What would you estimate for the length of a bass clarinet, assuming that it is modeled as a closed tube and that the lowest note that it can play is a D b whose frequency is 69 Hz

Explanation:

Se supone que si es 5.31 x 7.6 los límites son 38.98 ahora si fuera en suma mueves los puntos dos veces a la izquierda la sumatoria seria la siguiente .00531 + .0076 la respuesta seria

.00607

Answer:

The frequency is  [tex]f_n = 257.1 \ Hz[/tex]

Explanation:

From the question we are told that

    The third harmonic frequency of the tight guitar string is  [tex]f_3 = 540 \ Hz[/tex]

Let the original length be  L  

   Then the length at which it is fingered is  0.7 L

Generally the fundamental  is mathematically represented as

         [tex]f = \frac{v_s}{ 2L}[/tex]

Now when it finger at 70% it original length is

      [tex]f_n = \frac{v}{2 * (0.7 L)}[/tex]

      [tex]f_n = \frac{v}{1.4 L}[/tex]

Here v  the velocity of sound

  So  

         [tex]\frac{f_n}{f} = \frac{\frac{v}{1.4L} }{\frac{v}{2L} }[/tex]

Also the fundamental frequency for the original length can also be represented as

       [tex]f = \frac{f_3}{3}[/tex]

substituting values

          [tex]f = \frac{540}{3}[/tex]

          [tex]f = 180 \ Hz[/tex]

So

       [tex]\frac{f_n}{180} = \frac{\frac{v}{1.4L} }{\frac{v}{2L} }[/tex]

=>  [tex]f_n =\frac{180}{0.7}[/tex]

=>   [tex]f_n = 257.1 \ Hz[/tex]

The fundamental frequency, if it is fingered at a length of only 70% of its original length, will be 257.1  Hz.

Frequency is defined as the number of repetitions of a wave occurring waves in 1 second.

f is the frequency of tight guitar string = 540 Hz

Let's call the original length L.

The amount of time it is fingered is then 0.7 L.

In general, the fundamental frequency is expressed mathematically as;

[tex]\rm f = \frac{v_0}{2L} \\\\[/tex]

For the given conditions;

[tex]\rm f_n=\frac{v}{2 \times 0.7L} \\\\ \rm f_n=\frac{v}{1.4L}[/tex]

The ratio of the frequency is;

[tex]\rm \frac{f_n}{f} =\frac{\frac{v}{1.4L} }{\frac{V}{2L} }[/tex]

Also, the fundamental frequency for the original length can also be represented as;

[tex]\rm f= \frac{f'}{3} \\\\ f=\frac{540}{3} \\\\ \rm f=180\ Hz[/tex]

On putting the given data;

[tex]\rm \frac{f_n}{180} =\frac{\frac{v}{1.4L} }{\frac{V}{2L} }\\\\ \rm f_n=\frac{180}{0.7}\\\\\ \rm f_n=257.1\ Hz[/tex]

Hence the fundamental frequency, if it is fingered at a length of only 70% of its original length, will be 257.1  Hz.

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Answer:

therefore horizontal displacement changes increasing with linear velocity

Explanation:

Since the plane flies horizontally, the only speed that exists is

              v₀ₓ = 55.0 m / s

the time is the time it takes to reach the floor, which we can find because the speed on the vertical axis is zero

               y =y₀ + v₀ t - ½ g t2

               0 = I₀ + 0 - ½ g t2

               t = √ 2y₀o / g

time is that we use to calculate the x-axis displacement

 The distance it travels to reach the floor is

              x = v t

              x = 55 12

              x = 660 m

When the speed horizontally the time remains the same and 120

             x ’= v’ 12

therefore horizontal displacement changes increasing with linear velocity

Answer:

Net charge = E• b • L^3.

Explanation:

NB: here, the symbol representation of the flux is "p" = electric Field • Area(dot Product).

So, we will take a look at the flux through -x face, through x face and through -y face, through y face and through - z face and through z face.

(1). Starting from -z and z faces which are the back and front faces of the cube:

Thus, We have that the flux,p = 0 for -z and z.

(2). Recall that we are given that E = =(a+bx)i^+cj^.

Thus, p_-y = (a + bx)i + cj (-j) (L^2)

Where y = 0

p_-y = -cL^2.

Obviously for p_j, we will have cL^2 and y = L

(3). For p_-x = =(a + bx)i + cj (-i) (L^2).

p_-x = -aL^2

Where x = 0.

When x = L and p_x = (a + bL)L^2.

This, adding all together gives Net charge = E • b • L^3.

Answer:

please brainliest!!!

Explanation:

V1/√T1 =V2/√T2

V1 = 331m/s

T1 = 0°C = 273k

V2 = ?

T2 = 35°c = 308k

Answer:

118 minutes( 2 hours approximately )

Explanation:

Here, we are interested in calculating the orbital period of the satellite

Please check attachment for complete solution

Answer:

T = 7101 s = 118.35 mins = 1.9725 hrs

Explanation:

To solve the question, we apply the formula for gravitational acceleration

a = GM/r², where

a = acceleration due to gravity

G = gravitational constant

M = mass of the earth

r = distance between the satellite and center of the earth

Now, if we make r, subject of formula, we have

r = √(GM/a)

Recall also, that

a = v²/r, making v subject of formula

v = √ar

If we substitute the equation of r into it, we have

v =√a * √r

v =√a * √[√(GM/a)]

v = (GM/a)^¼

Again, remember that period,

T = 2πr/v, we already have v and r, allow have to do is substitute them in

T = 2π * √(GM/a) * [1 / (GM/a)^¼]

T = 2π * (GM/a³)^¼

T = 2 * 3.142 * [(6.67*10^-11 * 5.97*10^24) / (6.25³)]^¼

T = 6.284 * [(3.982*10^14) / 244.140]^¼

T = 6.284 * (1.63*10^12)^¼

T = 6.284 * 1130

T = 7101 s

T = 118.35 mins

T = 1.9725 hrs

Answer:

e)

Explanation:

In an RC series circuit, at any time, the sum of the voltages through the resistor and the capacitor must be constant and equal to the voltage of the DC voltage source, in order to be compliant with KVL.

At= 0, as the voltage through the capacitor can't change instantaneously, all the voltage appears through the resistor, which means that a current flows, that begins to charge the capacitor, up to a point that the voltage through the capacitor is exactly equal to the DC voltage, so no current flows in the circuit anymore, and the charge in the capacitor reaches to its maximum value.

Answer:

68.585m/sec , 779.1 N

Explanation:

To feel weightless, centripetal acceleration must equal g (9.8m/sec^2). The accelerations then cancel.

From centripetal motion.

F =( mv^2)/2

But since we are dealing with weightlessness

r = 480m

g = 9.8m/s^2

M also cancels, so forget M.

V^2 = Fr

V = √ Fr

V =√ (9.8 x 480) = 4704

= 68.585m/sec.

b) Centripetal acceleration = (v^2/2r) = (68.585^2/960) = 4704/960

= 4.9m/sec^2.

Weight (force) = (mass x acceleration) = 159kg x (g - 4.9)

159kg × ( 9.8-4.9)

159kg × 4.9

= 779.1N

A) The speed of the scooter at which the driver will feel weightlessness is;

v = 68.586 m/s

B) The apparent weight of both the driver and the scooter at the top of the hill is;

F_net = 779.1 N

We are given;

Mass; m = 159 kg

Radius; r = 480 m

A) Since it's motion about a circular hill, it means we are dealing with centripetal force.

Formula for centripetal force is given as;

F = mv²/r

Now, we want to find the speed of the scooter if the driver feels weightlessness.

This means that the centripetal force would be equal to the gravitational force.

Thus;

mg = mv²/r

m will cancel out to give;

v²/r = g

v² = gr

v = √(gr)

v = √(9.8 × 480)

v = √4704

v = 68.586 m/s

B) Now, he is travelling with speed of;

v = 68.586 m/s

And the radius is 2r

Let's first find the centripetal acceleration from the formula; α = v²/r

Thus; α = 4704/(2 × 480)

α = 4.9 m/s²

Now, since he has encountered a hill with a radius of 2r up the slope, it means that the apparent weight will now be;

F_app = m(g - α)

F_net = 159(9.8 - 4.9)

F_net = 779.1 N

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Answer:

Mass of shot (m) = 4 kg

Explanation:

Given:

Velocity (v) = 15 m/s

Mechanical kinetic energy (K.E) = 450 J

Find:

Mass of shot (m) = ?

Computation:

Mechanical kinetic energy (K.E) = 1/2mv²

Mechanical kinetic energy (K.E) = [1/2](m)(15)²

450 = [1/2](m)(15)²

900 = 225 m

Mass of shot (m) = 4 kg

Answer:

1. 7.08Kg

2. 311K

3. 0.268KJ/K

Explanation:

See attached file

The number of maxima of the standing wave pattern is two.

At the time when the receiver moves via one cycle so here two maximas should be considered. At the time when the two waves interfere by traveling in the opposite direction through the same medium so the standing wave pattern is formed.

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Answer:

The current is  [tex]I = 2042\ A[/tex]

Explanation:

From the question we are told that

    The length of the solenoid is  [tex]l = 2.2 \ m[/tex]

    The  radius is  [tex]r_i = 30 \ cm = 0.30 \ m[/tex]

    The number of turn is [tex]N = 1200 \ turns[/tex]

    The  magnetic field is  [tex]B = 1.4 \ T[/tex]

The  magnetic field produced  is mathematically represented as

         [tex]B = \frac{\mu_o * N * I }{l }[/tex]

making [tex]I[/tex] the subject

       [tex]I = \frac{B * l}{\mu_o * N }[/tex]

Where  [tex]\mu_o[/tex] is the permeability of free space with values [tex]\mu_o = 4\pi *10^{-7} N/A^2[/tex]

 substituting values

        [tex]I = \frac{1.4 * 2.2 }{4\pi *10^{-7} * 1200 }[/tex]

        [tex]I = 2042\ A[/tex]

Answer:

A.33.01

B.2.081

C.66%

Explanation:

See attached file pls

Answer:

Option A

Explanation:

Given that

Distance = Speed / Time

So, they are in inverse relation.

Such that when the time passes, the distance from the reacing point will become less and vice versa.

So, Yes! The more time that passes, the farther away a person riding a bike will be.

Answer:

The intensity is [tex]I = 500 mW/m^2[/tex]

Explanation:

From the question we are told that

    The  intensity of the unpolarized light is [tex]I_o = 1000 \ m W /m^2 = 1000 *10^{-3} \ W/m^2[/tex]

Generally the intensity of the light emerging from the polarizer is  mathematically represented as

          [tex]I = \frac{I_o}{2}[/tex]

substituting values

         [tex]I = \frac{1000 *10^{-3}}{2}[/tex]

         [tex]I = 500 *10^{-3} W/m^2[/tex]

         [tex]I = 500 mW/m^2[/tex]

Answer:

1/4F

Explanation:

We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.

So F α Qq

But if it is now half the initial charges, then

F α (1/2)Q *(1/2)q

F α (1/4)Qq

Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.

Thus the answer will be 1/4F

Answer:

B. reflects

Explanation:

Red objects appear red because they reflect red light.

Answer:

B

Explanation:

just did the quiz

Answer:

Billow clouds provide a visible signal to aviation interests of potentially dangerous turbulent sky since they indicate instability in air currents.

Explanation:

Billow clouds are created in regions that are not stable in a meteorological sense. They are frequently present in places with air flows, and have marked vertical shear and weak thermal separation and inversion (colder air stays on top of warmer air). Billow clouds are formed when two air currents of varying speeds meet in the atmosphere. They create a stunning sight that looks like rolling ocean waves. Billow clouds have a very short life span of minutes but they provide a visible signal to aviation interests of potentially dangerous turbulent sky since they indicate instability in air currents, which although may not affect us on the ground but is a concern to aircraft pilots. The turbulence due to the Billow wave is the only logical explanation for the loss of 500 m in altitude of the plane.

Answer:

speed when it reaches y = 4.00cm is

v = 14.9 g.m/s

Explanation:

given

q₁=q₂ =2.00 ×10⁻⁶

distance along x = 3.00cm= 3×10⁻²

q₃= 4×10⁻⁶C

mass= 10×10 ⁻³g

distance along y = 4×10⁻²m

r₁ = [tex]\sqrt{3^{2} +2^{2} }[/tex] = [tex]\sqrt{13}[/tex] = 3.61cm = 0.036m

r₂ = [tex]\sqrt{4^{2} + 3^{2} }[/tex] = [tex]\sqrt{25}[/tex] = 5cm = 0.05m

electric potential V = [tex]\frac{kq}{r}[/tex]

change in potential ΔV = [tex]V_{1} - V_{2}[/tex]

ΔV = [tex]\frac{2kq_{1} }{r_{1}} - \frac{2kq_{2} }{r_{2} }[/tex] , where [tex]q_{1} = q_{2}=[/tex]2.00μC

ΔV = [tex]2kq(\frac{1}{r_{1}} - \frac{1}{r_{2} })[/tex]

ΔV = 2 × 9×10⁹ × 2×10⁻⁶ × [tex](\frac{1}{0.036} - \frac{1}{0.05} )[/tex]

ΔV= 2.789×10⁵

[tex]\frac{1}{2}mv^{2}[/tex] = ΔV × q₃

[tex]\frac{1}{2}[/tex] ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶

v² = 223.12 g.m/s

v = 14.9 g.m/s

The speed of the charge q₃ when it starts from rest at y = 2 cm and reaches y = 4 cm is; v = 14.89 m/s

We are given;

Charge 1; q₁ = 2.00 μC = 2 × 10⁻⁶ C

Charge 2; q₂ = 2.00 μC = 2 × 10⁻⁶ C

Distance of charge 1 along x = 3 cm = 3 × 10⁻² m

Distance of charge 2 along x = -3 cm = -3 × 10⁻² m

Charge 3; q₃ = +4.00 μC  = 4 × 10⁻⁶ C

mass; m = 0.01 g

distance of charge 3 along y = 4 cm = 4 × 10⁻² m

q₃ starts from rest at y = 2 × 10⁻² m and reaches y = 4 × 10⁻² m.

Thus;

Distance of charge 1 from the initial position of q₃;

r₁ = √((3 × 10⁻²)² + ((2 × 10⁻²)²)

r₁ = 0.0361 m

Distance of charge 2 from the final position of q₃;

r₂ = √((3 × 10⁻²)² + ((4 × 10⁻²)²)

r₂ = 0.05 m

Now, formula for electric potential is;

V = kq/r

Where k = 9 × 10⁹ N.m²/s²

Thus,change in potential is;

ΔV = V₁ - V₂

Now, Net V₁ = 2kq₁/r₁

Net V₂ = 2kq₂/r₂

Thus;

ΔV = 2kq₁/r₁ - 2kq₂/r₂

ΔV = (2 × 9 × 10⁹)[(2 × 10⁻⁶/0.0361) - (2 × 10⁻⁶/0.05)]

ΔV = 277229.92 V

Now, from conservation of energy;

½mv² = q₃ΔV

Thus;

½ × 0.01 × v² = 4 × 10⁻⁶ × 277229.92

v² = 2 × 4 × 10⁻⁶ × 277229.92/0.01

v = √(221.783936)

v = 14.89 m/s

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Complete Question

Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperature of 2.81 K. Remember that Stefan's Law gives the Power (Watts) and Intensity is Power per unit Area (W/m2).

Answer:

The intensity is [tex]I = 3.535 *10^{-6} \ W/m^2[/tex]

Explanation:

From the question we are told that

    The temperature is  [tex]T = 2.81 \ K[/tex]

Now  According to Stefan's law

        [tex]Power(P) = \sigma * A * T^4[/tex]

Where  [tex]\sigma[/tex] is the Stefan Boltzmann constant with value  [tex]\sigma = 5.67*10^{-8} m^2 \cdot kg \cdot s^{-2} K^{-1}[/tex]

  Now the intensity of the cosmic background radiation emitted according to the unit from the question is mathematically evaluated as

        [tex]I = \frac{P}{A}[/tex]

=>      [tex]I = \frac{\sigma * A * T^4}{A}[/tex]

=>      [tex]I = \sigma * T^4[/tex]

substituting values

      [tex]I = 5.67 *10^{-8} * (2.81)^4[/tex]

       [tex]I = 3.535 *10^{-6} \ W/m^2[/tex]

Answer:

R = 42.67 ohms

Explanation:

It is given that, a thin-filament light bulb is connected to two 1.6 V batteries in series, the current is 0.075 A.

It means that when two batteries are connected in series, then the total voltage is 3.2 volts

Let R is the resistance of the glowing thin-filament bulb. So, using Ohm's law we get :

[tex]V=IR\\\\R=\dfrac{V}{I}[/tex]

So,

[tex]R=\dfrac{3.2}{0.075}\\\\R=42.67\ \Omega[/tex]

So, the resistance of the bulb is 42.67 ohms.

Answer:

Vf = 1.43 m/s

Explanation:

From Coulomb's Law, the electrostatic force between electron and proton is given as:

F = kq₁q₂/r²

F = Electrostatic force = ?

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

q₂ = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C

r = distance between electron and proton = 9 cm = 0.09 m

Therefore,

F = (9 x 10⁹ N.m²/C²)(1.6 x 10⁻¹⁹ C)(1.6 x 10⁻¹⁹ C)/(0.09 m)²

F = 2.84 x 10⁻²⁶ N

but, from Newton's second law:

F = 2.84 x 10⁻²⁶ N = ma

where,

m = mass of electron = 9.1 x 10⁻³¹ kg

a = acceleration of electron = ?

Therefore,

2.84 x 10⁻²⁶ N = (1.67 x 10⁻²⁷ kg)(a)

a = 2.84 x 10⁻²⁶ N/1.67 x 10⁻²⁷ kg

a = 17.03 m/s²

Now, we apply 3rd equation of motion to the motion of electron from a distance of 9 cm to 3 cm near to the proton:

2as = Vf² - Vi²

where,

s = distance traveled = 9 cm - 3 cm = 6 cm = 0.06 m

Vf = speed of electron when it is 3 cm from proton = ?

Vi = Initial speed of electron = 0 m/s

Therefore,

2(17.03 m/s²)(0.06 m) = Vf² - (0 m/s)²

Vf = √2.04 m²/s²

Vf = 1.43 m/s

Answer:

Explanation:

Magnetism is defined as the ability of a magnet to attract magnetic substance to itself. Such magnet has the ability of being magnetized. A magnet is known to possess poles which are the north poles and south poles. The presence of this poles is what makes them possess the properties of a magnet. An ordinary steel bar doesn't have the properties of a magnet unless it is magnetized and when you are trying to magnetize a steel bar, you are invariably introducing the magnetic poles.

According to the law of magnetism, like poles repel but unlike poles attract. From the above explanation, it can be concluded that the phenomenon of magnetism is best understood interns of existence of magnetic poles. This poles are called the north and the south poles.

Answer:

a) [tex] f = 8.62 \cdot 10^{8} Hz [/tex]

b) [tex] B = 1.9 \cdot 10^{-10} T [/tex]  

c) [tex] I = 4.30 \cdot 10^{-6} W/m^{2} [/tex]

Explanation:

a) The frequency (f) of the wave can be found as follows:

[tex] f = \frac{c}{\lambda} [/tex]

Where:

c: is the speed of light = 3x10⁸ m/s

λ: is the wavelength = 34.8 cm

[tex] f = \frac{3 \cdot 10^{8} m/s}{0.348 m} = 8.62 \cdot 10^{8} Hz [/tex]

b) The magnetic-flied amplitude (B) is:

[tex] B = \frac{E}{c} [/tex]      

Where:

E: is the electric field amplitude = 5.70x10⁻² V/m

[tex] B = \frac{E}{c} = \frac{5.70 \cdot 10^{-2} V/m}{3 \cdot 10^{8} m/s} = 1.9 \cdot 10^{-10} T [/tex]  

c) The intensity of the wave (I) is the following:

[tex] I = \frac{E*B}{2\mu_{0}} [/tex]

Where:

μ₀: is the permeability of free space =  1.26x10⁻⁶ m*kg/(s²A²)  

[tex] I = \frac{E*B}{2\mu_{0}} = \frac{5.70 \cdot 10^{-2} V/m*1.9 \cdot 10^{-10} T}{2*1.26 \cdot 10^{-6} m*kg/((s^{2}A^{2})} = 4.30 \cdot 10^{-6} W/m^{2} [/tex]

I hope it helps you!

The frequency of the wave is [tex]8.62\times 10^8\rm\;Hz[/tex], the magnetic-field amplitude is [tex]1.9\times 10^{-10}\rm\;T[/tex], and the intensity of the wave is [tex]4.298\rm\;W/m^2[/tex].

Given information:

A mobile phone emits electromagnetic radiation.

The wavelength of the wave is [tex]\lambda=34.8[/tex] cm.

The electric-field amplitude is  [tex]5.70\times10^{-2}[/tex] V/m.

Phone is at a distance of 210 m.

The speed of the electromagnetic wave is [tex]c=3\times 10^8[/tex] m/s.

(a)

Now, the frequency of the wave will be calculated as,

[tex]f=\dfrac{c}{\lambda}\\f=\dfrac{3\times 10^8}{0.348}\\f=8.62\times 10^8\rm\;Hz[/tex]

(b)

The magnetic-field amplitude can be calculated as,

[tex]B=\dfrac{E}{c}\\B=\dfrac{5.70\times10^{-2}}{3\times 10^8}\\B=1.9\times 10^{-10}\rm\;T[/tex]

(c)

[tex]\mu_0[/tex] is the permeability of the vacuum. [tex]\mu_0=1.26\times10^{-6} \rm\;\frac{kg-m}{(A^2s^2)}[/tex]

The intensity of the wave can be calculated as,

[tex]I=\dfrac{BE}{2\mu_0}\\I=\dfrac{1.9\times10^{-10 }\times5.7\times10^{-2}}{2\times1.26\times10^{-6}}\\I=4.298\rm\;W/m^2[/tex]

Therefore, the frequency of the wave is [tex]8.62\times 10^8\rm\;Hz[/tex], the magnetic-field amplitude is [tex]1.9\times 10^{-10}\rm\;T[/tex], and the intensity of the wave is [tex]4.298\rm\;W/m^2[/tex].

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Answer:

The y-value  is  z = 0.759 m

Explanation:

From the question we are told that

     The position of the first y-axis is  [tex]y_1 = 0.300 \ m[/tex]

     The current on the first wire is  [tex]I_ 1 = 26.0 \ A[/tex]

      The force per unit length on each wire is  [tex]\frac{F}{l} = 295 \mu N/m = 295 * 10^{-6} \ N/m[/tex]

Generally the force per unit length on first wire is mathematically represented as

                [tex]\frac{F}{l} = \frac{\mu_o * I_1 * I_2 }{2*\pi* y_1}[/tex]

Where  [tex]\mu _o[/tex] is the permeability of free space with value  [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]

    substituting values

                    [tex]295 *10^{-6} = \frac{ 4\pi * 10^{-7} * 26.0 * I_2 }{2 *3.142* 0.300}[/tex]

                [tex]I_2 = \frac{295 *10^{-6 } * 0.300 * 2* 3.142 }{ 4\pi * 10^{-7} * 26 }[/tex]

                 [tex]I_2 = 17.0 \ A[/tex]

Now the at the point where the magnetic field is zero the magnetic field of each wire are equal , let that point by z meters from the second wire on the y-axis  so

             [tex]\frac{\mu_o I_2}{2 * \pi * y_1} = \frac{\mu_o I_1}{2 * \pi * (y_1-z)}[/tex]

          [tex]I_2 (y_1 - z) = I_1 * y_1[/tex]

substituting values

         [tex]17.0 ( 0.300 - z) = 26 * 0.300[/tex]

         z = 0.759 m

Answer:

5.5x 10^-11 C

Explanation:

Pls see attached file

The net work done (W) on a particle by the external forces during the motion of the particle in terms of the initial and final kinetic energies is equal to [tex]W = K_f - K_i[/tex]

The net work done (W) can be defined as the work done in moving an object by a net force, which is the vector sum of all the forces acting on the object.

According to Newton's Second Law of Motion, the net work done (W) on an object or physical body is equal to the change in the kinetic energy possessed by the object or physical body.

Mathematically, the net work done (W) on an object or physical body is given by the formula:

[tex]W =\Delta K_E\\\\W = K_f - K_i[/tex]

Where:

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Answer:

2m/s

Explanation:

According to conservation of momentums, it states that the sum of collision of bodies before collision is equal to the sum of their momentum after collision. Both objects will move with the same velocity after collision.

According to the question, we were told that after they collide, glider B has a final velocity of 2 m/s. Since both bodies (Glider A and B) will move with the same velocity after collision according to the conservation of momentum, this means glider A will also have a final velocity of 2m/s like. Glider B.

Answer:

The de Broglie wavelength of electron βe = 2.443422 × 10⁻⁹ m

The de Broglie wavelength of proton βp = 5.70 × 10⁻¹¹ m

Explanation:

Thermal kinetic energy of electron or proton = KE

∴ KE = 3kbT/2

given that; kb = 1.38 x 10⁻²³ J/K , T = 1950 K

so we substitute

KE = ( 3 × 1.38 x 10⁻²³ × 1950 ) / 2

kE = 4.0365 × 10⁻²⁰ (  is the kinetic energy for both electron and proton at temperature T )

Now we know that

mass of electron M'e = 9.109 ×  10⁻³¹

mass of proton M'p = 1.6726 ×  10⁻²⁷

We also know that

KE = p₂ / 2m

from the equation, p = √ (2mKE)

{ p is momentum, m is mass }

de Broglie wavelength = β

so β = h / p = h / √ (2mKE)

h = Planck's constant = 6.626 ×  10⁻³⁴

∴ βe =  h / √ (2m'e × KE)

βe = 6.626 ×  10⁻³⁴ / √ (2 × 9.109 ×  10⁻³¹ × 4.0365 × 10⁻²⁰ )

βe = 6.626 ×  10⁻³⁴ / √  7.3536957 × 10⁻⁵⁰

βe = 6.626 × 10⁻³⁴  / 2.71176984642871 × 10⁻²⁵

βe = 2.443422 × 10⁻⁹ m

βp =  h / √ (2m'p ×KE)

βp = 6.626 ×  10⁻³⁴ / √ (2 × 1.6726 ×  10⁻²⁷ × 4.0365 × 10⁻²⁰ )

βp = 6.626 ×  10⁻³⁴ / √ 1.35028998 × 10⁻⁴⁶

βp =  6.626 ×  10⁻³⁴ / 1.16201978468527 ×  10⁻²³

βp = 5.702140 × 10⁻¹¹ m

Answer:

the smaller particle moves with speed of 8.706 m/s in the opposite direction to the bigger particle.

Explanation:

Speed of the original particle = 13 m/s

We designate particles as A and B

The final weights of the component particles are

Particle A = 1.4 N

particle B = 1.9 N

The speed of the larger piece (particle B) = 29 m/s

We know that weight is the product of a body's mass and acceleration due to gravity g which is equal to 9.81 m/s^2, therefore, masses of the particles are

particle A = 1.4/9.81 = 0.143 kg

Particle B = 1.9/9.81 = 0.194 kg

The momentum of a body is the product of its mass and its velocity i.e

P = mv

This means that the mass of the particle before splitting is  

0.143 kg + 0.194 kg = 0.337 kg

Momentum of the initial whole particle = mv

==> 0.337 x 13 = 4.381 kg-m/s

The bigger particle B remains horizontal, and has a momentum of

mv = 0.194 x 29 = 5.626 kg-m/s

According to the conservation of momentum, the total initial momentum of a system must be equal tot the total final momentum of the system.

Initial total momentum of the system = 4.381 kg-m/s (momentum of original particle before splitting)

Final total momentum of the system = Total momentum of the particles after splitting = 5.626 kg-m/s + ( 0.143 kg x [tex]V_{B}[/tex])

where  [tex]V_{B}[/tex]  is the velocity of smaller particle A

final total momentum of the system = 5.626 + 0.143[tex]V_{B}[/tex]

Equating the two momenta of the system, we'll have

4.381 = 5.626 + 0.143[tex]V_{B}[/tex]

4.381 - 5.626 = 0.143[tex]V_{B}[/tex]

-1.245 = 0.143[tex]V_{B}[/tex]

[tex]V_{B}[/tex]  = -1.245/0.143 = -8.706 m/s

The negative sign indicates that the smaller particle moves in the opposite direction to the bigger particle