A box of mass 115 kg sits on an inclined surface with an angle of 59. What is the component of the weight of the box along the surface?

Answer: 70.5 cm

Explanation:

The position on the meter stick, at which one would hang a third mass of 0.61 kg, to keep the meter stick in balance will be at the side of 0.31kg.

You will use the moment techniques.

That is,

Sum of the clockwise moment = sum of anticlockwise moments

Please find the attached file for the remaining explanation and solution.

Answer:

The  pressure at point 2 is [tex]P_2 = 254.01 kPa[/tex]

Explanation:

From the question we are told that

   The speed at point 1  is  [tex]v_1 = 3.57 \ m/s[/tex]

   The  gauge pressure at point 1  is  [tex]P_1 = 68.7kPa = 68.7*10^{3}\ Pa[/tex]

    The density of water is  [tex]\rho = 1000 \ kg/m^3[/tex]

Let the  height at point 1 be  [tex]h_1[/tex] then the height at point two will be

      [tex]h_2 = h_1 - 18.5[/tex]

Let the  diameter at point 1 be  [tex]d_1[/tex] then the diameter at point two will be

      [tex]d_2 = 2 * d_1[/tex]

Now the continuity equation is mathematically represented as  

         [tex]A_1 v_1 = A_2 v_2[/tex]

Here [tex]A_1 , A_2[/tex]  are the area at point 1 and 2

    Now given that the are is directly proportional to the square of the diameter [i.e [tex]A= \frac{\pi d^2}{4}[/tex]]

   which can represent as

             [tex]A \ \ \alpha \ \ d^2[/tex]

=>         [tex]A = c d^2[/tex]

where c is a constant

  so      [tex]\frac{A_1}{d_1^2} = \frac{A_2}{d_2^2}[/tex]

=>          [tex]\frac{A_1}{d_1^2} = \frac{A_2}{4d_1^2}[/tex]

=>        [tex]A_2 = 4 A_1[/tex]

Now from the continuity equation

        [tex]A_1 v_1 = 4 A_1 v_2[/tex]

=>     [tex]v_2 = \frac{v_1}{4}[/tex]

=>     [tex]v_2 = \frac{3.57}{4}[/tex]

       [tex]v_2 = 0.893 \ m/s[/tex]

Generally the Bernoulli equation is mathematically represented as

       [tex]P_1 + \frac{1}{2} \rho v_1^2 + \rho * g * h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho * g * h_2[/tex]

So  

         [tex]P_2 = \rho * g (h_1 -h_2 )+P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )[/tex]  

=>    [tex]P_2 = \rho * g (h_1 -(h_1 -18.3) + P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )[/tex]

substituting values

        [tex]P_2 = 1000 * 9.8 (18.3) )+ 68.7*10^{3} + \frac{1}{2} * 1000 ((3.57)^2 -0.893 ^2 )[/tex]

       [tex]P_2 = 254.01 kPa[/tex]

Answer:

Explanation:

Work done on a spring is expressed as [tex]W = 1/2 ke^{2}[/tex]

k is the elastic constant

e is the extension of the material

If 4 J of work is needed to stretch a spring from its natural length of 36 cm to a length of 47 cm, then;

Work done = 4J and the extension e = 47 cm - 36 cm; e = 11 cm

11cm = 0.11m

Substituting the given values into the equation above to get the elastic constant;

[tex]W = 1/2 ke^{2}\\4 = 1/2k(0.11)^{2} \\8 = 0.0121k\\k = 8/0.0121\\k = 661.16N/m[/tex]

a) In order to determine the amount of work needed work is needed to stretch the spring from 41 cm to 45 cm, wre will use the same formula as above.

[tex]W = 1/2ke^{2} \\e = 0.45 - 0.41\\e = 0.04 m\\ k = 661.16N/m[/tex]

[tex]W = 1/2 * 661.16 * 0.04^{2} \\W = 330.58*0.0016\\W = 0.53J (to\ 2d.p)[/tex]

b) According to hooke's law, F = ke where F is the applied force

We are to get the extension when a force of 15N is applied to the original length of the material.

e = F/k

e = 15/661.16

e = 0.02 m (to 1 d.p)

This means that the natural length of the spring will be stretched by 0.02 m when a force of 15N is applied to it.

Answer:

a) 6.57 m/s

b) 53.75 J

c) 6.37 m/s

d) -98.297 J

Explanation:

mass of player = [tex]m_{p}[/tex] = 117.5 kg

speed of player = [tex]v_{p}[/tex] = 6.5 m/s

mass of ball = [tex]m_{b}[/tex] = 0.43 kg

velocity of ball = [tex]v_{b}[/tex] = 26.5 m/s

Recall that momentum of a body = mass x velocity = mv

initial momentum of the player = mv = 117.5 x 6.5 = 763.75 kg-m/s

initial momentum of the ball = mv = 0.43 x 26.5 = 11.395 kg-m/s

initial kinetic energy of the player = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.5^{2}[/tex] =  2482.187 J

a) according to conservation of momentum, the initial momentum of the system before collision must equate the final momentum of the system.

for this first case that they travel in the same direction, their momenta carry the same sign

[tex]m_{p}[/tex][tex]v_{p}[/tex] + [tex]m_{b}[/tex][tex]v_{b}[/tex] = ([tex]m_{p}[/tex] +[tex]m_{b}[/tex])v

where v is the final velocity of the player.

inserting calculated momenta of ball and player from above, we have

763.75 + 11.395 = (117.5 + 0.43)v

775.145 = 117.93v

v = 775.145/117.93 = 6.57 m/s

b) the player's new kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.57^{2}[/tex] = 2535.94 J

change in kinetic energy = 2535.94 - 2482.187 = 53.75 J  gained

c) if they travel in opposite direction, equation becomes

[tex]m_{p}[/tex][tex]v_{p}[/tex] - [tex]m_{b}[/tex][tex]v_{b}[/tex] = ([tex]m_{p}[/tex] +[tex]m_{b}[/tex])v

763.75 - 11.395 = (117.5 + 0.43)v

752.355 = 117.93v

v = 752.355/117.93 = 6.37 m/s

d) the player's new kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.37^{2}[/tex]  = 2383.89 J

change in kinetic energy = 2383.89 - 2482.187 = -98.297 J

that is 98.297 J  lost

Answer:

The  speed  is  [tex]v = 214.35 \ m/s[/tex]

Explanation:

From the question we are told that

   The  linear mass density is  [tex]\mu = 3700 \ kg/m[/tex]

    The tension is  [tex]T = 1.7*10^8 \ N[/tex]

The  transverse wave speed is mathematically represented as  

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

substituting values

      [tex]v = \sqrt{\frac{1.7 *10^{8}}{3700} }[/tex]

      [tex]v = 214.35 \ m/s[/tex]

Answer:

D

Explanation:

Answer:

257 kN.

Explanation:

So, we are given the following data or parameters or information in the following questions;

=> "A jet transport with a landing speed

= 200 km/h reduces its speed to = 60 km/h with a negative thrust R from its jet thrust reversers"

= > The distance = 425 m along the runway with constant deceleration."

=> "The total mass of the aircraft is 140 Mg with mass center at G. "

We are also give that the "aerodynamic forces on the aircraft are small and may be neglected at lower speed"

Step one: determine the acceleration;

=> Acceleration = 1/ (2 × distance along runway with constant deceleration) × { (landing speed A)^2 - (landing speed B)^2 × 1/(3.6)^2.

=> Acceleration = 1/ (2 × 425) × (200^2 - 60^2) × 1/(3.6)^2 = 3.3 m/s^2.

Thus, "the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking" = The total mass of the aircraft × acceleration × 1.2 = 15N - (9.8 × 2.4 × 140).

= 140 × 3.3× 1.2 = 15N - (9.8 × 2.4 × 140).

= 257 kN.

The reaction N under the nose wheel B towards the end of the braking interval =  257 kN

Given data :

Landing speed of Jet = 200 km/h

Distance = 425 m

Total mass of aircraft = 140 Mg  with mass center at G

  Jet acceleration = 1 / (2 *425) * (200²  - 60² ) *  1 / (3.6)²

                              = 3.3 m/s²

Reaction N = Total mass of aircraft * jet acceleration* 1.2 = 15N - (9.8*2.4* 140).   ----- ( 1 )

∴ Reaction N = 140 * 3.3 * 1.2 = 15 N - ( 9.8*2.4* 140 )  

 Hence Reaction N = 257 KN

We can conclude that the The reaction N under the nose wheel B towards the end of the braking interval =  257 kN

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Answer:

The temperature of the filament when the flashlight is on is 2020 °C.

Explanation:

The resistivity varies linearly with temperature:

[tex] R = R_{0}[1 + \alpha*(T-T_{0})] [/tex]   (1)

Where:

T: is the temperature of the filament when the flashlight is on=?

T₀: is the initial temperature = 20 °C

α: is the temperature coefficient of resistance = 0.0045 °C⁻¹ 

R₀: is the resistance at T₀ = 1.5 Ω    

When V = 3.0 V, R is:

[tex]R = \frac{V}{I} = \frac{3.0 V}{0.20 A} = 15 \Omega[/tex]

By solving equation (1) for T we have:

[tex]T = \frac{R-R_{0}}{\alpha*R_{0}} + T_{0} = \frac{15-1.5}{0.0045*1.5} + 20 = 2020 ^{\circ} C[/tex]

Therefore, the temperature of the filament when the flashlight is on is 2020 °C.

I hope it helps you!            

Answer:

Explanation:

Kinetic energy is expressed as KE = 1/2 mv² where;

m is the mass of the body

v is the velocity of the body.

For the heavier shell;

m = 5kg

KE gained = 100J

Substituting this values into the formula above to get the velocity v;

100 = 1/2 * 5 * v²

5v² = 200

v² = 200/5

v² = 40

v = √40

v = 6.32 m/s

Note that after the explosion, both body fragments will possess the same velocity.

For the lighter shell;

mass = 2.0kg and v = 6.32m/s

KE of the lighter shell = 1/2 * 2 * 6.32²

KE of the lighter shell = 6.32²

KE of the lighter shell= 39.94m/s

Hence, the lighter one gains a kinetic energy of 39.94m/s.

The gain in the kinetic energy of the smaller fragment is 249.64 J.

The given parameters;

The velocity of the heavier fragment is calculated as follows;

[tex]K.E = \frac{1}{2} mv^2\\\\mv^2 = 2K.E\\\\v^2 = \frac{2K.E}{m} \\\\v= \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2 \times 100}{5} }\\\\v = 6.32 \ m/s[/tex]

Apply the principle of conservation of linear momentum to determine the velocity of the smaller fragment as;

[tex]m_1 u_1 + m_2 u_2 = v(m_1 + m_2)\\\\-6.32(5) \ + 2u_2 = 0(7)\\\\-31.6 + 2u_2 = 0\\\\2u_2 = 31.6\\\\u_2 = \frac{31.6}{2} \\\\u_2 = 15.8 \ m/s[/tex]

The gain in the kinetic energy of the smaller fragment is calculated as follows;

[tex]K.E_2 = \frac{1}{2} mu_2^2\\\\K.E_2 = \frac{1}{2} \times 2 \times (15.8)^2\\\\K.E_2 = 249.64 \ J[/tex]

Thus, the gain in the kinetic energy of the smaller fragment is 249.64 J.

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Answer:

I believe it is 91

Explanation:

Answer:

a) 24.4 Ω

b) 4.92 A

c) 495.9 W

d)

c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.

Explanation:

b)

The formula for power is:

P = IV

where,

P = Power of heater = 590 W

V = Voltage it takes = 120 V

I = Current Drawn = ?

Therefore,

590 W = (I)(120 V)

I = 590 W/120 V

I = 4.92 A

a)

From Ohm's Law:

V = IR

R = V/I

Therefore,

R = 120 V/4.92 A

R = 24.4 Ω

c)

For constant resistance and 110 V the power becomes:

P = V²/R

Therefore,

P = (110 V)²/24.4 Ω

P = 495.9 W

d)

If the resistance decreases, it will increase the current according to Ohm's Law. As a result of increase in current the power shall increase according to formula (P = VI). Therefore, correct option is:

c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.

Answer:

The energy coverted to heat is 200 kilojoules.

Explanation:

GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:

[tex]m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v[/tex]

Where:

[tex]m_{M}[/tex], [tex]m_{S}[/tex] - Masses of the small meteorite and the communication satellite, measured in kilograms.

[tex]v_{M}[/tex], [tex]v_{S}[/tex] - Speeds of the small meteorite and the communication satellite, measured in meters per second.

[tex]v[/tex] - Final speed of the satellite-meteorite system, measured in meters per second.

The final speed of the satellite-meteorite system is cleared:

[tex]v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}[/tex]

If [tex]m_{M} = 1\times 10^{-3}\,kg[/tex], [tex]m_{S} = 200\,kg[/tex], [tex]v_{M} = 20000\,\frac{m}{s}[/tex] and [tex]v_{S} = 0\,\frac{m}{s}[/tex], the final speed is now calculated:

[tex]v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}[/tex]

[tex]v = 0.1\,\frac{m}{s}[/tex]

Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

[tex]K_{S} + K_{M} - K - Q_{disp} = 0[/tex]

[tex]Q_{disp} = K_{S}+K_{M}-K[/tex]

Where:

[tex]K_{S}[/tex], [tex]K_{M}[/tex] - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

[tex]K[/tex] - Kinetic energy of the satellite-meteorite system, measured in joules.

[tex]Q_{disp}[/tex] - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

[tex]Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}][/tex]

Given that [tex]m_{M} = 1\times 10^{-3}\,kg[/tex], [tex]m_{S} = 200\,kg[/tex], [tex]v_{M} = 20000\,\frac{m}{s}[/tex], [tex]v_{S} = 0\,\frac{m}{s}[/tex] and [tex]v = 0.1\,\frac{m}{s}[/tex], the dissipated heat is:

[tex]Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right][/tex][tex]Q_{disp} = 200000\,J[/tex]

[tex]Q_{disp} = 200\,kJ[/tex]

The energy coverted to heat is 200 kilojoules.

Answer:

the phase relationship between two waves.

Explanation:

Coherence describes all properties of the correlation between physical quantities between waves. It is an ideal property of waves that determines their interference. In a situation in which there is a correlation or phase relationship between two waves. If the properties of one of the waves can be measure directly, then, some of the properties of the other wave can be calculated.

Answer:

vpg = 0.064 N

Explanation:

Upthrust = Volume of fluid displaced

upthrust liquid on the cube g=10ms−2

vpg =0.2 x 0.2 x 0.2 x0.8 x 10= 0.064N

vpg = 0.064 N

hope it helps.

Explanation:

a) d[phi]/dt = (dB/dt)*Acos(0) = (-0.20)*(pi(2.25*10^-2)^2) = -3.98*10^-4 Wb

E = (1/2r*pi)*(d[phi]/dt) = -2.8*10^-3 N/C

b) Clockwise because The induced magnetic field will be in the direction to oppose the change. Since the magnetic flux from the magnets is decreasing, the induced magnetic field will be in the same direction as the magnet's field.

Answer:

Explanation:

1 ) Average power supplied to an inductor is zero because the phase difference of potential and current is π / 2 .

So it is a wrong statement .

2 ) Step up transformer increases the voltage . At high voltage , lesser current is required to transport electrical energy . When current is reduced , the loss of energy due to heating effect is reduced .

3 ) voltage and current are in phase in resistance  in ac .

3 ) RMS stands for Root Mean Square .

Answer:

3.77x10^-5T

Explanation:

Magnetic field at center of the loop is given as

B=uo*I/2r =(4pi*10-7)*6/2*0.1

B=3.77*10-5Tor 37.7 uTi

Answer:

To verify that they're polarized, he could hold the two lenses perpendicular (90 degrees) to each other, one lens in front of the other, and point it at a light source. If no light passes through then the lenses are polarized

The test of Polarization of pair of sunglasses is , hold the two lenses perpendicular to each other, one lens in front of the other, and point it towards a light source. If no light passes through then the lenses are polarized.

For example, when fishing on a sunny day, you wouldn't see through the water. You would only see a reflection of the sun hitting the water.

To verify that  pair of sunglasses are  polarized, he could hold the two lenses perpendicular  to each other, one lens in front of the other, and point it towards a light source. If no light passes through then the lenses are polarized.

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Answer:

The impulse is 2145 kg-m/s

The final velocity is -8.34 m/s or 8.34 m/s in he opposite direction.

Explanation:

Force on the rail = 3900 N

Elapsed time of impact = 0.55 s

Impulse is the product of force and the time elapsed on impact

I = Ft

I is the impulse

F is force

t is time

For this case,

Impulse = 3900 x 0.55 = 2145 kg-m/s

If the initial velocity was 2.95 m/s

and mass of car plus driver is 190 kg

neglecting friction, the initial momentum of the car is given as

P = mv1

where P is the momentum

m is the mass of the car and driver

v1 is the initial velocity of the car

initial momentum of the car P = 2.95 x 190 = 560.5 kg-m/s

We know that impulse is equal to the change of momentum, and

change of momentum is initial momentum minus final momentum.

The final momentum = mv2

where v2 is the final momentum of the car.

The problem translates into the equation below

I = mv1 - mv2

imputing values, we have

2145 = 560.5 - 190v2

solving, we have

2145 - 560.5 = -190v2

1584.5 = -190v2

v2 = -1584.5/190 = -8.34 m/s

Answer:

The magnitude of the magnetic field is 55μT

Explanation:

Given;

number of turns of the solenoid per length, n = N/L = 1500 turns/m

current in the solenoid, I = 20 mA = 20 x 10⁻³ A

diameter of the solenoid, d = 4 cm = 0.04 m

The magnetic field at a point that is inside the solenoid;

B₁ = μ₀nI

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

B₁ = 4π x 10⁻⁷ x 1500 x 20 x 10⁻³

B₁ = 3.77 x 10⁻⁵ T

Given;

current in the wire, I = 2 A

distance of magnetic field from the wire, r = 1 cm = 0.01 m

The magnetic field at 1.0 cm from the wire;

[tex]B_2 = \frac{\mu_0I}{2\pi r} \\\\B_2 = \frac{4\pi*10^{-7}*2}{2\pi *0.01}\\\\B_2 = 4 *10^{-5} \ T[/tex]

The magnitude of the magnetic field;

[tex]B = \sqrt{B_1^2 +B_2^2} \\\\B = \sqrt{(3.77*10^{-5})^2 + (4*10^{-5})^2} \\\\B = 5.5 *10^{-5} \ T\\\\B = 55 \mu T[/tex]

Therefore, the magnitude of the magnetic field is 55μT

The magnitude of the magnetic field at a point that is inside the solenoid and 1.0 cm from the wire is [tex]5.5 \times 10^{-5}T[/tex]

Given the following parameters from the question  

The magnetic field at a point that is inside the solenoid is expressed according to the formula;  

Where;  

μ₀ is the permeability of free space = 4π x 10⁻⁷ m/A  

B₁ = 4π x 10⁻⁷ x 1500 x 20 x 10⁻³  

B₁ = 3.77 x 10⁻⁵ T

Next is to get the magnetic field strength in the second wire.

Substitute into the formula:

[tex]B_2=\dfrac{\mu_0 I}{2 \pi r} \\B_2=\frac{4\pi \times 10^{-7}\times 2}{2 \times 3.14\times 0.01} \\B_2 =4.0 \times 10^{-5}T[/tex]

Get the resultant magnetic field:

[tex]B = \sqrt{(0.00003771)^2+(0.00004)^2} \\B =5.5 \times 10^{-7}T[/tex]

Therefore the magnitude of the magnetic field at a point that is inside the solenoid and 1.0 cm from the wire is [tex]5.5 \times 10^{-5}T[/tex]

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Answer:

The moment of inertial of the wheel,  [tex]I = 8(\frac{1}{3}M_sL^2 ) + M_rL^2[/tex]

Explanation:

Given;

8 spokes of uniform diameter

mass of each spoke, = [tex]M_s[/tex]

length of each spoke, = L

mass of outer ring, = [tex]M_r[/tex]

The moment of inertial of the wheel will be calculated as;

[tex]I = 8I_{spoke} + I_{ring}[/tex]

where;

[tex]I_{spoke[/tex] is the moment of inertia of each spoke

[tex]I_{ring[/tex] is the moment of inertia of the rim

Moment of inertia of each spoke [tex]=\frac{1}{3}M_sL^2[/tex]

Moment of inertial of the wheel

[tex]I = 8(\frac{1}{3}M_sL^2 ) + M_rL^2[/tex]

Answer:

Since their wings and body develop the drag. When there is warm air then they expand their wings. Since,soaring birds and glider pilots have no engine, they always maintain their high speed to lift their weight in air for hours without expending power by convection

Explanation:

Answer:

Explanation:

Total mass m = 18 kg .

Spring are parallel to each other so total spring constant

= 4 x 24 = 96 N/cm = 9600 N/m

Time period of vibration

[tex]T=2\pi\sqrt{\frac{m}{k} }[/tex]

Putting the given  values

[tex]T=2\pi\sqrt{\frac{18}{9600} }[/tex]

= .27 s .

Answer:

a

The  speed of  wave is   [tex]v_1 = 129.1 \ m/s[/tex]

b

The new speed of the two waves is [tex]v = 129.1 \ m/s[/tex]

Explanation:

From the question we are told that

    The mass of the string is  [tex]m = 60 \ g = 60 *10^{-3} \ kg[/tex]

    The length is  [tex]l = 2.0 \ m[/tex]

    The tension is  [tex]T = 500 \ N[/tex]

Now the velocity of the first wave is mathematically represented as

     [tex]v_1 = \sqrt{ \frac{T}{\mu} }[/tex]

Where  [tex]\mu[/tex] is the linear density which is mathematically represented as

      [tex]\mu = \frac{m}{l}[/tex]

substituting values    

     [tex]\mu = \frac{ 60 *10^{-3}}{2.0 }[/tex]

     [tex]\mu = 0.03\ kg/m[/tex]

So

   [tex]v_1 = \sqrt{ \frac{500}{0.03} }[/tex]

   [tex]v_1 = 129.1 \ m/s[/tex]

Now given that the Tension, mass and length are constant the velocity of the second wave will same as that of first wave (reference PHYS 1100 )

Answer:

The current in the small radius loop must be 0.9677 A

Explanation:

Recall that the formula for the magnetic field at the center of a loop of radius R which runs a current I, is given by:

[tex]B=\mu_0\,\frac{I}{2\,R}[/tex]

therefore for the first loop in the problem, that magnetic field strength is:

[tex]B=\mu_0\,\frac{I}{2\,R} =\mu_0\,\frac{3}{2\,(0.093)} =16.129\,\mu_{0}\,[/tex]

with the direction of the magnetic field towards the plane.

For the second smaller loop of wire, since the current goes counterclockwise, the magnetic field will be pointing coming out of the plane, and will subtract from the othe field. In order to the addition of these two magnetic fields to be zero, the magnitudes of them have to be equal, that is:

[tex]16.129\,\,\mu_{0}=\mu_0\,\frac{I'}{2\,R'} =\mu_{0}\,\frac{I'}{2\,(0.03)} \\I'=16.129\,(2)\,(0.03)=0.9677\,\,Amps[/tex]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The wavelength is   [tex]\lambda= \frac{2 \pi }{k}[/tex]

Explanation:

From the question we are told that  

      The electric field is [tex]\= E = E_o sin (kx - wt )\r j[/tex]

       The magnetic field is  [tex]\= B = B_0 sin (kx -wt) \r k[/tex]

From the above equation

and  k is the wave number which is mathematically represented as

        [tex]k = \frac{2 \pi }{\lambda }[/tex]

=>     [tex]\lambda= \frac{2 \pi }{k}[/tex]

Where [tex]\lambda[/tex] is the wavelength

Answer:

gravitational potential energy at point A.

A) The gravitational potential energy at point A is; 705600 J

B) The velocity at point C, if the work done to move the roller coaster from point B to C is 264870 J is; v = 31.295 m/s

A) Formula for gravitational potential energy is;

PE = mgh

At point A;

mass; m = 900 kg

height; h = 80 m

Thus;

PE = 900 × 9.8 × 80

PE = 705600 J

B) Kinetic energy of the roller coaster at point C is given as;

KE = PE - W

We are given Workdone; W = 264870 J

Thus;

KE = 705600 - 264870

KE = 440730 J

Thus, velocity at point C is gotten from the formula of kinetic energy;

KE = ½mv²

v = √(2KE/m)

v = √(2 × 440730/900)

v = 31.295 m/s

Read more at; https://brainly.com/question/14295020

Answer:

Nails are made of iron. Iron consists of 26 protons and 26 electrons. protons are positively charged and electrons are negatively charged, so this force of attraction keeps the electrons together.

If electrons repel from each other, the positively charge protons and nucleus allow them to move in a definite orbit and prevent them flying out of the nail.

Answer:

E = 6.45 x 10⁻¹¹ J

Explanation:

First we need to find total number of photons detected in 1 hour. Therefore,

No. of Photons = n = (3.2 x 10⁸ photons/s)(1 h)(3600 s/1 h)

n = 11.52 x 10¹¹ photons

Now, the energy of these photons can be given by the formula:

E = nhc/λ

where,

E = Total Energy of the Photons = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of radiation = 3.55 mm = 3.55 x 10⁻³ m

Therefore,

E = (11.52 x 10¹¹)(6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(3.55 x 10⁻³ m)

E = 6.45 x 10⁻¹¹ J

Answer: it is easier to read in bright light than dim light.

Explanation:

The ray of light is the direction that is used by light in travelling through a medium. Rays that pass through a lens very close to the principle axis are more sharply focused than those that are very far from the axis.

Because of the fact that the rays are close to the principle axis, the spherical aberration helps us to understand the reason why it is easier for people to read in bright light than readin iin dim light.