On the moon, the gravitational acceleration is approximately one-sixth that on the surface of the earth. A 5-kg mass is "weighed" with a beam balance on the surface of the moon.
a) What is the expected reading?
b) If this mass is weighed with a spring scale that reads correctly for standard gravity on earth, what is the reading?

Answer:

V = 31.62 m V

Explanation:

Given,

Voltage across thick membrane = ?

Thickness of the membrane, d = 5.5 n m

Electric field = 5.75 M V/m

we know

[tex]E = \dfrac{V}{d}[/tex]

   V = E d

   V = 5.75 x 10⁶ x 5.5 x 10⁻⁹

   V = 31.62 x 10⁻³ V

   V = 31.62 m V

Hence, The Voltage across the membrane is equal to 31.62 m V.

The voltage across the membrane is dependent on the electric field and thickness of the membrane. The voltage across the membrane is 0.0316 V.

The voltage is defined as the difference in electric potential between two points.

Given that the electric field E is 5.75 mv/m and the thickness of the membrane d is 5.5 nm.

The voltage across the membrane is calculated as given below.

[tex]E = \dfrac {V}{d}[/tex]

[tex]V = Ed[/tex]

[tex]V = 5.75 \times 10^6 \times 5.5 \times 10^{-9}[/tex]

[tex]V = 0.0316 \;\rm V[/tex]

Hence we can conclude that the voltage across the membrane is 0.0316 V.

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Explanation:

The temperature of the solar nebula was decreasing as it moved away from its center. Therefore, only heavy elements could condense in the inner solar system and terrestrial planets could not form with light elements, such as gases. In the outer solar system, the Jovian planets formed mostly with gases, since temperatures were too low to allow rocky compositions.

Answer:

Explanation:

Given

Power drawn by toaster [tex]P_1=1840\ W[/tex]

Power drawn by Electric fan [tex]P_2=1460\ W[/tex]

Power drawn by lamp [tex]P_3=50\ W[/tex]

Voltage applied [tex]V=120\ V[/tex]

If appliances are applied in parallel then Voltage applied is same

Power is given by [tex]P=\frac{V^2}{R}[/tex]

Resistance of toaster [tex]R_1=\frac{120^2}{1840}=7.82\ \Omega [/tex]

Resistance of electric Fan [tex]R_2=\frac{120^2}{1460}=9.86\ \Omega [/tex]

Resistance of toaster [tex]R_3=\frac{120^2}{50}=288\ \Omega [/tex]

Current is given by

[tex]I=\frac{V}{R}[/tex]

[tex]I_1=\frac{120}{7.82}=15.34\ A[/tex]

[tex]I_2=\frac{120}{9.86}=12.17\ A[/tex]

[tex]I_3=\frac{120}{288}=0.416\ A[/tex]

Answer:

Explanation:

power of toaster, P1 = 1840 W

Power of electric frying pan, P2 = 1460 W

Power of lamp, P3 = 50 W

As they all are connected in parallel, so the voltage is same for all.

Let the current in toaster is i1.

P1 = V x i1

1840 = 120 x i1

i1 = 15.33 A

Let the current in frying pan is i2.

P2 = V x i2

1460 = 120 x i1

i1 = 12.17 A

Let the current in lamp is i3.

P3 = V x i3

50 = 120 x i3

i3 = 0.42 A

Answer:

 ΔP = 97.93 Pa ,     T₂-T₁ = 71.5° C

Explanation:

For this problem let's use Bernoulli's relationship, as point 1 we will take the plane and as point 2 the air

            P₁ + ½ ρ g v₁² + ρ g y₁ = P2 + ½ ρ v₂² + ρ g y₂

As the whole process occurs at the same height y₁ = y₂ = 11 km. We will consider that the air goes in the opposite direction to the plane

            P₂ –P₁ = ½ ρ (v₁² - v₂²)

Let's reduce the magnitudes to the SI system

           v₁ = 935 km / h (1000 m / 1 km) (1h / 3600s) = 259.72 m / s

           v₂ = 280 m / s

The density of air at 11000 m is

           Rho = 0.3629 kg / m

           P₂-P₁ = ½ 0.3629 (259.72 + 280)

           ΔP = 97.93 Pa

The variation of the temperature with the altitude is 0.65 per 100 m

            T₂ –T₁ = (0.65 / 100) 1000

             T₂-T₁ = 71.5° C

The temperature has decreased this value

Angular displacement of the cd is 3.25 rev

Explanation:

The question is incomplete. It is not given what is the maximum angular velocity of the cd.

Here we are going to assume that the maximum angular velocity is:

[tex]\omega = 30 rad/s[/tex]

The motion of the cd is an accelerated angular motion, therefore we can use the following suvat equation:

[tex]\theta = (\frac{\omega_0 + \omega}{2})t[/tex]

where:

[tex]\theta[/tex] is the angular displacement of the cd during the time interval t

[tex]\omega_0[/tex] is the initial angular velocity of the cd

[tex]\omega[/tex] is the final angular velocity

Here we have:

t = 1.36 s

[tex]\omega_0 = 0[/tex] (assuming the cd starts from rest)

Therefore, the angular displacement of the cd during this time is:

[tex]\theta=(\frac{0+30}{2})(1.36)=20.4 rad[/tex]

And since [tex]1 rev = 2 \pi rad[/tex], we can convert into number of revolutions completed:

[tex]\theta = 20.4 rad \cdot \frac{1}{2\pi rad/rev}=3.25 rev[/tex]

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Answer: a) 2.5 * 10^14, b) t = 1.2*10^-8 s, c) F = 2.2775 * 10^-15 N

Explanation: Since it starts from rest, initial velocity = 0, final velocity (v) = 3*10^6 m/s, distance covered (s) = 1.80cm = 1.80/100 = 0.018m

Since the force on the electron is constant, it acceleration will be constant too thus making newton's laws of motion valid.

Question a)

To get the acceleration, we use the formulae that

v² = u² + 2as

But u = 0

v² = 2as

(3*10^6)² = 2*a*(0.018)

9* 10^12 = 0.036*a

a = 9 * 10^12 / 0.036

a = 250 * 10^12

a = 2.5 * 10^14 m/s².

Question b)

To get the time, we use

v = u + at

But u = 0

v = at

3*10^6 = 2.5 * 10^14 * t

t = 3*10^6 / 2.5*10^14

t = 1.2*10^-8 s

Question c)

To get the force, we use the formulae below

F = ma

F = 9.11*10^-31 * 2.5 * 10^14

F = 22.775 * 10^-17

F = 2.2775 * 10^-15 N

a) The acceleration of the electron as it travels in a straight line to the grid is 2.5 × 10¹⁴m/s².

b) The time taken for the electron to reach the grid is 1.2 × 10⁻⁸ seconds.

c) The net force that is accelerating the electron is 2.28 × 10⁻¹⁶ Newton.

Given the data in the question

Acceleration; [tex]a = \ ?[/tex]

Time taken; [tex]t = \ ?[/tex]

Net force; [tex]F = \ ?[/tex]

a)

To determine the acceleration of the electron, we use the third equation of motion:

[tex]v^2 = u^2 + 2as[/tex]

Where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance traveled.

We substitute our given values into the equation

[tex](3.00 *10^6 m/s)^2 = 0^2 + [ 2 * a * 0.018m ]\\\\9.0 *10^{12} m^2/s^2 = a * 0.036m\\\\a = \frac{9.0 *10^{12} m^2/s^2}{0.036m} \\\\a = 2.5 * 10^{14}m/s^2[/tex]

Therefore, the acceleration of the electron as it travels in a straight line to the grid is 2.5 × 10¹⁴m/s².

b)

To the the time taken to reach the grid, we use the first equation of motion:

[tex]v = u + at[/tex]

Where v is the final velocity, u is initial velocity, a is the acceleration and t is the time taken

We substitute our values into the equation

[tex]3.00 * 10^6 m/s = 0 + [ ( 2.5*10^{14}m/s^2) * t\\\\3.00 * 10^6 m/s = ( 2.5*10^{14}m/s^2) * t\\\\t = \frac{3.00 * 10^6 m/s}{2.5*10^{14}m/s^2} \\\\t = 1.2 * 10^{-8}s[/tex]

Therefore, time taken for the electron to reach the grid is 1.2 × 10⁻⁸ seconds.

c)

To determine the net force of the electron, we use the expression from newton second law of motion:

[tex]F = m * a[/tex]

Where m is mass and a is the acceleration

We substitute our values into the equation

[tex]F = ( 9.11 * 10^{-31} kg ) * ( 2.5 * 10^{14}m/s^2)\\\\F = 2.28 * 10^{-16} kg.m/s^2\\\\F = 2.28 * 10^{-16}N[/tex]

Therefore, the net force that is accelerating the electron is 2.28 × 10⁻¹⁶ Newton.

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Explanation:

When the skydiver accelerates in the downward direction then tends to gain speed with each second. More is the resistance in air more will be the increase accompanied by the skydiver.

As a result, a point will come where air resistance force is balanced by gravitational force. Hence, the skydiver will attain terminal velocity.

So, air resistance for 1 [tex]m/s^{2}[/tex] = 0.45 [tex]m/s^{2}[/tex]

Air resistance for v [tex]m/s^{2}[/tex] = 0.45 v[tex]m/s^{2}[/tex]

As acceleration = change in velocity w.r.t time

                 a = [tex]\frac{dv}{dt}[/tex] = 0.45

            [tex]\frac{dv}{V}[/tex] = 0.45t

Now, we will integrate both the sides as follows.

            ln V = 0.45t

                V = [tex]e^{0.45t}[/tex]

Since,         [tex]F_{a} = F_{g}[/tex]   (in the given case)

so,                    ma = mg

On cancelling the common terms the equation will be as follows.

                     0.45 v = 9.8 [tex]m/s^{2}[/tex]

                         v = 21.77 [tex]m/s^{2}[/tex]

Thus, we can conclude that the skydivers terminal speed is 21.77 [tex]m/s^{2}[/tex].

We have that the  the rock's initial launch speed  is mathematically given as

From the question we are told

Generally the equation for the Motion   is mathematically given as

[tex]H=ut+0.5gt^2\\\\Therefore\\\\4.9=0*t+0.5*9.8*t^2\\\\t=1s\\\\Therefore\\\\sx=uxt\\\\20=ux*1\\\\[/tex]

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To determine the initial launch speed of the rock, we can use the equations of projectile motion. Since the rock is thrown horizontally, and it travels 20 m horizontally in 1 second, the initial launch speed can be calculated as 20 m/s.

To determine the initial launch speed of the rock, we can use the equations of projectile motion. Since the rock is thrown horizontally, its initial vertical velocity is zero. The only force acting on the rock in the vertical direction is gravity, which causes it to accelerate downward at a rate of 9.8 m/s². Using the formula:

-yf = yi + vit - 1/2gt²

where yi represents the initial height and yf represents the final height, we can solve for the initial downward velocity. Since the rock reaches the ground (yf = 0) and starts at a height of 4.9 m, the equation becomes:

0 = 4.9 + 0 - 1/2(9.8)t²

Simplifying the equation gives:

t² = 4.9/4.9

t = 1 second.

Since the horizontal distance traveled by the rock is 20 m, and it takes 1 second to reach the ground, the horizontal speed (initial launch speed) of the rock can be calculated using the formula:

v = d/t = 20 m/1 s = 20 m/s.

Therefore, the rock's initial launch speed is 20 m/s.

a) Not possible

b) Yes, it's possible (see graph in attachment)

Explanation:

a)

The average velocity of a body is defined as the ratio between the displacement and the time elapsed:

[tex]v=\frac{\Delta x}{\Delta t}[/tex]

where

[tex]\Delta x[/tex] is the displacement

[tex]\Delta t[/tex] is the time elapsed

In this problem, we want to have zero displacement and non-zero average velocity. From the equation above, we see that this is not possible. In fact, if the total displacement is zero,

[tex]\Delta x = 0[/tex]

And therefore as a consequence,

[tex]v=0[/tex]

which means that the average velocity is zero.

B)

Here we want to have zero displacement and non-zero velocity. In this case, it is possible: in fact, we are not talking  about average velocity, but we are talking about (instantaneous) velocity.

On a position-time graph, the instantaneous velocity is the slope of the graph. Look at the graph in attachment. We see that the position of the object first increases towards positive value, then it decreases (the object starts moving backward), then becomes negative, then it increases again until returning to the original position, x = 0.

In all of this, we notice that the total displacement of the object is zero:

[tex]\Delta x = 0[/tex]

However, we notice that the instantaneous velocity of the object at the various instants is not zero, because the slope of the graph is not zero.

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Zero displacement and nonzero average velocity can occur on an x-t graph, but it is not possible to have zero displacement and nonzero velocity simultaneously.

Zero displacement and nonzero average velocity can occur when an object moves back and forth over a certain distance and returns to its original position, but with a change in time. For example, if you consider a person running in a circular track, after each lap the person returns to the starting point, resulting in zero displacement. However, if the person completes the laps in different time intervals, their average velocity will be nonzero.

On the other hand, it is not possible to have zero displacement and nonzero velocity simultaneously. Velocity is defined as the rate of change of displacement with respect to time. If the displacement is zero, then the velocity must also be zero

In an x-t graph, zero displacement would be represented by the horizontal line at the x-axis, or the line parallel to the x-axis. Nonzero average velocity would be represented by a sloped line that is above or below the x-axis, indicating movement in either the positive or negative direction.

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The force between the two charges, according to Coulomb's law, is approximately 1.92 Newtons. Note that the force is attractive since one charge is positive and the other is negative.

The question refers to Coulomb's law, which is fundamental in Physics. The law describes the force between two electric charges. We apply the formula |F|=K|QQ′|/d^2 where the charges are Q1=-13.5nC (converted to C by multiplying by 10^-9) and Q2=35.5nC, the distance d=1.735mm (converted to meters by multiplying by 10^-3), and K=1/4πϵ0≈8.99x10^9 N.m²/C². Inserting these values into the formula we get |F| ≈ 1.92 N. The magnitude of the force is therefore approximately 1.92 Newtons. It's important to note that since one charge is positive and the other negative, the force will be attractive, not repulsive.

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Answer:

  q= 20 W                

Explanation:

Given that

Initial temperature ,T₁ = 2 ºC

Heat gains ,Q = 18 kJ

The final temperature ,T₂ = 20 ºC

time ,t= 15 min

We know that

1 min = 60 s

t= 15 x 60 = 900 s

The average rate of heat transfer is give as

[tex]q=\dfrac{Q}{t}[/tex]

[tex]q=\dfrac{18}{900}\ kW[/tex]

q=0.02 kW

q= 20 W

Therefore the rate of heat transfer will be 20 W.

A soda is initially at 2 °C and 15 minutes later, after gaining 18 kJ, its temperature is 20 °C. The average rate of heat transfer is 20 W.

Heat transfer describes the flow of heat (thermal energy) due to temperature differences and the subsequent temperature distribution and changes.

A soda is initially at 2 °C and 15 minutes later, after gaining 18 kJ, its temperature is 20 °C.

First, we will convert 15 min to seconds using the conversion factor 1 min = 60 s.

15 min × 60 s/1 min  = 900 s

Then, we can calculate the average rate of heat transfer using the following expression.

q = Q/t = 18 × 10³ J / 900 s = 20 W

where,

A soda is initially at 2 °C and 15 minutes later, after gaining 18 kJ, its temperature is 20 °C. The average rate of heat transfer is 20 W.

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Answer:

29.79 N/m

Explanation:

A spring connected in series behaves like a capacitor connected in series.

Note: Spring and capacitor are alike because they both store energy, While the former store mechanical energy, the later store electrical energy.

From the above, the effective spring connected in series is given by the formula below

1/Kt = 1/K1 + 1/K2 ........................ Equation 1

Where Kt = effective spring constant, K1 = spring constant of spring 1, K2 = spring constant of spring 2

Making K2 the subject of the equation,

K2 = KtK1/(K1-Kt)..................... Equation 2

Given: K1 = 61 N/m, Kt = 20 N/m.

Substitute into equation 2

K2 = (61×20)/(61-20)

K2 = 1220/41

K2 = 29.76 N/m.

Hence the spring constant in the second spring = 29.79 N/m

The spring constant for Spring #2 is 369.17 N/m.

To find the spring constant of Spring #2, we need to use the formula for the effective spring constant of springs in series. The formula is:

1/keff = 1/k1 + 1/k2

Given that k1 = 61.0 N/m and the effective spring constant (keff) is 20.0 N/m, we can plug these values into the formula and solve for k2:

1/20 = 1/61 + 1/k2

Now we can solve for k2:

1/k2 = 1/20 - 1/61

k2 = 1/(1/20 - 1/61)

k2 = 369.17 N/m

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Answer:

As potential energy depends on height

P.E=mgh

m and g remains same

just before release it has maximum height,thus option D it has maximum potential energy.

The ball has the most potential energy when it is at the top of the building D. before it is released.

Potential energy is stored energy that an object possesses due to its position, state, or condition. It can be gravitational, elastic, chemical, or electrical. The energy is converted to kinetic energy when the object moves or undergoes a change in its physical or chemical state.

The ball has the most potential energy when it is at the top of the building before it is released. Potential energy depends on an object's height and mass. When the ball is at the top of the building, it has the maximum height and therefore the most potential energy. As the ball falls, its potential energy decreases and is converted into kinetic energy.

As show the figure, peak of each blackbody curves shifts towards higher frequency. Then energy emitted by the black body is directly proportional to the frequency incident radiation.

Hence peak is shifted to higher energies.

Therefore the correct option is B: Toward higher energies.

The blackbody curve of a star moving towards Earth would see its peak shifted toward higher energies due to the Doppler Effect, creating a 'blue shift' in the light we observe.

The blackbody curve of a star moving toward Earth undergoes a phenomenon known as the Doppler Effect. When a star moves towards the observer, this effect causes the light from that star to be blue-shifted, which means the light moves towards higher energies, or shorter wavelengths. Therefore, the correct answer is (b) - the peak of the blackbody curve shifts towards higher energies.

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Answer:

Explanation:

Given

F_net(x)  = (3 x²) N

m v dv / dt = 3 x²

m v dv = 3 x² dx

integrating on both sides and taking limit from x = 2 to 4 m

m v² / 2 - 0 = 3 x³ / 3

mv² / 2 = 4³ - 2³

mv² / 2 = 64 -  8

3 x v² /2 = 56

v = 6.11 m / s

linear momentum

= m v

= 3 x 6.11

= 18.33 kgm/s

Answer:

[tex]p=m.v=37\ kg.m.s^{-1}[/tex]

Explanation:

Given:

WE know from the Newton's second law:

[tex]\frac{d}{dt} (p)=F[/tex]

where:

[tex]p=[/tex] momentum

[tex]F=[/tex] force

[tex]t=[/tex] times

Now put the values

[tex]\frac{d}{dt} (mv)=3\cdot x^2[/tex]

[tex]m.\frac{d}{dt} (v)=4\times x^2[/tex]

Now integrate both sides from final limit to initial:

[tex]m.v=\int\limits^4_2 {3x^2} \, dx[/tex]

[tex]m.v=[\frac{3x^3}{3} ]^4_2[/tex]

[tex]m.v=4^3-2^3[/tex]

[tex]p=m.v=56\ kg.m.s^{-1}[/tex]

Light takes 320 minutes to reach Earth from Pluto

Explanation:

We can solve this problem by applying the rule of three. We can do as follows:

- We call [tex]x[/tex] the distance between the Sun and the Earth

- Then the distance between the Earth and Pluto is 40 times this distance, so [tex]40x[/tex]

- Light takes about 8 minutes to cover the distance x between Sun and Earth

Therefore, calling T the time that light takes to cover distance between Earth and Pluto (40x), we can write:

[tex]\frac{x}{8}=\frac{40x}{T}[/tex]

And solving for T,

[tex]T=\frac{8\cdot 40 x}{x}=320 min[/tex]

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The acceleration of the car is 1.67 m/s². It takes approximately 11.23 seconds for the car to travel the length of the ramp. The traffic travels approximately 224.6 meters while the car is moving the length of the ramp.

(a) To find the acceleration of the car, we can use the equation:

v² = u² + 2as

Where v is the final velocity (20 m/s), u is the initial velocity (0 m/s), a is the acceleration, and s is the displacement (120 m).

Plugging in the values, we have:

20² = 0² + 2a(120)

Simplifying the equation, we get:

a = (20²)/(2 * 120)

Therefore, the acceleration of the car is approximately 1.67 m/s².

(b) Time can be calculated using the equation:

s = ut + 0.5at²

Where u is the initial velocity (0 m/s), a is the acceleration (1.67 m/s²), and s is the displacement (120 m).

Plugging in the values, we have:

120 = 0 + 0.5(1.67)t²

Simplifying the equation, we get:

t² = (120)/(0.5 * 1.67)

t = √(120/0.835)

t ≈ 11.23 s

Therefore, it takes approximately 11.23 seconds for the car to travel the length of the ramp.

(c) The distance the traffic travels while the car is moving the length of the ramp can be calculated using the equation:

d = v * t

Where v is the velocity of the traffic (20 m/s) and t is the time taken (11.23 s).

Plugging in the values, we have:

d = 20 * 11.23

Therefore, the traffic travels approximately 224.6 meters while the car is moving the length of the ramp.

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The final answers are [tex](a)~{a ~is~ 1.67 \, \text{m/s}^2} \], ~(b)~{t ~is~ 12s} ~and ~~(c)~d ~is~ 240 \, m[/tex].

To solve the problem, we will use the kinematic equations for uniformly accelerated motion. The car starts from rest, which means its initial velocity (u) is 0 m/s, and it reaches a final velocity (v) of 20 m/s over a distance (s) of 120 m.

a) To find the acceleration (a) of the car, we can use the third kinematic equation that relates initial velocity, final velocity, acceleration, and distance:

[tex]\[ v^2 = u^2 + 2as \]\\ Plugging in the known values, we get: \[ (20 \, \text{m/s})^2 = (0 \, \text{m/s})^2 + 2a(120 \, \text{m}) \] \[ 400 \, \text{m}^2/\text{s}^2 = 240a \, \text{m} \] \[ a = \frac{400 \, \text{m}^2/\text{s}^2}{240 \, \text{m}} \] \[ a = \frac{100}{60} \, \text{m/s}^2 \] \[ a = \frac{5}{3} \, \text{m/s}^2 \] \[ a \approx 1.67 \, \text{m/s}^2 \][/tex]

b) To find the time (t) it takes for the car to travel the length of the ramp, we can use the first kinematic equation:

[tex]\[ v = u + at \]\\ Since the initial velocity (u) is 0 m/s, the equation simplifies to: \[ 20 \, \text{m/s} = 0 + \left(\frac{5}{3} \, \text{m/s}^2\right)t \] \[ t = \frac{20 \, \text{m/s}}{\frac{5}{3} \, \text{m/s}^2} \] \[ t = \frac{20}{1} \cdot \frac{3}{5} \, \text{s} \] \[ t = 12 \, \text{s} \][/tex]

c) While the car is accelerating along the ramp, the traffic on the freeway is moving at a constant speed of 20 m/s. The distance (d) the traffic travels in the time (t) the car takes to travel the ramp is given by:

[tex]\[ d = vt \] \[ d = (20 \, \text{m/s})(12 \, \text{s}) \] \[ d = 240 \, \text{m} \][/tex]

So, the traffic on the freeway travels 240 meters while the car is moving the length of the ramp.

The final answers are:

[tex]\[ \boxed{a \approx 1.67 \, \text{m/s}^2} \] \[ \boxed{t = 12 \, \text{s}} \] \[ \boxed{d = 240 \, \text{m}} \][/tex]

To solve this problem we will apply the concepts related to wavelength as the rate of change of the speed of the wave over the frequency. Mathematically this is

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

Here,

v = Wave velocity

f = Frequency,

Replacing with our values we have that,

[tex]\lambda = \frac{340}{500}[/tex]

\lambda = 0.68m

The distance to move one speaker is half this

[tex]\lambda/2 = 0.34m[/tex]

Therefore the minimum distance will be 0.34m

The minimum distance that one of the speakers should be moved back away from the listener to produce destructive interference at the listener's ear is; Δ = 0.34 m

The formula for wavelength here as it relates to speed and frequency is given as;

λ = v/f

Where;

λ is wavelength

v is speed

f is frequency

We are given;

Frequency; f = 500 Hz

Speed of sound; v = 340 m/s

Thus;

λ = 340/500

λ = 0.68 m

Now, we are told that the line separating them creates a constructive interference at the listeners ear. Thus;

To calculate the minimum distance would one of the speakers be moved back away from the listener to produce destructive interference at the listener's ear we will use the formula;

formula for destructive path length is;

Δ = (m + ½)λ

And m here is zero

Thus;

Δ = ½λ

Δ = 0.68/2

Δ = 0.34 m

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Answer:

The correct option is that both the ends will remain neutral.

Explanation:

As the rod is grounded all of the prolonged charge will be converted to ground so the overall charge on the rod in absence of the charged ball will be equal to zero.Such that the both ends will not bear any charge.

The correct option is: There is negative charge spread evenly on both ends.

 When a conducting rod is brought into contact with a charged ball, charge transfer occurs due to the process of induction and conduction. Initially, when the charged ball (let's assume it is positively charged) is brought close to the rod, it will induce a negative charge on the end of the rod nearest to the ball (end A) and a positive charge on the far end (end B) due to the separation of charges within the conductor. This is because the positive charge on the ball will repel the positive charges within the rod, pushing them away, and attract the negative charges, pulling them towards the ball.

 Once the rod is actually touched to the ball, the negative charges on the rod (which were induced by the ball's positive charge) will be neutralized by the positive charges from the ball, leaving the rod with a net negative charge after the ball is moved away. This negative charge will be evenly distributed along the rod initially.

 However, since the rod is not grounded, the negative charge will not have a path to escape and will remain on the rod. Due to the repulsion between like charges (negative-negative), these charges will repel each other and spread out as far as possible from each other along the length of the rod. This means that the negative charge will be spread evenly on both ends of the rod, as this arrangement minimizes the repulsive forces between the charges and results in the lowest potential energy configuration.

 Therefore, the final arrangement of any charge on the rod, when the charged ball is far away, will be an even distribution of negative charge on both ends A and B of the rod. This is consistent with the principle of electrostatic equilibrium, where charges within a conductor will redistribute themselves to minimize repulsion and achieve the lowest energy state.

Answer:

Time: 0.35 s

Explanation:

The position vector of the particle is

[tex]r=(2.7m/s^2)t^2 i+(5.1 m/s^3)t^3j[/tex]

where the first term is the x-component and the 2nd term is the y-component.

The particle's velocity vector is given by the derivative of the position vector, so:

[tex]v=r'(t)=(2.7\cdot 2)t i + (5.1\cdot 3)t^2 j=5.4t i+15.3t^2j[/tex]

The particle's velocity has an angle with the x-axis of 45 degrees when the x and the y component have same magnitude. Therefore:

[tex]5.4t=15.3t^2\\5.4=15.3t\\t=\frac{5.4}{15.3}=0.35 s[/tex]

The time t at which the angle between the particle's velocity and the x-axis is 45°, given the vectors x = (2.7 m/s^2)t^2 i and y = (5.1 m/s^3)t^3 j, is found to be 0.35 seconds.

To solve this problem, first we need to find the velocity of the particle in both the x and y directions. The velocity in the x-direction, vx, is just the derivative of x with respect to time, dx/dt = (2.7 m/s2)(2t) = 5.4t m/s. The velocity in the y-direction, vy, is the derivative of y with respect to time, dy/dt = (5.1 m/s3)(3t2) = 15.3t2 m/s.

Now, the angle θ between the velocity and the x-axis can be found by taking the tangent inverse of the ratio of vy to vx: θ = tan-1(vy/vx). We want this angle to be 45°, so we set this equation equal to 45° and solve for t. θ = 45° = tan-1((15.3t2 m/s) / (5.4t m/s)) => tan(45) = 15.3t2 / 5.4t => 1 = 15.3t / 5.4 => t = 5.4 / 15.3 s= 0.35s.

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Answer:

0.12959085 J

Explanation:

k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]

q = Charge = 1.55 μC

d = Distance between charge = 0.5 m

Electric potential energy is given by

[tex]U=k\dfrac{q^2}{d}[/tex]

In this system with three charges which are equidistant from each other

[tex]U=k\dfrac{q^2}{d}+k\dfrac{q^2}{d}+k\dfrac{q^2}{d}[/tex]

[tex]\\\Rightarrow U=k\dfrac{3q^2}{d}\\\Rightarrow U=8.99\times 10^9\times \dfrac{3\times (1.55\times 10^{-6})^2}{0.5}\\\Rightarrow U=0.12959085\ J[/tex]

The potential energy of the system is 0.12959085 J

Answer:

greatest speed=0.99c

least speed=0.283c

Explanation:

To solve this problem, we have to go to frame of center of mass.

Total available energy fo π + and π - mesons will be difference in their rest energy:

[tex]E_{0,K_{0} }-2E_{0,\pi } =497Mev-2*139.5Mev\\[/tex]

                       =218 Mev

now we have to assume that both meson have same kinetic energy so each will have K=109 Mev from following equation for kinetic energy we have,

K=(γ-1)[tex]E_{0,\pi }[/tex]

[tex]K=E_{0,\pi}(\frac{1}{\sqrt{1-\beta ^{2} } } -1)\\\frac{1}\sqrt{1-\beta ^{2} }}=\frac{K}{E_{0,\pi}}+1\\ {1-\beta ^{2}=\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta ^{2}=1-\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta = +-\sqrt{\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\\\\beta =+-\sqrt{1-\frac{1}{(\frac{109Mev}{139.5Mev+1)^2}}[/tex]

[tex]u'=+-0.283c[/tex]

note +-=±

To find speed least and greatest speed of meson we would use relativistic velocity addition equations:

[tex]u=\frac{u'+v}{1+\frac{v}{c^{2} } } u'\\u_{max} =\frac{u'_{+} +v_{} }{1+\frac{v}{c^{2} } } u'_{+} \\u_{max} =\frac{0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } 0.828\\ u_{max} =0.99c\\u_{min} =\frac{u'_{-} +v_{} }{1+\frac{v}{c^{2} } } u'_{-}\\u_{min} =\frac{-0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } -0.828c\\u_{min} =0.283c[/tex]

Final answer:

The greatest and least speeds of the π mesons resulting from the decay of a high-speed K0 meson are calculated using the relativistic laws of velocity addition. The maximum speed is (85/86)c, and the minimum speed is (5/14)c.

Explanation:

The question revolves around a high-speed K0 meson decaying into two π mesons (π+ and π-). To determine the greatest and least speeds that the π mesons can have, we employ the relativistic addition of velocities. The formula used in this context is Vx=(Vx'+V)/(1+(Vx'V/c²)), where V is the speed of the K0 meson, Vx' the speed of π mesons in the K0 meson's rest frame, and c is the speed of light.

The maximum speed is achieved when the π meson is emitted in the same direction as the K0 meson's movement, while the minimum speed occurs when the π meson is emitted in the opposite direction. Considering the K0 meson's speed (0.9c) and the π mesons' speed in the rest frame (0.8c), the equations result in the maximum speed being (85/86)c and the minimum speed being (5/14)c respectively.

The speed equals half of its maximum speed at positions ±1.50 cm from the equilibrium for an amplitude of 3.00 cm.

Speed equals half of its maximum speed when the particle is at a position where it is halfway through its amplitude.

This occurs at x = ±0.5X from the equilibrium position, considering X as the amplitude of the motion.

For an amplitude of 3.00 cm, the speed equals half of its maximum speed at position ±1.50 cm.

Answer:

The constant force exerted on the bullet is 1590.87 N.

Explanation:

It is given that,

Mass of the bullet, m = 8.6 g

Initial speed of the bullet, u = 0          

Final speed of the bullet, v = 430.1 m/s

We need to find the force exerted on the bullet while it is traveling down the 0.5 m length of the barrel of the rifle. Let a is the acceleration of the bullet. So,

[tex]v^2-u^2=2ad[/tex]

[tex]v^2=2ad[/tex]

[tex]a=\dfrac{v^2}{2d}[/tex]

[tex]a=\dfrac{(430.1)^2}{2\times 0.5}[/tex]

[tex]a=184986.01\ m/s^2[/tex]

Let F is the force exerted. It is given by :

[tex]F=ma[/tex]

[tex]F=8.6\times 10^{-3}\times 184986.01[/tex]

F = 1590.87 N

So, the constant force exerted on the bullet is 1590.87 N. Hence, this is the required solution.                                                  

Explanation:

As the force is given as 5.30 kN or [tex]5.30 \times 1000 N[/tex]. Hence, mass will be calculated as follows.

           [tex]F_{w}[/tex] = mg

         [tex]5.30 \times 10^{3} = m \times 9.8 m/s^{2}[/tex]

                   m = 540.816 kg

(a)  At the top, centripetal force [tex]F_{c}[/tex] is acting upwards and the weight of the riders and car, [tex]F_{w}[/tex] will be acting downwards.

Therefore, force on the car by the boom will be calculated as follows.

                   [tex]F_{B} = F_{w} - F_{c}[/tex]

or,          [tex]F_{B} = mg - \frac{mv^{2}}{r}[/tex]

                         = [tex]5.30 \times 1000 N - \frac{540.816 kg \times (3.60)^{2}}{12}[/tex]

                         = 4715.919 N

Hence, the force [tex]F_{B}[/tex] on the car from the boom is 4715.919 N. This means that the car will be hanging on the boom and the boom will exert an upward force.

(b)   Now at the top, centripetal force [tex]F_{c}[/tex] will be acting upwards and the weight of cars and car riders will be acting in the downwards direction.

Hence, we will calculate the force on car by the boom as follows.

       [tex]F_{B} = F_{w} - F_{c}[/tex]

or,         [tex]F_{B} = mg - \frac{mv^{2}}{r}[/tex]

                        = [tex]5.30 \times 1000 N - \frac{540.816 kg \times (14.0)^{2}}{12}[/tex]

                        = -3533.33 N

Therefore, car will be pushing on the boom and the boom will exert a downward force.  

The average acceleration from t=0 to t=5.00 s is 0.5 m/s^2. instantaneous acceleration at t=0 is 0 m/s^2 and at t=5.00 s is 1.0 m/s^2. The velocity-time graph is parabolic, and the acceleration-time graph shows linear growth.

The velocity of a car as a function of time is given by vx(t) = \\(\alpha + \beta t^2\\), where \\(\alpha = 3.00 m/s\\) and \beta = 0.100 m/s^3. To find the average acceleration for the time interval from t = 0 to t = 5.00 s, we need the change in velocity over the time period. Using the given function, we can calculate the initial velocity at t = 0, which is v(0) = 3.00 m/s, and the final velocity at t = 5.00 s, which is v(5) = 3.00 + 0.100(5)^2 = 3.00 + 2.5 = 5.5 m/s. The average acceleration is then (5.5 m/s - 3.0 m/s) / (5.0 s - 0 s) = 0.5 m/s^2.

For instantaneous acceleration at any given time, we differentiate the velocity function with respect to time to obtain a(t) = dvx/dt = 2\beta t. Therefore, the instantaneous acceleration at t = 0 is a(0) = 2(0.100)(0) = 0 m/s^2, and at t = 5.00 s is a(5) = 2(0.100)(5) = 1.0 m/s^2.

When drawing the vx-t and ax-t graphs, for the velocity-time graph, the curve will start at 3.00 m/s and rise parabolically due to the \(t^2\) term in the function. For the acceleration-time graph, the line will start at 0 m/s^2 and increase linearly with time, passing through 1.0 m/s^2 at 5 seconds.

Answer:

The amplitude of the resultant wave will be 0.

Explanation:

Suppose the first wave has an amplitude of A. Its angle is given as wt.

The second way will also have the same amplitude as that of first.

After the reflection, a phase shift of π is added So the wave is given as

[tex]W_1=W_2=Acos(\omega t)\\W_1^{'}=Acos(\omega t+ \pi)[/tex]

Adding the two waves give

                               [tex]W_1'+W_2=Acos(\omega t+ \pi)+Acos(\omega t)\\W_1'+W_2=-Acos(\omega t)+Acos(\omega t)\\W_1'+W_2=0[/tex]

So the amplitude of the resultant wave will be 0.

Answer:

[tex]s=274.2857\ m[/tex] is the distance from the target before which the pilot must release the canister.

Explanation:

Given:

Time taken by the canister to hit the ground:

using equation of motion

[tex]h=u_y.t+\frac{1}{2} g.t^2[/tex]

where:

[tex]u_y=[/tex] initial vertical velocity of the canister = 0 (since the the object is dropped from a horizontally moving plane)

[tex]t=[/tex] time taken to hit the ground

[tex]90=0+0.5\times 9.8\times t^2[/tex]

[tex]t=4.2857\ s[/tex]

Now the horizontal distance travelled by the canister after dropping:

[tex]s=v_x\times t[/tex]

[tex]s=64\times 4.2857[/tex]

[tex]s=274.2857\ m[/tex] is the distance from the target before which the pilot must release the canister.

Answer:

No. The crown is made of lead

Explanation:

From Archimedes principle,

D/D' = W/U

Where D = density of the crown, D' = Density of water, W = weight of the crown in air, U = Upthrust of the crown in water.

make D the subject of the equation,

D = D'(W/U)................... Equation 1

W = mg

Where g = 9.8 m/s², m = 14.7 kg.

W = 14.7(9.8) = 144.06 N.

U = 9.8(14.7-13.4) = 12.74 N.

Constant: D' = 1000 kg/m³

Substitute into equation 1

D = 1000(144.06/12.74)

D = 11307.69 kg/m³

Density of the crown = 11307.69 kg/m³

And Density of Gold = 19300 kg/m³

From the above, The crown is not made of Gold

To determine if the crown is made of gold, we need to use Archimedes' principle and calculate its density. If the density is greater than the density of water, the crown is made of gold.

When a crown of mass 14.7kg is completely submerged in water, an accurate scale reading reads only 13.4kg. To determine if the crown is made of gold, we need to use Archimedes' principle. According to this principle, when an object is submerged in a fluid, it experiences an upward buoyant force equal to the weight of the fluid displaced by the object.

If the crown is made of gold, its density should be equal to or greater than the density of water. We can calculate the density of the crown by dividing its mass by the volume of water it displaces. If the density of the crown is greater than the density of water (1kg/L), then the crown is made of gold.

In this case, the mass of the crown is 14.7kg, and the mass of the water displaced is 13.4kg (since the scale reading is lower). By using the equation density = mass/volume, we can calculate the volume of water displaced by the crown and determine its density.

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Answer:

c. Friction will be in the opposite direction of the block's motion because molecular interactions between the block and the floor will deform the block in the opposite direction of motion.

Explanation:

When a block is put on the floor , there is interaction at molecular level between the molecules of block and floor . This mutual interaction

( attraction ) pulls them together. When the body is pushed forward, the distance between molecules is increased . Due to mutual attraction , block is deformed at the interface in the opposite direction of motion . This strain causes restoring stress which is also called friction . This elastic  stress creates force on the block in backward direction and force on the floor in forward direction .

Final answer:

The correct statement is C. Friction acts in the opposite direction of the block's motion due to molecular interactions between the block and the floor, and it always opposes the motion to slow down or stop the block.

Explanation:

The correct answer to the question, "Which statement best explains the direction of the force that friction applies on the moving block is that friction will be in the opposite direction of the block's motion. This occurs because friction always opposes the direction of motion to prevent or slow down the motion of the block. The encounter between the block and the floor's surface creates molecular interactions that resist the block's movement, and thus, the force of friction acts in the opposite direction to the motion of the block.

Moreover, it's essential to clarify that while friction does generate thermal energy due to the interactions at the interface between the block and the floor, this thermal energy does not add kinetic energy to the block or convert kinetic energy into potential energy in the context of motion. Instead, it's a byproduct of the frictional force's work against the block's motion.