To understand the experiment that led to the discovery of the photoelectric effect.
In 1887, Heinrich Hertz investigated the phenomenon of light striking a metal surface, causing the ejection of electrons from the metal. The classical theory of electromagnetism predicted that the energy of the electrons ejected should have been proportional to the intensity of the light. However, Hertz observed that the energy of the electrons was independent of the intensity of the light. Furthermore, for low enough frequencies, no electrons were ejected, no matter how great the intensity of the light became. The following problem outlines the methods used to investigate this new finding in physics: the photoelectric effect.
Suppose there is a potential difference between the metal that ejects the electrons and the detection device, such that the detector is at a lower potential than the metal. The electrons slow down as they go from higher to lower electric potential; since they must overcome this potential difference to reach the detector, this potential is known as the stopping potential. To reach the detector, the initial kinetic energy of an ejected electron must be greater than or equal to the amount of energy it will lose by moving through the potential difference.
(A) If there is a potential difference V between the metal and the detector, what is the minimum energy E_min that an electron must have so that it will reach the detector?
Express your answer in terms of V and the magnitude of the charge on the electron, e.
(B) Suppose that the light carries energy E_light. What is the maximum stopping potential V_0 that can be applied while still allowing electrons to reach the detector?
Express your answer in terms e, E_light, and Φ.

The magnitude of force acting on the bumper is 3760 N.

It states that the Work done in moving a body is equal to the change in kinetic energy of the body

Kinetic energy = 1/2 mv²

Given is a car's bumper designed to withstand 4.32 km/h or 1.2 m/s collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance.

The cushion collapses 0.180 m while bringing 940 kg car to rest from a initial speed of 1.2 m/s

Work done = Force x displacement

As the displacement of the bumper and force acted on it is in same direction, so the work done is

W = Fxcos0° = Fx

The body is coming to rest, so, final velocity is zero. Then, change in kinetic energy will be

ΔK.E = K.Ef - K.Ei

ΔK.E = m/2 (v² - u²)

According to work energy theorem, work done is

W = Fx  = m/2 (v² - u²)

Substitute the value and calculate the force,

F  = [940 x (0² - 1.2²)] / 2x0.180

F = 3760 N

Thus, the magnitude of force is 3760 N.

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The speed of light in material B is 1.6625 × 108 m/s.

The refractive index of a material is its optical density relative to that of a vacuum.

Material B has a refractive index of nB, and its speed of light is vB.

The speed of light in material A is given as 1.25 x 108 m/s.

The refractive indices of materials A and B have a ratio of nA/nB = 1.33.

We will use the formula:

nA/nB = vB/vA = nA/nB.

Therefore, nA/nB = vB/1.25 x 108 m/s.

This equation can be rearranged to give the speed of light in material B:

vB = nA/nB × 1.25 x 108 m/s.

Therefore, vB = 1.33 × 1.25 × 108 m/s.

We will perform this calculation:

vB = 1.6625 × 108 m/s.

Therefore, the speed of light in material B is 1.6625 × 108 m/s.

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Answer:

option B is correct

Explanation:

in geocentric model earth is at the center of the universe and sun,moon star etc orbited the earth  thats why option B is correct

Answer:

12.7m/s

Explanation:

Given parameters:

Mass of diver  = 77kg

Height of jump  = 8.18m

Unknown:

Final velocity  = ?

Solution:

To solve this problem, we apply the motion equation below:

             v²   = u²  + 2gH

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

H is the height

 Now insert the parameters and solve;

       v² = 0²  +  2 x 9.8 x 8.18

     v  = 12.7m/s

Answer:

Speed of A + B = 1 m/s

Explanation:

The speed of B = 0    because the distance does not change with time .

The speed of A =   1 m/s   ( the slope of the line)

Answer:

The circuit is missing attached below is the required circuit

answer :

a) Ic = 1.944 mA

  Rp = 288.66 kΩ

b) The Emitter-base Junction of the BJT is forward biased while its collector-base junction is reverse biased

Explanation:

Rc = 3.6 kΩ

VBE = 0.8 v

1) predict Ic and specify Rp to establish Vce at 5 V

we will apply Kirchhoff's voltage law to resolve this

solution attached below

b ) The BJT is said to be in Forward reverse bias because The Emitter-base Junction of the BJT is forward biased while its collector-base junction is reverse biased

The smallest angle at which the rope can bend up from the horizontal on either side of the tightrope walker to avoid breaking is approximately 82.8 degrees.

A tightrope walker is a skilled performer who walks or performs acrobatics on a thin, flexible rope, typically suspended at a considerable height.

Let's start by calculating the weight of the walker:

weight = mass x acceleration due to gravity = 60 kg x 9.8 m/s^2 = 588 N

To avoid breaking the rope, the tension in the rope cannot exceed 5200 N. Let's assume that the angle of the rope with respect to the horizontal is θ.

The tension in the rope can be calculated using the following formula:

tension = weight / (2 x cosθ)

Where

We can rearrange this formula to solve for the angle θ:

cosθ = weight / (2 x tension)

θ = arccos(weight / (2 x tension))

Plugging in the values for weight and tension gives:

θ = arccos(588 N / (2 x 5200 N)) = 82.8 degrees

Therefore, the smallest angle at which the rope can bend up from the horizontal on either side of the tightrope walker to avoid breaking is approximately 82.8 degrees.

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The centripetal force in Newtons provided by the friction between the blade of the skate and the ice is 490 N

The following data were obtained from the question:

The centripetal force can be obtained as illustrated below:

F = mv²/r

= (50 × 7²) / 5

= (50 × 49) / 5

= 2450 / 5

= 490 N

Thus, we can concluded that the centripetal force is 490 N

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An arrow pointing from the bottom of the ramp to the top, I assume it would be friction.

Given

m1 = 5 g

m2 = 10 g

Released at the same time

Procedure

Gravity causes everything to fall at the same speed. This is why balls that weigh different amounts hit the ground at the same time.

The answer would be "They will hit the ground at the same time"

The second reflection is created as the light frvl as it passes through the block. Multiple pictures are created via surface diffraction.

The act of light spreading out after passing through a narrow area or bending around an object is known as diffraction. The dispersion of light waves is a topic covered in physics classes. Sound, television, and water waves are all susceptible to diffraction.

Diffraction is the small bending of light that occurs when it travels around an object's edge. The degree of bending is influenced by the wavelength of light's size in relation to the opening's size. When the opening exceeds the wavelength of the light, the

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Answer:

Approximately \(0.293\; {\rm kg \cdot m^{2}}\).

Explanation:

Since the wheel started from rest, initial angular velocity will be \(\omega_{0} = 0\; {\rm rad \cdot s^{-1}}\). It is given that the angular velocity \(\omega_{1}\) is \(14.9\; {\rm rev \cdot s^{-1}}\) after \(t = 3.41\; {\rm s}\). Apply unit conversion and ensure that all angular velocity are measured in radians-per-second:

\(\begin{aligned} \omega_{1} &= 14.9\; {\rm rev \cdot s^{-1}} \times \frac{2\, \pi \; {\rm rad}}{1\; {\rm rev}} \\ &\approx 93.620\; {\rm rad \cdot s^{-1}}\end{aligned}\).

Change in angular velocity:

\(\begin{aligned} \Delta \omega = \omega_{1} - \omega_{0} \approx 93.620\; {\rm rad \cdot s^{-1}}\end{aligned}\).

Since the tangential force is constant and there is no friction on the wheel, the angular acceleration \(\alpha\) of this wheel will be constant. Since the change in velocity \(\Delta \omega \approx 93.620\; {\rm rad \cdot s^{-1}}\) was achieved within \(t = 3.41\; {\rm s}\), the average angular acceleration will be:

\(\begin{aligned} \alpha &= \frac{\Delta \omega}{t} \\ &\approx \frac{93.620\; {\rm rad \cdot s^{-1}}}{3.41\; {\rm s}} \\ &\approx 27.45\; {\rm rad \cdot s^{-2}}\end{aligned}\).

At a distance of \(r = 0.110\; {\rm m}\) from the axis of rotation, the tangential force \(F = 73.0\; {\rm N}\) will exert on the wheel a torque \(\tau\) of magnitude:

\(\begin{aligned} \tau &= F\, r \\ &= (73.0\; {\rm N})\, (0.110\; {\rm m}) \\ &\approx 8.030\; {\rm N \cdot m}\end{aligned}\).

Let \(I\) denote the moment of inertia of this wheel. The equation \(\alpha = (\tau / I)\) relates angular acceleration \(\alpha\) to moment of inertia \(I\!\) and net torque \(\tau\). Rearrange this equation to find the moment of inertia:

\(\begin{aligned}I &= \frac{\tau}{\alpha} \\ &\approx \frac{8.030\; {\rm N\cdot m}}{27.45\; {\rm rad \cdot s^{-2}}} \\ &\approx 0.293\; {\rm N \cdot m \cdot s^{2}} \\ &= 0.293 \; {\rm kg \cdot m^{2}}\end{aligned}\).

Note that the unit "radians" is typically ignored. Additionally, \(1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^{-2}}\).

Hence, the moment of inertia of this wheel is approximately \(0.293\; {\rm kg \cdot m^{2}}\).

Speeding up slowing down, and going around curves are all examples of acceleration .

A definition of acceleration is the speed at which velocity varies with respect to time.

Since acceleration has both a magnitude and a direction, it is a vector quantity. Additionally, it is the first derivative of velocity with respect to time or the second derivative of position with respect to time.

The acceleration is given by the formula ,

a = v/t

where , a =acceleration

v = velocity

t = time

the unit of acceleration is m/s² .

The overall change in velocity during the specified interval divided by the total amount of time required for the change is the definition of average acceleration over a particular period of time. It is represented as for a specific amount of time.

a = Δv/Δt

The definition of instantaneous acceleration is the ratio of velocity change during a specified period of time such that the period of time ends.

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Answer:

Water potential is the potential energy of water per unit volume relative to pure water in reference conditions. Water potential quantifies the tendency of water to move from one area to another due to osmosis, gravity, mechanical pressure and matrix effects such as capillary action.

Answer:

water potential is the potential energy of the water per unit volume.

Answer:

In 2004, the Stardust spacecraft made a close flyby of comet Wild-2, collecting comet and interstellar dust in a substance called aerogel. Two years later, the samples made it back to Earth in a return capsule that landed in the Utah desert.

Answer:

v₃ = 1.25[m/s]

Explanation:

In order to solve this problem, we must use the principle of conservation of the amount of movement and momentum.

That is, the amount of movement is preserved before and after the collision. Let's propose an equation, in which the elements to the left of the equal sign represent the moment before the collision and to the right the moment after the collision.

\(P=m*v\)

P = lineal momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

\((m_{1}*v_{1})+(m_{2}*v_{2})=(m_{1}+m_{2})*v_{3}\)

m₁ = mass of the first railroad = 2096 [kg]

m₂ = mass of the second railroad = 6280 [kg]

v₁ = velocity of the first railroad before the collision = 5 [m/s]

v₂ = velocity of the second railroad before the collision = 0 (initially at rest)

v₃ = velocity of the group after the collision [m/s]

Now replacing:

\((2096*5)+(6280*0)=(2096+6280)*v_{3}\\10480=8376*v_{3}\\v_{3}=1.25[m/s]\)

Hello,

I hope you and your family are doing well!

To find the angle of the applied force (α), you can rearrange the equation for the net force to solve for α. The equation for the net force is:

F_net = F * cos(α + β) - m * g * sin(β) - u * N

To solve for α, you can start by isolating the term containing α. You can do this by subtracting the other terms from both sides of the equation, which gives you:

F_net - F * cos(β) + m * g * sin(β) + u * N = F * cos(α + β)

Then, you can divide both sides of the equation by F * cos(β) to get:

[F_net - F * cos(β) + m * g * sin(β) + u * N] / (F * cos(β)) = cos(α + β)

Next, you can use the identity cos(a + b) = cos(a)cos(b) - sin(a)sin(b) to rewrite the right side of the equation:

[F_net - F * cos(β) + m * g * sin(β) + u * N] / (F * cos(β)) = cos(α)cos(β) - sin(α)sin(β)

Finally, you can rearrange this equation to solve for α:

α = atan2(sin(α) * cos(β) + cos(α) * sin(β), cos(α) * cos(β) - sin(α) * sin(β))

This equation gives the angle of the applied force (α) in terms of the other variables in the equation for the net force.

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Happy Holidays!

The radius of curvature of the bowl is 42 cm. It should be noted that this is a negative value, indicating that the mirror is concave (curved inward) rather than convex (curved outward).

The situation described can be modeled using the thin lens equation, which relates the object distance, image distance, and focal length of a lens or curved mirror. In this case, we can assume that the bowl acts as a concave mirror, with its reflecting surface forming a portion of a sphere.

We can also assume that the mirror is thin, meaning that the distance between the reflecting surface and the center of curvature is much greater than the radius of curvature.

Let us denote the radius of curvature of the mirror as R. When the child looks into the bowl, an erect image of himself is formed at a distance of 21 cm from the bowl. Using the thin lens equation, we can write:

\(1/f = 1/d_o + 1/d_i\)

where f is the focal length of the mirror, \(d_o\) is the object distance (i.e., the distance between the child and the mirror), and \(d_i\)is the image distance (i.e., the distance between the mirror and the image).

In this case, the object distance is negative (since the child is looking into the bowl from the same side as the object), and the image distance is positive (since the image is erect). Substituting the given values, we obtain:

\(1/f = -1/21 + 1/d_i\)

Next, when the child turns the bowl over, another erect image of himself is formed at a distance of 7 cm from the bowl. Using the same equation, we can write:

\(1/f = -1/d_o + 1/d_i'\)

where d_i' is the image distance for this case. In this case, the object distance is positive (since the child is looking at the bowl from the opposite side of the object), and the image distance is also positive (since the image is erect). Substituting the given values, we obtain:

\(1/f = 1/7 + 1/d_i'\)

Now we have two equations for f in terms of \(d_i\) and \(d_i\)', respectively. We can solve for \(d_i\)and \(d_i\)' by setting these two equations equal to each other and simplifying:

\(-1/21 + 1/d_i = 1/7 + 1/d_i'\\d_i' = 1/(1/7 - 1/21 + 1/d_i) = 28 cm\)

Now that we know both image distances, we can solve for the radius of curvature R using the mirror formula, which relates the focal length to the radius of curvature:

1/f = 2/R

Substituting the values of f and \(d_i\), we obtain:

\(1/R = 2/f = -1/21 + 1/d_i\)

Solving for R, we obtain:

R = -42 cm

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The capacitance C of a parallel-plate capacitor whose plates have an area A and are separated by a distance d is:

Where ε₀ is the vacuum permitivity:

Notice that the capacitance of the parallel-plate capacitor is given, as well as the area of its plates. Then, isolate d from the equation:

Replace A=100mm^2, C=100pF and the value of vacuum permitivity to find the separation between the plates:

Therefore, the correct choice is: 8.85x10^-6m.

Complete Question

A 940-g rock is whirled in a horizontal circle at the end of a 1.30-m-long string,  If the breaking strength of the string is 120 N, what's the minimum angle the string can make with the horizontal?

Answer:

The value is \(\theta = 4.41^o\)      

Explanation:

From the question we are told that

  The mass of the rock is  \(m_r = 940 \ g = 0.94 \ kg\)

   The length of the string is \(l = 1.30 \ m\)

    The breaking strength(i.e the maximum tension) on the string is \(T = 120 \ N\)

Gnerally the vertical component of the tension experienced by the string is mathematically represented as

     \(T_v = T sin(\theta)\)

Generally this vertical component  of tension is equivalent to the weight of the rock

So

     \(Tsin (\theta) = mg\)

=>  \(\theta = sin^{-1} [\frac{mg}{ T} ]\)

=>    \(\theta = sin^{-1} [\frac{0.940 *9.8 }{ 120} ]\)

=>    \(\theta = 4.41^o\)      

Answer:

First, we can find the velocity of the rock just before it hits the ground using the equation:

v^2 = 2gh

where v is the final velocity, g is the acceleration due to gravity (9.81 m/s^2), and h is the height the rock falls from (7.0 m).

v^2 = 2(9.81 m/s^2)(7.0 m) = 136.89 m^2/s^2

v = √136.89 m^2/s^2 = 11.7 m/s

The momentum change of the rock caused by its fall is:

Δp = mv = (18 kg)(11.7 m/s) = 211.2 kg m/s

The resulting change in the magnitude of earth's velocity can be found using the conservation of momentum equation:

m1v1 + m2v2 = (m1 + m2)v'

where m1 and v1 are the mass and velocity of the rock before it falls, m2 and v2 are the mass and velocity of the earth before the rock falls (which we can assume is negligible), and v' is the velocity of the rock-earth system after the rock falls.

We can rearrange this equation to solve for v':

v' = (m1v1)/(m1 + m2)

v' = (18 kg)(11.7 m/s)/(18 kg + 6.0 × 10^24 kg)

v' = 1.95 × 10^-24 m/s

This means that the change in the magnitude of earth's velocity is essentially zero (which makes sense, since the mass of the earth is so much greater than the mass of the rock).

The momentum change of the rock caused by its fall is 211.2 kg·m/s, and the resulting change in the magnitude of earth’s velocity is 3.5 × 10⁻²³ m/s.

What is Momentum?

Momentum is a physics concept that describes the amount of motion an object has.  The units of momentum are kg m/s. In a closed system, the total momentum before an event or interaction is equal to the total momentum after the event or interaction.

To solve this problem, we need to use the law of conservation of momentum, which states that the momentum of an isolated system remains constant.

The momentum change of the rock is given by the formula:

Δp = mv

where m is the mass of the rock, and v is the change in velocity.

First, we need to find the final velocity of the rock. We can use the formula for the velocity of a falling object:

v² = 2gh

where g is the acceleration due to gravity, and h is the height of the fall.

Plugging in the given values, we get:

v² = 2 × 9.8 m/s² × 7.0 m

v² = 137.2

v = √137.2

v = 11.7 m/s

The change in the magnitude of Earth's Velocity  can be found using the formula:

Δv = Δp / M

where M is the mass of the earth.

The mass of the earth is given as 6.0 × 10²⁴ kg.

Δp = mΔv

Δp = 18 kg × 11.7 m/s

Δp = 211.2 kg·m/s

Δv = Δp / M

Δv = 211.2 kg·m/s / 6.0 × 10²⁴ kg

Δv = 3.5 × 10⁻²³ m/s

Therefore, the momentum change of the rock caused by its fall is 211.2 kg·m/s, and the resulting change in the magnitude of earth’s velocity is 3.5 × 10⁻²³ m/s.

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The conservation of energy allows finding the result of the velocity of the body 2 when it reaches the ground is:

          v = 6.32 m / s

Given parameters

To find

The conservation of energy is one of the most important principles of physics, it establishes that if there is no friction force, the mechanical energy is conserved at all points, where the mechanical energy is the sum of the kinetic energy plus the different potential energies.

Let's write the energy at two points before we drop the system.

     Em₀ = U₁ + U₂

Where potential energy is

    U = m g y

They indicate that body 1 is on the ground and body 2 is at a height .

      y₂ = 8 m

      Em₀ = m₂ g y₂

Let's write the energy of the system for when body 2 is reaching the ground

       \(Em_f\) = K₂ + K₁ + U₂ + U₁

At this moment body 2 is reaching the ground, its height is zero and the height of body 1 is y₁ = 8 m. Since the two bodies are joined by the rope, they must have the same speed..

      \(Em_f\) = ½ (m₁ + m₂) v² + m₁ g y₁

As there is no friction, the energy is conserved.

      Em₀ = Em_f

      m₂ g y₂ = ½ (m₁ + m₂) v² + m₁ g y₁

The rope is inextensible, the heights are equal.

       y₁ = y₂ = y = 8.0 m

      v² = \(\frac{m2-m1}{m2+m1} \ 2g y\)  

Let's calculate.

       \(v^2 =\frac{22.9-13.6}{ 22.9+13.6} \ 2 \ 9.8 \ 8 \\ v = \sqrt{39.95}\)

      v = 6.32 m / s

In conclusion using the conservation of energy we can find the result of the velocity when the body 2 reaches the ground is:

          v = 6.32 m / s

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Answer:

D . The arrows on plates A and B would show the plated moving away from each other

Explanation:

The correct arrow in this instance will show that the plates A and B would be moving away from each other.

At the mid-ocean ridge, two plates are moving apart and pulling away from one another.

To determine the magnetic force on a straight wire carrying a current in a uniform magnetic field, we can use the formula for the magnetic force:

F = I * L * B * sin(θ)

where:

F is the magnetic force,

I is the current in the wire,

L is the length of the wire,

B is the magnitude of the magnetic field, and

θ is the angle between the wire and the magnetic field.

In this case, the values are:

I = 30 A (current in the wire)

L = 2.0 m (length of the wire)

B = 55 mT = 0.055 T (magnitude of the magnetic field)

θ = 20° (angle between the wire and the magnetic field)

Substituting the values into the formula:

F = 30 A * 2.0 m * 0.055 T * sin(20°)

Calculating sin(20°):

F = 30 A * 2.0 m * 0.055 T * 0.3420

F ≈ 1.5714 N

Therefore, the magnetic force on the 2.0-meter length of wire carrying a current of 30 A in a region with a uniform magnetic field of magnitude 55 mT and at an angle of 20° away from the wire is approximately 1.5714 N.

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A pair of cyan sunglasses will absorb or block light in the blue-green range and transmit light in the red-orange and yellow-green range.

What is Light?

Light is a form of electromagnetic radiation that is visible to the human eye. It is a type of energy that travels through space as waves and is characterized by its wavelength and frequency. Light can be produced by a variety of sources, including the sun, light bulbs, and electronic devices. It can also be manipulated through lenses, mirrors, and other optical devices to create different effects, such as magnification or dispersion.

Cyan (bluish-green) sunglasses will transmit or allow the passage of wavelengths of light that appear bluish-green to our eyes. This means that the lenses of these sunglasses are designed to absorb or block other wavelengths of visible light that correspond to other colors. The specific wavelengths of visible light that are absorbed or blocked by cyan sunglasses will depend on the exact properties of the dye or pigment used to color the lenses.

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