A 2450-Hz sound source and an observer are moving away from each other. The source is moving at 15.0m/s, and the observer is moving at 25.0m/s. Assuming the speed of sound to be 343 m/s, what is the frequency heard by the observer?

Answer:

Size of image: 1.8 cm inverted

Magnification of camera: -0.1

Explanation:

Given:

Assume:

Using sign convention, we have

[tex]u = -200\ cm\\v = 20\ cm\\h_o = 18\ cm[/tex]

The hole in the pinhole camera behaves as a convex lens. The other face opposite to the hole face behaves like a film where the image is formed to be seen. So, using the formula of magnification, we have

[tex]m = \dfrac{h_i}{h_o}=\dfrac{v}{u}\\\Rightarrow  \dfrac{h_i}{h_o}=\dfrac{v}{u}\\\Rightarrow  h_i=\dfrac{v}{u}h_o\\\Rightarrow  h_i=\dfrac{20}{-200}\times 18\\\Rightarrow  h_i=-1.8 cm[/tex]

This means the image of the bird measures 1.8 cm in length where negative sign in the calculation represents that the image formed is inverted.

Hence, the image of the bird on the film is 1.8 cm large.

Now, again using the formula of magnification, we have

[tex]m = \dfrac{h_i}{h_o}\\\Rightarrow m = \dfrac{-1.8}{18}\\\Rightarrow m =-0.1[/tex]

Hence, the magnification of the camera is -0.1.

In a pinhole camera, image size = object size * (camera depth / object distance). For a chicken 18 cm tall standing 2 meters away, the image size will be 0.198 cm. The magnification, or size of the image compared to the object, will be 0.011 or 1.1%

The size of the image formed by a pinhole camera is given by the formula: image size = object size * (camera depth / object distance). Substituting the values in this formula, image size = 18 cm * (22 cm / 2 m) = 0.198 cm. This tells us that the fierce chicken's image will be about 0.198 cm high on the film.

Next, the magnification of the camera is the ratio of the image size to the object size. So the magnification is 0.198 cm / 18 cm = 0.011. In other words, the image on the film is about 1.1% the size of the actual object.

Note:

These calculations are based on the simple model of a pinhole camera where only rays of light traveling in straight lines are considered. The actual picture may differ due to factors like the scattering of light.

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First, let's make some convertions:

[tex]268kg*\frac{1 lbm}{0.454kg}= 590.84 lbm[/tex]

[tex]133.7 ft^3*\frac{1m^3}{35.3147ft^3} = 3.78m^3[/tex]

a) weight of the fuel:

Newtons: The weight in newtons is equal to the mass in kilograms times the gravity in m/s^2.

[tex]W = m*g = 268kg*9.81m/s^2=2629.08 N[/tex]

lbf: The weight inlbf is equal to the mass in slugs times the gravity in ft/s^2.

[tex]W= m*g = 590.84 lbm *\frac{1 slug}{32.174lbm} *32.174ft/s^2 = 590.84 lbf[/tex]

b) density:

The density is the mass in kg of the fuel divided by its volume in m^3:

[tex]d = \frac{m}{v} =\frac{268kg}{3.78m^3} =70.9 kg/m^3[/tex]

c) specific volume:

The specific volume is the volume in ft^3 of the fuel divided by its mass in lbm:

[tex]v_{sp} = \frac{v}{m} =\frac{133.7 ft^3}{590.84 lbm} = 0.226 ft^3/lbm[/tex]

Final answer:

The pressure ratio between different diameter pipes can be calculated based on the square of their diameters.

Explanation:

To find the volume rate of flow (Q), we can use the principle of continuity that states the volume flow rate must be constant throughout the pipe. Q can be calculated using the area of cross-section (A) and velocity (v) with the formula Q = A1 x v1 = A2 x v2.

The ratio in water pressure between the larger and smaller water pipes can be determined by considering the cross-sectional areas of the pipes. As pressure is inversely proportional to the cross-sectional area of the pipe, the ratio in pressure is equal to the ratio of the squares of the pipe diameters. In this case, the ratio of pressures would be 36:1.

Answer:

[tex]acceleration=0.8 \frac{m}{s^{2} } \\distance=112.5 m[/tex]

Explanation:

We can find these answers following the equations of motion.

[tex]a=[/tex]Δ[tex]V/t[/tex]

Where Δ[tex]V[/tex] is the difference between the final speed and the initial speed. And [tex]t[/tex] is the time spent

We replace the terms:

[tex]a=\frac{18\frac{m}{s} -12\frac{m}{s} }{7.5s}[/tex]

We solve the difference:

[tex]a=\frac{6\frac{m}{s}}{7.5s}[/tex]

We divide the terms, so we can have the answer:

[tex]a=0.8 \frac{m}{s^{2}}[/tex]

2. To find the distance traveled by the truck, we use the equation:

[tex]x=V_0t+\frac{1}{2}at^{2}[/tex]

Where [tex]x[/tex] is the distance traveled, [tex]V_0[/tex] is the initial speed, [tex]a[/tex] is the acceleration and [tex]t[/tex] is the time.

We replace the terms:

[tex]x=(12\frac{m}{s}*7.5s)+\frac{1}{2}[0.8\frac{m}{s^{2} }*(7.5)^{2} ][/tex]

We multiply and solve the exponential:

[tex]x=90m+\frac{1}{2}(0.8\frac{m}{s^{2} }*56.25s^{2} )[/tex]

Then, we multiply the terms left:

[tex]x=90m+22.5m[/tex]

And add, so we can have the answer:

[tex]x=112.5m[/tex]

Answer:

(a) 5.6g

(b) 15.1g

Explanation:

Let us assume:

Part (a):

While Dr. John Paul Stapp accelerates the rocket to its top speed, we have

[tex]u = 0 m/s\\v = 282\ m/s\\t = 5.1\ s\\\therefore a = \dfrac{v-u}{t}\\\Rightarrow  a = \dfrac{282-0}{5.1}\\\Rightarrow  a =55.29\\\textrm{Dividing both sides by }g\\\Rightarrow \dfrac{a}{g}=\dfrac{55.29}{9.80}\\\Rightarrow \dfrac{a}{g}=5.6\\\Rightarrow a=5.6g[/tex]

Hence, the acceleration of the rocket in the direction of motion is 5.6g.

Part (b):

While Dr. John Paul Stapp accelerates the rocket from its top speed to rest, we have

[tex]u = 282 m/s\\v = 0\ m/s\\t = 1.9\ s\\\therefore a = \dfrac{v-u}{t}\\\Rightarrow  a = \dfrac{0-282}{1.9}\\\Rightarrow  a =-148.42\\\textrm{Dividing both sides by }g\\\Rightarrow \dfrac{a}{g}=\dfrac{-148.42}{9.80}\\\Rightarrow \dfrac{a}{g}=-15.1\\\Rightarrow a=-15.1g[/tex]

Here, the negative sign represents that the motion of acceleration of the rocket is opposite to the direction of motion.

Hence, the acceleration of the rocket in the direction opposite to that of motion is 15.1g.

Final answer:

The response provides the calculation for Dr. John Paul Stapp's acceleration in his direction of motion and the acceleration opposite to his motion in terms of multiples of g.

Explanation:

Acceleration:

(a) Acceleration in his direction of motion: To find the acceleration in Stapp's direction of motion, we can use the formula a = Δv / Δt. Plugging in the values, we get a = (282 m/s) / (5.1 s) = 55.3 m/s². Converting this to multiples of g, we have 55.3 m/s² / 9.81 m/s² ≈ 5.6 g.

(b) Acceleration opposite to his direction of motion: The deceleration can be calculated using the same formula a = Δv / Δt. Substituting the values, we have a = (282 m/s) / (1.9 s) = 148.4 m/s². Expressing this in multiples of g, we get 148.4 m/s² / 9.81 m/s² ≈ 15.1 g.

Answer:

(a) 1.73 s

(b) 14.75 m

(c) 3.36 s

(d) double

(e) 63.32 m

Explanation:

Vertical component of initial velocity, uy = 17 m/s

Horizontal component of initial velocity, ux = 18.3 m/s

(A) At highest point of trajectory, the vertical component of velocity is zero. Let the time taken is t.

Use first equation of motion in vertical direction

vy = uy - gt

0 = 17 - 9.8 t

t = 1.73 seconds

(B) Let the highest point is at height h.

Use III equation of motion in vertical direction

[tex]v^{2}=u^{2}-2gh[/tex]

0 = 17 x 17 - 2 x 9.8 x h

h = 14.75 m

(C) The time taken by the ball to return to original level is T.

Use second equation of motion i vertical direction.

[tex]h = ut + 0.5at^2[/tex]

h = 0 , u = 17 m/s

0 = 17 t - 0.5 x 9.8 t^2

t = 3.46 second

(D) It is the double of time calculated in part A

(E) Horizontal distance = horizontal velocity x total time

d = 18.3 x 3.46 = 63.32 m

Answer:

(a) 508.37 m

(b) 47.53 s

(c) 21.165 m

(d) 19.365 m

Explanation:

initial velocity, u = 77 km/h = 21.39 m/s

acceleration, a = - 0.45 m/s^2

(a) final velocity, v = 0

Let the distance traveled is s.

Use third equation of motion

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

[tex]0^{2}=21.39^{2}-2 \times 0.45 \times s[/tex]

s = 508.37 m

(b) Let t be the time taken to stop.

Use first equation of motion

v = u + at

0 = 21.39 - 0.45 t

t = 47.53 s

(c) Use the formula for the distance traveled in nth second

[tex]s_{n^{th}=u+\frac{1}{2}a\left ( 2n-1 \right )}[/tex]

where n be the number of second, a be the acceleration, u be the initial velocity.

put n = 1, u = 21.39 m/s , a = - 0.45m/s^2

[tex]s_{n^{th}=21.39-\frac{1}{2}\times 0.45\left ( 2\times 1-1 \right )}[/tex]

[tex]s_{n^{th}=21.165m[/tex]

(d)  Use the formula for the distance traveled in nth second

[tex]s_{n^{th}=u+\frac{1}{2}a\left ( 2n-1 \right )}[/tex]

where n be the number of second, a be the acceleration, u be the initial velocity.

put n = 5, u = 21.39 m/s , a = - 0.45m/s^2

[tex]s_{n^{th}=21.39-\frac{1}{2}\times 0.45\left ( 2\times 5-1 \right )}[/tex]

[tex]s_{n^{th}=19.365m[/tex]

Answer:

carrier power is 16.1 kW

Explanation:

given data

power transmit P = 20 kW

modulation index m = 0.7

to find out

carrier power

solution

we will apply here power transmit equation that is express as

[tex]P = 1 + \frac{m^2}{2} * carrier power[/tex]   ...............1

put here all value in equation 1  we get carrier power

[tex]P = 1 + \frac{m^2}{2} * carrier power[/tex]

[tex]20 = 1 + \frac{0.7^2}{2} * carrier power[/tex]

solve it we get

carrier power = 16.1

so carrier power is 16.1 kW

A) -0.15 s

First of all, we need to velocity of the swimmer in absence of current. This is given by:

[tex]v_0 = \frac{d}{t}[/tex]

where

d = 50.0 m

t = 25.0 s

Substituting,

[tex]v_0 = \frac{50.0}{25.0}=2.0 m/s[/tex]

In lane 1, the velocity of the current is

[tex]v_c = +1.2 cm/s = +0.012 m/s[/tex]

where the + sign means it is in the same direction as the swimmer. Therefore, the net velocity of the swimmer in lane 1 will be

[tex]v=v_0+v_c = 2.0 + 0.012 = = 2.012 m/s[/tex]

And so, the time the swimmer will take to cover the 50.0 m will be:

[tex]t=\frac{d}{v}=\frac{50.0}{2.012}=24.85 s[/tex]

So, the time would change by

[tex]\Delta t = 24.85 - 25.0 = -0.15 s[/tex]

which means that the swimmer will be 0.15 s faster.

B) +0.15 s

To solve this part, we just need to consider that the current goes in the opposite direction, so its velocity actually is:

[tex]v_c = -12 cm/s = -0.012 m/s[/tex]

where the negative sign indicates the opposite direction.

So, the net velocity of the swimmer in lane 8 is

[tex]v=v_0+v_c = 2.0 - 0.012 = = 1.988 m/s[/tex]

And so, the time the swimmer will take to cover the 50.0 m will be:

[tex]t=\frac{d}{v}=\frac{50.0}{1.988}=25.15 s[/tex]

So, the time would change by

[tex]\Delta t = 25.15 - 25.0 = +0.15 s[/tex]

which means that the swimmer will be 0.15 s slower.

Answer:

1010 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-68^2}{2\times 250}\\\Rightarrow a=-9.248\ m/s^2[/tex]

[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-68}{-9.248}\\\Rightarrow t=7.35\ s[/tex]

Time taken by the thunderbird to stop is 7.35 seconds

Time the thunderbird was at the pit is 5 seconds

[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{68^2-0^2}{2\times 420}\\\Rightarrow a=5.5\ m/s^2[/tex]

[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{68-0}{5.5}\\\Rightarrow t=12.36\ s[/tex]

Time taken to accelerate back to 68 m/s is 12.36 seconds

Total time to this point is 7.35+5+12.36 = 24.71 seconds

The Mercedes Benz is moving at a constant velocity hence it has no acceleration and we use the formula

Distance = Speed × Time

⇒Distance = 68 × 24.71 = 1680 m

The thunderbird has covered 250+420 = 670 m

So, the distance between them is 1680-670 = 1010 m

Final answer:

To find out how far the Thunderbird has fallen behind the Mercedes Benz, we calculate the time taken for its deceleration, pit stop, and acceleration, then determine how far the Mercedes traveled in that time. The Thunderbird falls behind by approximately 1346.4 meters.

Explanation:

The Thunderbird's driver decelerates to a stop, pauses for 5 seconds, then accelerates back to speed. The Mercedes Benz continues at a constant speed of 68.0 m/s throughout this time. We will calculate the time taken for the Thunderbird to decelerate and accelerate, and use this to determine how far the Mercedes has traveled in the same period, thus finding out how far the Thunderbird has fallen behind.

Deceleration Phase

First, we need to find the Thunderbird's deceleration. Using the formula v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity (68.0 m/s), a is the acceleration, and s is the distance (250 m), we solve for a to find the deceleration rate. Rearranging the formula gives us a = (v^2 - u^2) / 2s. Plugging in the numbers, we find that the deceleration rate is approximately -9.184 m/s^2.

To find out how long this deceleration took, we use the formula v = u + at, solving for t gives us t = (v - u) / a. This calculation reveals that the deceleration phase lasted approximately 7.4 seconds.

Acceleration Phase

Assuming the Thunderbird accelerates at the same rate but in the positive direction over 420 m, we'll use the same method to find out the acceleration time. The acceleration time is also approximately 7.4 seconds.

Total time for stop and acceleration equals deceleration time + pit stop time + acceleration time, which totals 19.8 seconds.

The Mercedes Benz, traveling at 68.0 m/s, would have traveled 1346.4 meters in this time.

Therefore, the Thunderbird has fallen behind by 1346.4 meters, considering both the stopping and accelerating phases and the pit stop time.

Answer:

The gravitational force is significantly constant.

Explanation:

The gravitational force is expressed as:

F= G*M*m/d^2

G= Gravitational constant

M= in this case it is the mass of the planet

m= mass of the rock

d= distance of the rocks from the ground (suppose both at the same height for better comparison)

At the same time, we know that the force that each rock experiences is equal to the product of the mass due to acceleration:

F=m*a

We can match both expressions for F:

G*M*m/d^2 = m*a

we simplify m, and we obtain that the acceleration is independent of the mass of the attracted bodies:

a= G*M/d^2

Answer:

The possible value of power fluctuation is 3.9811 W

Solution:

As per the question:

Power output, P = 10 W

Power fluctuation is from +3 dB to -3 dB

Therefore,

Power fluctuation in dB = 6 dB

Now,

P (in dB) = [tex]10log_{10}P'[/tex]

6 dB = [tex]10log_{10}P'[/tex]

[tex]0.6 = log_{10}P'[/tex]

[tex]P' = 10^{0.6} W = 3.9811 W[/tex]

Explanation:

Speed of the marathon runner, v = 9.51 mi/hr

Distance covered by the runner, d = 26.220 mile

Let t is the time taken by the marathon runner. We know that the speed of the runner is given by total distance divided by total time taken. Mathematically, it is given by :

[tex]v=\dfrac{d}{t}[/tex]

[tex]t=\dfrac{d}{v}[/tex]

[tex]t=\dfrac{26.220\ mi}{9.51\ mi/hr}[/tex]

t = 2.75 hours

Since, 1 hour = 60 minutes

t = 165 minutes

Since, 1 minute = 60 seconds

t = 9900 seconds

Hence, this is the required solution.

Answer:

The correct option is 'D': Equal and opposite force.

Explanation:

We know from Newton's third law of motion

"For every action there exists an equal and opposite reaction".

The statement is further explained as

If there exists a force from Body 1 to Body 2 then Body 2 must also exert an equal and opposite force on the Body 1.

In simple terms we can say that the action and the reaction forces exist in pairs and are always present if any one among them is present.

Answer:

Acceleration, a = [tex]97.52 m/s^{2}[/tex]

t = 0.08 s

Solution:

As per the question:

The velocity of the player, v = 7.9 m/s

distance, d = 0.32 m

Now, consider the direction of the initial velocity of the player as positive and the acceleration to be constant:

Also, the final velocity of the player, v' = 0 m/s as he finally stops

Using the third eqn of motion:

[tex]v'^{2} = v^{2} - 2ad[/tex]

[tex]0 = 7.9^{2} - 2a\times 0.32[/tex]

a = [tex]97.52 m/s^{2}[/tex]

Also, From eqn (1) of motion:

v' = v - at

0 = 7.9 - 97.52t

t = 0.08 s

Answer:

The hydro static force on the back of the dam is [tex]1.96\times10^{11}\ N[/tex]

Explanation:

Given that,

Width b= 1000 m

Depth d= 200 m

We need to calculate the average pressure

Using formula of  average pressure

[tex]P_{avg}=\rho\times g\times d_{avg}[/tex]

Put the value into the formula

[tex]P_{avg}=1000\times9.8\times100[/tex]

[tex]P_{avg}=980000\ Pa[/tex]

We need to calculate  the hydro static force on the back of the dam

Using formula of force

[tex]F = P_{avg}\times A[/tex]

Put the value into the formula

[tex]F = 980000\times1000\times200[/tex]

[tex]F=1.96\times10^{11}\ N[/tex]

Hence, The hydro static force on the back of the dam is [tex]1.96\times10^{11}\ N[/tex]

Answer:

upward acceleration is 211.04 m/s²

Explanation:

given data

velocity = 14.6 m/s

distance = 0.505 m

to find out

upward acceleration

solution

we have given distance and velocity and ball is going upward

so acceleration will be calculated by  velocity formula that is

v² - u² = 2×a×s   ............1

here v is velocity and u is initial velocity and s is distance and a is acceleration

and u = o because starting velocity zero

put here all there value in equation 1

v² - u² = 2×a×s

14.6² - 0 = 2×a×0.505

solve it we get a

a = 211.04

so upward acceleration is 211.04 m/s²

Answer:

The bones are 16925 years old

Explanation:

We have to use the radioactive decay law and know that the half life of carbon-14 is [tex]t_{\frac{1}{2}}=5730 \, years[/tex]. From this information we can know the decay rate of the carbon 14,

[tex]\lambda=\frac{ln(2)}{t_{\frac{1}{2}}}=1.21\times 10^{-4} s^{-1}[/tex]

Now to know the age of the bones we must directly use the radioactive decay law:

[tex]N(t)=N_0e^{-\lambda t}=0.129N_0[/tex]

Where the rightmost part of the equation comes from the statement that the activity found is just 12.9% of the activity that would be found in a similar live animal. This means that the number of carbon-14 atoms is just 12.9% of what it was at the moment the saber-toothed tiger died.

Solving for t we have:

[tex]t=-\frac{ln(0.129)}{\lambda}=16925 \, years[/tex]

Answer:

the radii of curvature is 30 cm.

Explanation:

given,

object is place at = 45 cm

image appears at = 90 cm

focal length = ?

refractive index = 1.5

radii of curvature = ?

[tex]\dfrac{1}{f} = \dfrac{1}{u} +\dfrac{1}{v}[/tex]

[tex]\dfrac{1}{f} = \dfrac{1}{45} +\dfrac{1}{90}[/tex]

f = 30 cm

using lens formula

[tex]\dfrac{1}{f} = (n-1)(\dfrac{1}{R_1} -\dfrac{1}{R_2})[/tex]

[tex]R_1 = R\ and\ R_2 = -R[/tex]

[tex]\dfrac{1}{f} = (n-1)(\dfrac{1}{R} +\dfrac{1}{R})[/tex]

[tex]R = (n -1)\ f [/tex]

[tex]R = 2(1.5 -1)\ 30[/tex]

R = 30 cm

hence, the radii of curvature is 30 cm.

Answer:

[tex]V=23.3kV[/tex]

Explanation:

Definition of the capacitance C, where a voltage V is applied and a charge Q is stored:

[tex]Q=C*V[/tex]

We solve to find V:

[tex]V=Q/C=0.14*10^{-3}C/6*10^{-9}F)=2.33*10^{4}V=23.3kV[/tex]

The equations of motion are used to find the motion characteristics of an object with time

A) The time it takes the motorcycle to reach the car is approximately 5.39 seconds

B) The distance the motorcycle travel from the moment it starts to accelerate to the moment it catches up with the car is approximately 155 meters

The reason the above values are correct is as follows:

The given parameters are;

The initial speed of the car and the motorcycle, v₁ = 18.0 m/s

The initial distance between the motorcycle following the car = 58.0 m

The time after which the motorcycle starts to accelerate, t₁ = 5.00 secs

The acceleration of the motorcycle = 4.00 m/s²

The time at which the motorcycle catches up with the car = t₂

Required:

A) The time it takes the motorcycle to accelerate before it catches up with the car, which is to find t₂ - t₁

Method:

The distance moved by motorcycle to reach the car = The distance moved by the car in the same time + 58.0 meters

Solution;

The appropriate equation of motion to use are [tex]s = u \cdot t + \dfrac{1}{2} \cdot a\cdot t^2[/tex] and [tex]d = u\cdot t[/tex]

Where;

s = The distance between the motorcycle and the car = 58.0 m

u = The initial velocity of the motorcycle = 18.0 m/s

t = Δt = t₂ - t₁ = The time the motorcycle accelerates

a = The acceleration of the motorcycle = 4.00 m/s²

d = The distance moved by the car

By the method, we have;

[tex]s = d + 58.0[/tex]

[tex]u \cdot t + \dfrac{1}{2} \cdot a\cdot t^2 = u \cdot t[/tex]

Plugging in the values into the equation gives;

[tex](18.0 \times t) + \dfrac{1}{2} \times 4 \times t^2 = (18.0 \times t) + 58.0[/tex]

Cancelling the like term, (18.0 × t), on both sides of the equation gives;

[tex]\dfrac{1}{2} \times 4 \times t^2 = 58.0[/tex]

[tex]2\cdot t^2 = 58.0[/tex]

[tex]t = \Delta t = \sqrt{\dfrac{58.0}{2} } = \sqrt{29} \approx 5.39[/tex]

The time it takes the motorcycle to reach the car, t ≈ 5.39 seconds

The time it takes the motorcycle to reach the car is approximately 5.39 seconds

B) Required:

The distance the motorcycle travels from the moment it starts to accelerate, t₁ to the time it catches up with the car, t₂

Method:

The distance travelled during the time interval t₂ - t₁ should be calculated

Solution:

The distance, s, travelled during the time interval [tex]t_2 - t_1 = \Delta t = t[/tex] is given as follows;

[tex]s = u \cdot t + \dfrac{1}{2} \cdot a \cdot t^2[/tex]

From part (A), u = 18.0 m/s t = [tex]\sqrt{29}[/tex] s, and a = 4.00 m/s², therefore;

[tex]s = 18.0 \times \sqrt{29} + \dfrac{1}{2} \times 4.00 \times (\sqrt{29} )^2 \approx 154.933[/tex]

Rounded to three significant figures, the distance the motorcycle travel, s = 155 meters

Learn more about kinetic equation motion here:

https://brainly.com/question/16793944

To solve the question, we use the equations of motion to equate the distances traveled by the car and motorcycle. After solving the quadratic equation, we find the time the motorcycle catches up with the car and the distance it traveled.

The question involves a situation where a motorcycle is following a car traveling at a constant speed on a highway. Initially, both the car and the motorcycle are traveling at the same speed of 18.0 m/s, and there is a distance of 58.0 m between them. After 5.00 seconds, the motorcycle begins to accelerate at 4.00 m/s² until it catches up with the car. To find the time t2-t1, which is the duration the motorcycle catches up to the car, and the distance the motorcycle covers from the moment it starts to accelerate until it catches up with the car, we use the equations of motion.

We know that the car travels with a constant velocity, so the distance it covers in time t after the motorcycle starts accelerating can be given by:

d_car = v_car × t

For the motorcycle, which starts to accelerate at time t1, the distance it covers can be given by:

d_moto = d_initial + v_initial × (t - t1) + ½ × a × (t - t1)²

Given that the initial distance d_initial is 58.0 m, v_initial is 18.0 m/s, and a is 4.00 m/s², we can equate the distances for both the car and motorcycle when the motorcycle catches up:

18.0 × t = 58.0 + 18.0 × (t - 5.00) + ½ × 4.00 × (t - 5.00)²

By solving this quadratic equation, we can find the value of t, and then t2-t1 would simply be t - 5.00 seconds. Finally, to find out the distance the motorcycle traveled, we substitute the value of t into the distance equation for the motorcycle.

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Answer:

equation of  motion for Bill is

[tex]y(t) = 4.9t^2[/tex]

equation of  motion for Ted is

[tex]y(t) = 2 + (-4.2)(t) + 4.9t^2[/tex]

Explanation:

Taking downward position positive and upward position negative

g = 9.8 m/s^2

equation of  motion for Bill is

[tex]y(t) = y_0 +v_0 t +\frac{1}{2}gt^2[/tex]

[tex]y(t) = 0 + 0(t) +\frac{1}{2}gt^2[/tex]

[tex]y(t) = \frac{1}{2}\times (9.8t)^2[/tex]

[tex]y(t) = 4.9t^2[/tex]

equation of  motion for Ted is

[tex]y_0 = 2m -1m = 2m[/tex]

[tex]y_0 = -4.2 m/s[/tex]

[tex]y(t) = y_0 +v_0 t +\frac{1}{2}gt^2[/tex]

[tex]y(t) = 2 + (-4.2)(t) +\frac{1}{2}gt^2[/tex]

[tex]y(t) = 2 + (-4.2)(t) +\frac{1}{2}\times (9.8t)^2[/tex]

[tex]y(t) = 2 + (-4.2)(t) + 4.9t^2[/tex]

Answer:

Answer:

For Bill:

[tex]x(t)=3-(4.9*t^{2})[/tex]

For Ted:

[tex]x(t)=1+(4.2*t)+(-4.9*t^{2} )[/tex]

Explanation:

For Bill:

[tex]Initial position=x_{0}=3[/tex]

[tex]Initial velocity=v_{0}=0[/tex]

Now using [tex]2^{nd}[/tex] equation of motion,we have

[tex]x-x_{0}=(v_{0} *t)+(1/2*g*t^{2})[/tex]

[tex]x_{0} =3[/tex]  ,[tex]v_{0}=0[/tex]

Thus,equation becomes

[tex]x-3=1/2*g*t^{2}[/tex]

[tex]x=3+(0.5*g*t^{2})[/tex]

Taking acceleration upward positive and downward negative.

[tex]g=-10[/tex] [tex]m/s^{2}[/tex]

[tex]x(t)=3-4.9*t^{2}[/tex]    for bill

For Ted

[tex]x_{0} =1[/tex]

[tex]v_{0}=4.2[/tex] [tex]m/s[/tex]

Using the same equation

[tex]x-x_{0}=(v_{0} *t)+(1/2*g*t^{2})[/tex]

[tex]x_{0}=1[/tex] [tex]m[/tex]

[tex]v_{0}=4.2[/tex] [tex]m/s[/tex]

Substitute values

[tex]x-1=(4.2*t)+(1/2*g*t^{2})[/tex]

[tex]g=-10[/tex] [tex]m/s^{2}[/tex]

Thus equation becomes

[tex]x(t)=1+(4.2*t)+(-4.9*t^{2})[/tex]   for Ted

Answer:

a)  Y component of the vector =15.54 m

b) Vector magnitude = 21.6 m

Explanation:

The given vector makes 44 degree angle with Y axis, as given. This is same as 90 -44 = 46 degrees with the horizontal or X axis.

b) X component of the given vector = [tex]A_{x}[/tex] = A cos 46 =15

⇒ A = 15/cos 16 = 21.6 m = Total vector magnitude.

a) Y component of the vector = 21.6 sin 46 = 15.54 m

b) A = 21.6 m

The y-component of vector A (Ay) is approximately 10.3 m, assuming a positive y-direction due to the angle being in the second quadrant. The magnitude of vector A, found using the Pythagorean theorem with its components, is approximately 18.0 m.

To find the y-component of vector A (Ay), we use the trigonometric relationship Ay = A sin(θ), where A is the magnitude of the vector and θ is the angle it makes relative to the x-axis. However, since the angle given is clockwise from the y-axis, we must first convert it to the counterclockwise angle from the x-axis, which would be θ = 90° + 44.0° = 134.0°.

Given that Ax = -15.0 m, and assuming θ is measured counterclockwise, we can find the magnitude of A using the Pythagorean theorem: A = √(Ax² + Ay²). To find Ay, we rearrange the equation for the y-component to Ay = A sin(θ) and solve for Ay using the magnitude we just found. Our direction angle θ is in the second quadrant, meaning Ay should be positive.

Since we only have Ax, we first find A as follows: A = √((-15.0 m)² + Ay²). To determine Ay, we need to look at signs and quadrants. Ax is negative, and because the angle is measured clockwise from the y-axis, Ay must be positive. Therefore, we can infer that Ay is approximately 10.3 m by implication of the Pythagorean theorem, knowing that the other possible value is negative, which does not fit the quadrant.

To find the magnitude of A, we use the derived components: A = √((-15.0 m)² + (10.3 m)²), yielding approximately 18.0m.

Answer:

398 m/s

Explanation:

The acceleration is given by:

a = K/(L - x)^2

K = 100 m^3/s^2

L = 0.169 m

This acceleration will result in a force:

F = m * a

F = m * K/(L - x)^2

This force will perform a work:

W = F * L

The ball will advance only until x = L - D/2

[tex]W = m * K \int\limits^{L - D/2}_0 {\frac{1}{(L - x)^2}} \, dx[/tex]

This work will be converted to kinetic energy

W = Ek

Ek = 1/2 * m * v^2

[tex]1/2 * m * v^2 = m * K \int\limits^{L - D/2}_0 {\frac{1}{(L - x)^2}} \, dx[/tex]

[tex]1/2 * v^2 = K \int\limits^{L - D/2}_0 {\frac{1}{(L - x)^2}} \, dx[/tex]

First we solve thr integral:

[tex]K \int\limits^{L - D/2}_0 {\frac{1}{(L - x)^2}} \, dx[/tex]

We use the replacement

u = L - x

du = -dx

And the limits

When x = L - D/2, u = D2/2, and when x = 0, u = L

[tex]-K \int\limits^{D/2}_L {u^-2}} \, du[/tex]

K / u evaluated between L and D/2

2*K / D - K / L

Then

1/2 * v^2 = 2*K / D - K / L

1/2 * v^2 = K * (2/D - 1/L)

v^2 = 2*K*(2/D - 1/L)

[tex]v = \sqrt{2*K*(2/D - 1/L)}[/tex]

[tex]v = \sqrt{2*100*(2/0.0025 - 1/0.169)} = 398 m/s[/tex]

Answer:

a) greatest voltage = 29.25 V

b) power = 16 W

Explanation:

The total resistance R of the three resistors in series is:

[tex]R = (6.7 + 30.4 + 16.3) \Omega = 53.4 \Omega[/tex]  

a) The greatest current I is the one that will burn the resistor with lower power rating, which is 9.12 W:

[tex]P_{max} = I_{max}^2 R = I_{max}^2 30.4\Omega = 9.12W\\I_{max} = 0.54 A[/tex]

The voltage is:

[tex]V_{max}=IR = 0.54*53.4V= 29.25 V[/tex]

b) When the current is 0.54 A, the power is:

[tex]P = RI^2=53.4*0.3 W = 16W[/tex]

Answer:

temperature change is 262.06°K

Explanation:

given data

mass = 0.07 kg

velocity = 258 m/s

to find out

what is its temperature change

solution

we know here

heat change Q is is equal to kinetic energy that is

KE = 0.5 × m× v²   ...........1

here m is mass and v is velocity

KE = 0.5 × 0.07 × 258²

KE = 2329.74 J

and we know

Q = mC∆t     .................2

here m is mass and ∆t is change in temperature and C is 127J/kg-K

so put here all value

2329.74 = 0.07 × 127 × ∆t

∆t = 262.06

so temperature change is 262.06°K

Final answer:

The temperature change of a 0.07-kg lead bullet that had been traveling at 258 m/s before coming to a stop is found to be approximately 260°C.

Explanation:

Calculating Temperature Change of a Lead Bullet

To calculate the temperature change of a 0.07-kg lead bullet that comes to a stop after traveling at 258 m/s, we must first determine how much kinetic energy the bullet had before stopping. The kinetic energy (KE) of the bullet can be calculated using the formula KE = 1/2 [tex]mv^2[/tex], where m is the mass of the bullet and v is its velocity. Plugging in the values gives us:

KE = 1/2 * 0.07 kg * (258 [tex]m/s)^2[/tex]

KE = 0.035 kg * 66564 [tex]m^2/s^2[/tex]

KE = 2330.74 J

This kinetic energy is converted to heat energy, which we will denote as Q. The temperature change (\ΔT) is then given by the formula Q = mc\ΔT, where m is the mass of the bullet and c is the specific heat capacity of lead.

Since c for lead is approximately 128 J/kg°C, we can rearrange the formula to solve for \Deltat:

ΔT = Q / (mc)

ΔT = 2330.74 J / (0.07 kg * 128 J/kg°C)

ΔT = 2330.74 J / 8.96 J/°C

ΔT = 259.96°C

Therefore, the temperature change of the bullet is approximately 260°C.

Answer:

Acceleration of the car, [tex]a=8\ m/s^2[/tex]

Explanation:

Given that,

Initial speed of the car, u = 0 (at rest)

Final speed of the car, v = 40 m/s

Distance covered, s = 100 m

We need to find the acceleration of the car. It can be calculated using third equation of motion as :

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

a is the acceleration of the car

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

[tex]a=\dfrac{(40)^2}{2\times 100}[/tex]

[tex]a=8\ m/s^2[/tex]

So, the acceleration of the car is [tex]8\ m/s^2[/tex]. Hence, this is the required solution

Answer:

a) v=2.743m/s

b) [tex]a_c = 8.363m/s^2[/tex]

c) T=2.543N

Explanation:

First, calculate the height of the ball at the starting point:

[tex]y' = 0.9cos(55)[/tex]

[tex]y' = 0.516[/tex]

At this point, just in the moment the ball is released, all the energy of the system is potencial gravitational energy. When it is at the bottom all the potencial energy is transformed into kinetic energy:

[tex]E_p=E_k\\mgh=\frac{mv^2}{2}[/tex]

Solving for v:

[tex]v=\sqrt{2gh}[/tex]

if h is the height loss: (l-y')

v=2.743m/s

The centripetal acceleration is the acceleration caused by the tension force exercised by the string, and is pointing outside of the trayectory path (at the lowest point, directly dawn):

[tex]a_c=\frac{v^2}{r}[/tex]

[tex]a_c = 8.363m/s^2[/tex]

To calculate tension, just make the free body diagram of forces in the ball, noticing the existence of the centripetal acceleration:

[tex]\sum{F_y}=ma_c=T-W\\T=ma_c+W\\T=m(a_c+g)\\T=0.14(8.363+9.8)\\T=2.543N[/tex]

Final answer:

Alcohol and mercury thermometers may not read the same at 60°C despite being calibrated at 0°C and 100°C because they have different rates of thermal expansion, with mercury being more linear and alcohol having a higher expansion coefficient.

Explanation:

Alcohol and mercury thermometers are calibrated at two fixed points: the freezing point of water (0°C) and the boiling point of water (100°C). However, these substances expand at different rates over the temperature range, which means the expansion is not linear. While both may be divided into 100 equal divisions between these two points, an alcohol thermometer and a mercury thermometer may not give the same reading at a temperature such as 60°C because their rates of thermal expansion differ. Mercury has a more linear expansion with temperature compared to alcohol, which has a higher coefficient of expansion, causing its physical expansion for an equal temperature increment to be larger. This inconsistency is often overlooked because it is somewhat small over the range of temperatures we normally encounter, but it becomes significant at more extreme temperatures.

Answer:

speed of sound = 321 m/s

Rule of thumb for kilometers: 1 km every 3 seconds

Explanation:

Hi!

If a thunder originates at a place a distance D from you, and it takes time T for its sound to reach you, then:

[tex]V = \frac{D}{T}[/tex]

Whre V is the speed of sound. Time T is the time elapsed between the moment you see the flash (becuase of the assumption of tha it takes no time for the light to reach you) and the moment you hear the thunder. Then

[tex]V = \frac{1mile}{5\;s} =\frac{1609.34 \;m}{5\;s} = 321\frac{m}{s}[/tex]

To calculate the rule in kilometers:

[tex]T = \frac{1\;km}{321*10^{-3} \frac{km}{s}} \approx 3\; s[/tex]

The estimated speed of sound derived from the rule of thumb is approximately 322 m/s. For the rule in kilometers, each five seconds would represent about 1.609 kilometers.

The rule-of-thumb that every five seconds between a lightning flash and the accompanying thunder equates to one mile of distance is based on the speed of sound. To convert this to meters per second we must know that 1 mile approximately equals 1609.34 meters. Therefore, if light travels nearly instantaneously, and one mile is covered in five seconds, the speed of sound can be estimated as about 1609.34 m/5 s or approximately 322 m/s.

For the rule in kilometers, we need to convert miles into kilometers. As 1 mile is approximately 1.609 kilometers, the rule of thumb in kilometers would be that each five seconds would represent 1.609 kilometers.

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