distant galaxy emits light that has a wavelength of 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm. (a) Decide whether this galaxy is approaching or receding from the earth. Give your reasoning. (b) Find the speed of the galaxy relative to the earth.

Answer:[tex]d=7.94\times 10^5\ m[/tex]

Explanation:

Given

Speed of Primary wave [tex]v_1=8\ km/s[/tex]

Speed of secondary wave [tex]v_2=4.5\ km/s[/tex]

difference in timing of two waves are [tex]77.2\ s[/tex]

Suppose both travel a distance of d km then

[tex]t_1=\frac{d}{8}\quad \ldots (i)[/tex]

[tex]t_2=\frac{d}{4.5}\quad \ldots (ii)[/tex]

Subtract (ii) from (i)

[tex]\frac{d}{4.5}-\frac{d}{8}=77.2[/tex]

[tex]d[\frac{1}{4.5}-\frac{1}{8}]=77.2[/tex]

[tex]d[0.0972]=77.2[/tex]

[tex]d=794.23\ km[/tex]

[tex]d=7.94\times 10^5\ m[/tex]

Final answer:

To calculate the distance to the earthquake's epicenter, we use the time difference between the arrival of P-waves and S-waves and their speeds. By setting up an equation and solving for the distance, the seismograph is found to be approximately 618,940 meters from the epicenter.

Explanation:

When an earthquake occurs, two types of waves are generated: P-waves (primary waves) and S-waves (secondary waves), each with distinctive speeds. To determine the distance to the epicenter of the earthquake, we use the formula d = v × t, where d is distance, v is velocity, and t is time. Given that the P-wave has a speed of 8.0 km/s and the S-wave has a speed of 4.5 km/s, and the time difference of arrival between the two waves is 77.2 seconds, we can calculate the distance from the seismograph to the earthquake's epicenter.

Let the distance be d, then:

To find the distance d, we solve the equation:

Now, to convert kilometers to meters:

Therefore, the seismograph is approximately 618,940 meters from the earthquake's epicenter.

Answer:

Explanation:

moment of inertia of the disc I = 1/2 m r²

= .5 x 8 x .25²

= .25 kg m²

The wok done by force will be converted into rotational kinetic energy

F x d = 1/2 I ω²

F is force applied , d is displacement , I is moment of inertia of disc and ω

is angular velocity of disc

41 x .9 = 1/2 x .25 ω²

ω² = .25

ω = 17.18  rad / s

The angular speed should be 17.18  rad / s

Since

moment of inertia of the disc I = 1/2 m r²

= .5 x 8 x .25²

= .25 kg m²

Now the work done by force should be converted into the rotational kinetic energy

F x d = 1/2 I ω²

here,

F is the force applied,

d is displacement,

I is moment of inertia of disc

and ω is angular velocity of disc

So,

41 x .9 = 1/2 x .25 ω²

ω² = .25

ω = 17.18  rad / s

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Answer:

It is somewhere in between

Explanation:

Wave length of sound from each of the  speakers = 340 / 1700 = 0.2 m = 20 cm

Distance between first speaker and the given point = 4 m.

Distance between second speaker and the given sound

D = √(4² + 2²)

D = √(16 + 4)

D = √20

D = 4.472 m

Path difference = 4.472 - 4 =

0.4722 m.

Path difference / wave length = 0.4772 / 0.2 = 2.386

This is a fractional integer which is neither an odd nor an  even multiple of half wavelength. Hence this point is of neither a perfect constructive nor a perfect destructive interference.

Answer:

291.509 Hz

Explanation:

Fundamental frequency, often referred to simply as the fundamental, is defined as the lowest frequency of a periodic waveform.

It is a vital concept in musical instruments and many aspects .

See the attached file for the solution to the given problem.

Answer:

Explanation:

In order to answer this problem you have to know the depth of the column, we say R, this information is important because allows you to compute some harmonic of the tube. With this information you can compute the depth of the colum of air, by taking tino account that the new depth is R-L.

To find the fundamental mode you use:

[tex]f_n=\frac{nv_s}{4L}[/tex]

n: mode of the sound

vs: sound speed

L: length of the column of air in the tube.

A) The fundamental mode id obtained for n=1:

[tex]f_1=\frac{v_s}{4L}[/tex]

B) For the 3rd harmonic you have:

[tex]f_3=\frac{3v_s}{4L}[/tex]

C) For the 2nd harmonic:

[tex]f_2=\frac{2v_s}{4L}[/tex]

Answer:

Explanation:

λ

given λ = 700 nm

a) for first maxima, d*sin(θ) =  λ

sin(theta) = λ/d

= 700*10^-9/(0.025*10^-3)

= 0.028

theta θ = sin^-1(0.028)

= 1.60 degrees

b) given R = 1 m,

delta_y = λ*R/d

= 700*10^-9*1/(0.025*10^-3)

= 0.028 m or 2.8 cm

c) for first maxima, d*sin(θ) = λ

sin(θ) = λ/d

= 700*10^-9/(2.5*10^-3)

= 0.00028

theta = sin^-1(0.00028)

= 0.0160 degrees

d) R = 25 mm = 0.025 m

δ_y = λ*R/d

= 700*10^-9*0.025/(2.5*10^-3)

= 7*10^-6 m or 7 micro m

e) No. But the position of maxima and minima will be shifted.

Answer:

[tex]\frac{dy}{dt}=k(y_t-R)[/tex]

Explanation:

According to Newton’s law of cooling, the rate of loss of heat from a body and the difference in the temperature of the body and its surroundings are proportional to each other.

[tex]\frac{dy}{dt}=k(y_t-R)[/tex]

Here, [tex]y_t[/tex] represents temperature at time t, R as the room temperature, t as the independent variable, y as the dependent variable.

The equation that represents Newton's Law of Cooling for this particular situation is dy/dt = k(yt-R)

According to Newton’s law of cooling refer, the rate of loss of heat from a body and also the difference in the temperature of the body and also its surroundings are proportional to each other.

dy/dt is = k(yt-R)

Therefore, yt conveys temperature at time t, R as the room temperature, t as the independent variable, y as the dependent variable.

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Answer:

Δx = 5.82mm

Explanation:

To find the distance between the eight maximum and the third one you use the following formula:

[tex]x_m=\frac{m\lambda D}{d}[/tex]   (1)

λ: wavelength = 452*10^-9 m

m: order of the fringes

D: distance to the scree = 0.49m

d: distance between slits = 0.190*10^-3 m

you use for m=8 and m=3, then you calculate x8 - x3:

[tex]x_8=\frac{8(452*10^{-9}m)(0.49m)}{0.190*10^{-3}m}=9.32*10^{-3}m\\\\x_3=\frac{3(452*10^{-9}m)(0.49m)}{0.190*10^{-3}m}=3.49*10^{-3}m\\\\\Delta x_{8,3}=5.82*10^{-3}m=5.82mm[/tex]

hence, the distance between these fringes is 5.82mm

First: CT Scan

Second: Fluoroscopy

Explanation:

Both correct

A CT scan can provide detailed images of a tumor and its surrounding area, which can be beneficial for a cancer patient. For ongoing throat issues, a Barium swallow study can be used, where the patient swallows a barium solution that is then visualized with an X-ray to identify abnormalities.

In the first scenario, the patient with an ongoing history of cancer might benefit most from a Computed Tomography (CT) scan. CT scans are capable of creating detailed pictures of organs, bones, and other tissues, making it an excellent tool for capturing the size and position of a tumor and surrounding blockages in the abdominal area. Furthermore, it can reveal whether the cancer has spread to other parts of the body.

In the second scenario, Joe's doctor chooses to use X-ray imaging to detect any abnormalities in his throat. The specific procedure used for this is known as a Barium swallow study. This involves swallowing a barium solution that coats the esophagus, enabling the X-ray to capture clear images of the region as the patient swallows.

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Answer:

in other to gain more power you would need to increase either current or voltage

Explanation:

In other to gain more power, you would need to increase either current or voltage,because power is directly proportional to voltage when current is kept constant.also power is directly proportional to current when voltage is kept constant

Answer:

Explanation:

Given that,.

A house hold power consumption is

475 KWh

Gas used is

135 thermal gas for month

Given that, 1 thermal = 29.3 KWh

Then,

135 thermal = 135 × 29.3 = 3955.5 KWh

So, total power used is

P = 475 + 3955.5

P =4430.5 KWh

Since 1 hr = 3600 seconds

So, the energy consumed for 1hr is

1KW = 1000W

P = energy / time

Energy = Power × time

E = 4430.5 KWhr × 1000W / KW × 3600s / hr

E = 1.595 × 10^10 J

So, using Albert Einstein relativity equation

E = mc²

m = E / c²

c is speed of light = 3 × 10^8 m/s

m = 1.595 × 10^10 / (3 × 10^8)²

m = 1.77 × 10^-7 kg

Then,

1 kg = 10^6 mg

m = 1.77 × 10^-7 kg × 10^6 mg / kg

m = 0.177mg

m ≈ 0.18 mg

Total monthly energy consumption of the house in kilowatt-hours (kWh) is calculated. Then, with the help of the equation E = mc², where c is the speed of light, the equivalent mass of energy in kilograms is found by converting energy to Joules and solving for mass.

The household uses 475475 kWh of electrical energy and 135135 therms for gas heating and cooking monthly. Considering that 1.00 therm is equal to 29.329.3 kWh, total energy consumption for the month in kWh can be calculated as (475475 + (135135 × 29.3)) kWh.

Einstein’s famous equation E = mc² relating energy (E) and mass (m) allows us to calculate the equivalent mass of this energy.

Here, c is the speed of light. We convert the energy into Joules (1 kWh = 3.6 × 10⁶ J) and then solve for m to find out the mass in kilograms that would need to be converted into energy each month. The question seems to have a typo in the values, you might need to correct them to calculate the accurate amount.

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Answer:

1.6 s

Explanation:

To find the time in which the potential difference of the inductor reaches 24V you use the following formula:

[tex]V_L=V_oe^{-\frac{Rt}{L}}[/tex]

V_o: initial voltage = 60V

R: resistance = 24-Ω

L: inductance = 42H

V_L: final voltage = 24 V

You first use properties of the logarithms to get time t, next, replace the values of the parameter:

[tex]\frac{V_L}{V_o}=e^{-\frac{Rt}{L}}\\\\ln(\frac{V_L}{V_o})=-\frac{Rt}{L}\\\\t=-\frac{L}{R}ln(\frac{V_L}{V_o})\\\\t=-\frac{42H}{24\Omega}ln(\frac{24V}{60V})=1.6s[/tex]

hence, after 1.6s the inductor will have a potential difference of 24V

To find the time when the potential difference across the inductor will be 24 V in an LR circuit, we need to use the formula for the time-dependent current in an RL circuit, and through mathematical manipulations, substitute into the equation for the induced emf in the inductor and solve for time.

The time for the potential difference across the inductor to be 24V in an LR circuit can be calculated using the formula for the time-dependent current in an RL circuit, which accounts for how current evolves over time. According to Faraday's law, a changing current in an inductor generates an opposition voltage, its magnitude is determined by L (inductance of the inductor) times dI/dt (the rate of change of current).

In our case we have the equation for the current in an RL circuit when turned on: I(t) = V/R*(1 - e^(-R*t/L)) and the induced emf in the inductor V(L) = L*dI/dt. Our goal is to find the time when V(L) equals 24V. This requires substituting I(t) into the equation for V(L), setting V(L) equal to 24V, and solving for t.

This process involves the principles of RL circuits, the characteristic time constant, and Faraday's law.

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Answer:

A) 27209506.5 N

B) 2393640 N

The force on the underwater vehicle is about 11.37 times the weight of the jetliner for comparison.

Explanation:

Force du to depth of water is

F = pghA

P = density of salt water = 1025 kg/m3

g = acceleration due to gravity 9.81 m/s2

h = depth of water 11000 m

A = area pressure acts

Area = ¶r^2 = 3.142 x 0.280^2 = 0.246 m^2

Therefore

F = 1025 x 9.81 x 11000 x 0.246

= 27209506.5 N

Weight of a jetliner with mass 2.44 × 10^5 kg is,

2.44×10^5 x 9.81 = 2393640 N

The force on the underwater vehicle is about 11.37 times the weight of the jetliner for comparison.

The force exerted by the water on the sea underwater vehicle's window at the depth of the Mariana Trench is approximately 2.73 x 10⁷ N. For comparison, the weight of a jetliner with a mass of 2.44 x 10⁵ kg is 2.39 x 10⁶ N. Thus, the force exerted on the underwater vehicle at the Mariana's depth is an order of magnitude greater than the weight of the jetliner.

To calculate the force that the water exerts on the observation window of an underwater vehicle, you first need to find the pressure at the depth of the Mariana Trench. The pressure is given by the formula P = ρgh, where P is pressure, ρ is density, g is acceleration due to gravity, and h is height (in this case, depth).

Substituting the given values, the pressure at a depth of 11,000 m is about 1.11 x 10⁸ Pa. The force exerted on the window can be found using F = PA, where A is the area of the window. With a radius of 0.280 m, the area of the window (A = πr²) is 0.246 m². Thus, the force applied by the water on the window is F = (1.11 x 10⁸ Pa) (0.246 m²) = 2.73 x 10⁷ N approx.

For the second part of the question, the weight (force) of the jetliner is given by the equation W = m × g. Using the given mass (2.44 x 10⁵ kg) and the acceleration due to gravity (9.8 m/s²), the weight of the jetliner is 2.44 x 10⁵ kg × 9.8 m/s² = 2.39 x 10⁶ N, which is an order of magnitude less than the force exerted on the underwater vehicle at the depth of the Mariana Trench.

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Answer:

The answer is A and C .

Explanation:

Reflection of object is reflected through the Normal (mirror) .

*incident angle = refracted angle

Answer:

a .

15.68

m/s

(b).  

21.72

m

Explanation:

To determine if the arrow will hit the target, we can analyze the projectile motion. By calculating the time of flight and horizontal displacement, we can determine if the arrow will hit the target. The velocity and direction of the arrow at impact can also be determined using the horizontal and vertical components of velocity.

To determine if the arrow will hit the target, we need to analyze the projectile motion of the arrow. First, we need to calculate the time of flight using the equation t = 2 * (v * sin(theta))/g, where v is the initial velocity of the arrow and theta is the launch angle. Next, we can calculate the horizontal displacement using the equation x = v * cos(theta) * t, where x is the distance traveled horizontally. Comparing the calculated horizontal displacement with the distance to the target, we can determine if the arrow will hit the target.

Given the initial velocity of the arrow (230 km/hr), launch angle (3.5 degrees), and distance to the target (50 m), we can convert the initial velocity to m/s (230 km/hr * (1000 m/km) * (1 hr/3600 s)) and use the formulas to calculate time of flight and horizontal displacement. If the horizontal displacement is less than or equal to the distance to the target, the arrow will hit the target.

To find the velocity and direction of the arrow at impact, we can use the equations v_x = v * cos(theta) and v_y = v * sin(theta), where v_x is the horizontal component of velocity and v_y is the vertical component of velocity. Since the angle of impact is not specified, the direction of the arrow at impact will depend on the particular situation.

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

The average distance calculated between a person's ears is 0.2 meters. The interaural time difference (ITD) is maximal when the sound source is directly to the side. For air (vs=338 m/s), Δ[tex]t_{max}[/tex] is 591.72 microseconds, and for seawater (vs= 1534 m/s), it is 130.38 microseconds. These calculations illustrate sound localization and the impact of medium on sound propagation.

Understanding Sound Localization and Interaural Time Difference

When a noise occurs directly to the right of a person, the sound waves reach the right ear before the left ear. We can calculate the maximum interaural time difference ( Δ[tex]t_{max}[/tex]) for sound reaching the ears using the given distance between the ears.

(a) The average distance between a person's ears was estimated as 0.2 meters (20 cm).

(b) To calculate Δ[tex]t_{max}[/tex] with the speed of sound in air (vs = 338 m/s), we can use the formula Δ[tex]t_{max}[/tex] = d / vs, where d is the distance between ears. Substituting the values, we get:

Δ[tex]t_{max}[/tex] = 0.2 m / 338 m/s = 0.0005917159763 seconds, or approximately 591.72 microseconds.

(c) Lastly, for sound traveling in seawater at room temperature where vs = 1534 m/s, we similarly get:

Δ[tex]t_{max}[/tex] = 0.2 m / 1534 m/s = 0.0001303763441 seconds, or approximately 130.38 microseconds.

This demonstrates the role of the medium in sound propagation and how it affects the interaural time difference.

Answer:

2.994 m/s

Explanation:

m = Mass of the car = 100 g

[tex]v_1[/tex] = Initial velocity = 3.33 m/s

h = Height = 0.108 m

g = Acceleration due to gravity = 9.81 m/s²

We know that energy in the system is conserved so

[tex]\dfrac{1}{2}mv_1^2=mgh+\dfrac{1}{2}mv_f^2\\\Rightarrow v_f=\sqrt{2\left({\dfrac{1}{2}v_1^2-gh}\right)}\\\Rightarrow v_f=\sqrt{2\left(\dfrac{1}{2}3.33^2-9.81\times 0.108\right)}\\\Rightarrow v_f=2.994\ m/s[/tex]

The final velocity of the car is 2.994 m/s

The final velocity of the car is approximately 6.05 m/s.

To find the final velocity of the car, we can use the principles of conservation of energy. In this case, the initial kinetic energy of the car is equal to its final potential energy. The initial kinetic energy can be calculated using the formula K.E. = 1/2 mv^2, where m is the mass of the car (100 g = 0.1 kg) and v is the initial velocity (3.33 m/s).

The final potential energy can be calculated using the formula P.E. = mgh, where m is the mass of the car, g is the acceleration due to gravity (9.8 m/s^2), and h is the change in altitude (0.108 m).

Setting the initial kinetic energy equal to the final potential energy, we can solve for the final velocity of the car:

1/2 mv^2 = mgh

Substituting the given values, we get:
1/2 (0.1 kg)(v^2) = (0.1 kg)(9.8 m/s^2)(0.108 m)

Simplifying and solving for v, we find that the final velocity of the car is approximately 6.05 m/s.

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Answer:

t = 5.94x10⁹ years.

Explanation:

The time of the explosion can be calculated using the decay equation:

[tex] N_{t} = N_{0}e^{-\lambda t} [/tex]

Where:

N(t): is the quantity of the element at the present time

N(0): is the quantity of the element at the time of explosion

λ: is the decay constant

t: is the time

Knowing that the present U-235/U-238 ratio is 0.00700 and that at the time of the explosion there were equal amount of U-235 and U-238, we have:

[tex]\frac{N_{U-235}}{N_{U-238}} = \frac{N_{U-235_{0}}e^{-\lambda_{U-235} t}}{N_{U-238_{0}}e^{-\lambda_{U-238} t}}[/tex]     (1)

The decay constant is equal to:

[tex] \lambda = \frac{ln(2)}{t_{1/2}} [/tex]  

For the U-235 we have:

[tex] \lambda_{U-235} = \frac{ln(2)}{0.700 \cdot 10^{9} y} = 9.90 \cdot 10^{-10} y^{-1} [/tex]

For the U-238 we have:

[tex] \lambda_{U-238} = \frac{ln(2)}{4.47 \cdot 10^{9} y} = 1.55 \cdot 10^{-10} y^{-1} [/tex]

By introducing the values of [tex]\lambda_{U-235}[/tex] and [tex]\lambda_{U-238}[/tex] into equation (1) we have:

[tex]0.00700 = \frac{e^{-9.90 \cdot 10^{-10} t}}{e^{-1.55 \cdot 10^{-10} t}}[/tex]        

[tex]0.00700 = e^{(-9.90 \cdot 10^{-10} + 1.55 \cdot 10^{-10}) t}[/tex]    

[tex]ln(0.00700) = (-9.90 \cdot 10^{-10} + 1.55 \cdot 10^{-10}) t[/tex]            

[tex]t = \frac{ln(0.00700)}{-9.90 \cdot 10^{-10} + 1.55 \cdot 10^{-10}} = 5.94 \cdot 10^{9} y[/tex]

Therefore, the star exploded 5.94x10⁹ years ago.

I hope it helps you!    

The explosion of the star that released the uranium that formed Earth happened approximately 6 billion years ago, determined using the half-lives of U-235 and U-238 and the current U-235/U-238 ratio.

The question regarding the age of the elements from a star explosion can be addressed using the concept of radioactive decay and isotope half-lives, specifically of Uranium-235 (U-235) and Uranium-238 (U-238). Using the current U-235/U-238 ratio of 0.00700 and knowing the half-lives of U-235 (0.700 × 109 years) and U-238 (4.47 × 109 years), we can calculate that the star exploded approximately 6 billion years ago based on the change in the U-235/U-238 ratio from an assumed equal initial amount. This estimate is consistent with the age of the solar system and the time it would take for such materials to coalesce into a planetary body like Earth.

Answer:

Explanation:

Given

Position of squirrel is given by

[tex]s(t)=t^3-12t^2+36t[/tex]

Velocity is given by

[tex]v(t)=\frac{ds(t)}{dt}=\frac{d(t^3-12t^2+36t)}{dt}[/tex]

[tex]v(t)=3t^2-12\times 2t+36[/tex]

[tex]v(t)=3t^2-24t+36[/tex]

(b) acceleration is given by

[tex]a(t)=\frac{da(t)}{dt}=\frac{d(3t^2-24t+36)}{dt}[/tex]

[tex]a(t)=6t-24[/tex]

(c)at [tex]s(3)=3^3-12(3)^2+36(3)[/tex]

[tex]s(3)=27\ m[/tex]

at [tex]s(4)=4^3-12(4)^2+36(4)[/tex]

[tex]s(4)=16\ m[/tex]

at [tex]t=3\ s[/tex] Position is [tex]27\ m[/tex] and at [tex]t=4\ s[/tex] position is [tex]16\ m[/tex]

therefore squirrel is moving down

Answer: 62.7

Explanation:

The specific heat capacity of water is 4.186 J you multiply 4.186 and 15.0 and you will get 62.7

Answer:

The required solution is 100890 Pa and 14.3lb/in²

Explanation:

See attached image

Answer:

1. p = 14.63 lb/in² or 100890.608 Pa

2. p = 74676 Pa or 10.83 lb/in²

3. P = 2450 W or 3.28 hp

4. [tex]p_{1}[/tex] = 490105 N/m²

Explanation:

1. Let's begin by listing out the given parameters:

density of mercury = 13.546 g/cm³ = 13546 kg/m³,

height of column = 76 cm = 0.76 m, acceleration due to gravity = 9.8m/s²

Using Pressure = density * acceleration due to gravity * height of column

p = ρ g h = 13546 * 9.8 * 0.76

p = 100890.608 Pa

To get the answer in lb/in², divide by 6895

p = 100890.608 ÷ 6895 = 14.632

p = 14.63 lb/in²

2. Let's list out the parameters given:

density of water = 62.43 lbm/ft³ = 62.43 * 16.018 = 1000kg/m³,

height of column = 25 ft = 25 ÷ 3.281 = 7.62 m,

acceleration due to gravity = 9.8m/s²

Using Pressure = density * acceleration due to gravity * height of column

p = ρ g h = 1000 * 9.8 * 7.62

p = 74676 Pa

To convert from Pa to lb/in², divide by 6895

p = 74676 ÷ 6895

p = 10.83 lb/in²

3. Let's list out the parameters given:

mass flow rate (ṁ) = 10 kg/s, [tex]h_{1}[/tex] = 5 m, [tex]h_{2}[/tex] = 30 m, Δh = 30 - 5 = 25 m, g = 9.8 m/s²

Using Power = Energy (Potential Energy) ÷ Time

Energy (Potential Energy) = m g h

Power = mgΔh ÷ t; m÷ t = ṁ

Substitute ṁ into the equation

P = ṁ g h = 10 * 9.8 * 25

P = 2450 W

To convert from W to hp, divide by 746

P = 2450 ÷ 746 = 3.284

P = 3.28 hp

4. Let's list out the parameters given:

height (Δh) = 50 m, ṁ = 1 kg/s, g = 9.8 m/s²,

p2 = 105N/m², ρ = 1000 kg/m³

Using Bernoulli's Equation,

p1 + ½ρ([tex]v_{1}[/tex])² + ρgh1 = p2 + ½ρ([tex]v_{2}[/tex])² + ρgh2

Assuming steady state flow; [tex]v_{2}[/tex] = [tex]v_{1}[/tex] ⇒ [tex]v_{2}[/tex] - [tex]v_{1}[/tex] = 0

[tex]p_{1}[/tex] - [tex]p_{2}[/tex]  = ½ρ([tex]v_{2}[/tex] - [tex]v_{1}[/tex])² + ρg([tex]h_{2}[/tex] - [tex]h_{1}[/tex])

[tex]p_{1}[/tex] - [tex]p_{2}[/tex] = ρgΔh

[tex]p_{1}[/tex] - 105 = 1000 * 9.8 * 50

[tex]p_{1}[/tex] = 490000 + 105 = 490105

[tex]p_{1}[/tex] = 490105 N/m²

Answer:

Her current biological age = 24.02 years

Explanation:

From time dilation equation, we know that;

t = t_o * [√(1-(v²/c²))]•L_o

Where;

t = dilated time

t_o = stationary time

v = the speed of the moving object

c = the speed of light in a vacuum

First, let's convert the rest time (t_o) from light years to years.

Thus;

t_o = [c/0.993c] * [42.2]

c will cancel out and we now have;

t_o = 42.5 years

Since t = t_o * [√(1-(v²/c²))]

Thus; t = 42.5 * [√(1-(0.993²c²/c²))]

t = 42.5/[√(1 - (0.993²))]

t = 42.5 * 0.1181

t = 5.02 years

Since the astronaut was 19 years old when the probe left the earth, thus;

Her current biological age now the probe has reached Capella, will be;

19 + 5.02 = 24.02 years

Answer:No it does not the chemical equation is unbalanced,this is because according to the law of conservation it states that mass cannot be created or destroyed ,the chemical equation N₂ + H₂ → NH₃ shows that energy has been created which is not possibly so the correct balanced chemical is equation is  1N₂ +3H₂ → 3NH₃

Explanation:

Answer:

The thickness of the paper is   [tex]t = 188\mu m[/tex]

Explanation:

From the question we are told that

   The length of the wedge-shaped air film  [tex]L = 12.5 cm = \frac{12.5}{100} = 0.125m[/tex]

   The wavelength of the light is  [tex]\lambda = 600nm = 600* 10^{-9}m[/tex]

  The spacing of the interference fringe is  [tex]D = 0.200mm = \frac{0.200}{1000} = 0.2*10^{-3} m[/tex]

For destructive interference the thickness is mathematically represented as

                [tex]t =\frac{\lambda * L}{2 * D }[/tex]

Substituting values

                [tex]t = \frac{600 * 10^{-9} * 0.125 }{2 * 0.2 *10^{-3}}[/tex]

                [tex]t = 188\mu m[/tex]

Answer:

a) v = 786.93 m/s

b) v = 122.40 m/s

Explanation:

a) To find the average exhaust speed (v) of the engine we can use the following equation:

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

Where:

F: is the thrust by the engine = 5.26 N

Δm: is the mass of the fuel = 12.7 g

Δt: is the time of the burning of fuel = 1.90 s

[tex]v = \frac{F*\Delta t}{\Delta m} = \frac{5.26 N*1.90 s}{12.7 \cdot 10^{-3} kg} = 786.93 m/s[/tex]

b) To calculate the final velocity of the rocket we need to find the acceleration.

The acceleration (a) can be calculated as follows:

[tex] a = \frac{F}{m} [/tex]

In the above equation, m is an average between the mass of the engine plus the rocket case mass and the mass of the engine plus the rocket case minus the fuel mass:

[tex]m = \frac{(m_{engine} + m_{rocket}) + (m_{engine} + m_{rocket} - m_{fuel})}{2} = \frac{2*m_{engine} + 2*m_{rocket} - m_{fuel}}{2} = \frac{2*25.0 g + 2*63.0 g - 12.7 g}{2} = 81.65 g[/tex]

Now, the acceleration is:

[tex] a = \frac{5.26 N}{81.65 \cdot 10^{-3} kg} = 64.42 m*s^{-2} [/tex]

Finally, the final velocity of the rocket can be calculated using the following kinematic equation:

[tex]v_{f} = v_{0} + at = 0 + 64.42 m*s^{-2}*1.90 s = 122.40 m/s[/tex]

I hope it helps you!

Final answer:

The average exhaust speed of the engine is calculated to be approximately 786.929 m/s, and the magnitude of the final velocity of the rocket if fired in outer space is roughly 94.916 m/s.

Explanation:

To calculate the average exhaust speed (v_ex), we can use the impulse-momentum theorem, which states that impulse is equal to the change in momentum of the system. Impulse is given by the product of the average force exerted by the engine (thrust) and the time interval during which the thrust is applied. If all the fuel is consumed, the change in mass (Δm) is the mass of the fuel.

Impulse = Thrust × Time = 5.26 N × 1.90 s = 9.994 N·s

The momentum change is equal to the mass of the fuel expelled times the average exhaust speed.

Δ(momentum) = Δm × v_ex

Substituting the impulse and solving for v_ex, we get:

v_ex = Impulse / Δm

v_ex = 9.994 N·s / 0.0127 kg = 786.929 m/s

Part B: Final Velocity of the Rocket in Space

The final velocity (V_final) of the rocket in space can be determined using the rocket equation, also known as Tsiolkovsky's rocket equation:

V_final = v_ex × ln(m_initial / m_final)

Where:

Calculating the final velocity:

V_final = 786.929 m/s × ln(0.088 kg / 0.0753 kg) ≈ 94.916 m/s

Answer:

volume flow rate:2.68m³/s

diameter :58.27cm

Explanation:

flow rate = Q = πr² v  = amount per second that flows through the pipe

Given:

pipe's diameter= 0.581

r= 0.581/2=>0.2905m

speed 'v'= 10.1 m/s

Q= (3.142)(0.2905)²(10.1)

so

volume flow rate= 2.68m³/s

->if no oil has been added or subtracted or compressed then Q is the same everywhere

therefore,

Q= πr²v

2.68 = π(d/2)²(5.85)

d= (2.68x4) /(3.142 x 5.85)

d= 0.5827m =>58.27cm

Answer:

The correct option is C

Explanation:

From the question we are told that

  The initial speed of the rocket is  [tex]v_i = 100 m/s[/tex]

    The speed of the rocket engine sound is [tex]V[/tex]

    The final speed of the rocket is  [tex]v_f = 200 \ m/s[/tex]

The speed of the sound at [tex]v_f[/tex] would still remain V this because the speed of sound wave is constant and is not dependent on the speed of the observer(The mountain ) or the speed of the source (The rocket ).

  A clear example when  lightning strikes you will first see (that is because it travels at  the speed of light which is greater than the speed of sound) but it would take some time before you hear the sound of the   lightning

Here we see that the speed of the lightning(speed of sound) does not affect the speed of the sound it generates  

To evaluate the final kinetic energy of a supply spacecraft under the given tractor beam force, you would need to integrate the force over the displacement. Direct calculation requires specific values for α, β, and displacement x, which are not provided.

The question involves calculating the final kinetic energy of a supply spacecraft under a specific tractor beam force, F(x)=αx3+β. This calculation would typically require the integration of the force over the displacement, as kinetic energy can be evaluated through the work-energy principle where work done by a force in moving an object is equal to the change in kinetic energy of the object. Here, without specific values or further context provided for α, β, or the displacement, x, a direct calculation cannot be made. However, in physics, especially in the study of mechanics and spacecraft dynamics, understanding how forces affect motion and energy forms the basis for analyzing and optimizing space missions. Evaluating the final kinetic energy would typically involve integrating the force function over the spacecraft's path, considering initial conditions, and any external forces or resistances.

Answer:[tex]r_i=0.016119\ m\approx 16.119\ mm[/tex]

Explanation:

Given

mass of first particle is [tex]m_1=3.12\times 10^{-3}\ kg[/tex]

mass of second particle is [tex]m_2=7.1\times 10^{-3}\ kg[/tex]

Charge on both the particle [tex]q=8.8\times 10^{-6}\ C[/tex]

Now final speed of first particle is [tex]v_1=131\ m/s[/tex]

Final separation between particles is [tex]r=0.15\ m[/tex]

As there is no external force therefore linear momentum is conserved

[tex]0+0=m_1v_1+m_2v_2[/tex]

[tex]0=3.12\times 10^{-3}\times 131+7.1\times 10^{-3}\times v_2[/tex]

[tex]v_2=-\dfrac{3.12\times 10^{-3}}{7.1\times 10^{-3}}\times 131[/tex]

[tex]v_2=-57.56\ m/s[/tex]

Conserving total energy

Initial Kinetic energy +Initial  Potential energy=Final Kinetic energy +Final Potential energy

[tex]\Rightarrow 0+\frac{kq^2}{r_i}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+\frac{kq^2}{r_f}[/tex]

[tex]\Rightarrow \frac{9\times 10^9\times 8.8^2\times 10^{-12}}{r_i}=\frac{1}{2}\times 3.12\times 10^{-3}\times 131^2+\frac{1}{2}7.1\times 10^{-3}\times (57.56)^2+\frac{9\times 10^9\times 8.8^2\times 10^{-12}}{0.15}[/tex]

[tex]\Rightarrow \frac{0.696}{r_i}=26.771+11.76+4.646[/tex]

[tex]\Rightarrow \frac{0.696}{r_i}=43.177[/tex]

[tex]r_i=0.016119\ m\approx 16.119\ mm[/tex]

Using the law of conservation of linear momentum and conservation of total energy, the obtained initial separation between the particles is 0.01612 m.

Given that the masses of the two particles are;

[tex]m_1 = 3.12\times 10^{-3}\,kg\\m_2 = 7.1\times 10^{-3}\,kg[/tex]

Also, the charges of both the particles are equal.

[tex]q_1 = q_2 = +8.8\times 10^{-6} \,C[/tex]

The final separation between the particles is;

[tex]r_f = 0.15\,m[/tex]

Also, the final speed of the first particle is;

[tex](v_1)_{final} = 131\,m/s[/tex]

There is no external force applied here; so by the law of conservation of linear momentum, we have;

[tex](m_1v_1)_{initial} +(m_2v_2)_{initial}=(m_1v_1)_{final} +(m_2v_2)_{final}[/tex]

But initially, the particles are at rest, so the initial velocities are zero.

[tex]0+0=(3.12\times 10^{-3}\,kg \, \times 131\,m/s ) +(7.1\times 10^{-3}\,kg\,)\,(v_2)_{final}\\[/tex]

[tex]\implies (v_2)_{final}=-\frac{3.12\times 10^{-3}\,kg \, \times 131\,m/s}{7.1\times 10^{-3}\,kg} =-57.57\,m/s[/tex]

Now, applying the law of conservation of total energy, we get;

[tex](KE)_{\,initial}\, + \,(PE)_{\,initial} = (KE)_{\,final}\, + \,(PE)_{\,final}[/tex]

But initially, the particles are at rest; so they have no initial kinetic energy.

They have electrostatic potential alone initially.

[tex]0+k\frac{q^2}{r_i} = \frac{1}{2}(m_1 v_1)_{initial} + \frac{1}{2}(m_2 v_2)_{final} + k\frac{q^2}{r_f}[/tex]

Substituting the known values, we get;

[tex](9\times 10^9 \, Nm^2/C^2)\frac{(8.8\times 10^{-6} \,C)^2}{r_i} =[ \frac{1}{2}(3.12\times 10^{-3}\,kg )\times (131\,m/s)^2] + [\frac{1}{2}(7.1\times 10^{-3}\,kg) \times (-57.57\,m/s)^2 + (9\times 10^9 \, Nm^2/C^2)\frac{(8.8\times 10^{-6} \,C)^2}{r_f}[/tex]

[tex]\implies\frac{0.696\,Nm^2}{r_i} =26.77\,J+11.76\,J+ 4.64\,J[/tex]

[tex]\implies r_i =\frac{0.696\,Nm^2}{43.17\,J}=0.01612\,m[/tex]

Learn more about conservation of momentum and energy in a collision here;

https://brainly.com/question/7247845

Answer:

T

Explanation:

Answer:

true

Explanation: