Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 4.0 cm . Two of the particles have a negative charge: q1

Answer:

The star will have a transverse speed of 315950.9 AU/year

Explanation:

d = 1/p

d = distance to star, measured in parsecs

p = parallax, measured in arcseconds = 0.5 arcsec/year

So, d = 1/0.5 = 2 parce

1 parsec = 3.26 light years

2 parce = 6.52 light years

⇒Transverse speed in AU / year = Distance/parallax

distance = 10pc = 2060000 AU

Transverse speed in AU / year =  2060000 Au/6.52 light years

Transverse speed  = 315950.9 AU/year

Therefore, A star (not Barnard's star) at a distance of 10 pc  observed to have a proper motion of 0.5 arcsec / year. Will have a transverse speed of 315950.9 AU/year

For the sinusoidal wave described by y = 0.25 sin (0.30x - 40t), the amplitude is 0.25 meters, angular frequency is 40 rad/s, the angular wave number is 0.30 rad/m and the wavelength is approximately 20.94 meters.

The given function for the sinusoidal wave is y = 0.25 sin (0.30x - 40t). We can extract the details of this wave from this function:

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Answer:

v = -4.22 x 10⁻⁴ m/s

Explanation:

given,

measured wavelength = 497.15 nm

Normally wavelength = 497.22 nm

Change in wavelength

Δ λ = 497.15 - 497.22

Δ λ = -0.07 nm

using Doppler's equation

[tex]\dfrac{\Delta \lambda}{\lambda}=\dfrac{v}{c}[/tex]

v is the speed of the star

c is the speed of light

[tex]\dfrac{-0.07\ nm}{497.22\ nm}=\dfrac{v}{3\times 10^8}[/tex]

      v = -4.22 x 10⁻⁴ m/s

Speed of the star moving is equal to v = -4.22 x 10⁻⁴ m/s

The speed of the star with respect to the Earth is -4.22× 10⁻⁴ m/s m/s. The negative sign indicates the star is moving away.

Given:

Observed wavelength (λ) = 497.15 nm

Rest wavelength (λ₀) = 497.22 nm

To calculate the speed of a star with respect to Earth. Doppler effect technique can be used.  The Doppler effect gives the relation between wavelength, speed, and speed of light.

The formula for the Doppler shift is given as:

Δλ / λ₀ = v / c

Δλ = λ - λ₀

Δλ = (497.15 x 10⁻⁹ m) - (497.22 x 10⁻⁹ m)

Δλ = -0.07 x 10⁻⁹ m

The speed of the star is evaluated as:

v = (Δλ / λ₀) x c

v = (-0.07 x 10⁻⁹ m / 497.22 x 10⁻⁹ m) x 299,792,458 m/s

v = -4.22× 10⁻⁴ m/s

Hence, the speed of the star with respect to the Earth is -4.22× 10⁻⁴ m/s m/s. The negative sign indicates the star is moving away.

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Answer

given,

storage coefficient, S = 6.8 x 10⁻⁴

thickness of aquifer, t = 50 m

porosity of the aquifer, n = 25 % = 0.25

Density of the water, γ = 9810 N/m³

Compressibilty  of water,β = 4.673 x 10⁻¹⁰ m²/N

We know,

   S = γ t(nβ + α)

where, α is the compressibility of the aquifer

   6.8 x 10⁻⁴  =9810 x 50 x (0.25 x 4.673 x 10⁻¹⁰+ α)

     α = 1.269 x 10⁻⁹ m²/N

Expansability of water

            = n t β γ

            = 0.25 x 50 x 4.673 x 10⁻¹⁰ x 9810

            = 5.73 x 10⁻⁵

Answer:

(a) 2.154×10⁻¹¹ A.

(b)  98300000.

Explanation:

(a)

Using Ohm's law,

V = IR ........................ Equation 1

Where V = Potential difference between the walls, I = current, R = Resistance between the walls.

Make I the subject of the equation

I = V/R...................... Equation 2

Given: V = 84 mV, = 0.084 V, R = 3.9×10⁹ Ω.

Substitute into equation 2

I = 0.084/(3.9×10⁹)

I = 2.154×10⁻¹¹ A.

(b)

I = q/t

q = It .................... Equation 1

Where q = quantity of electric charge, t = time.

Given: I = 2.154×10⁻¹¹ A, t = 0.73 s.

q = 2.154×10⁻¹¹×0.73

q = 1.572×10⁻¹¹ C.

The charge on an electron e = 1.6×10⁻¹⁹ C

n = q/e

where n = number of ions.

n = 1.572×10⁻¹¹/1.6×10⁻¹⁹

n = 9.83×10⁷

n = 98300000.

Using the concept of Newton's second law, as the elevator starts from rest and travels upward, the scale shows a value of 94.04 kg

Acceleration is the time derivative of velocity. Therefore acceleration can be given by;

[tex]a=a(t)=\frac{dv(t)}{dt} = \frac{d}{dt} (3t+0.2t^2) = 3+0.4t[/tex]

When t = 3s;

[tex]a= 3+(0.4\times 4)=4.6\,m/s^2[/tex]

As the elevator is moving upward the net force on the weighing scale is given using Newton's second law of motion;

[tex]F_{net}=m(a+g) =(64\,kg)\times (9.8\,m/s^2 +4.6\,m/s^2)=921.6\,N[/tex]

A weighing scale is usually caliberated to show the value in kilograms.

Therefore the weighing scale will show the reading;

[tex]m=\frac{F_{net}}{g} =\frac{921.6\,N}{9.8\,m/s^2}= 94.04\,kg[/tex]

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The apparent weight you would experience in the accelerating elevator is 925.12N. This higher reading is due to the addition of the upward acceleration of the elevator to the standard pull of gravity.

To solve this question, we first need to find the elevator's acceleration. The acceleration is the derivative of the velocity: v(t)=(3.0m/s²)t+(0.20m/s³)t² with respect to time t. The derivative is a(t) = 3.0m/s² +2×(0.20m/s³)×t.

Substitute t=4.0s into a(t) to get a(4.0s) = 3.0m/s² + 2×0.2×4.0 = 4.6m/s². This is the acceleration at t=4.0s.

The apparent weight, or reading on the scale, is given by the equation: F=m(g+a), where m is the mass (64 kg), g is the acceleration due to gravity (9.8 m/s²), and a is the acceleration of the elevator. Thus the apparent weight is F=64(9.8+4.6)=925.12 N.

This is your weight when the elevator is accelerating upward; you would feel heavier due to the addition of the elevator's upward acceleration to the natural gravitational pull.

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Answer:

His pitching speed is 38 m/s.

Explanation:

Hi there!

Please see the attached figure for a better understanding of the problem.

The position of the ball at any time t is given by the following vector:

r = (x0 + v0 · t, y0 + 1/2 · g · t²)

Where:

r = position vector of the ball at time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Let's place the origin of the frame of reference at the throwing point so that x0  and y0 = 0.

When the ball reaches the ground, its position vector will be r1 (see figure). Using the equation of the vertical component of the position vector, we can find the time at which the ball reaches the ground. At that time, the horizontal component of the position is 30 m and the vertical component is -3.0 m (see figure):

y = y0 + 1/2 · g · t²  (y0 = 0)

y = 1/2 · g · t²

-3.0 m = 1/2 · (-9.8 m/s²) · t²

-3.0 m / -4.9 m/s² = t²

t = 0.78 s

Now, knowing that at this time x = 30 m, we can find v0:

x = x0 + v0 · t  (x0 = 0)

x = v0 · t

30 m = v0 · 0.78 s

v0 = 30 m / 0.78 s

v0 = 38 m/s

His pitching speed is 38 m/s.

The pitching speed can be calculated using the horizontal distance and the height of the ledge. By considering the horizontal motion and the effects of gravity, the time taken can be determined. Substituting the given values into the appropriate equations, the pitching speed can be calculated as 38 m/s.

The pitching speed can be calculated using the horizontal distance and the height of the ledge. To find the initial velocity, we can use the equation:

Vox = d / t

where Vox is the horizontal component of velocity, d is the horizontal distance, and t is the time taken. Since the ball is thrown horizontally, the vertical component of velocity is zero, and the only force acting on the ball is gravity in the downward direction. Therefore, we can use the equation:

[tex]h = 0.5 * g * t^2[/tex]

where h is the height of the ledge, g is the acceleration due to gravity, and t is the time taken. By rearranging the equation, we can find the time taken:

t = sqrt(2 * h / g)

Substituting the values given in the question, we can calculate the pitching speed:

Vox = d / t = d * sqrt(g / (2 * h))

Using the values d = 30m, h = 3.0m, and g = 9.8m/s^2, we can find the pitching speed:

Vox = 30m * [tex]sqrt(9.8m/s^2[/tex]) = 38 m/s

Answer:

Part a: The value of yield strength at 0.2% offset is obtained from the engineering stress-strain curve which is given as 274 MPa.

Part b: The value of tensile strength  is obtained from the engineering stress-strain curve which is given as 417 MPa

Part c: The value of Young's modulus at given point is 172 GPa.

Part d: The percentage elongation is 18.55%.

Part e: The percentage reduction in area is 15.81%

Explanation:

From the complete question the data is provided for various Loads in ductile testing machine for a sample of d0=20 mm and l0=40mm. The plot is drawn between stress and strain whose values are calculated using following formulae. The corresponding values are attached with the solution.

The engineering-stress is given as

[tex]\sigma=\frac{F}{A}\\\sigma=\frac{F}{\pi \frac{d_0^2}{4}}\\\sigma=\frac{F}{\pi \frac{(20 \times 10^-3)^2}{4}}\\\sigma=\frac{F}{3.14 \times 10^{-4}}[/tex]    

Here F are different values of the load

Now Strain is given as

[tex]\epsilon=\frac{l-l_0}{l_0}\\\epsilon=\frac{\Delta l}{40}\\[/tex]

So the curve is plotted and is attached.

The value of yield strength at 0.2% offset is obtained from the engineering stress-strain curve which is given as 274 MPa.

The value of tensile strength  is obtained from the engineering stress-strain curve which is given as 417 MPa

Young's Modulus is given as

[tex]E=\frac{\sigma}{\epsilon}\\E=\frac{238 /times 10^6}{0.00138}\\E=172,000 MPa\\E=172 GPa[/tex]

The value of Young's modulus at given point is 172 GPa.

The percentage elongation is given as

[tex]Elongation=\frac{l_f-l_0}{l_0} \times 100\\Elongation=\frac{47.42-40}{40}\times 100\\Elongation=18.55 \%\\[/tex]

So the percentage elongation is 18.55%

The reduction in area is given as

[tex]Reduction=\frac{A_0-A_n}{A_0} \times 100\\Reduction=\frac{\pi \frac{d_0^2}{4}-\pi \frac{d_n^2}{4}}{\pi \frac{d_0^2}{4}}\times 100\\Reduction=\frac{{d_0^2}-{d_n^2}}{{d_0^2}} \times 100\\Reduction=\frac{{20^2}-{18.35^2}}{{20^2}} \times 100\\Reduction=15.81\%[/tex]

So the reduction in area is 15.81%

To determine the engineering properties, knowledge of additional values is needed. These properties include 0.2% offset yield strength, tensile strength, modulus of elasticity, % elongation, and % reduction in area, derived through respective formulas. The engineering stress-strain curve can be plotted with these details.

To calculate the engineering properties asked in your question some additional values such as the original length and diameter, the load at yield point, the maximum load sustained, and the length and diameter after fracture are required. However, the basic formulas for the calculations are as follows

Please note that the engineering stress-strain curve should be plotted after obtaining these values with stress on the Y-axis and strain on the X-axis. The curve typically starts from the origin, goes linearly upwards till the yield point (Proportional limit), followed by a non-linear portion (elastic limit), and reaches maximum at tensile strength, after which it falls down to the fracture point.

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Answer:

a. The apparent depth of the object is less than the real depth.

Explanation:

When we observe for any object lying at the bottom of the pool from the edge of the pool then we are actually viewing an object from an optically rarer medium into an optically denser medium.

The schematic shows the apparent view of the object due to the bending of the rays coming form the object to our eyes.

The rays when coming from a denser medium to a rarer medium they bend away from the normal of the interface.

The correct option is a. The apparent depth of the object is less than the real depth because of the refraction of light at the water-air interface.

Light bends away from the normal as it exits the water, making the object appear shallower than it actually is. When light travels from water to air, it bends away from the normal because water is denser than air.

This bending makes the object appear to be at a shallower depth than it actually is. The apparent depth of the object is therefore less than the real depth of the object.

To summarize, the correct statement is: The apparent depth of the object is less than the real depth. Option a is correct.

Answer : The volume of the tank is, 1.54 mL

Explanation :

To calculate the volume of gas we are using ideal gas equation:

[tex]PV=nRT\\\\PV=\frac{w}{M}RT[/tex]

where,

P = pressure of gas = 300 kPa = 2.96 atm

Conversion used : (1 atm = 101.325 kPa)

V = volume of gas = ?

T = temperature of gas = [tex]77^oC=273+77=350K[/tex]

R = gas constant = 0.0821 L.atm/mole.K

w = mass of gas = 4.6 kg  = 4600 g

M = molar mass of air = 28.96 g/mole

Now put all the given values in the ideal gas equation, we get:

[tex](2.96atm)\times V=\frac{4600g}{28.96g/mole}\times (0.0821L.atm/mole.K)\times (350K)[/tex]

[tex]V=1541.98L=1.54mL[/tex]        (1 L = 1000 mL)

Therefore, the volume of the tank is, 1.54 mL

Final answer:

For a step-up transformer with 22 primary turns and 800 secondary turns, to produce an output of 5300 V at 16 mA, the required input voltage is 145.75 volts and the input current needed is 0.5818 amperes.

Explanation:

Calculating Input Current and Voltage for a Step-Up Transformer

The student's question involves a step-up transformer with a known number of turns in the primary and secondary coils, a given secondary voltage, and a secondary current. To find the required input current and voltage, we can use the transformer equations that relate the primary and secondary sides of the transformer:

 Primary voltage (VP) / Secondary voltage (VS) = Number of turns in the primary coil (NP) / Number of turns in the secondary coil (NS)

 Primary current (IP) * Number of turns in the primary coil (NP) = Secondary current (IS) * Number of turns in the secondary coil (NS)

We're given:

 NP = 22 turns

 NS = 800 turns

 IS = 16 mA = 0.016 A

 VS = 5300 V

To find the input voltage VP:

VP = (NP / NS) * VS = (22 / 800) * 5300 V = 145.75 V

To find the input current IP:

IP = (NS / NP) * IS = (800 / 22) * 0.016 A = 0.5818 A

Therefore, the required input voltage is 145.75 volts, and the required input current is 0.5818 amperes.

Final answer:

The diameter of the iron ball bearing, which weighs 21.91 grams and is composed of iron atoms organized in a BCC structure, is roughly 1.62 cm.

Explanation:

To derive the diameter of a spherical iron ball bearing weighing 21.91 grams, given that iron atoms have an effective radius of 0.124 nm and are arranged in a Body-Centered Cubic (BCC) structure at room temperature, we need to calculate the volume of the iron ball and then find the diameter using the volume of a sphere formula. First, we will use the density of iron (7.9 g/cm³) to find the volume of the ball bearing:

V = mass / density = 21.91 g / 7.9 g/cm³ = 2.77342 cm³

Next, we use the volume of a sphere formula V = (4/3)πr³, where V is the volume and r is the radius, to find the diameter (d = 2r):

r³ = V / ((4/3)π) = 2.77342 cm³ / ((4/3)π) ≈ 0.52733 cm³

r ≈ 0.8092 cm

d = 2 * r ≈ 2 * 0.8092 cm ≈ 1.6184 cm

Therefore, the estimated diameter of the iron ball bearing is approximately 1.62 cm.

Answer:

The new position is 0.1865 m

Explanation:

As the context of the data is not available, thus following data is utilized from the question as attached above

x_relax=0.32 m

x_stiff=0.13 m

spring stiffness k=9 N/m

mass of block =0.073 kg

t=0.07 s

Velocity of the block is to be estimated thus

Force due to compression in spring is given as

F_s=k Δx

F_s=9(0.32-0.13)

F_s=1.71 N

Force on the block is given as

F_m=mg

F_m=0.073 x 9.8

F_m=0.71 N

Net Force

F=F_s-F_m

F=1.71-0.71 N

F=1 N

As Ft=Δp

So

Δp=1x0.07=0.07 kgm/s

Δp=p_final-p_initial

0.07=p_final-0

p_final=0.07 kgm/s

p_final=m*v_f

v_f=(p_final)/(m)

v_f=0.07/0.073

v_f=0.95 m/s

So now the velocity of the block is 0.95 m/s

time is 0.07 s

y_new=y_initial+y_travel

y_new=0.12+(0.95 x 0.07)

y_new=0.12+0.065

y_new=0.1865 m

So the new position is 0.1865 m

Final answer:

The new position of the bottom of the block at time t = 0.07 seconds is 1 m.

Explanation:

To find the new position of the bottom of the block at time t = 0.07 seconds, we can use the concept of average velocity. The average velocity is given by the change in position divided by the change in time. In this case, if we approximate the average velocity in the first time interval by the final velocity, we can say that the change in position is equal to the average velocity multiplied by the change in time. The new position can then be calculated by adding this change in position to the initial position.

Given that the initial position of the bottom of the block is at y = 0.12 m and the final velocity is approximated to be y = 1 m, we can calculate the change in position as:

Change in position = (Final velocity - Average velocity) * Change in time = (1 m - 0.12 m) * (0.07 s - 0 s) = 0.88 m

Therefore, the new position of the bottom of the block at time t = 0.07 seconds is y = 0.12 m + 0.88 m = 1 m.

Answer:

Explanation:

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Answer: the average speed of the rat from the information given above is 0.7m/s

Explanation:

position is given as

x(t) = pt² + qt

finding the diffencial of x(t) with respect to t, we have

d(x(t))/dt = 2pt + q

we substitute the p = 0.36m/s² and q= -1.10 m/s

d(x(t))/dt = 2(0.36)t + (-1.10)

so, at t= 1s

d(x(t))/dt = 2*(0.36)*1 - 1.1 = 0.72 - 1.1 = -0.38m/s

at t= 4s

d(x(t))/dt = 2*(0.36)*4 - 1.10 = 2.88 - 1.10 = 1.78 m/s

To find the average speed,

average speed = (V1 + V2)/ 2

average speed = (1.78 + (-0.38))/2 = 0.7m/s

The speed is defined as the distance per unit of time. The unit of speed is m/s. The speed is a scalar quantity which means it only depends on the magnitude.

According to the question, The average speed of the mouse is 0.7m/s

The solution of the question is as follows:-

The required equation is:-

[tex]x(t) = pt^2 + qt[/tex]

The Finding the differential of x(t) with respect to t, we have

[tex]\frac{dxt}{dt} = 2pt + q[/tex]

Put the value p = 0.36m/s² and q= -1.10 m/s

[tex]\frac{d(x(t)}{dt} = 2(0.36)t + (-1.10)[/tex]

so, at t= 1s

After solving it [tex]2*(0.36)*1 - 1.1 = 0.72 - 1.1 = -0.38m/s[/tex]

so,at t= 4s

After solving it =[tex]2*(0.36)*4 - 1.10 = 2.88 - 1.10 = 1.78 m/s[/tex]

The formula of average speed = [tex]\frac{(V1 + V2)}{2}[/tex]

[tex]= \frac{(1.78 + (-0.38))}{2} = 0.7m/s[/tex]

Hence, the average speed is 0.7m/s

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Answer:

d = 0.38 m

Explanation:

As we know that the person due to the airbag action, comes to a complete stop, in 36 msec or less, and during this time, is decelerated at a constant rate of 60 g, we can find the initial velocity (when airbag starts to work), as follows:

vf = v₀ -a*t  

If vf = 0, we can solve for v₀:

v₀ = a*t = 60*9.8 m/s²*36*10⁻³s = 21.2 m/s

With the values of v₀, a and t, we can find Δx, applying any kinematic equation that relates all of some of these parameters with the displacement.

Just for simplicity, we can use the following equation:

[tex]vf^{2} -vo^{2} = 2*a*d[/tex]

where vf=0, v₀ =21.2 m/s and a= -588 m/s².

Solving for  d:

[tex]d = \frac{-vo^{2}}{2*a} = \frac{(21.2m/s)^{2} }{2*588 m/s2} =0.38 m[/tex]

⇒ d = 0.38 m

Answer:

A person travels 39 cm in coming to a complete stop in 36 ms at a constant acceleration of 60 g.

Explanation:

Hi there!

The equation of position of an object moving in a straight line at constant acceleration is the following:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position at time t.

x0 = initial position.

v0 = initial velocity.

a = acceleration.

t = time.

So, let's see how much distance the person moves inside the car. Let's imagine that the person is initially at rest and suddenly is accelerated at 60 g (60 · 10 m/s² = 600 m/s²). In this case, x0 and v0 = 0 and the traveled distance will be:

x = 1/2 · 600 m/s² · (0.036)²

x = 0.39 m or 39 cm

Here, we have calculated the distance traveled by a person accelerated at 60 g from rest in 36 ms. Notice that the distance is the same if we calculate the traveled distance of a person that is brought to rest in 36 ms with an acceleration of 60 g.

A person travels 39 cm in coming to a complete stop in 36 ms at a constant acceleration of 60 g.

Answer:

changing the direction of the electric potential, we can get the particle to be in balance between the electric force, the weight and the thrust.

Explanation:

When the particle is removed from the wire, friction can be electrically charged, either with negative charges (extra electrons) or with positive charge by electron removal, in this case when the particle is between the condenser plates it experiences a force due to the electric field given by

          ΔV = E d

Where ΔV is the potential difference, d the distance between the plates and E the electric field.

In these cases we can use Newton's second law, where the acceleration is zero

                [tex]F_{e}[/tex] –W + B = 0

                 F_{e} = W –B

                 q E = mg - ρ_air g V

dodne B is the hydrostatic thrust

if  we know the density of the particular

                   ρ_particle = m / V

                   m = ρ_particle V

We replace

                 q E = g v (ρ_particle - ρ_air)

Therefore, by changing the direction of the electric potential, we can get the particle to be in balance between the electric force, the weight and the thrust.

Using kinematics equations, we can calculate the time for a rock to fall from the cliff on Earth. Without the value of acceleration due to gravity on the extrasolar planet, we can't quantify the time difference. A difference would exist if the gravitational accelerations of the two planets differ.

This question pertains to kinematics and physical laws regarding free fall. On earth, when the astronaut throws a rock straight up with a certain velocity, initially, gravity will slow down the rock and finally make it fall back toward the ground. This action is governed by the equation of motion: h = vit + 1/2gt² where h is the height, vi the initial velocity, t the time and g the acceleration due to gravity on earth (-9.81 m/s²).

In the extrasolar planet scenario, considering it is not mentioned, we could assume that the acceleration due to gravity might be similar to earth's. However, without specific details about the gravity, we can still calculate the time it takes for the rock to land from the 50m high cliff using the motion equation for a free-falling body. We can conclude that if the gravity in the two settings is different, then there would be a time difference. We can’t quantify the time difference without knowing the value of the gravity on the extrasolar planet.

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Answer

given,

Speed of the Aircraft,v = 175 mi/h

1 mi/h = 0.44704 m/s

175 mi/h = 78.232 m/s

time, t = 2.5 s

a) average acceleration = ?

[tex]a= \dfrac{v - u}{t}[/tex]

[tex]a= \dfrac{78.232 - 0}{2.5}[/tex]

a = 31.29 m/s²

b) Distance traveled by the Pane

using equation of motion

v² = u² + 2 a s

78.232² = 0² + 2 x 31.29 x s

s = 97.79 m

Distance moved by the plane is equal to 97.79 m

Answer:

W = 1.44 10⁻⁷ J

Explanation:

The expression for the job is

          W = ∫ F. dx

Where the point is the scalar product in this case the direction of the meteor and the depth is parallel, whereby the scalar product is reduced to the ordinary product

          W = 630 ∫ x³ dx

          W = 630 x⁴ / 4

Let's evaluate between the lower limit x = 0, w = 0 to the upper limite the point at x = 5.5 10⁻³ m

            W = 157.5 ((5.5 10⁻³)⁴ -0)

            W = 1.44 10⁻⁷ J

To solve this problem we will apply the concepts related to Newton's second law, which defines force as the product between mass and acceleration. Mathematically this can be described as,

[tex]F = ma[/tex]

Here,

m = Mass

a = Acceleration

Taking as reference the mass of the second boxcar, the force applied would be

[tex]F = m_2 a[/tex]

[tex]F = (16300kg)(0.569m/s)[/tex]

[tex]F = 9274.7N[/tex]

Therefore the boxcars pull on each other with a force of 9274.7N

The boxcars pull on each other with equal and opposite forces, which can be calculated using the formula F = ma. The force that the boxcars pull on each other is 16521.1 N.

When the locomotive pulls the first boxcar, it exerts a force on it. According to Newton's third law of motion, the first boxcar exerts an equal and opposite force on the locomotive. Similarly, when the first boxcar pulls the second boxcar, it exerts a force on the second boxcar, and the second boxcar exerts an equal and opposite force on the first boxcar. Therefore, the boxcars pull on each other with equal and opposite forces.

To calculate the magnitude of this force, we can use the formula F = ma, where F is the force, m is the mass, and a is the acceleration. In this case, the mass of the first and second boxcars together is 12,700 kg + 16,300 kg = 29,000 kg. Plugging in the values, we get F = (29,000 kg)(0.569 m/s²).

The force that the boxcars pull on each other is 16521.1 N (rounded to four significant figures).

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One of the characteristics of the luminous gas clouds is that they do not have direct affectation by some type of external electric or magnetic fields.

In addition, we must bear in mind that color is a variable that is depending on the gas in the mixture. Therefore its relationship with spectroscopy allows us to deduce that scientists take advantage of the wavelength spectrum to know the type of composition of one of the clouds. The speed of a cloud is measured by determining the Doppler shift of its spectral lines. From wine's law, wavelength of light emitting from the object depends on temperature of object

Therefore the correct option is D

Astronomers analyze starlight to obtain various pieces of information about stars, including their temperature, composition, and motion. The correct answer is (d) all of the above.

(a) Temperature: By examining the spectrum of starlight, astronomers can analyze the distribution of wavelengths or colors present in the light. The temperature of a star affects the intensity and distribution of light at different wavelengths.

(b) Composition: The spectrum of starlight also provides information about the chemical composition of stars. Different elements and molecules in a star's atmosphere absorb or emit light at specific wavelengths, creating characteristic absorption or emission lines in the spectrum.

(c) Motion: Through the analysis of starlight, astronomers can also determine the motion of stars. By studying the Doppler effect on spectral lines, which causes a shift in wavelength due to the motion of a star toward or away from Earth, astronomers can measure a star's radial velocity.

Therefore, by analyzing starlight, astronomers can gather information about a star's temperature, composition, and motion, making option (d) all of the above the correct choice

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Explanation:

The temperature of a fluid is proportional to the average kinetic energy of its molecules, since the room temperature is lower than the temperature in the boiling point, the energy that the water must overcome to become steam is greater. Therefore, the heat of vaporization will be greater.

Answer:

E. Energy

Explanation:

Electron volt is a unit of energy commonly used in various branches of physics.

It is defined as the energy gained by an electron when the electrical potential of the electron increases by one volt.

The electron volt = 1.602 × 10^−12 erg, or 1.602 × 10^−19 joule

Answer:

The electron-volt is a unit of A. charge. B. electric potential. C. electric field. D. electric force. E. energy.

The answer is option E (energy)

Explanation:

There are many forms of energy which are divided into Potential Energy, Kinetic energy and the major energy sources are nonrenewable and renewable sources. Energy makes change; it does things for us. Since energy is a fundamental physical quantity, it is a property of matter that can be converted into work, heat or radiation. Energy can be converted from one form to another, but it cannot be created, nor can it be destroyed.

Energy conversion is essential for energy utilization. Energy can be measured in many different units which includes joules, calories, electron-volts, kilowatt-hours, and so many more.

The electron-volt is a unit used to measure the energy of subatomic particles. The electron-volt, symbol eV, can be defined as the amount of energy gained by the charge of a single electron (a charged particle carrying unit electronic charge) moved across an electric potential difference of one volt. One electron-volt, eV is equal to 1.602176634×10−19 J. Where J is in joules.

Answer:

48 m

Explanation:

As she travels at the rate of 5m/s due north, the amount of time it would take for her to cross the 80m wide river would be

t = 80 / 5 = 16 seconds

This is also the time it takes for the river to push her to the east side at the rate of 3m/s. So after 16 seconds, she would reach the opposite point at a horizontal distance from her starting of

s = 16*3 = 48 m

Answer:

Part a: The schematic diagram is attached.

Part b: The pressure at the end is 1 bar.

Part c: Weight of 0.5kmol of water is 88.2 N.

Part d: The specific volume is 0.001 m^3/kg

Explanation:

The schematic is given in the diagram attached.

Pressure is given using the ideal gas equation as

Here

                      [tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\frac{1\times V_1}{300}=\frac{P_2\times 2V_1}{600}\\P_2=\frac{600}{600}\\P_2=1 bar[/tex]

So the pressure at the end is 1 bar.

Mass of 0.5kmol is given as follows

                              [tex]Mass=n_{moles} \times Molar \, Mass\\Mass=0.5 \times 10^3 \times 18 \times 10^{-3}\\Mass=9.0 kg[/tex]

Weight is given as

                           [tex]W=mxg\\W=9 \times 9.8\\W=88.2 \, N[/tex]

So weight of 0.5kmol of water is 88.2 N.

Specific volume is given as

                        [tex]v=\frac{Volume}{Mass}\\v=\frac{0.009}{9}\\v=0.001 m^3/kg[/tex]

So the specific volume is 0.001 m^3/kg

A. Draw a schematic of the system with a boundary around the piston-cylinder device. B. The final pressure can be determined using the ideal gas law equation. C. The weight of H2O can be calculated using the relationship between moles, molecular mass, and mass. D. The specific volume of H2O can be determined by dividing the volume by the mass.

A. To study the air in the piston-cylinder device as a closed system, we consider the device itself as the system and draw a boundary around it, including the air inside and excluding the surroundings. The schematic would show a cylindrical container with a piston separating the initial and final air volumes.

B. To determine the final pressure of the air, we can use the ideal gas law. The equation is PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature. Since the volume doubles and the temperature increases to 600 K, we can set up the equation (1 bar)(2V) = n(R)(600 K), and solve for the final pressure.

C. To determine the weight of H2O in the piston-cylinder device, we use the relationship between number of moles, molecular mass, and mass. The weight of H2O in N can be calculated as (0.5 kmol)(molecular mass of H2O)(Acceleration due to gravity).

D. The specific volume of H2O can be determined by dividing the volume (0.009 m3) by the mass of H2O (which we can calculate from the number of moles and molecular mass).

Answer:

D) result of the calculation is zero

Explanation:

Coulomb's Law is valid for only point-like particles. Since the ring is not a point-like, then we have to choose an infinitesimal portion (ds) of the ring, apply the Coulomb's Law to this portion and then integrate over the ring to find the total force.

The small portion (dq) will have the same charge density as the ring itself. Furthermore, the length of the infinitesimal portion is equal to the radius times the corresponding angle, dθ.

[tex]\lambda = \frac{Q}{2\pi R} = \frac{dq}{Rd\theta}\\dq = \frac{Qd\theta}{2\pi}[/tex]

Therefore, the force between the charge at the center and the small portion is

[tex]dF = \frac{1}{4\pi\epsilon_0}\frac{qdq}{R^2} = \frac{1}{4\pi\epsilon_0}\frac{qQd\theta}{2\pi R^2}[/tex]

Since force is a vector, we have separate its x- and y-components,

[tex]dF_x = \frac{1}{4\pi\epsilon_0}\frac{qQd\theta}{2\pi R^2}\cos(\theta)\\dF_y = \frac{1}{4\pi\epsilon_0}\frac{qQd\theta}{2\pi R^2}\sin(\theta)[/tex]

Now, we can integrate both of them over the ring.

[tex]F_x = \int\limits^{2\pi}_0 dF_x = \frac{1}{4\pi\epsilon_0}\frac{qQ}{2\pi R^2}\int\limits^{2\pi}_0\cos(\theta)d\theta = 0\\F_y = \int\limits^{2\pi}_0 dF_y = \frac{1}{4\pi\epsilon_0}\frac{qQ}{2\pi R^2}\int\limits^{2\pi}_0\sin(\theta)d\theta = 0[/tex]

Since the integration from 0 to 2π for sine and cosine functions results as zero.

Therefore, the force on the charge at the center of a uniformly distributed ring is equal to zero.

The total force exerted on a charge q placed at the center of a circle with a uniformly distributed charge Q along the circumference is zero due to the symmetry of the charge distribution.

When a charge Q is spread uniformly along the circumference of a circle with radius R, and a point particle with charge q is placed at the center of this circle, we must apply Coulomb's law to calculate the force exerted on the charge q. Thanks to the symmetry of the charge distribution, the forces exerted by individual segments of the charged circumference on the central charge will cancel each other out in every direction. Hence, while the distance from the charge q to any point on the circle is R, the resulting total force on charge q will be zero due to symmetry.

It's important not to confuse the circumference with other distances, such as the diameter (2R) or the circumferential length (2πR), as these are not relevant for calculating the force on the central charge in this symmetric setup. Therefore, the correct answer is that the result of the calculation is zero (Option D), because the uniform distribution of charge Q around the circle results in an equilibrium of forces.

Answer:

Angular velocity, [tex]\omega=3747.33\ rev/min[/tex]

Explanation:

In this case, we need to find the angular speed needed for a centrifuge to produce an acceleration of 759 times the gravitational acceleration.

Radius of the circular path, r = 4.83 cm

The acceleration acting on the particle in circular path is given by :

[tex]a=r\omega^2[/tex]

[tex]\omega[/tex] is the angular speed in rad/s

[tex]\omega=\sqrt{\dfrac{a}{r}}[/tex]

[tex]\omega=\sqrt{\dfrac{759\times 9.8}{4.83\times 10^{-2}}}[/tex]

[tex]\omega=392.42\ rad/s[/tex]

or

[tex]\omega=3747.33\ rev/min[/tex]

So, there are 3747.33 revolutions per minute that is needed. Hence, this is the required solution.

Answer:

Q_out,shell = 9Q

Explanation:

Given:

- Q_in,shell = -10 Q

- Q_sphere = +1Q

Find:

How much charge is on the outer surface?

Solution:

The electric field in the material of both the sphere and shell must be zero. The only way for this to occur is if the charge inside the

inner surface of the shell is such that its charge plus the solid's charge is zero. The rest of the excess charge from the shell moves to  the outside of the shell.

Hence,

                     Q_out,shell + Q_in,shell + Q_sphere = 0

                     Q_out,shell -10 Q + 1 Q = 0

                    Q_out,shell = 9Q

Final answer:

The charge on the outer surface of the hollow conducting sphere, which initially had a charge of -10Q and contains an inner sphere with a charge of +1Q, would be -9Q, as determined by Gauss' Law and the conservation of charge.

Explanation:

The student is asking about the charge distribution on concentric conducting spheres when one sphere is placed inside another and they each have different charges. According to Gauss' Law, when a charge is placed inside a conducting shell, it induces an equal and opposite charge on the inner surface of the shell to maintain an electric field of zero inside the material of the conductor. In the given scenario, the inner solid sphere has a charge of +1Q and the outer hollow sphere has a charge of -10Q.

By Gauss' Law, since the electric field inside a conductor must be zero, we know that the inner surface of the hollow outer sphere must have a charge of -1Q to cancel out the electric field from the +1Q charge of the inner solid sphere.

Considering charge conservation, if the outer sphere initially had a total charge of -10Q and now there is -1Q on the inner surface, the outer surface of the hollow sphere must have the remainder, which is -10Q + 1Q = -9Q. Therefore, the charge on the outer surface of the outer hollow sphere is -9Q.

Answer:

a) Earth

Explanation:

When constructing an accurate scale model of the solar system on a football field with the Sun at one end and Neptune at the other, the planet closest to the center of the field would be Earth.

From the given options the earth is the closest planet to the center of the field. The center of the field can be considered as the sun. In reality we have mercury as the closest planet in our  solar system.