You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.17 mm and place your screen 8.95 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.57 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength λ expressed in nanometers?

Answer:

contaminated drinking water

deaths of sea creatures that are used as a food source

limits to potential economic activities such as a fishing

Explanation:

A watershed is a large area that comprises of drainage area of all the surrounding water bodies meeting at a common affluence point before draining into sea or ocean or any other large water body. Pollution in this area can pollute the small water streams flowing through it, thereby polluting the larger water body into which it drains.  

Thus, the water extracted for drinking from such area will be contaminated. Pollution in larger water body can cause death of water creature and hence pose a threat to fishing.  

Answer:

loss of areas for tourism

contaminated drinking water

deaths of sea creatures that are used as a food source

limits to potential economic activities such as fishing

Answer:

Explanation:

Young modulus ε = 1.4 × 10¹⁰ Pa

ΔL = 1% of the original length = 0.01 x where x is the original length

cross sectional area = 3.0 cm² =( 3 .0 / 10000) m²= 0.0003 m²

ε = Stress / strain

stress = ε × strain

stress = F /A

F force = ε × A × ( ΔL / L) =  1.4 × 10¹⁰ Pa × 0.0003 m² × 0.01 = 4.2 × 10⁴ N

b) F net = F max - mg ( weight) = 84000 - ( 70 × 9.8 m/s² ) ( F is double since the stress on the two leg is equally distributed)

f net = ma = 84000 - ( 70 × 9.8 m/s² )

a =  (84000 - ( 70 × 9.8 m/s² )) = 1190.2 m/s²

v = u + at

where final velocity equal zero

- u = -at since it coming downwards

u = at = 1190.2 m/s² × 0.03s = 35.706 m/s

using conservation of energy

1/2 mv² = mgh

1/2v²/ g = h

h = 0.5 × (35.706 m/s )² / 9.8 = 65.04 m

Answer:

3241.35J

Explanation:

No. Of rods = 5

Mass = 2.89kg

Length (L) = 0.827m

W = 507rpm

Kinetic energy of rotation = ½I*ω²

For each rod, the moment of inertia (I) = ML² / 3

I = ML² / 3

I = [2.89*(0.827)²] / 3

I = 1.367 / 3 = 0.46kgm²

ω = 507 rev/min. Convert rev/min to rev/sec.

507 * 2Πrads/60s = 53.09rad/s

ω = 53.09rad/s

k.e = ½ I * ω²

K.E = ½ * 0.46 * (53.09)²

K.E = 648.27.

But there five (5) rods, so kinetic energy is equal to

K.E = 5 * 648.27 = 3241.35J

The rotational kinetic energy of a propeller with five blades, each modeled as a uniform rod, can be determined using formulas for moment of inertia and rotational kinetic energy.

The kinetic energy of rotating objects depends on their moment of inertia and angular velocity. Given that each propeller blade is modeled as a uniform rod rotating about its end, we can calculate the moment of inertia (I) of all five blades using the formula I=5*(1/3*m*L^2), where m is the mass and L is the length of each rod. After we find the moment of inertia, we can determine the angular velocity (ω) in radian per second, given that the propeller rotates at 507 rpm, by using the conversion factor ω=(507*2*pi)/60. Finally, we calculate the rotational kinetic energy (K) using the formula K=1/2*I*ω^2.

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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]

Answer:

47.76°

Explanation:

Magnitude of dipole moment = 0.0243J/T

Magnetic Field = 57.5mT

kinetic energy = 0.458mJ

∇U = -∇K

Uf - Ui = -0.458mJ

Ui - Uf = 0.458mJ

(-μBcosθi) - (-μBcosθf) = 0.458mJ

rearranging the equation,

(μBcosθf) - (μBcosθi) = 0.458mJ

μB * (cosθf - cosθi) = 0.458mJ

θf is at 0° because the dipole moment is aligned with the magnetic field.

μB * (cos 0 - cos θi) = 0.458mJ

but cos 0 = 1

(0.0243 * 0.0575) (1 - cos θi) = 0.458*10⁻³

1 - cos θi = 0.458*10⁻³ / 1.397*10⁻³

1 - cos θi = 0.3278

collect like terms

cosθi = 0.6722

θ = cos⁻ 0.6722

θ = 47.76°

Answer:

The initial angle between the dipole moment and the magnetic field is 47.76⁰

Explanation:

Given;

magnitude of dipole moment, μ = 0.0243 J/T

magnitude of magnetic field, B = 57.5 mT

change in kinetic energy, ΔKE = 0.458 mJ

ΔKE = - ΔU

ΔKE = - (U₂ -U₁)

ΔKE = U₁ - U₂

U₁ -U₂ = 0.458 mJ

[tex](-\mu Bcos \theta_i )- (-\mu Bcos \theta_f) = 0.458 mJ\\\\-\mu Bcos \theta_i + \mu Bcos \theta_f = 0.458 mJ\\\\\mu Bcos \theta_f -\mu Bcos \theta_i = 0.458 mJ\\\\\mu B(cos \theta_f - cos \theta_i ) = 0.458 mJ[/tex]

where;

θ₁ is the initial angle between the dipole moment and the magnetic field

[tex]\theta_f[/tex] is the final angle which is zero (0) since the  dipole moment is aligned with the magnetic field

μB(cos0 - cosθ₁) = 0.458 mJ

Substitute the given values of μ and B

0.0243 x 0.0575 (1 - cosθ₁) = 0.000458

0.00139725 (1 - cosθ₁) = 0.000458

(1 - cosθ₁) = 0.000458 / 0.00139725

(1 - cosθ₁) = 0.327787

cosθ₁ = 1 -  0.327787

cosθ₁ = 0.672213

θ₁ = cos⁻¹ (0.672213)

θ₁ = 47.76⁰

Thus, the initial angle between the dipole moment and the magnetic field is 47.76⁰

Answer:

a) The work done on the astronaut by the force from the helicopter is [tex]W_{h}=16087.68\ J[/tex].

b) The work done on the astronaut by the gravitational force is  [tex]W_{g}=-15082.2\ J[/tex] .

Explanation:

We are told that the mass of the astronaut is [tex]m=81\ kg[/tex], the displacement is [tex]\Delta x=19\ m[/tex], the acceleration of the astronaut is  [tex]|\vec{a}|=\frac{g}{15}[/tex]  and the acceleration of gravity is [tex]g=9.8\ \frac{m}{s^{2}}[/tex] .

We suppose that in the vertical direction the force from the helicopter [tex]F_{h}[/tex] is upwards and the gravitational force [tex]F_{g}[/tex] is downwards. From the sum of forces we can get the value of [tex]F_{h}[/tex]:

                                               [tex]F_{h}-F_{g}=m.a[/tex]

                                              [tex]F_{h}-mg=m.\frac{g}{15}[/tex]

                                              [tex]F_{h}=mg(1+\frac{1}{15})[/tex]

                              [tex]F_{h}=(\frac{16}{15}).81\ kg.\ 9.8\ \frac{m}{s^{2}}\ \Longrightarrow\ F_{h}=846.72\ N[/tex]

We define work as the product of the force, the displacement of the body and the cosine of the angle [tex]\theta[/tex] between the direction of the force and the displacement of the body:

                                                 [tex]W=F.\Delta x.\ cos(\theta)[/tex]

a) The work done on the astronaut by the force from the helicopter

                                                  [tex]W_{h}=F_{h}.\Delta x[/tex]

                                             [tex]W_{h}=846.72\ N.\ 19\ m[/tex]

                                                  [tex]W_{h}=16087.68\ J[/tex]

b) The work done on the astronaut by the gravitational force

                                                   [tex]W_{g}=-F_{g}.\Delta x[/tex]

                                                    [tex]W_{g}=-mg\Delta x[/tex]

                                            [tex]W_{g}=-81\ kg.\ 9.8\ \frac{m}{s^{2}}.\ 19\ m[/tex]

                                                  [tex]W_{g}=-15082.2\ J[/tex]

Answer:

a) Work done on the astronaut by the force from the helicopter = 16.104 kJ

b) Work done on the astronaut by the gravitational force = -15.082 kJ

Explanation:

mass of the astronaut, m = 81 kg

height, h = 19 m

acceleration of the astronaut, a = g/15

Since the astronaut is lifted up, using the third law of motion:

T - mg = ma

T = mg + ma

T = (81*9.81) + 81*(9.81/15)

T = 847.584 N

Work done on the astronaut by the helicopter

Work done = Tension * height

W = T* h

W = 847.584 * 19

Work done, W = 16104.096 Joules

W = 16.104 kJ

b) Work done on the astronaut by the gravitational force on her

[tex]W = -f_{g} h[/tex]

[tex]f_{g} = mg = 81 * 9.8\\f_{g} = 793.8 N[/tex]

[tex]W = -793.8 * 19\\W =- 15082.2 J[/tex]

W = -15.082 kJ

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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The linear charge density on the wire is approximately 0.002 N/Cm.

The linear charge density on the wire can be calculated using the formula:

λ = E × r / 2πk

where λ is the linear charge density, E is the electric field, r is the distance from the wire, and k is the Coulomb's constant. Plugging in the known values of E = 20.0 N/C and r = 2.00 cm = 0.02 m, we can solve for λ:

λ = (20.0 N/C) × (0.02 m) / (2π × 8.99 × 10^9 Nm^2/C^2) ≈ 0.002 N/Cm

Therefore, the linear charge density on the wire is approximately 0.002 N/Cm.

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Answer:

[tex]2.08\cdot 10^9 A[/tex]

Explanation:

The magnetic dipole moment of a circular coil with a current is given by

[tex]\mu = IA[/tex]

where

I is the current in the coil

[tex]A=\pi r^2[/tex] is the area enclosed by the coil, where

[tex]r[/tex] is the radius of the coil

So the magnetic dipole moment can be rewritten as

[tex]\mu = I\pi r^2[/tex] (1)

Here we can assume that the magnetic dipole moment of Earth is produced by charges flowing in Earth’s molten outer core, so by a current flowing in a circular path of radius

[tex]r=3500 km = 3.5\cdot 10^6 m[/tex]

Here we also know that the Earth's magnetic dipole moment is

[tex]\mu = 8.0\cdot 10^{22} J/T[/tex]

Therefore, we can re-arrange eq (1) to find the current that the charges produced:

[tex]I=\frac{\mu}{\pi r^2}=\frac{8.00\cdot 10^{22}}{\pi (3.5\cdot 10^6)^2}=2.08\cdot 10^9 A[/tex]

Answer:

The angular velocity is 4.939rad/sec

Explanation:

θ = 61°

N = 118 rev/min

The formula for angular velocity is given as;

w = θ/t

where;

w = angular velocity

θ = angle

t = time in rad/sec

1 rev/ min = 0.1047 radian/second.

Therefore 118 rev/min = 118*0.1047rad/sec

                                     =12.35rad/sec

Substituting into the formula, we have

w = 61/12.35

w = 4.939rad/sec

Answer:

The temperature change in Celsius is 46°C.

The temperature change in Fahrenheit is 82.8°F.

Explanation:

A degree of Celsius scale is equal to that of kelvin scale; therefore,

[tex]\Delta C = \Delta K = 283K-237K\\\\ \boxed{\Delta C = 46^oC}[/tex]

A degree in Fahrenheit is 1.8 times the Celsius degree; therefore

[tex]\Delta F = 1.8(46^o)[/tex]

[tex]\boxed{\Delta F = 82.8^oF}[/tex]

Hence, the temperature change in Celsius is 46°C, and the temperature change in Fahrenheit is 82.8°F.

Final answer:

The temperature change of 46 degrees from Kelvin to Celsius in Granville, North Dakota, is equivalent to a change of 82.8 degrees Fahrenheit.

Explanation:

On January 21, 1918, the temperature in Granville, North Dakota, changed from 237 K to 283 K within 12 hours. To convert the change in temperature to the Celsius scale, we first recognize that 0 degrees Celsius is equivalent to 273.15 K. Therefore, the initial temperature in Celsius would have been 237 K - 273.15 K = -36.15  extdegree C, and the final temperature would have been 283 K - 273.15 K = 9.85  extdegree C. The change in the Celsius scale is then 9.85  extdegree C - (-36.15  extdegree C) = 46  extdegree C

To convert the change in temperature to the Fahrenheit scale, we start with the change in Celsius and use the conversion formula (F = C  imes \frac{9}{5} + 32). Since we are looking for the change, we only need to convert the difference in Celsius to Fahrenheit. Therefore, the change in Fahrenheit is 46  extdegree C  imes \frac{9}{5} = 82.8  extdegree F.

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:

49 N

Explanation:

The diagram of the bar is obtained online and attached to this solution.

The free body diagram is also attached.

Since the weight of the bar acts at the middle of the bar, the torque due to the weight of the bar is given by

τ = mgx

where x = (L/2) cos 35° = 0.45 × cos 35° = 0.3686 m

τ = (7)(9.8)(0.3686) = 25.29 Nm

The force acting on pin A = torque ÷ (length × sin 35°) = 25.29 ÷ (0.9 × sin 35°)

= 25.29 ÷ 0.5162 = 48.99 N = 49 N

Hope this Helps!!!

Answer:

The force supported by the pin at A is 69.081 N

Explanation:

The diagram is in the figure attach. The angular acceleration using the moment expression is:

[tex]-mg(\frac{Lcos\theta }{2} )=I\alpha \\\alpha =\frac{-3g}{2L} cos\theta[/tex]

Where

L = length of the bar = 900 mm = 0.9 m

[tex]\alpha =\frac{-3*9.8cos35}{2*0.9} =-13.38rad/s^{2}[/tex]

The acceleration in point G is equal to:

[tex]a_{G} =a_{A} +\alpha kr_{G/A} -w^{2} r_{G/A}[/tex]

Where

aA = acceleration at A = 0

w = angular velocity of the bar = 3 rad/s

rG/A = position vector of G respect to A = [tex]\frac{L}{2} cos\theta i-\frac{L}{2} cos\theta j[/tex]

[tex]a_{G} =(\frac{L}{2}\alpha sin\theta -\frac{w^{2}Lcos\theta }{2} )i+(\frac{L}{2}\alpha cos\theta +\frac{w^{2}Lsin\theta }{2} )j=(\frac{0.9*(-13.38)*sin35}{2} -(\frac{3^{2}*0.9*cos35 }{2} )i+(\frac{0.9*(-13.38)*cos35}{2} +\frac{3^{2} *0.9*sin35)}{2} )j=-6.76i-2.61jm/s^{2}[/tex]

The force at A in x is equal to:

[tex]A_{x} =ma_{G} =7*(-6.76)=-47.32N[/tex]

The force at A in y is:

[tex]A_{y} =ma_{G} +mg=(7*(-2.61))+(7*9.8)=50.33N[/tex]

The magnitude of force A is equal to:

[tex]A=\sqrt{A_{x}^{2}+A_{y}^{2} } =\sqrt{(-47.32^{2})+50.33^{2} } =69.081N[/tex]

Explanation:

Given that,

Mass of the hockey puck, m₁ = 0.45 kg

Initial peed of the hockey puck, u₁ = 5.25 m/s (east)

Mass of other puck, m₁ =  0.85 kg

Initial speed of other puck, u₂ = 0 (at rest)

Let v₁ and v₂ are the final speeds of both pucks after the collision respectively. Using the conservation of momentum as :

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\m_1v_1+m_2v_2=0.45\times 5.25+0.85\times 0\\\\m_1v_1+m_2v_2=2.36\\\\0.45v_1+0.85v_2=2.36.........(1)[/tex]

The coefficient of restitution for elastic collision is equal to 1.          

[tex]C=\dfrac{v_2-v_1}{u_1-u_2}\\\\1=\dfrac{v_2-v_1}{u_1-u_2}\\\\1=\dfrac{v_2-v_1}{5.25-0}\\\\v_2-v_1=5.25.......(2)[/tex]      

On solving equation (1) and (2) we get :

[tex]v_1=-1.611\ m/s\\\\v_2=3.63 m/s[/tex]

Hence, this is the required solution.

Velocity changes with time and this is called acceleration.

The question is incomplete but I will try to explain the meaning of acceleration to you. The term acceleration refers to the change of velocity with time.

a = Δv/t

Hence, velocity changes with time and this is called acceleration.

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The increase of the pressure of the system would lead to an increase of acceleration.

As pressure equals to the force upon action. Here F is the force and the A the area. As Newton's second law       F = ma  We substitute      P A = m a

Hence the pressure is directly proportional to acceleration.

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Answer:

R = 1.81 10² km

Explanation:

Let's start by looking for the power in the visible range emitted this is 10W, the energy of that power is one second is

         P = E₁ / t

         E₁ = P t

         E₁ = 10 J

 Let's find the energy of a photon with Planck's equation

          E = h f

          c = λ f

we substitute

         E = h c /λ

         E = 6.63 10⁻³⁴ 3 10⁸/580 10⁻⁹

         E = 3.42 10⁻¹⁹ J

we can use a direct proportions rule to find the number of photons in the energy E₁

          #_photon = E₁ / E

          #_photon = 10 / 3.42 10⁻¹⁹

          #_photon = 2.92 10¹⁹ photons

This number of photons is distributed on the surface of a sphere. Let's find what the distance is so that there are 500 photons in 3 mm = 0.003 m.

the area of ​​the sphere is

            A = 4π R²

area of ​​the circle is

           A´ = π r²

as the intensity is constant over the entire sphere

         P = #_photon / A = 500 / A´

       # _photon / 4π R² = 500 / π r²

       R² = #_photon r² / 4 500

       r = d / 2 = 0.003 / 2 = 0.0015 m

       R² = 2.92 10¹⁹ 0.0015 2/2000

       R = √ (3,285 10¹⁰)  

        R = 1.81 10⁵ m

       R = 1.81 10² km

Answer:

T1 = 112.07[lb]

T2 = 487.3 [lb]

Explanation:

To solve this problem we must perform a static balance analysis, for this we perform a free body diagram. In this free body diagram we use the angles mentioned in the description of the problem.

Performing a sum of forces on the X-axis equal to zero, we can find an equation that relates the tension of the T1 & T2 cables.

Then we perform a summation of forces on the Y-axis, in which we can find another equation. In this new equation, we replace the previous one and we can find the tension T2.

T1 = 112.07[lb]

T2 = 487.3 [lb]

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 tension T₁ in string 1 at the moment that the monkey is halfway between the ends of the bar is 34.68 N

Explanation:

Given;

mass of the uniform horizontal bar, m₁ = 2.5 kg

length of the bar, L = 3.0 m

mass of the monkey, m₂ = 1.25 kg

distance from the right end of the second string, d = 0.74 m

For a body to remain in rotational equilibrium, the net external torque acting on it due to applied external forces must be equal.

∑τ = 0

For vertical equilibrium of bar-string system, in which T₁ and T₂ are the tension on both ends of the string;

T₁ + T₂ = (m₁ + m₂)g

T₁ + T₂ = (2.5 + 1.25) 9.8

T₁ + T₂ = 36.75 N

For rotational equilibrium when the monkey is halfway between the ends of the bar, take moment about the left end of the string.

(m₁ + m₂) L /2 = T₂(L - d)

(m₁ + m₂)0.5L = T₂( L - d)

(2.5 + 1.25)0.5 x 3 = T₂ ( 3 - 0.74)

4.6875 = T₂ (2.26)

T₂ = (4.6875) / (2.26)

T₂ = 2.074 N

Thus, T₁ = 36.75 N - T₂

T₁ = 36.75 N  - 2.074 N

T₁ = 34.68 N

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

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:

The magnitude of their velocities will be the same but their direction will be reversed.

Final answer:

In a completely elastic collision between two particles of identical mass and equal but opposite velocities, both particles simply swap their velocities as a result of the collision, conserving both momentum and kinetic energy.

Explanation:

In a completely elastic head-on collision between two particles with identical mass and equal but opposite speeds, the outcome is quite straightforward. Since the collision is elastic, both conservation of momentum and conservation of kinetic energy apply.

Before the collision, let's assume particle 1 with mass m has velocity v, and particle 2 also with mass m has velocity -v. The total momentum before collision would be m*v + m*(-v) = 0, and the total kinetic energy would be (1/2)*m*v² + (1/2)*m*(-v)².

For identical masses and equal and opposite velocities, the particles simply swap velocities after collision. Therefore, the first particle will have a velocity of -v and the second a velocity of v post-collision, as momentum and kinetic energy are conserved. They essentially 'bounce' off each other and move in the opposite directions with the same speed they had before the collision.

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.

The mass of the ion is 5.96 X 10⁻²⁵ kg

Explanation:

The electrical energy given to the ion Vq will be changed into kinetic energy [tex]\frac{1}{2}mv^2[/tex]

As the ion moves with velocity v in a magnetic field B then the magnetic Lorentz force Bqv will be balanced by centrifugal force [tex]\frac{mv^2}{r}[/tex].

So,

[tex]Vq = \frac{1}{2}mv^2[/tex]

and

[tex]Bqv = \frac{mv^2}{r}[/tex]

Right from these eliminating v, we can derive

[tex]m = \frac{B^2r^2q}{2V}[/tex]

On substituting the value, we get:

[tex]m = \frac{(0.4)^2X (0.305)^2 X1.602X 10^-^1^9}{2X 2000}\\\\[/tex]

m = 5.96 X 10⁻²⁵ kg.

The mass of the ion is found using the principles of kinetic and magnetic forces, is approximately 9.65 x 10⁻²⁷ kg.

To find the mass of the ion, we use the principles of energy and magnetic force. When a singly charged positive ion is accelerated through a voltage (V), it gains kinetic energy equal to the electrical potential energy provided by the voltage:

Kinetic Energy (KE) = qV

Since the ion starts from rest, this kinetic energy is also given by:

Kinetic Energy (KE) = (1/2)mv²

Equating these two expressions for kinetic energy gives:

qV = (1/2)mv²

Solving for the velocity (v), we get:

v =√(2qV/m)

When the ion enters the magnetic field (B) and bends into a circular path, the centripetal force required is provided by the magnetic force:

Magnetic Force = qvB

The centripetal force is also given by:

Centripetal Force = mv²/r

Equating these two expressions for the forces, we get:

qvB = mv²/r

Solving for the mass (m) of the ion:

m = qBr/v

We already have v = √(2qV/m), substituting this into the equation above to eliminate v, we get:

m = (qBr) / √(2qV/m)

Square both sides to simplify and solve for m:

m² = (q²B²r²) / (2qV)

m = (qB²r²) / (2V)

Using the given values (q = 1.602 x 10⁻¹⁹ C as charge of a singly charged positive ion, V = 2000 V, B = 0.400 T, and r = 0.305 m), we get:

m = (1.602 x 10⁻¹⁹ C x (0.400 T)² x (0.305 m)²) / (2 x 2000 V)

Calculating this gives the mass of the ion approximately equal to:

m ≈ 9.65 x 10⁻²⁷ kg

Answer:

Maximum speed for a point on the string at anti node will be 22.6 m/sec

Explanation:

We have given length of string L = 3.5 m

For 7th harmonic length of the string [tex]L=\frac{7\lambda }{2}[/tex]

So [tex]\lambda =\frac{2L}{7}[/tex]

Speed of the wave in the string is 150 m/sec

Frequency corresponding to this wavelength [tex]f=\frac{v}{\lambda }=\frac{7v}{2L}[/tex]

So angular frequency will be equal to [tex]\omega =2\pi f=2\pi \times \frac{7v}{2L}=2\times 3.14\times \frac{7\times 150}{2\times 3.5}=942rad/sec[/tex]

Maximum speed is equal to [tex]v_m=A\omega =0.024\times 942=22.60m/sec[/tex]

So maximum speed for a point on the string at anti node will be 22.6 m/sec

The speed of waves on the string is 32 m/s.

The speed of waves on a string can be calculated using the formula:

velocity = frequency × wavelength

In this case, the frequency is given as 80.0 Hz and the distance between adjacent antinodes is 20.0 cm. Since the wavelength is twice the distance between adjacent antinodes, we have:

wavelength = 2 × 20.0 cm = 40.0 cm = 0.4 m

Plugging these values into the formula, we get:

velocity = 80.0 Hz × 0.4 m = 32 m/s

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The net force is  [tex]F_{net}= 6.44 *10^{-4} N[/tex]

Explanation:

Generally the net force is a force that come up due to the unequal centripetal force(A difference in centripetal force ) and it is mathematically represented  as

                     [tex]F_{net} = \Delta F_{cen}[/tex]

and the difference in centripetal force  [tex]\Delta F_{cen}[/tex] is mathematically represented as

               [tex]\Delta F_{cen} = \Delta m* rw^2[/tex]

Which the difference in mass multiplied by the centripetal acceleration

  Substituting 10 mg = [tex]10 *10^{-3}g[/tex] for [tex]\Delta m[/tex] , 12 cm = [tex]\frac{12}{100} = 0.12m[/tex] for radius

  and  70,000 rpm = [tex]70,000 *[\frac{2 \pi rad}{1 rev}][\frac{1 min}{60s} ] = 7326.7 rad/s[/tex]

              [tex]F_{net} = \Delta F_{cen} = 10*10^{-3} * 0.12 * 7326.7[/tex]

                     [tex]F_{net}= 6.44 *10^{-4} N[/tex]

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! ]

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.

Final answer:

Using the conservation of momentum for an inelastic collision, Ender's mass is calculated to be approximately 51.43 kg when the given velocities and Shen's mass are considered.

Explanation:

The subject of this question is physics, specifically dealing with the concept of conservation of momentum during collisions. We are asked to find Ender's mass given the velocities and mass involved in a collision in space. Since the collision here is inelastic (Ender and Shen hold on to each other), the total momentum before the collision must be equal to the total momentum after the collision.

According to the conservation of momentum, m1 × v1 + m2 × v2 = (m1 + m2) × v3, where m1 and m2 are the masses of Ender and Shen respectively, and v1, v2, and v3 are their velocities.

We know Shen's mass (m2) is 45 kg, Ender's velocity (v1) is 12 m/s, Shen's velocity (v2) is 9 m/s, and the combined velocity after the collision (v3) is 6.4 m/s. Using the momentum conservation formula, we can solve for Ender's mass (m1) as follows:

m1 × 12 m/s + 45 kg × 9.0 m/s = (m1 + 45 kg) × 6.4 m/s

Expanding this and rearranging the terms, we get:

12m1 = 6.4m1 + 6.4 × 45

12m1 - 6.4m1 = 6.4 × 45

m1(12 - 6.4) = 6.4 × 45

m1 = (6.4 × 45) / (12 - 6.4)

m1 = 288 / 5.6

m1 = 51.43 kg

Therefore, Ender's mass is approximately 51.43 kg.