What is the maximum wavelength of incident light that can produce photoelectrons from silver? The work function for silver is Φ=2.93 eV.

Answers

Answer 1

Answer:

The maximum wavelength of incident light that can produce photoelectrons from silver is 423.5 nm.

Explanation:

Given;

work function of silver, Φ = 2.93 eV = 2.93 x 1.602 x 10⁻¹⁹ J = 4.6939 x 10⁻¹⁹ J

Apply Einstein Photo electric effect;

E = K.E + Ф

Where;

E is the energy of the incident light

K.E is the kinetic of electron

Ф is the work function of silver surface

For the incident light to have maximum wavelength, the kinetic energy of the electron will be zero.

E = Ф

hf = Ф

[tex]h\frac{c}{\lambda} = \phi[/tex]

where;

c is speed of light = 3 x 10⁸ m/s

h is Planck's constant, = 6.626 x 10⁻³⁴ J/s

λ is the wavelength of the incident light

[tex]\lambda = \frac{hc}{\phi}\\\\\lambda =\frac{6.626*10^{-34} *3*10^8}{4.6939*10^{-19}} \\\\\lambda = 4.235 *10^{-7} \ m\\\\\lambda = 423.5 *10^{-9} \ m\\\\\lambda = 423.5 \ nm[/tex]

Therefore, the maximum wavelength of incident light that can produce photoelectrons from silver is 423.5 nm.


Related Questions

An LR circuit consists of a 35-mH inductor, a resistance of 12 ohms, an 18-V battery, and a switch. What is the current 5.0 ms after the switch is closed

Answers

Answer:

Current, I = 1.23 A

Explanation:

Given that,

Inductance, L = 35 mH

Resistance, R = 12 ohms

Potential difference, V = 18 V

We need to find current 5 ms after the switch is closed. Current in LR circuit is given by :

[tex]I=I_o(1-e^{-t/\tau })[/tex] ....(1)

Here,

[tex]I_o[/tex] is final current

[tex]I_o=\dfrac{V}{R}\\\\I_o=\dfrac{18}{12}=1.5\ A[/tex]

[tex]\tau[/tex] is time constant

[tex]\tau=\dfrac{L}{R}\\\\\tau=\dfrac{35\times 10^{-3}}{12}\\\\\tau=0.00291\ s[/tex]

So, equation (1) becomes :

[tex]I=1.5\times (1-e^{-5\times 10^{-3}/0.00291})\\\\I=1.23\ A[/tex]

So, after 5 ms the current in the circuit is 1.23 A.

Four charges each of magnitude 15 µC are arranged on the corners of a square of side 5 cm. What is the total potential energy of the system?

Answers

Answer:

-105J

Explanation:

See attached file

A 590-turn solenoid is 12 cm long. The current in it is 36 A . A straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us).
What is the magnitude of the force on this wire assuming the solenoid's field points due east?

Answers

Complete Question

A 590-turn solenoid is 12 cm long. The  current in it is 36 A . A 2 cm straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us).

What is the magnitude of the force on this wire assuming the solenoid's field points due east?

Answer:

The force is  [tex]F = 0.1602 \ N[/tex]

Explanation:

From the question we are told that

   The number of turns is  [tex]N = 590 \ turns[/tex]

   The  length of the solenoid is  [tex]L = 12 \ cm = 0.12 \ m[/tex]

   The current is  [tex]I = 36 \ A[/tex]

   The  diameter is  [tex]D = 4.5 \ cm = 0.045 \ m[/tex]

   The  current carried by the wire is  [tex]I = 27 \ A[/tex]

    The  length of the wire is  [tex]l = 2 cm = 0.02 \ m[/tex]

Generally the magnitude of the force  on this wire assuming the solenoid's field points due east is mathematically represented as

           [tex]F = B * I * l[/tex]

Here  B  is the magnetic field which is mathematically represented as

          [tex]B = \frac{\mu_o * N * I }{L}[/tex]

Here   [tex]\mu _o[/tex] is permeability of free space with value  [tex]\mu_ o = 4\pi *10^{-7} \ N/A^2[/tex]

substituting values

         [tex]B = \frac{4 \pi *10^{-7} * 590 * 36 }{ 0.12}[/tex]

           [tex]B = 0.2225 \ T[/tex]

So

      [tex]F = 0.2225 * 36 * 0.02[/tex]

      [tex]F = 0.1602 \ N[/tex]

Some stove tops are smooth ceramic for easy cleaning. If the ceramic is 0.630 cm thick and heat conduction occurs through an area of 1.45 ✕ 10−2 m2 at a rate of 500 J/s, what is the temperature difference across it (in °C)? Ceramic has the same thermal conductivity as glass and concrete brick.

Answers

Answer:

The temperature difference [tex]\Delta T = 258.6 \ ^ o\ C[/tex]

Explanation:

From the question we are told that

   The  thickness is [tex]\Delta x = 0.630 cm = 0.0063 m[/tex]

    The  area is  [tex]A = 1.45 *10^{-2 } \ m^2[/tex]

     The rate is  [tex]P = 500 J/s[/tex]

       The  thermal conductivity is  [tex]\sigma = 0.84J[\cdot s \cdot m \cdot ^oC ][/tex]

Generally the rate heat conduction mathematically represented as

       [tex]P = \sigma * A * \frac{\Delta T}{\Delta x }[/tex]

=>    [tex]\Delta T = \frac{P * \Delta x }{\sigma * A }[/tex]

=>     [tex]\Delta T = \frac{ 500 * 0.00630 }{ 0.84 * 1.45 *10^{-2} }[/tex]

=>    [tex]\Delta T = 258.6 \ ^ o\ C[/tex]

A car travels at 45 km/h. If the driver breaks 0.65 seconds after seeing the traffic light turn yellow, how far will the car continue to travel before it begins to slow?

Answers

Answer:

8.1 m

Explanation:

Convert km/h to m/s.

45 km/h × (1000 m/km) × (1 h / 3600 s) = 12.5 m/s

Distance = speed × time

d = (12.5 m/s) (0.65 s)

d = 8.125 m

A long bar slides on two contact points and is in motion with velocity ν. A steady, uniform, magnetic field B is present. The induced current through resistor R is:

Answers

Answer:

The induced current in the resistor is I = BLv/R

Explanation:

The induced emf ε in the long bar of length, L in a magnetic field of strength, B moving with a velocity, v is given by

ε = BLv.

Now, the current I in the resistor is given by

I = ε/R where ε = induced emf in circuit and R = resistance of resistor.

So, the current I = ε/R.

substituting the value of ε the induced emf, we have

I = ε/R

I = BLv/R

So, the induced current through the resistor is given by I = BLv/R

The sun generates both mechanical and electromagnetic waves. Which statement about those waves is true?
OA. The mechanical waves reach Earth, while the electromagnetic waves do not.
OB. The electromagnetic waves reach Earth, while the mechanical waves do not.
OC. Both the mechanical waves and the electromagnetic waves reach Earth.
OD. Neither the mechanical waves nor the electromagnetic waves reach Earth.

Answers

Answer: The correct answer for this question is letter (B) The electromagnetic waves reach Earth, while the mechanical waves do not. The sun generates both mechanical and electromagnetic waves. Space, between the sun and the earth is a nearly vacuum. So mechanical wave can not spread out in the vacuum.

Hope this helps!

Answer:

The electromagnetic waves reach Earth, while the mechanical waves do not

How much work is needed to pump all the water out of a cylindrical tank with a height of 10 m and a radius of 5 m

Answers

Answer:

Explanation:

volume of water being lifted

= π r² h , where r is radius of cylinder and h is height of cylinder

= 3.14 x5² x 10

= 785 m³

mass of water = 785 x 10³ kg

mass of this much of water is lifted so that its centre of mass is lifted by height

10 / 2 = 5m .

So work done = mgh , m is mass of water , h is displacement of centre of mass and g is acceleration due to gravity

= 785 x 10³ x 9.8 x 5

= 38.465 x 10⁶ J  

A resistor made of Nichrome wire is used in an application where its resistance cannot change more than 1.35% from its value at 20.0°C. Over what temperature range can it be used (in °C)?

Answers

Answer:

Pls seeattached file

Explanation:

A resistor made of Ni chrome wire is used in an application where its resistance cannot be more than 1.35 % so its temperature range will be from 33.75 to -33.75 °C.

What is Resistance?

Electrical resistance, or resistance to electricity, is a force that opposes the flow of current. Ohms are used to expressing resistance values.

When there is an electron difference between two terminals, electricity will flow from high to low. In opposition to that flow is resistance. As resistance rises, the current declines. On the other side, when the resistance falls, the current rises.

According to the question,

R = R₀ (1 + α ΔT)

(1 + 0.0135)R₀ = R₀(1 + α ΔT)

ΔT = (1 + 0.0135) / α

= 0.0135 / 0.0004

= 33.75 °C.

ΔT = [(1 - 0.0135) -1]/0.004

= -33.75 °C

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If the
refractive index of benzere is 2.419,
what is the speed of light in benzene?

Answers

Answer:

[tex]v=1.24\times 10^8\ m/s[/tex]

Explanation:

Given that,

The refractive index of benzene is 2.419

We need to find the speed of light in benzene. The ratio of speed of light in vacuum to the speed of light in the medium equals the refractive index. So,

[tex]n=\dfrac{c}{v}\\\\v=\dfrac{c}{n}\\\\v=\dfrac{3\times 10^8}{2.419}\\\\v=1.24\times 10^8\ m/s[/tex]

So, the speed of light in bezene is [tex]1.24\times 10^8\ m/s[/tex].

Two long, parallel wires are separated by a distance of 2.60 cm. The force per unit length that each wire exerts on the other is 4.30×10^−5 N/m, and the wires repel each other. The current in one wire is 0.520 A.Required:a. What is the current in the second wire? b. Are the two currents in the same direction or in opposite directions?

Answers

Answer:

10.75 A

The current is in opposite direction since it causes a repulsion force between the wires

Explanation:

Force per unit length on the wires = 4.30×10^−5 N/m

distance between wires = 2.6 cm = 0.026 m

current through one wire = 0.52 A

current on the other wire = ?

Recall that the force per unit length of two wires conducting and lying parallel and close to each other is given as

[tex]F/l[/tex] = [tex]\frac{u_{0}I_{1} I_{2} }{2\pi r }[/tex]

where [tex]F/l[/tex] is the force per unit length on the wires

[tex]u_{0}[/tex] = permeability of vacuum = 4π × 10^−7 T-m/A

[tex]I_{1}[/tex] = current on the first wire = 0.520 A

[tex]I_{2}[/tex] = current on the other wire = ?

r = the distance between the two wire = 0.026 m

substituting the value into the equation, we have

4.30×10^−5 = [tex]\frac{4\pi *10^{-7}*0.520*I_{2} }{2\pi *0.026}[/tex] =  [tex]\frac{ 2*10^{-7}*0.520*I_{2} }{0.026}[/tex]

4.30×10^−5 = 4 x 10^-6 [tex]I_{2}[/tex]

[tex]I_{2}[/tex] = (4.30×10^-5)/(4 x 10^-6) = 10.75 A

The current is in opposite direction since it causes a repulsion force between the wires.

What is the magnitude of the applied electric field inside an aluminum wire of radius 1.4 mm that carries a 4.5-A current

Answers

Answer:

Explanation:

From the question we are told that

    The radius is  [tex]r = 1.4 \ mm = 1.4 *10^{-3} \ m[/tex]

     The  current is  [tex]I = 4.5 \ A[/tex]

Generally the electric field is mathematically represented as

         [tex]E = \frac{J}{\sigma }[/tex]

Where [tex]\sigma[/tex] is the conductivity of  aluminum with value [tex]\sigma = 3.5 *10^{7} \ s/m[/tex]

J is the current density which mathematically represented as  

      [tex]J = \frac{I}{A}[/tex]

Here A is the cross-sectional area which is mathematically represented as  

       [tex]A = \pi r^2[/tex]

       [tex]A = 3.142 * (1.4*10^{-3})^2[/tex]

       [tex]A = 6.158*10^{-6} \ m^2[/tex]

So

    [tex]J = \frac{ 4.5 }{6.158*10^{-6}}[/tex]

    [tex]J = 730757 A/m^2[/tex]

So

       [tex]E = \frac{ 730757}{3.5*10^{7} }[/tex]

       [tex]E = 0.021 \ N/C[/tex]

Suppose you are planning a trip in which a spacecraft is to travel at a constant velocity for exactly six months, as measured by a clock on board the spacecraft, and then return home at the same speed. Upon return, the people on earth will have advanced exactly 120 years into the future. According to special relativity, how fast must you travel

Answers

Answer:

I must travel with a speed of 2.97 x 10^8 m/s

Explanation:

Sine the spacecraft flies at the same speed in the to and fro distance of the journey, then the time taken will be 6 months plus 6 months

Time that elapses on the spacecraft = 1 year

On earth the people have advanced 120 yrs

According to relativity, the time contraction on the spacecraft is gotten from

[tex]t[/tex] = [tex]t_{0} /\sqrt{1 - \beta ^{2} }[/tex]

where

[tex]t[/tex] is the time that elapses on the spacecraft = 120 years

[tex]t_{0}[/tex] = time here on Earth = 1 year

[tex]\beta[/tex] is the ratio v/c

where

v is the speed of the spacecraft = ?

c is the speed of light = 3 x 10^8 m/s

substituting values, we have

120 = 1/[tex]\sqrt{1 - \beta ^{2} }[/tex]

squaring both sides of the equation, we have

14400 = 1/[tex](1 - \beta ^{2} )[/tex]

14400 - 14400[tex]\beta ^{2}[/tex] = 1

14400 - 1 = 14400[tex]\beta ^{2}[/tex]

14399 = 14400[tex]\beta ^{2}[/tex]

[tex]\beta ^{2}[/tex] =  14399/14400 = 0.99

[tex]\beta = \sqrt{0.99}[/tex] = 0.99

substitute β = v/c

v/c = 0.99

but c = 3 x 10^8 m/s

v = 0.99c = 0.99 x 3 x 10^8 = 2.97 x 10^8 m/s

A deep-space vehicle moves away from the Earth with a speed of 0.870c. An astronaut on the vehicle measures a time interval of 3.10 s to rotate her body through 1.00 rev as she floats in the vehicle. What time interval is required for this rotation according to an observer on the Earth

Answers

Answer:

t₀ = 1.55 s

Explanation:

According to Einstein's Theory of Relativity, when an object moves with a speed comparable to speed of light, the time interval measured for the event, by an observer in  motion relative to the event is not the same as measured by an observer at rest.

It is given as:

t = t₀/[√(1 - v²/c²)]

where,

t = time measured by astronaut in motion = 3.1 s

t₀ = time required according to observer on earth = ?

v = relative velocity = 0.87 c

c = speed of light

3.1 s = t₀/[√(1 - 0.87²c²/c²)]

(3.1 s)(0.5) = t₀

t₀ = 1.55 s

Answer:

The time interval required for this rotation according to an observer on the Earth = [tex]6.29sec[/tex]

Explanation:

Time interval required for this rotation according to an observer on the Earth is given as [tex]\delta t[/tex]

where,

[tex]t_o = 3.1\\\\v = 0.87[/tex]

[tex]\delta t = \frac{t_o}{\sqrt{1-\frac{v^2}{c^2}}}\\\\\delta t = \frac{3.1}{\sqrt{1-(\frac{0.87c}{c})^2}}\\\\\delta t = 6.29sec[/tex]

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A series LR circuit contains an emf source of 19 V having no internal resistance, a resistor, a 22 H inductor having no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4 s after the switch is closed, what is the resistance of the resistor

Answers

Answer: R = 394.36ohm

Explanation: In a LR circuit, voltage for a resistor in function of time is given by:

[tex]V(t) = \epsilon. e^{-t.\frac{L}{R} }[/tex]

ε is emf

L is indutance of inductor

R is resistance of resistor

After 4s, emf = 0.8*19, so:

[tex]0.8*19 = 19. e^{-4.\frac{22}{R} }[/tex]

[tex]0.8 = e^{-\frac{88}{R} }[/tex]

[tex]ln(0.8) = ln(e^{-\frac{88}{R} })[/tex]

[tex]ln(0.8) = -\frac{88}{R}[/tex]

[tex]R = -\frac{88}{ln(0.8)}[/tex]

R = 394.36

In this LR circuit, the resistance of the resistor is 394.36ohms.

A sinusoidal electromagnetic wave is propagating in a vacuum in the +z-direction. If at a particular instant and at a certain point in space the electric field is in the +x-direction and has a magnitude of 4.00 V/m, what is the magnitude of the magnetic field of the wave at this same point in space and instant in time?

Answers

Answer:

B = 1.33 10⁻⁸ T , the magnetic field must be in the y + direction

Explanation:

In an electromagnetic wave the electric and magnetic fields are in phase

         c = E / B

         B = E / c

let's calculate

          B = 4.00 / 3 10⁸

          B = 1.33 10⁻⁸ T

To determine the direction we use that the electric and magnetic fields and the speed of the wave are perpendicular.

 If the wave advances in the + Z direction and the electric field is in the + x direction, the magnetic field must be in the y + direction

Matter's resistance to a change in motion is called _____ and is directly proportional to the mass of an object

Answers

Answer:

Matter's resistance to a change in motion is called INERTIA and is directly proportional to the mass of an object.

Explanation:

3. What are the first steps that you should take if you are unable to get onto the Internet? (1 point)
O Check your router connections then restart your router.
O Plug the CPU to a power source and reboot the computer.
O Adjust the display properties and check the resolution.
Use the Control Panel to adjust the router settings.​

Answers

Answer:

Check your router connections then restart your router.

Explanation:

Answer:

Check your router connections then restart your router.

Explanation:

Most internet access comes from routers so the problem is most likely the router.

What is the minimum thickness of coating which should be placed on a lens in order to minimize reflection of 566 nm light? The index of refraction of the coating material is 1.46 and the index of the glass is 1.71.

Answers

Answer:

The  thickness is   [tex]t = 1.415 *10^{-7 } \ m[/tex]

Explanation:

From the question we are told that

    The wavelength is  [tex]\lambda = 566 \ nm = 566 *10^{-9} \ m[/tex]

     The  index of refraction of glass is  [tex]n_g = 1.71[/tex]

     The index of refraction of the coating is  [tex]n= 1.46[/tex]

Generally the condition for destructive interference is  

         [tex]2 t = (m + \frac{1}{2} ) * \frac{\lambda }{n }[/tex]

Here m is the order of the interference pattern and given from the question that we are considering minimizing  reflection  m = 0

t = thickness of the coating

substituting values

         [tex]2 t = (0 + \frac{1}{2} ) * \frac{ 566 *10^{-9}}{ 1.46 }[/tex]

    =>    [tex]t = 1.415 *10^{-7 } \ m[/tex]

           

10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ????? How far will this pulse travel in the same time if the tension is doubled?

Answers

Answer: Tension = 47.8N, Δx = 11.5×[tex]10^{-6}[/tex] m.

              Tension = 95.6N, Δx = 15.4×[tex]10^{-5}[/tex] m

Explanation: A speed of wave on a string under a tension force can be calculated as:

[tex]|v| = \sqrt{\frac{F_{T}}{\mu} }[/tex]

[tex]F_{T}[/tex] is tension force (N)

μ is linear density (kg/m)

Determining velocity:

[tex]|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }[/tex]

[tex]|v| = \sqrt{0.00874 }[/tex]

[tex]|v| =[/tex] 0.0935 m/s

The displacement a pulse traveled in 1.23ms:

[tex]\Delta x = |v|.t[/tex]

[tex]\Delta x = 9.35.10^{-2}*1.23.10^{-3}[/tex]

Δx = 11.5×[tex]10^{-6}[/tex]

With tension of 47.8N, a pulse will travel Δx = 11.5×[tex]10^{-6}[/tex]  m.

Doubling Tension:

[tex]|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }[/tex]

[tex]|v| = \sqrt{2.0.00874 }[/tex]

[tex]|v| = \sqrt{0.01568}[/tex]

|v| = 0.1252 m/s

Displacement for same time:

[tex]\Delta x = |v|.t[/tex]

[tex]\Delta x = 12.52.10^{-2}*1.23.10^{-3}[/tex]

[tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex]

With doubled tension, it travels [tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex] m

Water flows at speed v in a pipe of radius R. At what speed does the water flow through a constriction in which the radius of the pipe is R/3

Answers

Answer:

   v₂ = 9 v

Explanation:

For this exercise in fluid mechanics, let's use the continuity equation

           v₁ A₁ = v₂ A₂

where v is the velocity of the fluid, A the area of ​​the pipe and the subscripts correspond to two places of interest.

The area of ​​a circle is

           A = π R²

let's use the subscript 1 for the starting point and the subscript 2 for the part with the constraint

     

In this case v₁ = v and the area is

            A₁ = π R²

in the second point

           A₂= π (R / 3)²

we substitute in the continuity equation

           v π R² = v₂ π R² / 9

            v = v₂ / 9

           

            v₂ = 9 v

To get an idea of the order of magnitude of inductance, calculate the self-inductance in henries for a solenoid with 1500 loops of wire wound on a rod 13 cm long with radius 2 cm

Answers

Answer:

The self-inductance in henries for the solenoid is 0.0274 H.

Explanation:

Given;

number of turns, N = 1500 turns

length of the solenoid, L = 13 cm = 0.13 m

radius of the wire, r = 2 cm = 0.02 m

The self-inductance in henries for a solenoid is given by;

[tex]L = \frac{\mu_oN^2A}{l}[/tex]

where;

[tex]\mu_o[/tex] is permeability of free space = [tex]4\pi*10^{-7} \ H/m[/tex]

A is the area of the solenoid = πr² = π(0.02)² = 0.00126 m²

[tex]L = \frac{4\pi *10^{-7}(1500)^2*(0.00126)}{0.13} \\\\L = 0.0274 \ H[/tex]

Therefore, the self-inductance in henries for the solenoid is 0.0274 H.

The highest mountain on mars is olympus mons, rising 22000 meters above the martian surface. If we were to throw an object horizontaly off the mountain top, how long would it take to reach the surface? (Ignore atmospheric drag forces and use gMars=3.72m/s^2

a. 2.4 minutes
b. 0.79 minutes
c. 1.8 minutes
d. 3.0 minutes

Answers

Answer:

  t = 1.81 min ,     the correct answer is c

Explanation:

This is a missile throwing exercise

The object is thrown horizontally, so its vertical speed is zero (voy = 0), let's use the equation

             y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

the final height is y = 0 and the initial height is y₀ = 22000 m

            0 = y₀ + 0 - ½ g t²

             

            t = √y 2y₀ / g

let's calculate

           t = √(2  22000 / 3.72)

           t = 108.76 s

let's reduce to minutes

           t = 108.76 s (1 min / 60 s)

           t = 1.81 min

The correct answer is c

A child is trying to throw a ball over a fence. She gives the ball an initial speed of 8.0 m/s at an angle of 40° above the horizontal. The ball leaves her hand 1.0 m above the ground and the fence is 2.0 m high. The ball just clears the fence while still traveling upwards and experiences no significant air resistance. How far is the child from the fence?

Answers

Answer:

the child is 1.581 m far from the fence

Explanation:

The diagrammatic illustration that give a better view of what the question denote can be seen in the image attached below.

From the image attached below, let assume that the release point is the origin, then equation of the motion (x) is as follows:

[tex]x - x_o = u_xt[/tex]

[tex]\mathtt{x = u_xt \ \ \ since (x_o = 0)}[/tex]  ---- (1)

the equation of the motion y is :

[tex]\mathtt{y - y_o =u_yt - 0.5 gt^2}[/tex]

[tex]\mathtt{y = u_yt-4.9t^2 \ \ \ since (y_o =0)}[/tex]

[tex]\mathtt{ 1= (u \ sin 40^0)t -4.9 \ t^2 }[/tex]

[tex]\mathtt{1 = 8 sin 40^0 t - 4.9 t^2}[/tex]

[tex]\mathtt{1 = 5.14t - 4.9t^2}[/tex]

[tex]\mathtt{4.9t^2 - 5.14t +1 = 0}[/tex]

By using the quadratic formula, we have;

[tex]\mathtt{ \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}} }[/tex]

where;

a = 4.9,   b = -5.14     c = 1

[tex]= \mathtt{ \dfrac{ -(-5.14) \pm \sqrt{(-5.14)^2 - 4(4.9)(1)}}{2(4.9)}} }[/tex]

[tex]= \mathtt{ \dfrac{ 5.14 \pm \sqrt{26.4196 -19.6}}{9.8}} }[/tex]

[tex]= \mathtt{ \dfrac{ 5.14 \pm \sqrt{6.8196}}{9.8}} }[/tex]

[tex]= \mathtt{ \dfrac{ 5.14+ \sqrt{6.8196}}{9.8} \ \ OR \ \ \dfrac{ 5.14- \sqrt{6.8196}}{9.8}} }[/tex]

[tex]= \mathtt{ \dfrac{ 5.14+ 2.6114}{9.8} \ \ OR \ \ \dfrac{ 5.14- 2.6114}{9.8}} }[/tex]

[tex]= \mathtt{ \dfrac{ 7.7514}{9.8} \ \ OR \ \ \dfrac{ 2.5286}{9.8}} }[/tex]

[tex]= \mathbf{ 0.791 \ \ OR \ \ 0.258} }[/tex]

In as much as the ball is traveling upward, then we consider t= 0.258sec.

From equation (1)

[tex]\mathtt{x = u_x(0.258)}[/tex]

[tex]\mathtt{x = ucos 40^0 (0.258)}[/tex]

[tex]\mathtt{x = 8 \ cos 40^0 (0.258)}[/tex]

[tex]\mathbf{x = 1.581 \ m}[/tex]

Thus, the child is 1.581 m far from the fence

A nearsighted person has a far point that is 4.2 m from his eyes. What focal length lenses in diopters he must use in his contacts to allow him to focus on distant objects?

Answers

Answer:

-0.24diopters

Explanation:

The lens is intended that makes an object at infinity appear to be 4.2 m away, so do=infinity, dI = - 4.2m (minus sign because image is on same side of lens as object)

So 1/do +1/di = 1/f

1/infinity + 1/-4.2 = 1/f

1/f = 1/-4.2 = -0.24diopters

A 26-g rifle bullet traveling 220 m/s embeds itself in a 3.8-kg pendulum hanging on a 2.7-m-long string, which makes the pendulum swing upward in an arc, Determine the vertical and horizontal component of the pendulum's maximum displacement

Answers

Answer:

displacements are 0.776m, 0.114m

Explanation:

We were given mass of 26-g rifle bullet , then we can convert to Kg since

Momentum is conserved here.

The initial momentum before impact = (Mi * Vi)

Where Mi= initial given mass

Vi=initial velocity given

= 0.026 * 220 = 5.72 kgm/s

The final momentum after impact is (Mf * Vf )

Mf= final mass

5.72=( 3.82* Vf )

= 5.72/ 3.82

= 1.497 m/s

the speed of the pendulum bob with bullet afterwards= 1.497 m/s

the total energy after the collision is the addition of the kinetic energy of the bob+bullet and the potential energy of the bob and bullet, potential energy can be taken as zero.

M = 3.82 kg the mass of the bob containing the bullet

E(total) = ¹/₂MV² = 1/2 * (3.82kg)*(1.497m/s)² = 4.280J

When the Bob got to highest point the kinetic energy is zero and the potential energy is due to the increase in height of the bob, and the addition of the potential and kinetic energies still equal the total energy from before

E(total) = Mgh + 0 = Mgh = 4.280J

solving for h and substituting,

h = 4.280 J/(9.8m/s^2*3.82kg) = 0.114 m

Since the height is found,we the angle of the pendulum at the top of the swing can also be determined

A = arccos[(2.7 - 0.114) / 2.7] or A = 16.71degrees

Since A is known, the displacement along the horizontal axis can be calculated as

x = 2.7* sin(A) = 0.776m

therefore, displacement is 0.776m, 0.114m

the vertical and horizontal component of the pendulum's maximum displacement are displacement is 0.776m, 0.114m

Kinetic and
A brick is resting on a smooth wooden board that is at a 30° angle. What is one way to overcome the static friction that is holding the brick in place?
11 point)
O raise the board to a higher angle
O press down on the brick in a direction that is perpendicular to the board
O roughen up the texture of the wooden board
o lower the board so it's level with the ground

Answers

Answer:

raise the board to a higher angle

Explanation:

Static friction is the force opposite to the applied force.

Static friction is dependent on the angle of inclination, it means as the angle of incline increases, the force of friction will increases as normal force will decrease.

So, if the board will be raised to a higher angle, it will increase the angle of incline and will overcome the static friction and block will be able slide.

Hence, the correct option is "raise the board to a higher angle".

When a mercury thermometer is heated, the mercury expands and rises in the thin tube of glass. What does this indicate about the relative rates of expansion for mercury and glass

Answers

Answer:

This means that mercury has a higher or faster expansion rate than glass

Explanation:

This is because When a container expands, the reservoir in the glass expands at the same rate as the glass. Thus, if there is something in a glass and both expand at the same rate, they have no change - but if the contents expand faster, they will fill the container to a higher level, and if the contents expand slower, they will fill the container to a lower level (relative to the new size of the container).

The positron has the same mass as an electron, with an electric charge of +e. A positron follows a uniform circular motion of radius 5.03 mm due to the force of a uniform magnetic field of 0.85 T. How many complete revolutions does the positron perform If it spends 2.30 s inside the field? (electron mass = 9.11 x 10-31 kg, electron charge = -1.6 x 10-19 C)

Answers

Answer:

5.465 × 10^10 revolutions

Explanation:

Formula for Magnetic Field = m. v/ q . r

M = mass of electron = mass of positron = 9.11 x 10^-31 kg,

radius of the positron = 5.03 mm

We convert to meters.

1000mm = 1m

5.03mm = xm

Cross multiply

x = 5.03/1000mm

x = 0.00503m

q = Electric charge = -1.6 x 10^-19 C

Magnetic field (B) = 0.85 T

Speed of the positron is unknown

0.85 = 9.11 x 10^-31 kg × v/ -1.6 x 10^-19 C × 0.00503

0.85 × 1.6 x 10^-19 C × 0.00503 = 9.11 x 10^-31 kg × v

v = 0.85 × -1.6 x 10^-19 C × 0.00503/9.11 x 10^-31 kg

v = 6.8408 ×10-22/ 9.11 x 10^-31 kg

v = 750911086.72m/s

Formula for complete revolutions =

Speed × time / Circumference

Time = 2.30s

Circumference of the circular path = 2πr

r =0.00503

Circumference = 2 × π × 0.00503

= 0.0316044221

Revolution = 750911086.72 × 2.30/0.0316044221

= 1727095499.5/0.0316044221

= 546541562294 revolutions

Approximately = 5.465 × 10^10 revolutions

A high school physics student claims her muscle car can achieve a constant acceleration of 10 ft/s/s. Her friend develops an accelerometer to confirm the feat. The accelerometer consists of a 1 ft long rod (mass=4 kg) with one end attached to the ceiling of the car, but free to rotate. During acceleration, the rod rotates. What will be the angle of rotation of the rod during this acceleration? Assume the road is flat and straight.

Answers

Answer: Ф = 17.2657 ≈ 17°

Explanation:

we simply apply ET =0 about the ending of the rod

so In.g.L/2sinФ - In.a.L/2cosФ = 0

g.sinФ - a.cosФ = 0

g.sinФ = a.cosФ

∴ tanФ = a/g

Ф =  tan⁻¹ a / g

Ф = tan⁻¹ ( 10 / 32.17405)

Ф = tan⁻¹ 0.31080948777

Ф = 17.2657 ≈ 17°

Therefore the angle of rotation of the rod during this acceleration is 17.2657 ≈ 17°

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