A beam of light, with a wavelength of 4170 Å, is shined on sodium, which has a work function (binding energy) of 4.41 × 10 –19 J. Calculate the kinetic energy and speed of the ejected electron.

Answer:

8 minutes

Explanation:

The time it takes for light to travel from the Sun to Earth can be found using the formula time = distance/speed. Given the speed of light is known to be 300,000 km/s and the distance is 150,000,000 km, it will take light about 500 seconds to travel from the Sun to the Earth.

To calculate the time it takes for light to travel from the Sun to Earth, you need to use the formula for time which is distance/speed. The speed of light is 300,000 km/s and the distance from the Sun to Earth is 150,000,000 km.

So, Time = Distance / Speed = 150,000,000 km / 300,000 km/s = 500 seconds

Therefore, it takes approximately 500 seconds for light to travel from the Sun to Earth.

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Answer:

[tex]v_{f} \approx 6.647\,\frac{m}{s}[/tex]

Explanation:

The final speed of the cyclist is determined by applying the Principle of Energy Conservation:

[tex]\frac{1}{2}\cdot m\cdot v_{o}^{2} + m\cdot g\cdot h_{o} = \frac{1}{2}\cdot m\cdot v_{f}^{2} + m\cdot g\cdot h_{f}[/tex]

[tex]\frac{1}{2} \cdot v_{o}^{2} + g\cdot (h_{o}-h_{f}) = \frac{1}{2}\cdot v_{f}^{2}[/tex]

[tex]v_{f}^{2}=v_{o}^{2} + 2\cdot g \cdot (h_{o}-h_{f})[/tex]

[tex]v_{f} = \sqrt{v_{o}^{2}+2\cdot g \cdot (h_{o}-h_{f})}[/tex]

[tex]v_{f} = \sqrt{v_{o}^{2}-2\cdot g \cdot \Delta s \cdot \sin \theta}[/tex]

[tex]v_{f} = \sqrt{(9\,\frac{m}{s} )^{2}-2\cdot (9.807\,\frac{m}{s} )\cdot (12\,m)\cdot \sin 9^{\textdegree}}[/tex]

[tex]v_{f} \approx 6.647\,\frac{m}{s}[/tex]

Answer:

the final speed of the 10.85 m/s.

Explanation:

Given that,

Slope with respect to horizontal, [tex]\theta=9^{\circ}[/tex]

Distance travelled, d = 12 m

Initial speed of the cyclist, u = 9 m/s

We need to find the final speed of the cyclist. Let h is the height of the sloping surface such that,

[tex]h=d\times sin\thetah\\\\=12\times sin(9) \\\\h = 1.877 m[/tex]

v is the final speed of the cyclist. It can be calculated using work energy theorem as

[tex]\dfrac{1}{2}m(v^2-u^2)\\\\=mgh\dfrac{1}{2}(v^2-u^2)\\\\=gh\dfrac{1}{2}\times (v^2-(9.0)^2)\\\\=9.8\times 1.87\\\\v = 10.85 m/s[/tex]

Thus,the final speed of the 10.85 m/s.

Answer:

The initial angular acceleration is [tex]16.8 s^{-2}[/tex]

Explanation:

From the parallel axis theorem the moment of inertia about 25cm mark is

[tex]I = \dfrac{1}{12}ML^2+MD^2[/tex]

since [tex]L = 1m[/tex] & [tex]D = 0.25m[/tex], we have

[tex]I = \dfrac{1}{12}M(1)^2+M(0.25m)^2[/tex]

[tex]I = \dfrac{7}{48} M[/tex]

Now, the gravitational force (equal to the weight of the object) acts as a torque [tex]\tau[/tex] on the center of mass of the rod, which induces angular acceleration [tex]\alpha[/tex]according to

[tex]\tau = I\alpha[/tex]

since [tex]\tau = Mg D[/tex]

[tex]MgD =I \alpha[/tex]

[tex]MgD = \dfrac{7}{48} M \alpha[/tex]

solving for [tex]\alpha[/tex] we get

[tex]\boxed{\alpha = \dfrac{48}{7} gD}[/tex]

putting in [tex]g= 9.8m/s^2[/tex] and [tex]D = 0.25m[/tex] we get:

[tex]\boxed{\alpha = 16.8\: s^{-2}.}[/tex]

which is the initial angular acceleration.

The initial angular acceleration of the stick is 16.812 radians per square second.

First, we calculate the resulting moment of inertia by Steiner theorem:

[tex]I_{O} = I_{g} + M\cdot r^{2}[/tex] (1)

Where:

If we know that [tex]I_{g} = \frac{1}{12} \cdot M\cdot L^{2}[/tex], [tex]L = 1\,m[/tex] and [tex]r = 0.25\,m[/tex], then the formula for the moment of inertia is:

[tex]I_{O} = \frac{1}{12}\cdot M + \frac{1}{16}\cdot M[/tex]

[tex]I_{O} = \frac{7}{48}\cdot M[/tex] (2)

We know that initial angular acceleration ([tex]\alpha[/tex]), in radians per square second, is solely due to gravity. By the second Newton's law and the D'Alembert principle, we derive an expression for the initial angular acceleration of the stick:

[tex]\Sigma M = M\cdot g\cdot r = I_{O}\cdot \alpha[/tex] (3)

Where:

By (2) and (3) we have the following simplified expression:

[tex]M\cdot g \cdot r = \frac{7}{48} \cdot M\cdot \alpha[/tex]

[tex]\alpha = \frac{48}{7}\cdot g \cdot r[/tex] (4)

If we know that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]r = 0.25\,m[/tex], then the initial angular acceleration of the stick is:

[tex]\alpha = \frac{48}{7}\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.25\,m)[/tex]

[tex]\alpha = 16.812\,\frac{rad}{s^{2}}[/tex]

The initial angular acceleration of the stick is 16.812 radians per square second.

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Answer:in a spring tide the sun , moon and the earth are in a straight line which causes regular higher tide and low tide to increase.

Explanation:

In a spring tide, the Moon, Earth, and Sun are all in allignment, which causes the greatest gravitational pull on the water of Earth. Basically, when the Moon and Sun align towards and in the same direction at the Earth, their gravitational pulls join together making the pull stronger. (see image below)

What this very strong pull does is make high tides higher than usual, and low tides lower than usual. This is because the pull on tides is very strong and more water is being pulled to form higher high tides, making there less water for there to be in low tides.

(image is from oceanservice.gov)

Answer:

a. yes

Explanation:

The initial speed of the circular saw is:

[tex]\dot n_{o} = \frac{6000}{60} \,\frac{rev}{s}[/tex]

[tex]\dot n_{o} = 100\,\frac{rev}{s}[/tex]

Deceleration rate needed to stop the circular saw is:

[tex]\ddot n = -\frac{100\,\frac{rev}{s} }{2\,s}[/tex]

[tex]\ddot n = - 50\,\frac{rev}{s^{2}}[/tex]

The number of turns associated with such deceleration rate is:

[tex]\Delta n = \frac{\dot n^{2}}{2\cdot \ddot n}[/tex]

[tex]\Delta n = \frac{\left(100\,\frac{rev}{s} \right)^{2}}{2\cdot \left(50\,\frac{rev}{s^{2}} \right)}[/tex]

[tex]\Delta n = 100\,rev[/tex]

Since the measured number of revolutions is lesser than calculated number of revolution, the circular saw meets specifications.

Answer:

5.213ft

Explanation:

Z² = x² + y²

x = √(z² - y²)

y = 52ft, dx = 6ft, z = 105ft, dz = ?

d(z² = x² + y²)

2zdz = 2xdx

dz = xdx/z

But x = √(z² - y²)

dz = √(z² - y²)/z * dx

dz = [√(105² - 52²)/105] * 6

dz = √(8521)/ 17.5

dz = 5.213ft

Answer:

a) P ( X ≤ 49 ) = 0.9149

b) P ( X ≥ 48 ) = 0.2119

c) 0.9876

Explanation:

Solution:-

- Lets define a random variable X: the maximum speed of a moped.

- The random variable follows normal distribution with the following parameters:

                    X ~ Norm ( u , σ^2 )

Where,

              u = Mean = 46.6 km/h

              σ = Standard deviation 1.75 km/h

Hence,

                    X ~ Norm ( 46.6 , 1.75^2 ).

a) The probability that the maximum speed of mopeds is atmost 49 km/h?

- To evaluate the probability of P ( X ≤ 49 ). We will find the standard Z-score value using the following formula:

                       [tex]Z-score = \frac{x-u}{s} \\\\ Z-score = \frac{49-46.6}{1.75} \\\\Z-score = 1.37142[/tex]

- Now use the standard normal tables to determine the required probability of:

                     P ( Z < 1.37142 ) = 0.9149

Hence,

                     P ( X ≤ 49 ) = 0.9149  

b) The probability that the maximum speed of mopeds is at-least 48 km/h?

- To evaluate the probability of P ( X ≥ 48 ). We will find the standard Z-score value using the following formula:

                       [tex]Z-score = \frac{x-u}{s} \\\\ Z-score = \frac{48-46.6}{1.75} \\\\Z-score = 0.8[/tex]

- Now use the standard normal tables to determine the required probability of:

                     P ( Z ≥ 0.8 ) = 0.2119

Hence,

                     P ( X ≥ 48 ) = 0.2119                          

c) The probability that the maximum speed of mopeds differs from mean by at-most 2.5 standard deviation.

- The required probability is the standard error from the mean value "u" of 2.5 standard deviation.

- We don't need to evaluate the test statistics as we are already given the standard error about mean.

- Using, the standard normal Z-score: The required probability is:

                P ( -2.5 < Z < 2.5 ) = 2*P ( Z < 2.5 ) - 1

                                              = 2*0.9938 - 1

                                              = 0.9876

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Answer:

a) αA = 4.35 rad/s²

αB = 1.84 rad/s²

b) t = 3.7 rad/s²

Explanation:

Given:

wA₀ = 240 rpm = 8π rad/s

wA₁ = 8π -αA*t₁

The angle in B is:

[tex]\theta _{B} =4\pi =\frac{1}{2} \alpha _{B} t_{1}^{2} =\frac{1}{2} (\frac{r_{A} }{r_{B} } )^{3} \alpha _{A} t_{1}^{2}=\frac{1}{2} (\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}^{2}[/tex]

[tex]\alpha _{A} =8\pi (\frac{0.2}{0.15} )^{3} =59.57rad[/tex]

[tex]w_{B,1} =\alpha _{B} t_{1}=(\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}=0.422\alpha _{A} t_{1}[/tex]

The velocity at the contact point is equal to:

[tex]v=r_{A} w_{A} =0.15*(8\pi -\alpha _{A} t_{1})=1.2\pi -0.15\alpha _{A} t_{1}[/tex]

[tex]v=r_{B} w_{B} =0.2*(0.422\alpha _{A} t_{1})=0.0844\alpha _{A} t_{1}[/tex]

Matching both expressions:

[tex]1.2\pi -0.15\alpha _{A} t_{1}=0.0844\alpha _{A} t_{1}\\\alpha _{A} t_{1}=16.09rad/s[/tex]

b) The time during which the disks slip is:

[tex]t_{1} =\frac{\alpha _{A} t_{1}^{2}}{\alpha _{A} t_{1}} =\frac{59.574}{16.09} =3.7s[/tex]

a) The angular acceleration of each disk is

[tex]\alpha _{A}=\frac{\alpha _{A} t_{1}}{t_{1} } =\frac{16.09}{3.7} =4.35rad/s^{2} (clockwise)[/tex]

[tex]\alpha _{B}=(\frac{0.15}{0.2} )^{3} *4.35=1.84rad/s^{2} (clockwise)[/tex]

The angular acceleration of each disk and the time during which the disks slip cannot be determined with the given information.

(a) To determine the angular acceleration of each disk, we can use the formula: angular acceleration = (final angular velocity - initial angular velocity) / time. For disk A, the initial angular velocity is 240 rpm counterclockwise and the final angular velocity is 0 rpm. The time during which disk A stops is not given, so we cannot determine its angular acceleration at this time. For disk B, the initial angular velocity is 0 rpm and the final angular velocity is also 0 rpm. Since disk B is at rest initially and then starts moving, it undergoes an angular acceleration that brings it to rest. However, the specific time during which disk B slips is not given, so we cannot determine its angular acceleration during slipping.

(b) The time during which the disks slip is not provided in the question, so we cannot determine the exact time. However, we know that disk B makes 2 revolutions before reaching its final angular velocity, so during this period, disk B is slipping.

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Answer:

The answer is -x direction.

Explanation:

According to the right hand rule for electromagnetism, when the thumb points up, index finger points forward and the middle finger is perpendicular to the index finger, the thumb points in the direction of the magnetic force, the index finger in the direction of the charge movement and the middle finger in the direction of the electromagnetic lines. If the electric field is pointing in the -z direction which is into the screen and the magnetic field is in the +y direction which is upwards, then the magnetic lines are in the -x direction.

I hope this answer helps.

Answer:

a) L = 1.17 m

b) width of central maxima = 1.28 mm

Explanation:

Given:-

- The wavelength of light, λ = 588.0 nm

- The slit of width, a = 0.74 mm

Find:-

(a) At what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be 0.93 mm from the central maximum

(b) Calculate the width of the central maximum.

Solution:-

- The results of Young's single slit experiment are given in form of a relation as the angle of separation between fringes ( θ ) as function of fringe order ( m ) and wavelength ( λ ).

- Destructive interference produces the dark fringes ( minimum ).  Dark fringes in the diffraction pattern of a single slit are found at angles θ for which:

                        a*sin ( θ ) = m*λ

Where,

            m : The order number for the minimum ( dark fringe ).

- We are to investigate for the first (m = 1 )) dark fringe (minima) which is y = 0.93 mm from central order ( m = 0 ) for which the screen must be placed at a distance L:

                        a*y/L = m*λ

                        L = a*y / m*λ      

                        L = (0.74*0.93) / (1*588*10^-9)    

                        L = 1.17 m

- The distance L of screen should be 1.17 m away from slit.

- The central maximum - central bright fringe. The maxima lie between the minima and the width of the central maximum is simply the distance between the 1st order minima from the centre of the screen on both sides of the centre.

                                tanθ ≈ θ ≈ y/L = w*λ

                                y = w*λ*L

The width of the central maximum is simply twice this value:

- Width of central maximum = 2λLw = 2*588*10^-6*1170*0.93  = 1.28 mm                                

Answer:

a) 1.17 m

b) 0.929mm

Explanation:

(a) to find the distance to the screen you use

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

m: order of the fringe

lambda: wavelength of the light = 588*10^{-9} m

D: distance to the screen

d: distance between slits = 0.74*10^-3 m

by doing D the subject of the formula and replacing the values of the other parameters you obtain:

[tex]D=\frac{dy}{m\lambda}=\frac{(0.74*10^{-3}m)(0.93*10^{-3})}{(1)(588*10^{-9}m)}=1.17m[/tex]

the distance to the screen is 1.17m

(b) to find the width of the central maximum you calculate the position of the first dark fringe:

[tex]y=\frac{(1)(588*10^{-9}m/2)(1.17m)}{0.74*10^{-3}m}=4.64*10^{-4}m=0.464mm[/tex]

[tex]y=\frac{m(\lambda/2)D}{d}[/tex]

2y = 2(0.464mm)=0.929mm is the width of the central maximum

Answer:

8.049 MW

Explanation:

The expression for gravitational potential energy is given as

Ep = mgh............. Equation 1

Ep = gravitational potential energy, m = mass of water, h = height, g = acceleration due to gravity.

Given: m = 58.4×10³ kg, h = 20.1 m, g = 9.81 m/s²

Substitute into equation 1

Ep =  58.4×10³(20.1)(9.81)

Ep = 1.6098×10⁷ J.

If one half the gravitational potential energy of the water were converted to electrical energy

Electrical energy = Ep/2

Electrical energy = (1.6098×10⁷)/2

Electrical energy = 8.049×10⁶ J

In one seconds,

The power generated = 8.049×10⁶ W

Power generated = 8.049 MW

Answer:

[tex]W=173.48J[/tex]

Explanation:

information we know:

Total force: [tex]F=45N[/tex]

Weight: [tex]w=100N[/tex]

distance: [tex]4m[/tex]

vertical component of the force: [tex]F_{y}=12N[/tex]

-------------

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: [tex]F_{x}=Fcos\theta[/tex]

vertical component: [tex]F_{y}=Fsen\theta[/tex]

but from the given information we know that [tex]F_{y}=12N[/tex]

so, equation these two [tex]F_{y}=Fsen\theta[/tex] and [tex]F_{y}=12N[/tex]

[tex]Fsen\theta =12N[/tex]

and we know the force [tex]F=45N[/tex], thus:

[tex]45sen\theta=12[/tex]

now we clear for [tex]\theta[/tex]

[tex]sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466[/tex]

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

[tex]F_{x}=Fcos\theta[/tex]

[tex]F_{x}=45cos(15.466)\\F_{x}=43.37N[/tex]

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

[tex]W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J[/tex]

the work done is W=173.48J

To calculate the work done by the force in moving the crate, we use the formula Work = Force x Distance x cos(theta). We are given the force, distance, and the vertical component of the force, so we can find the horizontal component using trigonometry. Substituting the values into the formula gives us the work done as 174 J.

To calculate the work done by a force, we use the formula:

Work = Force x Distance x cos(theta)

In this case, the force is 45 N, the distance is 4 m, and the vertical component of the force is 12 N. We need to find the horizontal component of the force, which can be calculated using the given information.

Given that the vertical component of the force is 12 N, we can use the trigonometric relationship between the horizontal and vertical components of a force to find the horizontal component as follows:

sin(theta) = vertical component / total force

sin(theta) = 12 N / 45 N

sin(theta) = 0.267

theta = sin^(-1)(0.267)

theta = 15.1 degrees

Once we have the horizontal component of the force, we can calculate the work done as follows:

Work = Force x Distance x cos(theta)

Work = 45 N x 4 m x cos(15.1 degrees)

Work = 180 J x cos(15.1 degrees)

Work = 180 J x 0.966

Work = 174 J

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Answer: you'll see cyan color on the screen

Explanation:

Saturating the red cone causes them to stop functioning, hence you can't perceive the red part of white light. White light is made up of three main colors which are blue, red and green. When one can no longer perceive the red part of light, one is left with the grean and blue part. The green and blue part of light will superimpose to give a cyan color.

Answer:

Cyan color

Explanation:

When an individual stares at a red object and immediately look at a white area afterward, there will be an image formation almost immediately that is the same size and shape as the previous one. The only difference is there will be a color change to blue-green, or cyan.

This is due to the eyes generally using the red, blue and green cone cells to perceive white light. Since the red cone cells are fatigued due to long view of it and cone cells being saturated , the blue and green color which are the remaining color cone cells use to perceive white light is viewed. This gives rise to the blue green or cyan color.

Answer:

C. 2.3A

Explanation:

V/Ω=A

9V / 4Ω = 2.25 ≅ 2.3A

The current flowing through the circuit is 2.3 A. So, option C is correct.

Ohm's law states that, the voltage applied across a circuit is directly proportional to the steady current flowing through the circuit and also proportional to the resistance of the circuit.

Here,

Voltage applied in the circuit, V = 9 V

Resistance of the bulb, R = 4 Ω

According to Ohm's law,

V [tex]\alpha[/tex] I

V [tex]\alpha[/tex] R

So, V = IR

Therefore, current flowing through the circuit,

I = V/R

I = 9/4

I = 2.25 A    approximately, 2.3 A

Hence,

The current flowing through the circuit is 2.3 A.

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Answer:

Note that the emf induced is

emf = B d v cos (A)

---> v = emf / [B d cos (A)]

where

B = magnetic field

d = distance of two rails

v = constant speed

A = angle of rails with respect to the horizontal

Also, note that

I = emf/R

where R = resistance of the bar

Thus,

I = B d v cos (A) / R

Thus, the bar experiences a magnetic force of

F(B) = B I d = B^2 d^2 v cos (A) / R, horizontally, up the incline.

Thus, the component of this parallel to the incline is

F(B //) = F(B) cos(A) = B I d = B^2 d^2 v cos^2 (A) / R

As this is equal to the component of the weight parallel to the incline,

B^2 d^2 v cos^2 (A) / R = m g sin (A)

where m = the mass of the bar.

Solving for v,

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]   [ANSWER, the constant speed, PART A]

******************************

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]

Plugging in the units,

m/s = [ [ohm * kg * m/s^2] / [T^2 m^2] ]

Note that T = kg / (s * C), and ohm = J * s/C^2

Thus,

m/s = [ [J * s/C^2 * kg * m/s^2] / [(kg / (s * C))^2 m^2] ]

= [ [J * s/C^2 * kg * m/s^2] / [(kg^2 m^2) / (s^2 C^2)]

As J = kg*m^2/s^2, cancelling C^2,,

= [ [kg*m^2/s^2 * s * kg * m/s^2] / [(kg^2 m^2) / (s^2)]

Cancelling kg^2,

= [ [m^2/s^2 * s * m/s^2] / [(m^2) / (s^2)]

Cancelling m^2/s^2,

= [s * m/s^2]

Cancelling s,

=m/s   [DONE! WE SHOWED THE UNITS ARE CORRECT! ]

Explanation:

Given that,

Mass of the truck, m = 2268 kg

Speed of the truck, v = 30 m/s

The friction force has an average magnitude of 700 N, f = -700 N

(a) The initial kinetic energy of the truck is given by :

[tex]K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 2268\times (30)^2\\\\K=1.02\times 10^6\ J[/tex]

(b) Finally, the stops due to an effective friction force that the road, final speed is 0. Let d is the stopping distance of the truck. using third equation of motion to find it as :

[tex]v^2-u^2=2ad\\\\d=\dfrac{-u^2}{2a}[/tex]

Since, f = ma

[tex]d=\dfrac{-u^2m}{2f}\\\\d=\dfrac{-(30)^2\times 2268}{-2\times 700}\\\\d=1458\ m[/tex]

So, the stopping distance of the truck is 1458 meters.

Answer:

d. Planets formed from the disk of gas and dust that surrounded the Sun, and such disks are common around young stars.

Explanation:

A star passes through various stages before it becomes a full fledged star with its own planetary system. The same is with Sun as well. It was born out of a Nebula. Nebula is a cloud of dust and gases. The dust and gases start accumulating to form what we call as a protostar.

A lot of material at this stage is thrown out from the young star, this material forms a disk around the star. This disk is known as proto-planetary disk. The gases and dust of this disk then coalesce together to make planets and other objects of the star system.

We have observed such disks in the Orion nebula.

The solar nebula theory suggests that planets are formed from the disk of gas and dust that surrounds young stars. Such disks are commonly observed around stars, leading to the inference that planet formation is a universal occurrence.

The solar nebula theory, or nebular hypothesis, explains the formation of our solar system from a nebular cloud of gas and dust. According to this theory, planets are formed from the disk of gas and dust that surrounds a young star, also known as protoplanetary disk. This leads to the implication that planets are common because such disks are commonly found around young stars.

When a star condenses from a nebula, it forms a hot, spinning disk of gas and dust around it. Over time, particles within this disk begin to collide and stick together, gradually forming planetesimals and eventually, planets. Since we observe many stars surrounded by these disks, we can infer that planet formation is a common process in the universe.

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Explanation:

Case 1,

A force F is pushing perpendicular on an object a distance L/2 from the rotation axis. Torque is given by :

[tex]\tau_1=Fd\sin \theta\\\\\tau_1=\dfrac{FL}{2}\sin (90)\\\\\tau_1=\dfrac{FL}{2}\ ..........(1)[/tex]

Case 2,

The same force is pushing at an angle of 30 degrees a distance L from the axis. New torque is given by :

[tex]\tau_2=Fd\sin \theta\\\\\tau_2=FL\sin (30)\\\\\tau_2=\dfrac{FL}{2}\ ..........(2)[/tex]

From equation (1) and (2), it is clear that the force in both cases is same.

Question:

a.  It would decrease by a factor of 14 .

b. It would increase by a factor of 4.

c. It would remain unchanged.

d. It would increase by a factor of 2.

e. It would decrease by a factor of 12 .

Answer:

The correct option is;

d. It would increase by a factor of 2

Explanation:

Here we have

[tex]\frac{1}{2}kx^2 = \frac{1}{2}mv^2 + F_kx[/tex]

The formula for maximum mass is given by the following relation;

[tex]v_{max1} =A\sqrt{\frac{k}{m} }[/tex]

Where:

[tex]v_{max}[/tex] = Maximum speed of mass on spring

k = Spring constant

m = Mass attached to the spring

A = Amplitude of the oscillation

Therefore, when k is increased by a factor of 4 we have;

[tex]v_{max2} =A\sqrt{\frac{4 \times k}{m} } = 2 \times A\sqrt{\frac{ k}{m} }[/tex]

Therefore, [tex]v_{max2} = 2 \times v_{max1}[/tex], that is the velocity will increase by a factor of 2.

Answer:

A) The new amplitude = 0.048 m

B) Period T = 0.6 seconds

Explanation: Please find the attached files for the solution

The number of revolutions is 301

Explanation:

Angular acceleration, α = 3.5 rad/s

Time, t = 9s

Number of revolutions, n = ?

We know,

[tex]\alpha = \frac{w}{t}[/tex]

where,

ω = angular velocity

t = time

It can also be written as:

[tex]\alpha = \frac{\pi n}{t X 30}[/tex]

On substituting the value, we get:

[tex]3.5 = \frac{\pi X n}{9 X 30} \\\\n = 300.8[/tex]

Therefore, the number of revolutions is 301

The  answer is 22 complete revolutions.

The number of complete revolutions the disk makes in 9 seconds is given by the formula for angular displacement due to constant angular acceleration:

[tex]\[ \theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2 \][/tex]

where:

-[tex]\( \theta \)[/tex] is the final angular displacement in radians,

-[tex]\( \theta_0 \)[/tex] is the initial angular displacement, which is zero since the disk starts from rest,

-[tex]\( \omega_0 \)[/tex] is the initial angular velocity, which is also zero since the disk starts from rest,

- [tex]\( \alpha \)[/tex] is the angular acceleration, which is 3.5 [tex]rad/s\( ^2 \)[/tex],

- [tex]\( t \)[/tex] is the time in seconds, which is 9 seconds.

Since the disk starts from rest, [tex]\( \theta_0 = 0 \)[/tex] and [tex]\( \omega_0 = 0 \)[/tex], the equation simplifies to:

[tex]\[ \theta = \frac{1}{2} \alpha t^2 \][/tex]

Substituting the given values:

[tex]\[ \theta = \frac{1}{2} \times 3.5 \, \text{rad/s}^2 \times (9 \, \text{s})^2 \][/tex]

[tex]\[ \theta = \frac{1}{2} \times 3.5 \times 81 \][/tex]

[tex]\[ \theta = 1.75 \times 81 \][/tex]

\[ \theta = 141.75 \, \text{radians} \]

To find the number of complete revolutions, we need to convert the angular displacement from radians to revolutions. Since there are [tex]\( 2\pi \)[/tex] radians in one revolution, we divide the angular displacement by [tex]\( 2\pi \)[/tex]:

[tex]\[ \text{Number of revolutions} = \frac{\theta}{2\pi} \][/tex]

[tex]\[ \text{Number of revolutions} = \frac{141.75}{2\pi} \][/tex]

[tex]\[ \text{Number of revolutions} = \frac{141.75}{6.2831853071796} \][/tex]

[tex]\[ \text{Number of revolutions} \approx 22.57 \][/tex]

Answer: 3 m/s²

Explanation:

m = 3 kg

F = 9 N

a = ?

F = m × a

9 = 3 × a

[tex]\frac{9}{3}[/tex] = a

a = 3 m/s²

Answer:

Potential difference across resistor will be 87.66 volt

Explanation:

We have given number of electrons [tex]n=4.1\times 10^{21}[/tex]

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

So total charge [tex]Q=4.1\times 10^{21}\times 1.6\times 10^{-19}=656C[/tex]

Time is given t = 5 min

1 minute = 60 sec

So 5 minute = 5×60 = 300 sec

So current [tex]i=\frac{Q}{t}=\frac{656}{300}=2.1866A[/tex]

Resistance is given R = 40 ohm

Sp from ohm's law potential difference across resistor v = iR = 2.1866×40 = 87.466 volt

The potential difference across the 40 Ω resistor when 4.1 × 1021 electrons have passed through it over a time period of 5 minutes is calculated using Ohm's law, which results in 87.6 Volts.

The subject of this question is based on Ohm's law, an important concept in Physics. Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

First, calculate the total charge (Q) passed through the resistor. This can be done by multiplying the number of electrons by the fundamental charge of one electron (Q = electrons x electron charge). Thus, Q = 4.1 × 1021 electrons x 1.602 × 10-19 C = 656.82 Coulombs.

Next, use the formula to obtain the electric current (I) passed through the resistor. Since current (in Amperes) is equal to the total charge (in Coulombs) divided by the time (in seconds) and time here is given in minutes, we first need to convert minutes to seconds. So, 5 min = 5 x 60 = 300 seconds. Hence, I = Q / t = 656.82 C / 300 s = 2.19 Ampere.

Lastly, we can now find the potential difference (V) across the resistor. According to Ohm's law, V = I * R. Thus, V = 2.19 Ampere * 40 Ω = 87.6 Volts.

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Answer:

  h ’= 3.75 inch

Explanation:

For this exercise we use the definition of magnification which is the height of the image over the height of the object

                 m = h ’/ h

                  h’= m h

Let's calculate

               h ’= 5  3/4

              h ’= 3.75 inch

Answer: Alloy A is best chosen for the sheet forming operation

Explanation:

taking both alloy A and B into consideration;

For alloy A;

the true strain ration, R₀ = width strain / thickness strain

   R₀ = in (0.65/0.75) / in (0.092/0.1) = 1.716

at 45⁰  and 90⁰, the true strain ratio becomes;

R45⁰ = in (0.63/0.75) / in (0.093/0.1) = 2.4025

R90⁰ = in (0.67/0.75) / in (0.087/0.1) = 0.8099

where R(avg) = (R₀+2Ras+Rao) / A = (1.716+2(2.4025)+0.8099) = 3.0340

R(avg) = 3.0340

For alloy B;

R₀ = in (0.7/0.75) / in (0.082/0.1) = 0.3477

R45⁰ = in (0.69/0.75) / in (0.083/0.1)  = 0.4475

R90⁰ = in (0.71/0.75) / in (0.078/0.1) = 0.2206

R(avg) = (R₀+2Ras+Rao) / A = (0.3477+2(0.4475)+0.2206) = 0.365825

R(avg) = 0.365825

comparing both we have that,

R(avg) for allow A > R(avg) for allow B

∴ Alloy A is the best to be selected for sheet formation operation.

cheers i hope this helps!!!1

Answer: 0.88 m/s

Explanation:

given

Force constant of the spring, k = 965 N/m

Compression of the spring, x = 3.7 cm = 0.037m

Mass of the block, m = 1.7 kg

To solve this, we would be using law of conservation of energy, where initial energy is all kinetic and elastic energy is elastic but potential.

E(f) = E(i)

1/2kx² = 1/2mv²

kx² = mv²

965 * 0.037² = 1.7 * v²

1.321 = 1.7 v²

v² = 1.321 / 1.7

v² = 0.777

v = 0.88 m/s

Thus, the initial speed of the block is v = 0.88 m/s

Answer:

Check the explanation

Explanation:

Atomization, also called the spraying technique or procedure, refers to a process in which molten metals are broken down into little or tiny drops of liquid through the use of high-speed fluids (liquid as water, gas as air or inert gas) or fluids with centrifugal force, and then solidified into a powdered state.

Kindly check the attached image below to get the step by step solution to the above question.

Answer:

Answer: The thickness of film is equal to 1/2 of wavelength in the film

Explanation:

Consider a thin film surrounded with two other transparent materials. Index of refraction of the top material is 1.2; index of refraction of a film 1.3 and of the bottom material 1.4 For normal incidence this film will look bright for light which wavelength satisfies the following condition a. Thickness of a film is equal to of wavelength of light in bottom material b. Thickness of a film is equal ½ of wavelength in the film c. Thickness of a film is equal ¼ of wavelength in the film. d. Thickness of a film is equal to ½ of wavelength of light in upper material e. Thickness of a film is equal ¼ of wavelength of light in upper material f. Thickness of a film is equal 4 of wavelength in bottom material

For constructive interference

     Zt = ∧/n

    t=∧/n

        2

thickness = 1/2 x wavelength in the film

The thickness of film is equal to 1/2 of wavelength in the film

The condition for a thin film to look bright is when its thickness is ½ of the wavelength in the film itself, due to constructive interference resulting from the phase shift upon reflection.

The question deals with thin film interference, a phenomenon where light waves reflected off the top and bottom surfaces of a thin film interfere with each other. For a thin film to appear bright for light of a certain wavelength, the condition for constructive interference must be satisfied. This often involves the film's thickness being a multiple or a fraction of the wavelength of light in the film itself or in the surrounding media.

In the context of thin films surrounded by materials with different indices of refraction, the correct condition for the film to look bright is when its thickness is equal to ½ of the wavelength in the film itself. This accounts for the phase shift that occurs upon reflection, leading to constructive interference when the optical path difference is a multiple of the wavelength.

It's important to note that the index of refraction of the material influences the effective wavelength of light within it. Therefore, the wavelength in the material can be determined by dividing the wavelength in vacuum by the material's index of refraction.

Answer:

The separation distance between the two charges must be 82704.2925 m

Explanation:

Given:

Two negative charges that are both q = -3.8 C

Force of 19 N

Question: How far apart are the two charges, s = ?

First, you need to get the electrostatic force of this two negative charges:

[tex]F=\frac{kq}{s^{2} } \\s=\sqrt{\frac{kq}{F} }[/tex]

Here

k = electric constant of the medium = 9x10⁹N m²/C²

Substituting values:

[tex]s=\sqrt{\frac{9x10^{9}*(-3.8)^{2} }{19} } =82704.2925m[/tex]