Consider two identical small glass balls dropped into two identical containers, one filled with water and the other with oil. Which ball will reach the bottom of the container first? Why?

Answer:

If the particle is an electron [tex]E_y = 3.311 * 10^3 N/C[/tex]

If the particle is a proton, [tex]E_y = 6.08 * 10^6 N/C[/tex]

Explanation:

Initial speed at the origin, [tex]u = 3 * 10^6 m/s[/tex]

[tex]\theta = 38^0[/tex] to +ve x-axis

The particle crosses the x-axis at , x = 1.5 cm = 0.015 m

The particle can either be an electron or a proton:

Mass of an electron, [tex]m_e = 9.1 * 10^{-31} kg[/tex]

Mass of a proton, [tex]m_p = 1.67 * 10^{-27} kg[/tex]

The electric field intensity along the positive y axis [tex]E_y[/tex], can be given by the formula:

[tex]E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\[/tex]

If the particle is an electron:

[tex]E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\[/tex]

[tex]E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\[/tex]

[tex]E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C[/tex]

If the particle is a proton:

[tex]E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\[/tex]

[tex]E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\[/tex]

[tex]E_y = 6.08 * 10^6 N/C[/tex]

Answer:

1.1ohms

Explanation:

According to ohms law E = IR

If potential difference of a battery is 2.2 V when it is connected across a resistance of 5 ohm and if suddenly the voltage Falls to 1.8V then the current in the 5ohms resistor I = V/R = 1.8/5

I = 0.36A (This will be the load current).

Before we can calculate the value of the internal resistance, we need to know the voltage drop across the internal resistance.

Voltage drop = 2.2V - 1.8V = 0.4V

Then we calculate the internal resistance using ohms law.

According to the law, V = Ir

V= voltage drop

I is the load current

r = internal resistance

0.4 = 0.36r

r = 0.4/0.36

r = 1.1 ohms

Answer:

   θ = 4.78º

with respect to the vertical or 4.78 to the east - north

Explanation:

This is a velocity compound exercise since it is a vector quantity.

The plane takes a direction, the air blows to the west and the result must be to the north, let's use the Pythagorean theorem to find the speed

                  v_fly² = v_nort² + v_air²

                  v_nort² = v_fly² + - v_air²

Let's use trigonometry to find the direction of the plane

        sin θ = v_air / v_fly

        θ = sin⁻¹ (v_air / v_fly)

let's calculate

        θ = sin⁻¹ (10/120)

         θ = 4.78º

with respect to the vertical or 4.78 to the north-east

Answer:

[tex]v_B=3.78\times 10^5\ m/s[/tex]

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is [tex]6.63\times 10^{-27}\ kg[/tex]

Speed of nucleus at A is [tex]v_A=6.2\times 10^5\ m/s[/tex]

Potential at point A, [tex]V_A=1.5\times 10^3\ V[/tex]

Potential at point B, [tex]V_B=4\times 10^3\ V[/tex]

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

[tex]\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s[/tex]

So, the speed at point B is [tex]3.78\times 10^5\ m/s[/tex].

Answer:

9.6 Ns

Explanation:

Note: From newton's second law of motion,

Impulse = change in momentum

I = m(v-u).................. Equation 1

Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.

Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)

Substitute into equation 1

I = 2.4[2.5-(-1.5)]

I = 2.4(2.5+1.5)

I = 2.4(4)

I = 9.6 Ns

The magnitude of impulse will be "9.6 Ns".

According to the question,

Mass,

Final velocity,

Initial velocity,

By using Newton's 2nd law of motion, we get

→ Impulse, [tex]I = m(v-u)[/tex]

By substituting the values, we get

                     [tex]= 2.4[2.5-(1.5)][/tex]

                     [tex]= 2.4(2.5+1.5)[/tex]

                     [tex]= 2.4\times 4[/tex]

                     [tex]= 9.6 \ Ns[/tex]

Thus the above answer is right.    

Learn more about Impulse here:

https://brainly.com/question/15495020

Complete question:

Two conductors made of the same material are connected across the same potential difference. Conductor A has seven times the diameter and seven times the length of conductor B. What is the ratio of the power delivered to A to power delivered to B.

Answer:

The ratio of the power delivered to A to power delivered to B is 7 : 1

Explanation:

Cross sectional area of a wire is calculated as;

[tex]A = \frac{\pi d^2}{4}[/tex]

Resistance of a wire is calculated as;

[tex]R = \frac{\rho L}{A} \\\\R = \frac{4\rho L}{\pi d^2} \\\\[/tex]

Resistance in wire A;

[tex]R = \frac{4\rho _AL_A}{\pi d_A^2}[/tex]

Resistance in wire B;

[tex]R = \frac{4\rho _BL_B}{\pi d_B^2}[/tex]

Power delivered in wire;

[tex]P = \frac{V^2}{R}[/tex]

Power delivered in wire A;

[tex]P = \frac{V^2_A}{R_A}[/tex]

Power delivered in wire B;

[tex]P = \frac{V^2_B}{R_B}[/tex]

Substitute in the value of R in Power delivered in wire A;

[tex]P_A = \frac{V^2_A}{R_A} = \frac{V^2_A \pi d^2_A}{4 \rho_A L_A}[/tex]

Substitute in the value of R in Power delivered in wire B;

[tex]P_B = \frac{V^2_B}{R_B} = \frac{V^2_B \pi d^2_B}{4 \rho_B L_B}[/tex]

Take the ratio of power delivered to A to power delivered to B;

[tex]\frac{P_A}{P_B} = (\frac{V^2_A \pi d^2_A}{4\rho_AL_A} ) *(\frac{4\rho_BL_B}{V^2_B \pi d^2_B})\\\\ \frac{P_A}{P_B} = (\frac{V^2_A d^2_A}{\rho_AL_A} )*(\frac{\rho_BL_B}{V^2_B d^2_B})\\\\[/tex]

The wires are made of the same material, [tex]\rho _A = \rho_B[/tex]

[tex]\frac{P_A}{P_B} = (\frac{V^2_A d^2_A}{L_A} )*(\frac{L_B}{V^2_B d^2_B})\\\\[/tex]

The wires are connected across the same potential; [tex]V_A = V_B[/tex]

[tex]\frac{P_A}{P_B} = (\frac{ d^2_A}{L_A} )* (\frac{L_B}{d^2_B} )[/tex]

wire A has seven times the diameter and seven times the length of wire B;

[tex]\frac{P_A}{P_B} = (\frac{ (7d_B)^2}{7L_B} )* (\frac{L_B}{d^2_B} )\\\\\frac{P_A}{P_B} = \frac{49d_B^2}{7L_B} *\frac{L_B}{d^2_B} \\\\\frac{P_A}{P_B} =\frac{49}{7} \\\\\frac{P_A}{P_B} = 7\\\\P_A : P_B = 7:1[/tex]

Therefore, the ratio of the power delivered to A to power delivered to B is

7 : 1

Answer:

0.15 m

Explanation:

First calculating the center of mass from the saw horse

[tex]\frac{3.15}{2} -1=0.575 m[/tex]

from the free body diagram we can write

Taking moment about the saw horse

55.9×9.81×y=14.5×0.575×9.81

y= 0.15 m

So, the painter walk from the saw horse on the right until the board begins to tip is 0.15 m far.

Answer:

v = 349.23 m/s

Explanation:

It is required to find the orbital speed for a satellite [tex]3.5\times 10^8\ m[/tex] from the center of mass.

Mass of Mars, [tex]M=6.4\times 10^{23}\ kg[/tex]

The orbital speed for a satellite is given by the formula as follows :

[tex]v=\sqrt{\dfrac{GM}{r}} \\\\v=\sqrt{\dfrac{6.67\times 10^{-11}\times 6.4\times 10^{23}}{3.5\times 10^8}} \\\\v=349.23\ m/s[/tex]

So, the orbital speed for a satellite is 349.23 m/s.

Answer:

d) Both blocks have had equal losses of energy to friction

Explanation:

As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces

Moreover, the block A is twice as fast than block B at the time of crossing the finish line

So based on the above information,  it contains the losses of identical friction

And we also know that

Friction energy loss is

[tex]= \mu \times m \times g \times D[/tex]

It would be the same for both the blocks

hence, the option d is correct

The correct answer will be both blocks have had equal losses of energy to friction.

Friction is defined as when any object is slides on a surface by means of any external force then the force in the opposite direction generated between the surface and the body restrict the motion of the body this force is called as the friction.

As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces.

Moreover, the block A is twice as fast as block B at the time of crossing the finish line.

So based on the above information,  it contains the losses of identical friction.

And we also know that

Friction energy loss is

[tex]E_f=\mu m g D[/tex]

It would be the same for both the blocks

Hence both blocks have had equal losses of energy to friction.

To know more about friction, follow

https://brainly.com/question/24386803

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

[tex]T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}[/tex]

k = 1.4

[tex]T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K[/tex]

Work done is given as;

[tex]W = \frac{1}{2} *m*(v_i^2 - v_e^2)[/tex]

inlet velocity is negligible;

[tex]v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s[/tex]

Therefore, the exit velocity is 629.41 m/s

Answer:

  c.  the net work a spring does on a body is zero when the body returns to its initial position

Explanation:

A force is conservative when the net work done over any path that returns to the initial position is zero. Choice C matches that definition.

An ideal spring of the kind used in physics problems has the characteristic that it applies the same force at the same distance always. So any work required to extend or compress the spring is reversed when the reverse motion takes place.

Answer:

The magnitude of the electric force on the particle in terms of the variables given is, F = qE

The magnitude of the magnetic force on the particle in terms of the variables given is, F = q (v x B)

Explanation:

Given;

a charged particle, q

magnitude of electric field, E

magnitude of magnetic field, B

The magnitude of the electric force on the particle in terms of the variables given;

F = qE

The magnitude of the magnetic force on the particle in terms of the variables given;

F = q (v x B)

where;

v is the constant velocity of the charged particle

Answer:

The magnitude of the electric force acting on a charged particle moving through an electric field = |qE|

The magnitude of the magnetic force of a charged particle moving at a particular velocity through a magnetic field = |qv × B|

Explanation:

The electric force acting on a charged particle, q, moving through an electric field, E, is given as a product of the charge on the particle (a scalar quantity) and the electric field (a vector quantity).

Electric force = qE

The magnitude of the electric force = |qE|

That is, magnitude of the product of the charge and the electric field vector.

The magnetic force acting on a charged particle, q, moving with a velocity, v, through a magnetic field, B is a vector product of qv [a product of the charge of the particle (a scalar quantity) and the velocity of the particle (a vector quantity)] and B (a vector quantity).

It is given mathematically as (qv × B)

The magnitude of the magnetic force is the magnitude of the vector product obtained.

Magnitude of the magnetic force = |qv × B|

Hope this Helps!!!

Answer:

a

  [tex]KE = 7.17 *10^{7} \ J[/tex]

b

 [tex]t = 6411.09 \ s[/tex]

Explanation:

From the question we are told that

    The radius of the flywheel is  [tex]r = 1.50 \ m[/tex]

      The mass of the flywheel is [tex]m = 430 \ kg[/tex]

          The rotational speed of the flywheel is [tex]w = 5,200 \ rev/min = 5200 * \frac{2 \pi }{60} =544.61 \ rad/sec[/tex]

      The power supplied by the motor is  [tex]P = 15.0 hp = 15 * 746 = 11190 \ W[/tex]

     Generally the moment of inertia of the flywheel is  mathematically represented as

       [tex]I = \frac{1}{2} mr^2[/tex]

substituting values

       [tex]I = \frac{1}{2} ( 430)(1.50)^2[/tex]

       [tex]I = 483.75 \ kgm^2[/tex]

The kinetic energy that is been stored is  

       [tex]KE = \frac{1}{2} * I * w^2[/tex]

substituting values

        [tex]KE = \frac{1}{2} * 483.75 * (544.61)^2[/tex]

        [tex]KE = 7.17 *10^{7} \ J[/tex]

Generally power is mathematically represented as

          [tex]P = \frac{KE}{t}[/tex]

=>      [tex]t = \frac{KE}{P}[/tex]

substituting the value

        [tex]t = \frac{7.17 *10^{7}}{11190}[/tex]

        [tex]t = 6411.09 \ s[/tex]

Answer:

The two value of the wavelength for the out of tune guitar is  

[tex]\lambda _2 = (6.48,6.52) \ cm[/tex]

Explanation:

From the question we are told that

     The wavelength of the note is [tex]\lambda = 6.50 \ cm = 0.065 \ m[/tex]

     The difference in beat frequency is [tex]\Delta f = 17.0 \ Hz[/tex]

Generally the frequency of the note played by the guitar that is in tune is  

        [tex]f_1 = \frac{v_s}{\lambda}[/tex]

Where [tex]v_s[/tex] is the speed of sound with a constant value [tex]v_s = 343 \ m/s[/tex]

       [tex]f_1 = \frac{343}{0.0065}[/tex]

      [tex]f_1 = 5276.9 \ Hz[/tex]

The difference in beat is mathematically represented as

       [tex]\Delta f = |f_1 - f_2|[/tex]

Where [tex]f_2[/tex] is the frequency of the sound from the out of tune guitar

     [tex]f_2 =f_1 \pm \Delta f[/tex]

substituting values

      [tex]f_2 =f_1 + \Delta f[/tex]

      [tex]f_2 = 5276.9 + 17.0[/tex]  

     [tex]f_2 = 5293.9 \ Hz[/tex]

The wavelength for this frequency is

      [tex]\lambda_2 = \frac{343 }{5293.9}[/tex]

     [tex]\lambda_2 = 0.0648 \ m[/tex]

    [tex]\lambda_2 = 6.48 \ cm[/tex]

For the second value of the second frequency

     [tex]f_2 = f_1 - \Delta f[/tex]

     [tex]f_2 = 5276.9 -17[/tex]

      [tex]f_2 = 5259.9 Hz[/tex]

The wavelength for this frequency is

   [tex]\lambda _2 = \frac{343}{5259.9}[/tex]

   [tex]\lambda _2 = 0.0652 \ m[/tex]

   [tex]\lambda _2 = 6.52 \ cm[/tex]

This question involves the concepts of beat frequency and wavelength.

The possible wavelengths of the out-of-tune guitar are "6.48 cm" and "6.52 cm".

The beat frequency is given by the following formula:

[tex]f_b=|f_1-f_2|\\\\[/tex]

f₂ = [tex]f_b[/tex] ± f₁

where,

f₂ = frequency of the out-of-tune guitar = ?

[tex]f_b[/tex] = beat frequency = 17 Hz

f₁ = frequency of in-tune guitar = [tex]\frac{speed\ of\ sound\ in\ air}{\lambda_1}=\frac{343\ m/s}{0.065\ m}=5276.9\ Hz[/tex]

Therefore,

f₂ = 5276.9 Hz ± 17 HZ

f₂ = 5293.9 Hz (OR) 5259.9 Hz

Now, calculating the possible wavelengths:

[tex]\lambda_2=\frac{speed\ of\ sound}{f_2}\\\\\lambda_2 = \frac{343\ m/s}{5293.9\ Hz}\ (OR)\ \frac{343\ m/s}{5259.9\ Hz}\\\\[/tex]

λ₂ = 6.48 cm (OR) 6.52 cm

Learn more about beat frequency here:

https://brainly.com/question/10703578?referrer=searchResults

Answer:

Explanation:

Given that:

The mass of the cell is 9.0 x 10^-14 kg

The charges of the cell is -2.5pC and the other -3.30 pC

[tex]q_1=-2.5\times10^{-12}C \ \ and \ \ q_2=-3.75\times10^{-12}C[/tex]

Radius is  3.75 × 10-6 m

The final distance is twice the radius

i.e [tex]2*(3.75 \times 10^{-6}) = 7.5*10^{-6}m[/tex]

The formula for the velocity of the cell is

[tex]mv^2=\frac{q_1q_2}{4\pi \epsilon 2 r} \\[/tex]

[tex]v=\sqrt{\frac{q_1q_2}{4\pi \epsilon 2 r} }[/tex]

[tex]=\sqrt{\frac{(-2.5\times10^{-12})(-3.3\times10^{-12}}{4(3.14)(8.85\times10^{-112}(2\times3.75\times10^{-6})(9\times10^{-14})} } \\\\=\sqrt{\frac{(-8.25\times10^{-24})}{(7503.03\times10^{-32})} } \\\\=\sqrt{109955.5779} \\\\=331.60m/s[/tex]

The maximum acceleration of the cells as they move toward each other and just barely touch is

[tex]ma= \frac{q_1q_2}{4\pi \epsilon (2r)^2} \\\\a= \frac{q_1q_2}{4\pi \epsilon (2r)^2(m)}[/tex]

[tex]=\frac{(-2.5\times10^{-12})(-3.3\times10^{-12})}{4(3.14)(8.85\times10^{-12})(2\times3.75\times10^{-6})^2(9\times10^{-14})}[/tex]

[tex]=\frac{(-8.25\times10^{-24})}{(56272.725\times10^{-38})} \\\\=1.47\times10^{10}m/s^2[/tex]

The answers obtained are;

1. The initial speed of each of the red blood cells is [tex]v= 331.66\,m/s[/tex].

2. The maximum acceleration of the cells is [tex]a=1.47\times 10^{10}\,m/s^2[/tex].

The answer is explained as shown below.

We have, the mass of the red blood cell;

Also, the charges of the cells are;

The distance between the charges when they barely touch will be two times the radius of each charge.

1. As both the cells are negatively charged they will repel each other.

Now, substituting all the known values in the equation, we get;

[tex]v^2=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{7.5\times10^{-6}\,m\times(9\times 10^{-14}\,kg)} =110000\,m^2/s^2[/tex]

2. Also as the force between them is repulsive, there must be an acceleration to make them barely touch each other.

Substituting the known values, we get;

[tex](9\times 10^{-14}\,kg)\times a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2}[/tex]

[tex]\implies a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2\times(9\times 10^{-14}\,kg) }[/tex]

Find out more information about moving charges here:

https://brainly.com/question/14632877

Answer:

Explanation:

(a)

The true atmospheric pressure will has more value than the reading in the barometer. If Parm is the atmospheric

pressure in the tube then the resulting vapour pressure is

Patm - pgh = Prapor

The final reading ion the barometer is

pgh = Palm - Proper

Hence, the true atmospheric pressure is greater.

you can find the answer in this book

physics principles with Applications, Global Edition Problem 67P: Chapter: CH 13 Problem:67p

Answer:

blue  θ₂ = 22.26º

red    θ₂ = 22.79º

Explanation:

When a light beam passes from one material medium to another, it undergoes a deviation from the path, described by the law of refraction

         n₁ sin θ₁ = n₂ sin θ₂

where n₁ and n₂ are the incident and transmitted media refractive indices and θ are the angles in the media

let's apply this equation to each wavelength

λ = blue

in this case n₁ = 1, n₂ = 1,645

       sin θ₂ = n₁/ n₂ sin₂ θ₁

let's calculate

       sin θ₂ = 1 / 1,645 sint 38.55

       sin θ₂ = 0.37884

       θ₂ = sin⁻¹ 0.37884

       θ₂ = 22.26º

λ = red

n₂ = 1,609

         sin θ₂ = 1 / 1,609 sin 38.55

         sin θ₂ = 0.3873

         θ₂ = sim⁻¹ 0.3873

         θ₂ = 22.79º

the refracted rays are between these two angles

Complete question is;

After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands? Assume the leg can be treated as a uniform rod x = 0.98 m long that pivots freely about the hip.

Answer:

Tangential speed of foot just before it lands is; v = 5.37m/s

Explanation:

Let U (potential energy) be zero on the ground.

So, initially, U = mgh

where, h = 0.98/2 = 0.49m (midpoint of the leg)

Now just before the leg hits the floor it would have kinetic energy as;

K = ½Iω²

where ω = v/r and I = ⅓mr²

So, K = ½(⅓mr²)(v/r)²

K = (1/6) × (mr²)/(v²/r²)

K = (1/6) × mv²

From principle of conservation of energy, we have;

Potential energy = Kinetic energy

Thus;

mgh = (1/6) × mv²

m will cancel out to give;

gh = (1/6)v²

Making v the subject, we have;

v = √6gh

v = √(6 × 9.81 × 0.49)

v = √28.8414

v = 5.37m/s

Answer:

about 1 meter

Explanation:

The distance past the edge that the man will get before the plank starts to tip is; 0.4285 m

We are given;

Mass of plank; m = 30 kg

Length of plank; L = 6m

Mass of man; M = 70 kg

Since the plank has 2 supports which are the deck of the ship, then it means that, we can take moments about the right support before the 2m stick out of the plank.

Thus;

Moment of weight of plank about the right support;

τ_p = mg((L/2) - 2)

τ_p = 30 × 9.8((6/2) - 2)

τ_p = 294 N.m

Moment of weight of man about the right support;

τ_m = Mgx

where x is the distance past the edge the man will get before the plank starts to tip.

τ_m = 70 × 9.8x

τ_m = 686x

Now, moment of the board is counterclockwise while that of the man is clockwise. Thus;

τ_m = τ_p

686x = 294

x = 294/686

x = 0.4285 m

Read more at; https://brainly.com/question/22150651

Answer:

A) M

Explanation:

The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:

Box with mass M

[tex]\Sigma F = F - F' = M\cdot a[/tex]

Box with mass 2M

[tex]\Sigma F = F' - F'' = 2\cdot M \cdot a[/tex]

Box with mass 3M

[tex]\Sigma F = F'' = 3\cdot M \cdot a[/tex]

On the third equation, acceleration can be modelled in terms of F'':

[tex]a = \frac{F''}{3\cdot M}[/tex]

An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.

[tex]F' = 2\cdot M \cdot a + F''[/tex]

[tex]F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''[/tex]

[tex]F' = \frac{5}{3}\cdot F''[/tex]

Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:

[tex]F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)[/tex]

[tex]F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''[/tex]

[tex]F = 2\cdot F''[/tex]

[tex]F'' = \frac{1}{2}\cdot F[/tex]

Afterwards, F' as function of the external force can be obtained by direct substitution:

[tex]F' = \frac{5}{6}\cdot F[/tex]

The net forces of each block are now calculated:

Box with mass M

[tex]M\cdot a = F - \frac{5}{6}\cdot F[/tex]

[tex]M\cdot a = \frac{1}{6}\cdot F[/tex]

Box with mass 2M

[tex]2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F[/tex]

[tex]2\cdot M \cdot a = \frac{1}{3}\cdot F[/tex]

Box with mass 3M

[tex]3\cdot M \cdot a = \frac{1}{2}\cdot F[/tex]

As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.

Answer:

88.29 N

Explanation:

mass of the ball = 4.5 kg

weight of the ball will be = mass x acceleration due to gravity(9.81 m/s^2)

weight W = 4.5 x 9.81 = 44.145 N

centrifugal forces Tc act on the ball as it swings.

At the top point of the vertical swing,

Tension on the rope = Tc - W.

At the bottom point of the vertical swing,

Tension on the rope = Tc + W

therefore,

difference in tension between these two points will be;

Net tension = tension at bottom minus tension at the top

= Tc + W - (Tc - W) = Tc + W -Tc + W

= 2W

imputing the value of the weight W, we have

2W = 2 x 44.145 = 88.29 N

Answer:

v = 4.08m/s₂

Explanation:

Answer:

46648 J

Explanation:

mass m= 85 Kg

velocity v = 56 m/s

distance covered s =40 m

According to Question,

frictional drag force to him that matched his weight

[tex]\Rightarrow F_d =mg\\=85\times9.81=833 N[/tex]

Therefore, work done by practometer against the drag force = heat was generated by this frictional drag force in J

W=Q= F_d×s

=833×56 = 46648 J

Answer:

μ_k = 0.1773

Explanation:

We are given;

Initial velocity;u = 20 m/s

Final velocity;v = 0 m/s (since it comes to rest)

Distance before coming to rest;s = 115 m

Let's find the acceleration using Newton's second law of motion;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging relevant values;

a = (0² - 20²)/(2 × 115)

a = -400/230

a = -1.739 m/s²

From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:

F_k = −ma - - - (1)

We also know that F_k is defined by;

F_k = μ_k•N

Where;

μ_k is coefficient of kinetic friction

N is normal force which is (mg)

Since gravity acts in the negative direction, the normal force will be positive.

Thus;

F_k = μ_k•mg - - - (2)

where g is acceleration due to gravity.

Thus,equating equation 1 and 2,we have;

−ma = μ_k•mg

m will cancel out to give;

-a = μ_k•g

μ_k = -a/g

g has a constant value of 9.81 m/s², so;

μ_k = - (-1.739/9.81)

μ_k = 0.1773

The coefficient of kinetic friction between the hockey puck and ice is equal to 0.178

Given the following data:

Scientific data:

To determine the coefficient of kinetic friction between the hockey puck and ice:

First of all, we would calculate the acceleration of the hockey puck by using the third equation of motion.

[tex]V^2 = U^2 + 2aS\\\\0^2 =20^2 + 2a(115)\\\\-400=230a\\\\a=\frac{-400}{230}[/tex]

Acceleration, a = -1.74 [tex]m/s^2[/tex]

Note: The negative signs indicates that the hockey puck is slowing down or decelerating.

From Newton's Second Law of Motion, we have:

[tex]\sum F_x = F_k + F_n =0\\\\F_k =- F_n\\\\\mu mg =-ma\\\\\mu = \frac{-a}{g}\\\\\mu = \frac{-(-1.74)}{9.8}\\\\\mu = \frac{1.74}{9.8}[/tex]

Coefficient of kinetic friction = 0.178

Read more: https://brainly.com/question/13821217

Answer:

t = 17.68s

Explanation:

In order to calculate the total elapsed time that skydiver takes to reache the ground, you first calculate the distance traveled by the skydiver in the first 5.00s. You use the following formula:

[tex]y=y_o-v_ot-\frac{1}{2}gt^2[/tex]            (1)

y: height for a time t

yo: initial height = 1000m

vo: initial velocity = 0m/s

g: gravitational acceleration = 9.8m/s^2

t: time = 5.00 s

You replace the values of the parameters to get the values of the new height of the skydiver:

[tex]y=1000m-\frac{1}{2}(9.8m/s^2)(5.00s)^2\\\\y=877.5m[/tex]

Next, you take this value of 877.5m as the initial height of the second part of the trajectory of the skydiver. Furthermore, use the value of 7.00m/s as the initial velocity.

You use the same equation (1) with the values of the initial velocity and new height. We are interested in the time for which the skydiver arrives to the ground, then y = 0

[tex]0=877.5-7.00t-4.9t^2[/tex]       (2)

The equation (2) is a quadratic equation, you solve it for t with the quadratic formula:

[tex]t_{1,2}=\frac{-(-7.00)\pm \sqrt{(-7.00)^2-4(-4.9)(877.5)}}{2(-4.9)}\\\\t_{1,2}=\frac{7.00\pm 131.33}{-9.8}\\\\t_1=12.68s\\\\t_2=-14.11s[/tex]

You use the positive value of t1 because it has physical meaning.

Finally, you sum the times of both parts of the trajectory:

total time = 5.00s + 12.68s = 17.68s

The total elapsed time taken by the skydiver to arrive to the ground from the airplane is 17.68s

Answer: Part 1: Propellant Fraction (MR) = 8.76

Part 2: Propellant Fraction (MR) = 1.63

Explanation: The Ideal Rocket Equation is given by:

Δv = [tex]v_{ex}.ln(\frac{m_{f}}{m_{e}} )[/tex]

Where:

[tex]v_{ex}[/tex] is relationship between exhaust velocity and specific impulse

[tex]\frac{m_{f}}{m_{e}}[/tex] is the porpellant fraction, also written as MR.

The relationship [tex]v_{ex}[/tex] is: [tex]v_{ex} = g_{0}.Isp[/tex]

To determine the fraction:

Δv = [tex]v_{ex}.ln(\frac{m_{f}}{m_{e}} )[/tex]

[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]

Knowing that change in velocity is Δv = 9.6km/s and [tex]g_{0}[/tex] = 9.81m/s²

Note: Velocity and gravity have different measures, so to cancel them out, transform km in m by multiplying velocity by 10³.

Part 1: Isp = 450s

[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]

ln(MR) = [tex]\frac{9.6.10^{3}}{9.81.450}[/tex]

ln (MR) = 2.17

MR = [tex]e^{2.17}[/tex]

MR = 8.76

Part 2: Isp = 2000s

[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]

ln (MR) = [tex]\frac{9.6.10^{3}}{9.81.2.10^{3}}[/tex]

ln (MR) = 0.49

MR = [tex]e^{0.49}[/tex]

MR = 1.63

Answer:

The beam used is a negatively charged electron beam with a velocity of

v = E / B

Explanation:

After reading this long statement we can extract the data to work on the problem.

* They indicate that when the beam passes through the plates it deviates towards the positive plate, so the beam must be negative electrons.

* Now indicates that the electric field and the magnetic field are contracted and that the beam passes without deviating, so the electric and magnetic forces must be balanced

           [tex]F_{e} = F_{m}[/tex]

           q E = qv B

           v = E / B

this configuration is called speed selector

They ask us what type of beam was used.

The beam used is a negatively charged electron beam with a velocity of v = E / B

Answer:

  r₂ = 1,586 m

Explanation:

For this problem we are going to solve it by parts, let's start by finding the sound intensity when we are 25 m

         β = 10 log (I / I₀)

where Io is the sensitivity threshold 10⁻¹² W / m²

          I₁ / I₀ = [tex]e^{\beta/10}[/tex]

          I₁ = I₀  e^{\beta/10}

let's calculate

          I₁ = 10⁻¹² e^{25/10}

          I₁ = 1.20 10⁻¹¹ W / m²

the other intensity in exercise is

          I₂ = 10⁻¹² e^{80/10}

          I₂ = 2.98 10⁻⁹ W / m²

now we use the definition of sound intensity

          I = P / A

where P is the emitted power that is a constant and A the area of ​​the sphere where the sound is distributed

         P = I A

the area a sphere is

         A = 4π r²

we can write this equation for two points of the found intensities

          I₁ A₁ = I₂ A₂

where index 1 corresponds to 25m and index 2 to the other distance

          I₁ 4π r₁² = I₂ 4π r₂²

          I₁ r₁² = I₂ r₂²

           r₂ = √ (I₁ / I₂) r₁

let's calculate

           r₂ = √ (1.20 10⁻¹¹ / 2.98 10⁻⁹) 25

           r₂ = √ (0.40268 10⁻²) 25

           r₂ = 1,586 m

Answer:

The velocity of the bird is [tex]v_f = 4.87 \ m/s[/tex]

Explanation:

From the question we are told that  

    The mass of the bird  is [tex]m_1 = 300 \ g = 0.3 \ kg[/tex]

    The initial speed of the bird is  [tex]u_1 = 6.2 \ m/s[/tex]

     The mass of the insect is [tex]m_2 = 10 \ g = 0.01 \ kg[/tex]

       The speed of the insect is [tex]u_ 2 =-35 \ m/s[/tex]

The negative sign is because it is moving in opposite direction  to the bird

According to the principle of linear momentum conservation

       [tex]m_1 u_1 + m_2 u_2 = (m_1 + m_2 )v_f[/tex]

substituting values

        [tex](0.3 * 6.2 ) + (0.01 * (-35)) = (0.3 + 0.01 )v_f[/tex]

    [tex]1.51 = 0.31 v_f[/tex]

     [tex]v_f = 4.87 \ m/s[/tex]

The Final velocity of Bird =  4.87 m/s

Mass of the bird = 300 g = 0.3 kg

Velocity of bird = 6.2 m/s

Momentum of Bird = Mass of bird [tex]\times[/tex] Velocity of Bird = 0.3 [tex]\times[/tex] 6.2 =  1.86 kgm/s

Mass of the insect = 10 g = 0.01 kg

Velocity of insect =   - 35 m/s

Momentum of the Insect = Mass of Insect [tex]\times[/tex] Velocity of Insect = - 0.35  kgm/s

According to the law of conservation of momentum We can write that

In the absence of external forces on the system , the momentum of system remains conserved in that particular direction.

The bird opens the mouth and enjoys the free lunch  hence

Let the final velocity of bird is [tex]v_f[/tex]

Initial momentum of the system = Final momentum of the system

1.86 -0.35 = [tex]v_f[/tex] ( 0.01 + 0.3 )

1.51 =  [tex]v_f[/tex] 0.31

[tex]v_f[/tex] = 4.87 m/s

The Final velocity of Bird =  4.87 m/s

For more information please refer to the link below

https://brainly.com/question/18066930