A professor of physics cannot see clearly at a distance shorter than 1 m. What should be the focal length of an eyeglass lens that would assist him in reading a newspaper while holding it at a desired distance of 25 cm

Answer:

[tex]M=7.05*10^{-4}[/tex]

Explanation:

From the question we are told that:

Coil one turns N_1=1550 Turns/m

Radius [tex]r=0.0240m[/tex]

Turns 2 [tex]N_2=200N[/tex]

Generally the equation for area is mathematically given by

[tex]A=\pi*r^2[/tex]

[tex]A=\pi*0.024^2[/tex]

[tex]A=\1.81*10^{-3} m^2[/tex]

Therefore

The mutual inductance of this system is

[tex]M=\mu*N_1*N_2*A[/tex]

[tex]M=(4 \pi*10^{-7})*1550*200*1.81*10^{-3}[/tex]

[tex]M=7.05*10^{-4}[/tex]

Answer:

1) ans: The ratio of load to effort in a simple machine is called Mechanical Advantage.

2) ans: Frictuon produced in Simple machine affect the mechanical advantage of a machine.

3) ans: The machine whose efficiency is 100% is called ideal machine.

4) ans: The work done by the machine is called output work.

ans: The work done in the machine is called input work.

5) ans: The turning effect of force is called moment.

Answer:

The fraction of kinetic energy to the total energy is [tex]\frac{K}{T}=\frac{8}{9}[/tex].

Explanation:

displacement is one third of the amplitude.

Let the amplitude is A.

x= A/3

The kinetic energy of the body executing SHM is

[tex]K = 0.5 mw^2(A^2 - x^2)\\\\K = 0.5 m w^2 \left ( A^2 -\frac{A^2}{9} \right )\\\\K = 0.5 mw^2\times \frac{8A^2}{9}......(1)[/tex]

The total energy is

[tex]T =0.5 mw^2A^2 ..... (2)[/tex]

Divide (1) by (2)

[tex]\frac{K}{T}=\frac{8}{9}[/tex]

Answer:

75

Explanation:

i am not sure but if 200N boy is being held back then the force that's holding him back must be equal to or greater than his weight. if 125N is already exerted then the tension will be:

T=200-125

= 75

Answer:

a)  [tex]F=475.7Hz[/tex]

b)  [tex]F'=410.899Hz[/tex]

Explanation:

From the question we are told that:

Velocity of eagle [tex]V_1=35m/s[/tex]

Frequency of eagle [tex]F_1=440Hz[/tex]

Velocity of Black bird [tex]V_2=10m/s[/tex]

Speed of sound [tex]s=343m/s[/tex]

a)

Generally the equation for Frequency is mathematically given by

 [tex]F=f_0(\frac{v-v_2}{v-v_1})[/tex]

 [tex]F=440(\frac{343-10}{343-35})[/tex]

 [tex]F=475.7Hz[/tex]

b)

Generally the equation for Frequency is mathematically given by

 [tex]F'=f_0(\frac{v+v_2}{v+v_1})[/tex]

 [tex]F'=440(\frac{343+10}{343+35})[/tex]

 [tex]F'=410.899Hz[/tex]

Answer:

[tex]r=200m[/tex]

Explanation:

From the question we are told that:

Charges:

[tex]Q_1=8.0mC[/tex]

[tex]Q_2=2.0mC[/tex]

[tex]Q_3=8.mC[/tex]

Distance [tex]d=300m[/tex]

Generally the equation for Force is mathematically given by

[tex]F=\frac{kq_1q_2}{r^2}[/tex]

Therefore

[tex]F_{32}=F_{31}[/tex]

[tex]\frac{q_2}{(300-r)^2}=\frac{q_1}{r^2}[/tex]

[tex]\frac{2*10^{-3}}{(300-r)^2}=\frac{8*10^{-3}}{r^2}[/tex]

[tex]r=2(300-r)[/tex]

[tex]r=200m[/tex]

Answer:

1.671L

Explanation:

In the given case the pressure is constant therefore it is an isobaric process, the process in which the pressure remains constant is called as isobaric process.

Work done= external pressure× change in volume.

288= 2×( final volume-intial volume)×101.32

144÷101.32=(finalvolume-0.250)

1.421=final volume-0.250

final volume=1.671L

Answer:

F = 118 N

Explanation:

Assume Ann and Bob lift at their respective ends of the table

Sum moments about Bob's position to zero.

Let F be Ann's upward force

F[2.25] - 18.5(9.80)[2.25 / 2] - 8.33(9.80)[0.750] = 0

F = 117.86133333...  

Given:

The weight of table will be:

→ [tex]W_t = m_t g[/tex]

        [tex]= 18.5\times 9.8[/tex]

        [tex]= 181.3 \ N[/tex]

Now,

→ [tex]\sum M_A_{nn} = -81.634\times 0.750-181.3\times 1.125+R_{bob}\times 1.125[/tex]

or,

→ [tex]R_{bob} = \frac{61.2255+203.9625}{2.25}[/tex]

          [tex]= \frac{265.188}{2.25}[/tex]

          [tex]= 117.9 \ N[/tex]

Thus the above answer is right.

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Answer:

(a) v = 32 t

h = 16 t^2

g = 32 ft/s^2

(b) 64 ft/s

Explanation:

height, h = 64 feet

g = - 32 ft/s^2

(a) Let the time is t .

Let the velocity after time t is v.

Use first equation of motion

v = u + at

- v = 0 - 32 t

v = 32 t

Let the distance is h from the top.

Use second equation of motion

[tex]h = u t + 0.5 at^2 \\\\- h = 0 - 0.5\times 32 \times t^2\\\\h = 16 t^2[/tex]

The acceleration is constant for entire motion.

(b) Let the velocity is v as it hits the ground. Use third equation of motion

[tex]v^2 = u^2 + 2 a s \\\\v^2 = 0 + 2 \times 32\times 64\\\\v = 64 feet/s[/tex]

Answer:

[tex]L=9.76*10^{16}m[/tex]

Explanation:

From the question we are told that:

Time on earth [tex]T_e= 13yrs[/tex]

Time on ship [tex]T_s= 7.9yrs[/tex]

Therefore

[tex]r=\frac{t_e}{t_s}[/tex]

[tex]r=\frac{13}{7.9}[/tex]

[tex]r=1.65[/tex]

Generally the equation for Constant Velocity is mathematically given by

[tex]V=C\sqrt{1-\frac{1}{r^2}}[/tex]

[tex]V=3*10^8\sqrt{1-\frac{1}{1.64^2}}[/tex]

[tex]V=2.38*10^8m/s[/tex]

Therefore

[tex]L=V*t[/tex]

Where

[tex]t=(13*365.25*24*3600)s[/tex]

[tex]t=4.1*10^8[/tex]

[tex]L=2.38*10^8m/s*4.1*10^8[/tex]

[tex]L=9.76*10^{16}m[/tex]

Answer:

b. 0.20 m/s.

Explanation:

Given;

initial mass, m = 0.2 kg

maximum speed,  v = 0.3 m/s

The total energy of the spring at the given maximum speed is calculated as;

K.E = ¹/₂mv²

K.E = 0.5 x 0.2 x 0.3²

K.E = 0.009 J

If the mass is changed to 0.4 kg

¹/₂mv² = K.E

mv² = 2K.E

[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2\times 0.009}{0.4} } \\\\v = 0.21 \ m/s\\\\v \approx 0.20 \ m/s[/tex]

Therefore, the maximum speed is 0.20 m/s

Answer:

The correct answer is "2205.72 N".

Explanation:

Given:

m₁ = 161.5 kg

m₂ = 72.3 kg

By taking moment about point A, we get

⇒ [tex]m_1 g Cos \theta.\frac{l}{2 Cos \theta} + m g Cos \theta.\frac{l}{ Cos \theta} -T Cos(90-2 \theta).\frac{l}{2 Cos \theta} = 0[/tex]

By substituting the values, we get

⇒ [tex]\frac{161.5\times 9.8\times 13}{2}+72.3\times 9.8\times 13-T.2 Sin \theta.l =0[/tex]

⇒ [tex]10287.55+9211.02-T 2\times 0.34\times 13=0[/tex]

⇒                                            [tex]T\times 8.84=19498.57[/tex]

⇒                                                       [tex]T= 2205.72 N[/tex]

Answer:

Time to move out of the way = 1.74 s

Explanation:

Time to move out of the way is one fourth of period = 6.95/4 = 1.74 seconds.

Time to move out of the way = 1.74 s

Answer:

[tex]\omega_1=150rads/sec[/tex]

Explanation:

From the question we are told that:

Number of Revolution [tex]N=15=30\pi[/tex]

Deceleration [tex]d= -120 rads/2[/tex]

Generally the equation for  initial angular speed [tex]\omega_1[/tex] is mathematically given by

 [tex]\omega_2^2=\omega_1^2 +2(d)(N)[/tex]

 [tex]0=\omega_1^2 +2(-120)(20 \pi)[/tex]

 [tex]\omega_1^2=7200 \pi[/tex]

 [tex]\omega_1=150rads/sec[/tex]

Answer:

the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m

Explanation:

Given the data in the question;

Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J

Mass of proton = 1.673 × 10⁻²⁷ kg

Charge of proton = 1.602 × 10⁻¹⁹ C

distance d = 2 m

we know that

Kinetic Energy = Charge of proton × Potential difference ΔV

so

Potential difference ΔV = Kinetic Energy / Charge of proton

we substitute

Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )

Potential difference ΔV = 20287.14 V

Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;

E = Potential difference ΔV / distance d

we substitute

E = 20287.14 V / 2 m

E = 10143.57 V/m or 1.01 × 10⁴ V/m

Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m

• Point charge (Q) = 10 μC = 10 × 10⁻⁶ C

• Potential (V) = 1000 V

• Distance (r) = ?

[tex]\implies V = \dfrac{KQ}{r} \\ [/tex]

[tex]\implies r= \dfrac{KQ}{V}[/tex]

[tex]\implies r= \dfrac{9 \times {10}^{9} \times 10 \times {10}^{ - 6} }{1000} \\ [/tex]

[tex]\implies r= \dfrac{90 \times {10}^{3} }{1000} \\ [/tex]

[tex]\implies r= \dfrac{90 \times {10}^{3} }{ {10}^{3} } \\ [/tex]

[tex]\implies r= 90 \times {10}^{3} \times {10}^{ - 3} \\ [/tex]

[tex]\implies\bf r= 90\:m \\ [/tex]

Hence,the option B) 90 m is the correct answer.

Answer:

The final speed of electron=[tex]5.93\times 10^6m/s[/tex]

Explanation:

We are given that

Initial velocity, u=0

Potential, V=100 V

We have to find the final speed.

Mass of electron, [tex]m=9.1\times 10^{-31} kg[/tex]

Charge on electron, q=[tex]1.6\times 10^{-19}C[/tex]

We know that

[tex]qV=\frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]

Using the formula

[tex]1.6\times 10^{-19}\times 100=\frac{1}{2}\times 9.1\times 10^{-31} v^2-0[/tex]

[tex]v^2=\frac{2\times 1.6\times 10^{-19}\times 100}{9.1\times 10^{-31}}[/tex]

[tex]v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 100}{9.1\times 10^{-31}}}[/tex]

[tex]v=5.93\times 10^6m/s[/tex]

Hence, the final speed of electron=[tex]5.93\times 10^6m/s[/tex]

Answer:

Explanation:

Let the mass of objects be m₁ and m₂ .

Total kinetic energy = 1/2 m₁ v² + 1/2 m₂ v²= 1/2 ( m₁ + m₂ ) v²

Total kinetic energy after collision= 1/2 ( m₁ + m₂ ) v² / 4  =  1/2 ( m₁ + m₂ ) v² x .25

final KE / initial KE = 1/2 ( m₁ + m₂ ) v² x .25 / 1/2 ( m₁ + m₂ ) v²

= 0.25

b )

Applying law of conservation of momentum to the system . Let m₁ > m₂

m₁ v -  m₂ v = ( m₁ + m₂ ) v / 2

m₁ v -  m₂ v = ( m₁ + m₂ ) v / 2

m₁  -  m₂  = ( m₁ + m₂ )  / 2

2m₁ - 2 m₂ = m₁ + m₂

m₁ = 3m₂

m₁ / m₂ = 3 / 1

Answer:

(a) The ratio is 1 : 4.

(b) The ratio is 1 : 3.

Explanation:

Let the mass of each object is m and m'.

They initially move with velocity v opposite to each other.

Use conservation of momentum

m v - m' v = (m + m') v/2

2 (m - m')  = (m + m')

2 m - 2 m' = m + m'

m = 3 m' .... (1)

(a) Let the initial kinetic energy is K and the final kinetic energy is K'.

[tex]K = 0.5 mv^2 + 0.5 m' v^2 \\\\K = 0.5 (m + m') v^2..... (1)[/tex]

[tex]K' = 0.5 (m + m') \frac{v^2}{4}.... (2)[/tex]

The ratio is

K' : K = 1 : 4

(b) m = 3 m'

So, m : m' = 3 : 1

Answer:

40 m

Explanation:

Applying,

v² = u²+2as............... Equation 1

Where v = final velocity, u = initial velocity, a = deceleration, s = distance

make s the subject of the equation

s = (v²-u²)/2a............ Equation 2

From the question,

Given: v = 0 m/s (comes to rest), u = 20 m/s, a = -5 m/s² (begins to slow)

Substitute these values into equation 2

s = (0²-20²)/[2(-5)]

s = -400/-10

s = 40 m

Hence the bus was 40 m from the stop

Answer:

(a) Altitude = 1.95 x 10⁶ m = 1950 km

(b) g = 5.9 m/s²

Explanation:

(a)

The time period of the satellite is given by the following formula:

[tex]T^2 = \frac{4\pi^2r^3}{GM_E}[/tex]

where,

T = Time period = (125 min)([tex]\frac{60\ s}{1\ min}[/tex]) = 7500 s

r = distance of satellite from the center of earth = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

[tex]M_E[/tex] = Mass of Earth = 6 x 10²⁴ kg

Therefore,

[tex](7500\ s)^2 = \frac{4\pi^2r^3}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(6\ x\ 10^{24}\ kg)}\\\\r^3 = \frac{(7500\ s)^2(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(6\ x\ 10^{24}\ kg)}{4\pi^2}\\\\r = \sqrt[3]{5.7\ x\ 10^{20}\ m^3} \\[/tex]

r = 8.29 x 10⁶ m

Hence, the altitude of the satellite will be:

[tex]Altitude = r - radius\ of\ Earth\\Altitude = 8.29\ x\ 10^6\ m - 6.34\ x\ 10^6\ m[/tex]

Altitude = 1.95 x 10⁶ m = 1950 km

(b)

The weight of the satellite will be equal to the gravitational force between satellite and Earth:

[tex]Weight = Gravitational\ Force\\\\M_sg = \frac{GM_EM_s}{r^2}\\\\g = \frac{GM_E}{r^2}\\\\g = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(6\ x\ 10^{24}\ kg)}{(8.23\ x\ 10^6\ m)^2}[/tex]

g = 5.9 m/s²

Answer:

Please find the complete question in the attached file.

Explanation:

[tex]V=12\ V\\\\I=1.2\ A\\\\T=5\times 60=300\ second\\\\[/tex]

Calculating the electrical energy dissipated:

[tex]w=p\cdot t=V\cdot I \cdot t\\\\[/tex]

              [tex]=12\times 1.2 \times 300 \ J\\\\=4320\ J[/tex]

[tex]\Delta T=20^{\circ}\ C\\\\W=m\cdot c\cdot \Delta T\\\\4320=m(4186 \times 20)\\\\m=\frac{4320}{4186 \times 20}=51.6 \ grams=0.516 \ kg\\\\[/tex]

Answer:

Explanation:

Moment of inertia of sphere I = 2 /5 x m R²

m is mass and R is radius of the sphere .

I = 2/5 x 1800 x 0.5²

= 180 kg m².

For rotational motion starting from rest

θ = 1/2 α t²

Here θ = π /2 = 1.57 radians

t = 12 s

1.57 radians  = 1/2 x α x 12²

Angular acceleration α = .022 rad /s²

Torque applied = I α

=  180 kg m² x  .022 rad /s²

= 3.96 N.m

b ) Torque = F x r

F = Torque / r

= 3.96 / .5 = 7.92 N .

c )

Average weight of a cellophone is 1.3  N .

So the force applied is a little more  than weight of a cellophone.

Answer:

C is the right answer

Explanation:

fitness logs is a great way to track your progress. You can easily look back and see how you have progressed over time. In addition, it can help you plan and prepare for future workouts, as well as identify patterns of what seems to work well for you and when you have the most success

hope it was useful for you

Answer:

the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm

Explanation:

Given the data in the question;

wavelength of light; λ = 463 nm = 463 × 10⁻⁹ m

Index of refraction; n = 1.35

Now, the thinnest thickness of the soap film can be determined from the following expression;

[tex]t_{min[/tex] = ( λ / 4n )

so we simply substitute in our given values;

[tex]t_{min[/tex] = ( 463 × 10⁻⁹ m ) / 4(1.35)

[tex]t_{min[/tex] = ( 463 × 10⁻⁹ m ) / 5.4

[tex]t_{min[/tex] = ( 463 × 10⁻⁹ m ) / 4(1.35)

[tex]t_{min[/tex] = 8.574 × 10⁻⁸ m

[tex]t_{min[/tex] = 85.74 × 10⁻⁹ m

[tex]t_{min[/tex] = 85.74 nm

Therefore, the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm

Answer:

Approximately [tex]63\; \rm m[/tex].

Explanation:

Convert the initial speed of this truck to standard units:

[tex]\begin{aligned} v &= 72\; \rm km \cdot h^{-1} \times \frac{1\; \rm h}{3600\; \rm s} \times \frac{1000\; \rm m}{1\; \rm km} \\ &= 20\; \rm m \cdot s^{-1}\end{aligned}[/tex].

Calculate the initial kinetic energy of this truck:

[tex]\begin{aligned}\text{KE} &= \frac{1}{2} \, m \cdot v^{2} \\ &= \frac{1}{2} \times 2600\; \rm kg \times (20\; \rm m \cdot s^{-1})^{2} \\ &= 5.2 \times 10^{5} \; \rm J\end{aligned}[/tex].

The kinetic energy of the truck would be [tex]0[/tex] when the truck is not moving. Hence, the friction on the truck would need to do [tex](-5.2 \times 10^{5}\; \rm J)[/tex] of work on the truck to bring the truck to a stop.

Calculate the displacement required to achieve [tex](-5.2 \times 10^{5}\; \rm J)[/tex] of work at [tex]8200\; \rm N[/tex]:

[tex]\begin{aligned}x &= \frac{W}{F} \\ &= \frac{-5.2 \times 10^{5}\; \rm J}{8200\; \rm N} \approx -63\; \rm m\end{aligned}[/tex].

This result is negative because the direction of friction is opposite to the direction of the motion of the truck.

Hence, the truck would come to a stop after skidding for approximately [tex]63\; \rm m[/tex].

Answer:

In order to lift off the ground, the air in the balloon must be heated to 710.26 K

Explanation:

Given the data in the question;

P = 1.01 × 10⁵ Pa

V = 480 m³

ρ = 1.29 kg/m³

M = 381 kg

we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol

let F represent the force acting upward.

Now in a condition where the hot air balloon is just about to take off;

F - Mg - m[tex]_g[/tex]g = 0

where M is the mass of the balloon and its occupants, m[tex]_g[/tex] is the mass of the hot gas inside the balloon.

the force acting upward F = Vρg

so

Vρg - Mg - m[tex]_g[/tex]g = 0

solve for m[tex]_g[/tex]

m[tex]_g[/tex] = ( Vρg - Mg ) / g

m[tex]_g[/tex] =  Vρg/g - Mg/g

m[tex]_g[/tex] =  ρV - M ------- let this be equation 1

Now, from the ideal gas law, PV = nRT

we know that number of moles n = m[tex]_g[/tex] / μ

where μ is the molecular mass of air

so

PV = (m[tex]_g[/tex]/μ)RT

solve for T

μPV = m[tex]_g[/tex]RT

T = μPV / m[tex]_g[/tex]R -------- let this be equation 2

from equation 1 and 2

T = μPV / (ρV - M)R

so we substitute in our values;

P = 1.01 × 10⁵ Pa

V = 480 m³

ρ = 1.29 kg/m³

M = 381 kg

we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol

T = [ (29 × 10⁻³) × (1.01 × 10⁵) × 480 ] / [ (( 1.29 × 480 ) - 381)8.31 ]

T =  1405920 / 1979.442

T =  710.26 K

Therefore, In order to lift off the ground, the air in the balloon must be heated to 710.26 K

The temperature required for the air to be heated is 710.26 K.

Given data:

The mass of a hot air-balloon is, m = 381 kg.

The pressure of air outside the balloon is, [tex]P = 1.01 \times 10^{5} \;\rm Pa[/tex].

The density of air is, [tex]\rho = 1.29 \;\rm kg/m^{3}[/tex].

The volume of heated balloon is, [tex]V = 480 \;\rm m^{3}[/tex].

The condition where the hot air balloon is just about to take off is as follows:

[tex]F-mg - m'g =0[/tex]

Here,

m' is the mass of hot gas inside the balloon and g is the gravitational acceleration and F is the force acting on the balloon in upward direction. And its value is,

[tex]F = V \times \rho \times g[/tex]

Solving as,

[tex](V \times \rho \times g)-mg - m'g =0\\\\ m'=(V \rho )-m[/tex]

Now, apply the ideal gas law as,

PV = nRT

here, R is the universal gas constant and n is the number of moles and its value is,

[tex]n=\dfrac{m'}{M}[/tex]

M is the molecular mass of gas. Solving as,

[tex]PV = \dfrac{m'}{M} \times R \times T\\\\\\T=\dfrac{P \times V\times M}{m'R}\\\\\\T=\dfrac{P \times V\times M}{(V \rho - m)R}[/tex]

Since, the standard value for the molecular mass of air is, [tex]M = 29 \times 10^{-3} \;\rm kg/mol[/tex]. Then solve for the temperature as,

[tex]T=\dfrac{(1.01 \times 10^{5}) \times 480\times 381}{(480 \times (1.29) - 381)8.31}\\\\\\T = 710.26 \;\rm K[/tex]

Thus, we can conclude that the temperature required for the air to be heated is 710.26 K.

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Answer:

12.74 V

Explanation:

We are given that

Current, I1=54 A

Potential difference, V1=9.18V

I2=2.10 A

V2=12.6 V

We have to find the battery's emf.

[tex]E=V+Ir[/tex]

Using the formula

[tex]E=9.18+54r[/tex] ....(1)

[tex]E=12.6+2.10r[/tex]  .....(2)

Subtract equation (1) from (2)

[tex]0=3.42-51.9r[/tex]

[tex]3.42=51.9r[/tex]

[tex]r=\frac{3.42}{51.9}=0.0659ohm[/tex]

Using the value of r in equation (1)

[tex]E=9.18+54(0.0659)[/tex]

[tex]E=12.74 V[/tex]

Answer:

30 V

Explanation:

Given that:

The uniform electric field = 50 N/C

Voltage = 80 V

distance = 1.0 m

The potential difference of the electric field = Δ V

E_d = V₁ - V₂

50 × 1 = 80V - V₂

50 - 80 V = - V₂

-30 V = - V₂

V₂ = 30 V