720 - 710 = 10 N   where the positive direction is chosen as down

Fw is the weight of the person and Fn the force exerted on the person by the elevator,

The acceleration of the person the becomes F = m a = m * 10 N and will be downward agreeing with our choice of coordinate axes.

Answer:

Train wheels and rails are both made of steel, and the steel-steel friction coefficient is around 0.25. As a result, the stopping time and distance will be three to four times that of a car.

Answer:

The first person will gain more velocity as a result of recoil.

Explanation:

Let us recall that from Newton's third law of motion, action and reaction are equation and opposite. A consequence of this law is the proposition that ''momentum can neither be created nor destroyed.''

Hence, when two people who have the same mass, throw two different objects at the same velocity but the first object is heavier than the second, the first object possesses greater momentum than the second object hence the first person will gain more velocity as a result of recoil.

Answer:

a) The energy efficiency of the out-of-condition professor is 9.082 %.

b) The food calories needed by the well-conditioned athlete is 181.644 kilocalories.

Explanation:

a) The energy efficiency of the food metabolization ([tex]\eta[/tex]), no unit, is defined by following formula:

[tex]\eta = \frac{W}{E}\times 100\,\%[/tex] (1)

Where:

[tex]W[/tex] - Useful work, in joules.

[tex]E[/tex] - Food energy, in joules.

If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]E = 2.092\times 10^{6}\,J[/tex], the energy efficiency of the food metabolization is:

[tex]\eta = \frac{1.90\times 10^{5}\,J}{2.092\times 10^{6}\,J} \times 100\,\%[/tex]

[tex]\eta = 9.082\,\%[/tex]

The energy efficiency of the out-of-condition professor is 9.082 %.

b) If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]\eta = 25\,\%[/tex], then the quantity of food energy is:

[tex]E = \frac{W}{\eta}\times 100\,\%[/tex]

[tex]E = 1.90\times 10^{5}\,J\times \frac{100\,\%}{25\,\%}[/tex]

[tex]E = 7.60\times 10^{5}\,J[/tex]

[tex]E = 181.644\,kcal[/tex]

The food calories needed by the well-conditioned athlete is 181.644 kilocalories.

Answer:

The initial angular speed of the CD is equal to 14.73 rad/s.

Explanation:

Given that,

Angular displacement, [tex]\theta=12\ rad[/tex]

Final angular speed, [tex]\omega_f=17\ rad/s[/tex]

The acceleration of the CD,[tex]\alpha =3\ rad/s^2[/tex]

We need to find the initial angular speed of the CD. Using third equation of kinematics to find it such that,

[tex]\omega_f^2=\omega_i^2+2\alpha \theta\\\\\omega_i^2=\omega_f^2-2\alpha \theta[/tex]

Put all the values,

[tex]\omega_i^2=(17)^2-2\times 3\times 12\\\\\omega_i=\sqrt{217}\\\\\omega_i=14.73\ rad/s[/tex]

So, the initial angular speed of the CD is equal to 14.73 rad/s.

Answer:

13.3 pounds.

Explanation:

For a spring of constant K, with an attached object of mass M, the period can be written as:

T = 2*π*√(M/K)

Where π = 3.14

First, we know that the period is 4 seconds, then we have:

4s = (2*π)*√(M/K)

We know that if the mass is reduced by 10lb, the period becomes 2s.

Then the new mass of the object will be: (M - 10lb)

Then the period equation becomes:

2s = (2*π)*√((M-10lb)/K)

So we have two equations:

4s = (2*π)*√(M/K)

2s = (2*π)*√((M-10lb)/K)

We want to solve this for M.

First, we need to isolate K in one of the equations.

Let's isolate K in the first one:

4s = (2*π)*√(M/K)

(4s/2*π) = √(M/K)

(2s/π)^2 = M/K

K = M/(2s/π)^2 = M*(π/2s)^2

Now we can replace it in the other equation.

2s = (2*π)*√((M-10lb)/K)

First, let's simplify the equation:

2s/(2*π) = √((M-10lb)/K)

1s/π =  √((M-10lb)/K)

(1s/π)^2 =  ((M-10lb)/K

K*(1s/π)^2 = M - 10lb

Now we can use the equation: K =  M*(π/2s)^2

then we get:

K*(1s/π)^2 = M - 10lb

(M*(π/2s)^2)*(1s/π)^2 = M - 10lb

M/4 = M - 10lb

10lb = M - M/4

10lb = (3/4)*M

10lb*(4/3) = M

13.3 lb = M

Answer:

V = k Q / R       potential at distance R for a charge Q

R = k Q / V

R = 9 * 10E9 * 15 * 10E-9 / 30 = 9 * 15 / 30 = 4.5 m

Note: Our equation says that if R if infinite then V must be zero.

Answer:

True

Explanation:

This is because of the point where the forces are applied by our muscles and

the angle they have about the bones. Take for example the  diagram I uploaded.

If we do a free body diagram and a sum of torques, we would get that:

[tex]F_{muscle}sin \theta r1 - mg r2 = 0[/tex]

In this case, mg is the same in magnitude as the force made by the hand to hold the ball, so:

[tex]F_{muscle}sin \theta r_{1} - F_{hand} r_{2} = 0[/tex]

If we solve the equation for the force of the muscle we would get that:

[tex]F_{muscle}=\frac{F_{hand}r_{2}}{r_{1}sin \theta}[/tex]

Since r2 is greater than r1 and the sin function can only return values that are less than 1, this means that the force of the muscle is much greater than the force used by the hand to hold the weight.

Let's use some standard values to prove this, let's say that r1=10cm, r2=35cm and theta=60 degrees. When inputing the values into the equation we get:

[tex]F_{muscle}=\frac{F_{hand}(35cm)}{(10cm)sin (60^{o})}[/tex]

which yields:

[tex]F_{muscle}=4.04 F_{hand}[/tex]

so in this example, the force made by the muscle is 4 times as big as the force exerted by the hand.

Answer:

Explanation:

When a falcon is hovering, the force of up thrust is balanced by the weight.

When it begins to fall towards the ground, the weight acts downwards, kinetic friction is upwards, drag is upwards, normal force is upwards, thrust is upwards.

Answer:

The correct solution is "122.2211".

Explanation:

Given:

deceleration,

a = 22 ft/sec²

Initial velocity,

[tex]V_i=50 \ m/h[/tex]

Now,

[tex]V_i=50 \ m/h\times 5280 \ ft/m\times hr/3600 \ s[/tex]

    [tex]=73.333 \ ft/sec[/tex]

Now,

Final velocity,

[tex]V_f=0[/tex]

Initial velocity,

[tex]V_{initial} = 73.333 \ ft/sec[/tex]

hence,

⇒ [tex]V_f^2=V_i^2+2aD[/tex]

By putting the values, we get

      [tex]0=(73.333)^2+2\times( -22) D[/tex]

  [tex]44D=(73.333)^2[/tex]

      [tex]D=\frac{(73.333)^2}{44}[/tex]

          [tex]=122.2211[/tex]

Answer:

c). Easier setup

Explanation:

As per the question, 'easier setup' cannot be characterized as the advantage of using Access because it comprises of plenty of steps that must be followed in the sequential order to establishing a database or carrying transactions based on time. However, there are plenty of advantages of using Microsoft access like 'enhanced and increased storage of data,' 'hassle free database systems,' 'easy importing of data,' 'highly economical,' 'capability to allow multiple users,' 'extra features for reporting,' and much more. Hence, option c is the correct answer.

[tex]4.4×10^{-7}\:\text{m}[/tex]

Explanation:

The minimum energy needed to kick out an electron from a metal's surface is when the energy of the incident radiation is equal to the metal's work function [tex]\phi[/tex]:

[tex]E = h\nu - \phi = \dfrac{hc}{\lambda} - \phi = 0[/tex]

or

[tex]\dfrac{hc}{\lambda} = \phi[/tex]

Solving for the wavelength [tex]\lambda[/tex],

[tex]\lambda = \dfrac{hc}{\phi}[/tex]

[tex]\:\:\:\:\:=\dfrac{(6.62×10^{-34}\:\text{J-s})(3.0×10^8\:\text{m/s})}{4.5×10^{-19}\:\text{J}}[/tex]

[tex]\:\:\:\:\:= 4.4×10^{-7}\:\text{m}[/tex]

Note that as the radiation's wavelength increases, its energy decreases. So a radiation whose wavelength is longer than this maximum will lose its ability to kick out an electron from this metal.

The maximum wavelength, in nm, of light that can eject an electron from the metal, given the data is 441.73 nm.

To find the wavelength, the given values are,

Energy (E) = 4.50×10¯¹⁹ J

The distance between two consecutive crests and troughs is called the wavelength of a wave.

Here, for the wavelength,

Energy (E) = 4.50×10¯¹⁹ J

Planck's constant (h) = 6.626×10¯³⁴ Js

Speed of light (v) = 3×10⁸ m/s

The wavelength of the light can be obtained as illustrated below:

E = hv / λ

Cross multiply λ,

E × λ = hv

Divide both sides by E,

λ = hv / E

Substituting all the values,

λ = (6.626×10¯³⁴ × 3×10⁸) / 4.50×10¯¹⁹

λ = 0.000000441733 m

λ = 441.73nm

λ - The maximum wavelength of light.

Thus, the wavelength of the light that can eject an electron from the metal is 441.73 nm

Learn more about wavelength,

https://brainly.com/question/13047641

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Complete Question

The barometer of a mountain hiker reads 980 mbars at the beginning of a hiking trip and 790 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. Assume an average air density of 1.20kg/m^2

Answer:

[tex]h=1614m[/tex]

Explanation:

From the question we are told that:

Initial Pressure [tex]P_1=980mbar=>98000Pa[/tex]

Final Pressure [tex]P_2=790mbar=>79000Pa[/tex]

Density [tex]\rho=1.20kg/m^2[/tex]

Generally the equation for Height climbed is mathematically given by

[tex]h=\frac{P_1-P_2}{\rho*g}[/tex]

[tex]h=\frac{P_1-P_2}{1.20*9.81}[/tex]

[tex]h=1614m[/tex]

Answer:

A) C_{eq} = 15 10⁻⁶  F,  B)   U₃ = 3 J,  U₄ = 0.5 J

Explanation:

In a complicated circuit, the method of solving them is to work the circuit in pairs, finding the equivalent capacitance to reduce the circuit to simpler forms.

In this case let's start by finding the equivalent capacitance.

A) Let's solve the part where C1 and C3 are. These two capacitors are in serious

         [tex]\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_3}[/tex]            (you has an mistake in the formula)

         [tex]\frac{1}{C_{eq1}} = (\frac{1}{30} + \frac{1}{15}) \ 10^{6}[/tex]

         [tex]\frac{1}{C_{eq1}}[/tex] = 0.1   10⁶

         [tex]C_{eq1}[/tex] = 10 10⁻⁶ F

capacitors C₂, C₄ and C₅ are in series

          [tex]\frac{1}{C_{eq2}} = \frac{1}{C_2} + \frac{1}{C_4} + \frac{1}{C_5}[/tex]

          [tex]\frac{1}{C_{eq2} } = (\frac{1}{15} + \frac{1}{30} + \frac{1}{10} ) \ 10^6[/tex]

          [tex]\frac{1}{C_{eq2} }[/tex] = 0.2 10⁶

          [tex]C_{eq2}[/tex] = 5 10⁻⁶ F

the two equivalent capacitors are in parallel therefore

          C_{eq} = C_{eq1} + C_{eq2}

          C_{eq} = (10 + 5) 10⁻⁶

          C_{eq} = 15 10⁻⁶  F

B) the energy stored in C₃

The charge on the parallel voltage is constant

is the sum of the charge on each branch

         Q = C_{eq} V

         Q = 15 10⁻⁶ 6

         Q = 90 10⁻⁶ C

the charge on each branch is

         Q₁ = Ceq1 V

         Q₁ = 10 10⁻⁶ 6

          Q₁ = 60 10⁻⁶ C

         Q₂ = C_{eq2} V

         Q₂ = 5 10⁻⁶ 6

         Q₂ = 30 10⁻⁶ C

now let's analyze the load on each branch

Branch C₁ and C₃

In series combination the charge is constant    Q = Q₁ = Q₃

          U₃ = [tex]\frac{Q^2}{2 C_3}[/tex]

          U₃ =[tex]\frac{ 60 \ 10^{-6}}{2 \ 10 \ 10^{-6}}[/tex]

          U₃ = 3 J

In Branch C₂, C₄, C₅

since the capacitors are in series the charge is constant Q = Q₂ = Q₄ = Q₅

          U₄ = [tex]\frac{30 \ 10^{-6}}{ 2 \ 30 \ 10^{-6}}[/tex]

          U₄ = 0.5 J