Two point charges of +2.50 x 10^-5 C and -2.50 x 10^-5 C are separated by 0.50m. Which of the following describes the force between them?

A): 90 N, repulsive
B): 90 N, attractive
C): 23 N, attractive
D): 45 N, attractive

If one could answer this ASAP, it will be greatly appreciated

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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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.

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.

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:

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:

[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:

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:

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:

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 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:

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:

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:

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:

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

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]

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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:

[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:

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

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:

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 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

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.

Learn more about acceleration:https://brainly.com/question/12134554

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.

Find out more information about acceleration.

brainly.com/question/4279093.

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: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:

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:

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]

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:

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:

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.