A horizontal force of 480 n is applied to a stationary wooden box in one direction and a 600 n horizontal force is applied in the opposite direction. What is the additional force is needed for the box to remain stationary

Answer:

a) m=1, y₁ = 0.08 m , θ₁ = 4.57º ,  b) m=2,  y₂ = 0.16 m , θ₂ = 9.09º ,  c)  m=3,   y₃ = 0.24 m ,   θ₃ = 13.5º

Explanation:

After reading your strange statement, I understand that this is an interference problem, I transcribe the data to have it more clearly. Number of slits 2, distance between slits 5000 nm, wavelength 400 nm, distance to the screen 1 m.

They ask us to calculate the angles for the first, second and third interference, they also ask us to write down the distance from the central maximum.

The expression for constructive interference for two slits is

             d sin θ = m λ

where d is the distance between the slits, λ is the wavelength used, m is an integer representing the order of interference

Let's use trigonometry to find the distance from the central maximum

         tan θ = y / L

in all interference experiments the angle is small,

         tan θ = sin θ / cos θ = sin θ

         sint θ = y / L

let's replace

        d y / L = m λ

        y = m λ L / d

let's calculate

distance to the first maximum m = 1

          y₁ = 1  400 10⁻⁹ 1/5000 10⁻⁹

          y₁ = 0.08 m

distance to second maximum m = 2

          y₂ = 2  400 10⁻⁹ 1/5000 10⁻⁹

          y₂ = 0.16 m

distance to the third maximum m = 3

          y₃ = 3  400 10⁻⁹ 1/5000 10⁻⁹

          y₃ = 0.24 m

with these values ​​we can search for each angle

           tan θ = y / L

           θ = tan⁻¹ y / L

for m = 1

            θ₁ = tan⁻¹ (0.08 / 1)

            θ₁ = 4.57º

for m = 2

            θ₂ = tan⁻¹ (0.16 / 1)

            θ₂ = 9.09º

for m = 3

            θ₃ = tan⁻¹ (0.24 / 1)

            θ₃ = 13.5º

Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after the objects collide and bounce apart?

a-velocity

b-mass

c-momentum

d-direction

Answer:

b. Mass

Explanation:

This question has to do with the principle of the law of conservation of momentum which states that the momentum of a system remains constant if no external force is acting on it.

As the question states, two objects collide with each other and eventually bounce apart, so their momentum may not be conserved but the mass of the objects is constant for each non-relativistic motion. Because of this, the mass of each object prior to the collision would be the same as the mass after the collision.

Therefore, the correct answer is B. Mass.

This question involves the concept of the law of conservation of momentum.

Jerome should always keep the "mass" of each object constant after the objects collide and bounce apart.

The law of conservation of momentum states that the momentum of a system of objects must remain constant before and after the collision has taken place.

Mathematically,

[tex]m_1u_1+m_2u_2=m_1v_1+m_v_2[/tex]

where,

m₁ = mass of the first object

m₂ = mass of the second object

u₁ = velocity of the first object before the collision

u₂ = velocity of the second object before the collision

v₁ = velocity of the first object after the collision

v₂ = velocity of the second object after the collision

Hence, it is clear from the formula that the only thing unchanged before and after the collision is the mass of each object.

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The attached picture illustrates the law of conservation of momentum.

Answer:

The correct answer is option 3 .

Please check the answer once :)

Answer:

F₁ = 100 N

Explanation:

The pressure must be equally transmitted from the piston to the narrow arm. Therefore,

P₁ = P₂

F₁/A₁ = F₂/A₂

F₁/F₂ = A₁/A₂

where,

F₁ = Force Required to be applied to piston = ?

F₂ = Force to push cork at narrow arm = 16 N

A₁ = Area of wider arm = πd₁²/4

A₂ = Area of narrow arm = πd₂²/4

Therefore,

F₁/16 N = (πd₁²/4)/(πd₂²/4)

F₁ = (16 N)(d₁²/d₂²)

but, it is given that the diameter of wider arm is 2.5 times the diameter of the narrow arm.

d₁ = 2.5 d₂

Therefore,

F₁ = (16 N)[(2.5 d₂)²/d₂²]

F₁ = (16 N)(6.25)

F₁ = 100 N

Answer:

The length of the pendulum depends on acceleration due to gravity (g) which varies in different Earth's location beacuse Earth is not perfectly spherical.

Explanation:

The period of oscillation is calculated as;

[tex]T = 2\pi\sqrt{\frac{l}{g} }[/tex]

where;

L is the length of the pendulum bob

g is acceleration due to gravity

If we make L the subject of the formula in the equation above, we will have;

[tex]T = 2\pi\sqrt{\frac{l}{g}}\\\\\sqrt{\frac{l}{g} } = \frac{T}{2\pi} \\\\\frac{l}{g} = (\frac{T}{2\pi} \)^2\\\\\frac{l}{g} =\frac{T^2}{4\pi^2}\\\\L = \frac{gT^2}{4\pi^2}[/tex]

The length of the pendulum depends on acceleration due to gravity (g).

Acceleration due to gravity is often assumed to be the same everywhere on Earth, but it varies because Earth is not perfectly spherical. The variation of acceleration due to gravity (g) as a result of Earth's geometry, will also cause the length of the pendulum to vary.

Answer:

f = 357.29Hz

Explanation:

In order to calculate the fundamental frequency in the closed tube, you use the following formula:

[tex]f_n=\frac{nv}{4L}[/tex]       (1)

n: order of the mode = 1

v: speed of sound = 343m/s

L: length of the tube = 24cm = 0.24m

You replace the values of the parameters in the equation (1):

[tex]f_1=\frac{(1)(343m/s)}{4(0.24m)}=357.29Hz[/tex]

The fundamental frequency of in the tube is 357.29Hz

Answer:

T = 7.61*10^6 s

Explanation:

In order to calculate the Mercury's period. in its orbit around the sun, you take into account on the Kepler's law. You use the following formula:

[tex]T=\sqrt{\frac{4\pi^2r^3}{GM_s}}[/tex]         (1)

T: period of Mercury

r: distance between Mercury and Sun

Ms: mass of the sun = 1.98*10^30 kg

G: Cavendish's constant = 6.674*10^-11 m^3 kg^-1 s^-2

You replace the values of all parameters in the equation (1):

[tex]T=\sqrt{\frac{4\pi^2(5.79*10^{10}m)^3}{(6.674*10^{-11}m^3kg^{-1}s^{-2})(1.98*10^{30}kg)}}\\\\T=7.61*10^6s*\frac{1h}{3600s}*\frac{1d}{24h}=88.13\ days[/tex]

The period of Mercury is 7.61*10^6 s, which is approximately 88.13 Earth's days

Answer:

see that the entire spectrum of the visible is between the integers from 3 to 5 so only three wavelengths are reflected with constructive interference

Explanation:

This is an interference problem in thin films, the refractive index of water is 1.33 and the refractive index of oil is 1.5

Let's analyze the light beam path emitted by the diver.

* when the beam passes from the water to the oil with the highest refractive index, it has a phase change of 180º

* also the wavelength of light in a material medium changes

      λ_n =  λ / n

where  λ_n is the wavelength in the material and  λ the wavelength in the vacuum air and n the refractive index.

If we include these aspects, the constructive interference equation is

       2t = (m + ½)  λ_n

       2nt = (m + ½)  λ

let's apply this equation to our case

            λ = 2nt / (m + ½)

The incidence of replacement of the oil with respect to water is

        n = n_oil / n_water = 1.5 / 1.33

        n = 1,128

let's calculate

        λ = 2 1,128 t / (m + ½)

        λ = 2,256 t / (m + ½)

In your statement you do not include the value of the oil layer that is the thin film, suppose a value to finish the calculation

          t = 0.001 mm = 1 10⁻⁶ m

the formula remains

        λ = 2,256 10⁻⁶ / (m + ½)

Let's find what values ​​of m we have to cut light in the visible range (400 to 700) 10⁻⁹ m

     m + ½ = 2,256 10⁻⁶ / λ

     m = 2,256 10⁻⁶ / λ - ½

light purple lan = 400 10⁻⁹m

     m = 2,256 10-6 / 400 10⁻⁹ - ½

     m = 5.64 - 0.5

     m = 5.14

     m = 5

red light  λ = 700 10⁻⁹m

      m = 2,256 1-6 / 700 10⁻⁹ - ½

      m = 3.22 - 0.5

      m = 2.72

      m = 3

we see that the entire spectrum of the visible is between the integers from 3 to 5 so only three wavelengths are reflected with constructive interference

Answer:

a.     E = 1.02*10^-27 J

b.     E = 6.39*10^-9eV

Explanation:

a. In order to calculate the energy of the radio photon, you use the following formula:

[tex]E=hf[/tex]             (1)

h: Planck's constant = 6.626*10^-34 Js

f: frequency of the photon = 1545kHz = 1.545*10^6 Hz

Then, by replacing you obtain the energy of the photon:

[tex]E=(6.626*10^{-34}Js)(1.545*10^6s^{-1})=1.02*10^{-27}J[/tex]

b. In electron volts, the energy of the photon is:

[tex]E=1.02*10^{-27}J*\frac{6.242*10^{18}eV}{1J}=6.39*10^{-9}eV[/tex]

Answer:

The power exerted by the student is 51.2 W

Explanation:

Given;

extension of the elastic band, x = 0.8 m

time taken to stretch this distance, t = 0.5 seconds

the spring constant, k = 40 N/m

Apply Hook's law;

F = kx

where;

F is the force applied to the elastic band

k is the spring constant

x is the extension of the elastic band

F = 40 x 0.8

F = 32 N

The power exerted by the student is calculated as;

P = Fv

where;

F is the applied force

v is velocity = d/t

P = F x (d/t)

P = 32 x (0.8 /0.5)

P = 32 x 1.6

P = 51.2 W

Therefore, the power exerted by the student is 51.2 W

Answer:

D. All of the above

Explanation:

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James Marcia is a psychologist known mainly for his research and theories in human identity. Identity according to him is the sum total of a person's beliefs, values, and ideologies that shape what a person actually becomes and is known for. Occupation and Ideologies primarily determine identity. The four stages of Identity status include, Identity diffusion, foreclosure,  moratorium, and achievement.

Answer:

The only force acts on a projectile is gravitational force {Fg}, therefore its acceleration a=Fg/m will always directed towards the direction of force i.e. vertically downwards. Therefore it will always be perpendicular to the x direction or here we can say that a is always perpendicular to Vx}.

Explanation:

The vector r indicates the instantaneous displacement of a projectile from the origin. At the instant when the projectile is at r , its velocity and acceleration vectors are v and a . Which statement is correct?

Answer:

[tex]F=1.8\times 10^{-3}\ N[/tex]

Explanation:

We have,

Length of wires is 11 m

Separation between wires is 0.033 m

Current in both the wires is 5.2 A

It is required to find the magnitude of force between two wires. The force between wires is given by :

[tex]F=\dfrac{\mu_o I_1I_2l}{2\pi r}\\\\F=\dfrac{4\pi \times 10^{-7}\times 5.2\times 5.2\times 11}{2\pi \times 0.033}\\\\F=1.8\times 10^{-3}\ N[/tex]

So, the magnitude of force between wires is [tex]1.8\times 10^{-3}\ N[/tex]

Answer:

C. explicit and implicit

Explanation:

E20

Answer:

The tube should be held vertically, perpendicular to the ground.

Explanation:

As the power lines  of ground are equal, so its electrical field is perpendicular to the ground and the equipotential surface is cylindrical. Therefore, if we put the position fluorescent tube parallel to the ground so the both ends of the tube lie on the same equipotential surface and the difference is zero when its  potential.

And the ends of the tube must be on separate equipotential surfaces to optimize potential. The surface near the power line has a greater potential value and the surface farther from the line has a lower potential value, so the tube must be placed perpendicular to the floor to maximize the potential difference.

Answer:

a) P = 10.27 kW

b) Pmax = 10.65 kW

c) E = 5.47 MJ

Explanation:

Mass of the loaded car, m = 950 kg

Angle of inclination of the shaft, θ = 28°

Acceleration due to gravity, g = 9.8 m/s²

The speed of the car, v = 2.35 m/s

Change in time, t = 14.0 s

a) The power that must be provided by the winch motor when the car is moving at constant speed.

P = Fv

The force exerted by the motor, F = mg sinθ

P = mgv sinθ

P = 950 * 9.8 *2.35* sin28°

P = 10,271.3 W

P = 10.27 kW

b) Maximum power that the motor must provide:

[tex]P = mv\frac{dv}{dt} + mgvsin \theta\\dv/dt = \frac{2.35 - 0}{14} \\dv/dt = 0.168 m/s^2\\P = (950*2.35*0.168) + (950*9.8*2.35* sin28)\\P = 374.74 + 10271.3\\P = 10646.04 W\\10.65 kW[/tex]

c) Total energy transferred:

Length of the track, d = 1250 m

[tex]E = 0.5 mv^2 + mgd sin \theta\\E = (0.5 * 950 * 2.35^2) + (950 * 9.8 * 1250 * sin 28)\\E = 2623.19 + 5463475.31\\E = 5466098.50 J\\E = 5.47 MJ[/tex]

Answer:

The total momentum is  [tex]p__{T }} =(2400 -4 v_2) \ Dalton \cdot m/s[/tex]

Explanation:

The diagram illustration this  system is shown on the first uploaded image (From physics animation)

From the question we are told that

     The mass of the first object is [tex]M_1 = 12 \ Dalton[/tex]

      The speed of the first mass is [tex]v_1 = 200 \ m/s[/tex]

      The mass of the second object is  [tex]M_2 = 4 \ Dalton[/tex]

      The speed of the second object is  assumed to be  [tex]- v_2[/tex]

The total momentum of the system is the combined momentum of both object which is mathematically represented as

           [tex]p__{T }} = M_1 v_1 + M_2 v_2[/tex]

   substituting values

            [tex]p__{T }} = 12 * 200 + 4 * (-v_2)[/tex]

            [tex]p__{T }} =(2400 -4 v_2) \ Dalton \cdot m/s[/tex]

Answer:

5.02 m

Explanation:

Applying the formula of maximum height of a projectile,

H = U²sin²Ф/2g...................... Equation 1

Where H = maximum height, U = initial velocity, Ф = angle, g = acceleration due to gravity.

Given: U = 46 ft/sec = 14.021 m/s, Ф = 45°

Constant: g = 9.8 m/s²

Substitute these values into equation 1

H = (14.021)²sin²45/(2×9.8)

H = 196.5884×0.5/19.6

H = 5.02 m.

Hence the ball goes 5.02 m high

The ball reaches the maximum height of 54 feet

The question is about projectile motion,

the ball is shot at an angle α = 45°, and

the initial velocity u = 46 ft/s.

Under the projectile motion, the maximum height H is given by:

[tex]H=\frac{u^2sin^2\alpha }{2g} [/tex]

where, g = 9.8 m/s²

substituting the given values we get:

[tex]H=\frac{46^2sin^{2}(45)}{2*9.8}\\ \\ H=\frac{46*46*(1/2)}{2*9.8}\\ \\ H=54 feet[/tex]

Hence, the maximum height is 54 feet.

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Answer:

P₃₅₀ = 0.28 watt

Explanation:

First we find the resistance of the wire at 20°C:

R₀ = ρL/A

where,

ρ = resistivity = 1 x 10⁻⁶ Ωm

L = Length of wire = 100 cm = 1 m

A = cross-sectional area of wire = πr² = π(0.5 x 10⁻³ m)² = 0.785 x 10⁻⁶ m²

Therefore,

R₀ = (1 x 10⁻⁶ Ωm)(1 m)/(0.785 x 10⁻⁶ m²)

R₀ = 1.27 Ω

Now, from Ohm's Law:

V = I₀R₀

where,

V = Potential Difference = ?

I₀ = Current Passing at 20°C = 0.5 A

Therefore,

V = (0.5 A)(1.27 Ω)

V = 0.64 volts

Now, we need to find the resistance at 350°C:

R₃₅₀ = R₀(1 + αΔT)

where,

R₃₅₀ = Resistance at 350°C = ?

α = temperature coefficient of resistance = 0.4 x 10⁻³ °C⁻¹

ΔT = Difference in Temperature = 350°C - 20°C = 330°C

Therefore,

R₃₅₀ = (1.27 Ω)[1 + (0.4 x 10⁻³ °C⁻¹)(330°C)]

R₃₅₀ = 1.44 Ω

Now, for power at 350°C:

P₃₅₀ = VI₃₅₀

where,

P₃₅₀ = Power dissipation at 350°C = ?

V = constant potential difference = 0.64 volts

I₃₅₀ = Current at 350°C = V/R₃₅₀ (From Ohm's Law)

Therefore,

P₃₅₀ = V²/R₃₅₉

P₃₅₀ = (0.64 volts)²/(1.44 Ω)

P₃₅₀ = 0.28 watt

Answer:

The motion of the lighter child would look faster than that of the heavier child, but both have the same period of oscillation.

Explanation:

Oscillation is a type of simple harmonic motion which involves the to and fro movement of an object. The oscillation takes place at a required time called the period of oscillation.

Since the swings are similar, the period of oscillation of the two children are the same and they would complete one oscillation in the same time. Though the oscillation of the lighter child seems faster than that of the heavy child, their masses does not affect the period of oscillation.

When a heavy object oscillates, its mass increases the drag or damping force, but not the period of oscillation. Thus, it oscillate slowly.

Answer:

a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon

Explanation:

F= ma

v²=u² -2aS

(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)

a=1.36×10⁹m/s²

recall

F=ma

F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²

F= 2.55 × 10⁻¹⁹N

the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon

F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)

= 1.38 × 10⁸

Answer:

Explanation:

From the question, the kinectic energy of the train will be equal to the energy stored in the spring.

Kinetic energy = 1/2 mv² and energy stored in a spring E = 1/2 ke².

Equating both we will have;

1/2 mv² = 1/2ke²

mv² = ke²

m is the mass of the train

v is the velocity of then train

k is the spring constant

e is the extension caused by the spring.

Given m = 30000kg, v = 4 m/s, e = 4 - 2.4 = 1.6m

Substituting this values into the formula will give;

30000*4² =  k*1.6²

[tex]k = \frac{30,000*16}{1.6^2}\\ \\k = \frac{480,000}{2.56}\\ \\k = 187,500Nm^{-1}[/tex]

The value of the spring constant is 187,500N/m

Answer:

8.9 seconds

Explanation:

The height of the object at time t is:

y = h + vt − 4.9t²

where h is the initial height, and v is the initial velocity.

Given h = 30 and v = 40:

y = 30 + 40t − 4.9t²

When y = 0:

0 = 30 + 40t − 4.9t²

4.9t² − 40t − 30 = 0

Solving with quadratic formula:

t = [ -(-40) ± √((-40)² − 4(4.9)(-30)) ] / 2(4.9)

t = [ 40 ± √(1600 + 588) ] / 9.8

t = 8.9

It takes 8.9 seconds for the object to land.

Answer:

4L

Explanation:

Data provided in the question according to the question is as follows

Length = L

Gravity = G

For friend

Length = ?

Growth = 4G

Moreover,

[tex]T_1 = T_2[/tex]

Based on the above information ,

Now we have to apply the simple pendulum formula which is shown below:

[tex]T = 2\pi \frac{L}{G}[/tex]

Now equates these equations in both sides

[tex]2\pi \frac{L}{G} = 2\pi \frac{L}{4G}[/tex]

So after solving this, the length of the pendulum is 4L

Answer:

the length of a pendulum on your friend s planet should be 4 times than that on earth

Explanation:

We know that time period of simple pendulum is given by

[tex]T= 2\pi\sqrt{\frac{L}{g} }[/tex]

L= length of pendulum

g= acceleration due to gravity

therefore, Let T_1 and T_2 be the time period of the earth and other planet respectively.

[tex]\frac{T_1}{T_2} =\sqrt(\frac{L_1}{L_2}\times\frac{g_2}{g_1})[/tex]

ATQ

T_1=T_2=T,   g_2=4g_1

Putting the values in above equation and solving we get

[tex]\frac{L_1}{L_2} =\frac{1}{4}[/tex]

Answer:

The effective (rms) current when the circuit is in resonance is 6 A

Explanation:

Given;

resistance of the resistor, R = 20 ohms

capacitance of the capacitor, C = 0.75 microfarads

inductance of the inductor, L =  0.12 H

effective rms voltage, [tex]V_{rms}[/tex] = 120

At resonance, the impedance Z = R, Since the capacitive reactance (Xc) is equal to inductive reactance (XL).

The effective (rms) current, = [tex]V_{rms}[/tex] / R

                                              = 120 / 20

                                              = 6 A

Therefore, the effective (rms) current when the circuit is in resonance is 6 A

Answer:

a) E = 2.7x10⁶ N/C

b) F = 54 N

Explanation:

a) The electric field can be calculated as follows:

[tex] E = \frac{Kq}{d^{2}} [/tex]

Where:

K: is the Coulomb's constant = 9x10⁹ N*m²/C²

q: is the charge

d: is the distance

Now, we need to find the electric field due to charge 1:

[tex] E_{1} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*50 \cdot 10^{-6} C}{(0.5 m)^{2}} = 1.8 \cdot 10^{6} N/C [/tex]

The electric field due to charge 2 is:

[tex]E_{2} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*(-25) \cdot 10^{-6} C}{(0.5 m)^{2}} = -9.0 \cdot 10^{5} N/C[/tex]

The electric field at a point midway between them is given by the sum of E₁ and E₂ (they are in the same direction, that is to say, to the right side):

[tex]E_{T} = E_{1} + E_{2} = 1.8 \cdot 10^{6} N/C + 9.0 \cdot 10^{5} N/C = 2.7 \cdot 10^{6} N/C to the right side[/tex]                                                                                                

Hence, the electric field at a point midway between them is 2.7x10⁶ N/C to the right side.  

b) The force on a charge q₃ situated there is given by:

[tex]E_{T} = \frac{F_{T}}{q_{3}} \rightarrow F_{T} = E_{T}*q_{3}[/tex]

[tex] F = 2.7 \cdot 10^{6} N/C*20 \cdot 10^{-6} C = 54 N [/tex]

Therefore, the force on a charge q₃ situated there is 54 N.  

I hope it helps you!

(a) The electric field at a point midway between [tex]q_1[/tex] and [tex]q_2[/tex] is obtained to be [tex]2.7\times 10^6 \,N/C[/tex].

(b) The electrostatic force on the third charge [tex]q_3[/tex] situated between [tex]q_1[/tex] and [tex]q_2[/tex] is obtained as 54 N.

The answer can be explained as follows.

Given that the two charges are;

(a) At the midpoint; [tex]r = 0.5\,m[/tex].

We know that the electric field due to charge [tex]q_1[/tex].

Where, [tex]k=9\times 10^9\,Nm^2/C[/tex]

The electric field due to charge [tex]q_2[/tex] is given by;

Therefore, the net electric field in the midpoint is given by;

The direction is towards the right side.

(b) Now, there is another charge [tex]q_3=20\times 10^{-6}[/tex] in the midpoint.

So the force on the charge is ;

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Answer:

The kinetic energy of a spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.

Explanation:

Let us first consider the initial characteristics of the angular motion of the disk

moment of inertia = [tex]I[/tex]

angular speed = ω

For the second case, we consider the characteristics to now be

moment of inertia = [tex]5I[/tex]  (five times larger)

angular speed = ω/5  (five times smaller)

Recall that the kinetic energy of a spinning body is given as

[tex]KE = \frac{1}{2}Iw^{2}[/tex]

therefore,

for the first case, the K.E. is given as

[tex]KE = \frac{1}{2}Iw^{2}[/tex]

and for the second case, the K.E. is given as

[tex]KE = \frac{1}{2}(5I)(\frac{w}{5} )^{2} = \frac{5}{50}Iw^{2}[/tex]

[tex]KE = \frac{1}{10}Iw^{2}[/tex]

this is one-tenth the kinetic energy before its spinning characteristics were changed.

This implies that the kinetic energy of the spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.

A spinning disk's kinetic energy will change to one-tenth if its moment of inertia was five times larger but its angular speed was five times smaller.

Relation between Kinetic energy and Moment of Inertia:

Now, let's consider moment of inertia =  I  and angular speed = ω

It is asked that what would be change in Kinetic energy if

moment of inertia =   (five times larger)

angular speed = ω/5  (five times smaller)

The kinetic energy of a spinning body is given as:

[tex]K.E.=\frac{1}{2} I. w^2[/tex]

On substituting the values, we will get:

[tex]K.E.= \frac{1}{2} (5I) (\frac{w}{5} )^2 \\\\K.E. =\frac{1}{10} I. w^2[/tex]

Kinetic energy will be one-tenth to the kinetic energy before its spinning characteristics were changed.

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brainly.com/question/12337396

Answer:

Mass of the rope = 2.8 kg

Explanation:

The speed of waves travelling through a rope with linear density (μ) and under tension T is given as v = √(T/μ)

The speed of waves in the rope is also calculated as

v = (d/t)

d = L = length of the rope = 15 m

t = time taken for the wave to move through the rope = 0.545 s

Speed = v = (15/0.545) = 27.523 m/s

Speed = v = √(T/μ)

T = tension in the rope = 140 N

μ = linear density = ?

27.523 = √(140/μ)

27.523² = (140/μ)

(140/μ) = 757.512

μ = (140/757.512) = 0.1848155556 = 0.1848 kg/m

Linear density = μ = (m/L)

m = mass of the rope = ?

L = length of the rope = 15 m

0.1848 = (m/15)

m = 0.1848 × 15 = 2.77 kg = 2.8 kg to 1 d.p.

Hope this Helps!!!

Answer:

Higher than the coefficient of kinetic friction.

Explanation:

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