Donna drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 8 hours. When Donna drove home, there was no traffic and the trip only took 4 hours. If her average rate was 45 miles per hour faster on the trip home, how far away does Donna live from the mountains?

Answer:

The plane needs 1,56 seconds to clear the intersection.

Explanation:

This is a case of uniformly accelerated rectilinear motion.

[tex]V_0^{2} = V_f^{2} - 2ad[/tex]

[tex]V_0=\sqrt{V_0^{2} } = ?[/tex]

Vf=50 m/s

[tex]V_f^{2}  = (50 m/s)^{2} = 2500  m^{2}/s^{2}[/tex]

a = -5.4 [tex]m/s^{2}[/tex] (Negative because is decelerating)

d = displacement needed to clear the intersection. It should be the width of the intersection plus the lenght of the plane.

d= 59,7m + 25 m = 84.7 m

Calculating [tex]V_0[/tex]:

[tex]V_0^{2} = V_f^{2} - 2ad[/tex]

[tex]V_0^{2}= 2500 \frac{m^{2} }{s^{2} } - 2(-5.4\frac{m}{s^{2} })(84.7 m)[/tex]

[tex]V_0^{2}= 3,414.76 \frac{m^{2} }{s^{2} }[/tex]

[tex]V_0= \sqrt{3,414.76} = 58.44 \frac{m}{s}[/tex]

Otherwise:

[tex]t = \frac{V_f-V_0}{a} =\frac{50\frac{m}{s} - 58.44\frac{m}{s}  }{-5.4 \frac{m}{s^{2} } } = 1.56 s[/tex]

The time needed for the plane to clear the intersection is approximately 1.88 s.

Given data:

- Length of the plane (L) = 59.7 m

- Width of the intersection (d) = 25.0 m

- Deceleration (a) = 5.4 m/s²

- Final speed (v) = 50 m/s

The plane decelerates through the intersection. The final speed v and the deceleration a are given. The total distance the plane covers while decelerating through the intersection is the sum of the length of the plane and the width of the intersection:

[tex]\[ \text{Total distance} = L + d = 59.7 \, \text{m} + 25.0 \, \text{m} = 84.7 \, \text{m} \][/tex]

Determine the initial speed [tex](v_0)[/tex]

Using the kinematic equation:

[tex]\[ v^2 = v_0^2 + 2a \Delta x \][/tex]

We rearrange to solve for [tex]\( v_0 \)[/tex]:

[tex]\[ v_0^2 = v^2 - 2a \Delta x \][/tex]

[tex]\[ v_0^2 = (50 \, \text{m/s})^2 - 2 \times 5.4 \, \text{m/s}^2 \times 84.7 \, \text{m} \]\[ v_0^2 = 2500 - 2 \times 5.4 \times 84.7 \]\[ v_0^2 = 2500 - 913.56 \]\[ v_0^2 = 1586.44 \]\[ v_0 = \sqrt{1586.44} \]\[ v_0 \approx 39.83 \, \text{m/s} \][/tex]

Calculate the time (t)

Using the kinematic equation:

[tex]\[ v = v_0 + at \][/tex]

Rearrange to solve for t:

[tex]\[ t = \frac{v - v_0}{a} \]\[ t = \frac{50 \, \text{m/s} - 39.83 \, \text{m/s}}{5.4 \, \text{m/s}^2} \]\[ t = \frac{10.17 \, \text{m/s}}{5.4 \, \text{m/s}^2} \]\[ t \approx 1.88 \, \text{s} \][/tex]

Answer:

89.16pounds

Explanation:

The equation that defines this problem is as follows

W=k/X^2

where

W=Weight

K= proportionality constant

X=distance from the center of Earth

first we must find the constant of proportionality, with the first part of the problem

k=WX^2=131x3960^2=2054289600pounds x miles^2

then we use the equation to calculate the woman's weight with the new distance

W=2054289600/(4800)^2=89.16pounds

Answer:

a=8.1832m/s^2

Explanation:

Vo=initial speed=0m/s

Vf=final speed=26.8m/s

t=time=3.275s

the vehicle moves with a constant acceleration therefore we can use the following equation

A=aceleration=(Vf-Vi)/t

A=(26.8m/s-0m/s)/3.275s=8.1832m/s^2

the magnitude of the car´s aceleration is 8.1832m/s^2

Final answer:

To calculate the angle of the second displacement, typically one would use vector addition and trigonometry, but the question, as presented, does not provide sufficient information to determine this angle.

Explanation:

To find the angle of the second displacement for the particle, we need to consider the information given: the first displacement is 11 m at an angle of 82° from the positive x-axis, and the resulting displacement after the second move is 8.7 m at 135° to the positive x-axis, using the counterclockwise direction as positive.

Using vector addition, the resulting displacement's direction can help us determine the second displacement angle. However, without more information about the specifics of the second displacement (like its magnitude), we cannot calculate the exact angle or vector using typical methods such as vector components or the law of cosines.

Given the problem as stated, assuming standard vector addition, a common approach would involve drawing a vector diagram and applying trigonometry; but in this case, we seem to be missing details to perform accurate calculations.

Answer:

The moist filter paper is in charge of preventing evaporation and ensuring the proper saturation of the air of the chromatography chamber.

Explanation:

Thin-layer chromatography (TLC) is a chromatography technique used to separate non-volatile mixtures.

A strip of moist filter paper is put into the chromatography chamber so that its bottom touches the solvent and the paper lies on the chamber wall and reaches almost to the top of the container.

The container is closed and left for a few minutes to let the solvent vapors ascend the moist filter paper and saturate the air in the chamber.

The moist filter paper is in charge of preventing evaporation and ensuring the proper saturation of the air of the chamber.

Answer:

A)t=1.375s

B)t=11s

Explanation:

for this problem we will assume that the east is positive while the west is negative, what we must do is find the relative speed between the wave and the powerboat, and then with the distance find the time for each case

ecuations

V=Vw-Vp  (1)

V= relative speed

Vw= speed of wave

Vp=Speesd

t=X/V(2)

t=time

x=distance=44m

A) the powerboat moves to west

V=18-(-14)=32m/s

t=44/32=1.375s

B)the powerboat moves to east

V=18-14=4

t=44/4=11s

Final answer:

The frequency at which the boat encounters a wave crest is 0.09 Hz regardless of the direction the boat is traveling.

Explanation:

To find the frequency at which the boat encounters a wave crest, we need to determine the time it takes for one wave crest to pass by the boat. The speed of the boat relative to the ocean floor doesn't affect the frequency, only the speed of the wave. The formula to calculate the frequency of a wave is:

Frequency = Wave Speed / Wavelength

Wave speed = Speed of the wave + Speed of the boat (since the boat is traveling in the opposite direction)
Frequency = (Speed of the wave + Speed of the boat) / Wavelength
Frequency = (18 m/s + (-14 m/s)) / 44 m
Frequency = 4 m/s / 44 m = 0.09 Hz

Wave speed = Speed of the wave - Speed of the boat (since the boat is traveling in the same direction)
Frequency = (Speed of the wave - Speed of the boat) / Wavelength
Frequency = (18 m/s - 14 m/s) / 44 m
Frequency = 4 m/s / 44 m = 0.09 Hz

Answer:

The height of the ball after 5 seconds is 162 ft.

Explanation:

First, replace the variable t with how many seconds the ball has dropped, which in this case is 5.

[tex]h = 512 - 14 {t}^{2} \\ h = 512 - 14 \times {5}^{2} [/tex]

Solve.

[tex]h = 512 - 14 \times {5}^{2} \\ h = 512 - 14 \times 25 \\ h = 512 - 350 \\ h = 162[/tex]

For a duration of 5 seconds, the ball had managed to drop 350 feet, with 162 feet left to go to touch ground level.

Answer : The cell emf for this cell is 0.077 V

Solution :

The balanced cell reaction will be,  

Oxidation half reaction (anode):  [tex]Zn(s)\rightarrow Zn^{2+}+2e^-[/tex]

Reduction half reaction (cathode):  [tex]Zn^{2+}+2e^-\rightarrow Zn(s)[/tex]

In this case, the cathode and anode both are same. So, [tex]E^o_{cell}[/tex] is equal to zero.

Now we have to calculate the cell emf.

Using Nernest equation :

[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}{diluted}}{[Zn^{2+}{concentrated}]}[/tex]

where,

n = number of electrons in oxidation-reduction reaction = 2

[tex]E_{cell}[/tex] = ?

[tex][Zn^{2+}{diluted}][/tex] = 0.0111 M

[tex][Zn^{2+}{concentrated}][/tex] = 4.50 M

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

[tex]E_{cell}=0-\frac{0.0592}{2}\log \frac{0.0111M}{4.50M}[/tex]

[tex]E_{cell}=0.077V[/tex]

Therefore, the cell emf for this cell is 0.077 V

The cell emf, or electromotive force, for a voltaic cell, like the Zn2+-Zn kind described, can be calculated with the Nernst equation. The emf is determined by the standard electrode potential as well as the concentrations of the redox species in each half-cell.

A voltaic cell runs on a spontaneous redox reaction occurring indirectly, with the oxidant and reductant redox couples contained in separate half-cells. In your example, both half-cells are Zn2+-Zn electrodes where the reaction is Zn2+ + 2e− → Zn (s) with a standard electrode potential, E°, of -0.763 V. The cell emf, or electromotive force, is calculated through the Nernst Equation:

Ecell = E° - (RT/nF) * ln(Q)

where Q is the reaction quotient, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred in the redox reaction, and F is Faraday's constant. Since we're dealing with a concentration cell, Q is given by the ratio of the concentrations of the reduced and oxidized species. Depending on the temperature and assuming that Zn2+ is reduced at the cathode and oxidized at the anode, the cell emf can be calculated accordingly.

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Answer:

209 m

Explanation:

The y-component of a vector is the magnitude times the sine of the angle.

y = 253 sin 55.8°

y = 209

Answer:

The answer to your question is: 209 m

Explanation:

Data

length = 253 m

d = 55.8°

y - component = ?

Formula

To solve this problem we need to use a right triangle a the trigonometric functions (sine, cosine, tangent, etc)

Now we have to choose among the trigonometric functions which one to use that relates the opposite side and the hypotenuse.

And that one is sineФ = os/h

we clear os = sineФ x h

     os = sine 55.8 x253

     os =  209 m

soh, cah, toa

Answer:

[tex]-0.44 m/s^2[/tex]

Explanation:

First of all, we need to calculate the distance covered by the locomotive during the reaction time, which is

t = 0.46 s

During this time, the locomotive travels at

v = 13 m/s

And the motion is uniform, so the distance covered is

[tex]d_1 = vt = (13)(0.46)=6.0 m[/tex]

The locomotive was initially 200 m from the crossing, so the distance left to stop is now

[tex]d=200 - 6.0 = 194.0 m[/tex]

And now the locomotive has to slow down to a final velocity of [tex]v=0[/tex] in this distance. We can find the minimum deceleration needed by using the suvat equation:

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

where

v = 0 is the final velocity

u = 13 m/s is the initial velocity

a is the deceleration

d = 194.0 m is the distance to stop

Solving for a,

[tex]a=\frac{v^2-u^2}{2d}=\frac{0^2-13^2}{2(194)}=-0.44 m/s^2[/tex]

The text looks incomplete. Here the complete question found on google:

"The acceleration of a bus is given by ax(t)=αt, where α = 1.28m/s3 is a constant. Part A If the bus's velocity at time t1 = 1.13s is 5.09m/s , what is its velocity at time t2 = 2.02s ? If the bus's position at time t1 = 1.13s is 5.92m , what is its position at time t2 = 2.02s ?"

A) 6.88 m/s

The velocity of the bus can be found by integrating the acceleration. Therefore:

[tex]v(t) = \int a(t) dt = \int (\alpha t)dt=\frac{1}{2}\alpha t^2+C[/tex] (1)

where

[tex]\alpha = 1.28 m/s^3[/tex]

C is a constant term

We know that at [tex]t_1 = 1.13 s[/tex], the velocity is [tex]v=5.09 m/s[/tex]. Substituting these values into (1), we can find the exact value of C:

[tex]C=v(t) - \frac{1}{2}\alpha t^2 = 5.09 - \frac{1}{2}(1.28)(1.13)^2=4.27 m/s[/tex]

So now we can find the velocity at time [tex]t_2 = 2.02 s[/tex]:

[tex]v(2.02)=\frac{1}{2}(1.28)(2.02)^2+4.27=6.88 m/s[/tex]

B) 11.2 m

To find the position, we need to integrate the velocity:

[tex]x(t) = \int v(t) dt = \int (\frac{1}{2}\alpha t^2 + C) dt = \frac{1}{3}\alpha t^3 + Ct +D[/tex] (2)

where D is another constant term.

We know that at [tex]t_1 = 1.13 s[/tex], the position is [tex]v=5.92 m[/tex]. Substituting these values into (2), we can find the exact value of D:

[tex]D = x(t) - \frac{1}{6}\alpha t^3 -Ct = 5.92-\frac{1}{6}(1.28)(1.13)^3 - (4.27)(1.13)=0.79 m[/tex]

And so now we can find the position at time [tex]t_2 = 2.02 s[/tex] using eq.(2):

[tex]x(2.02)=\frac{1}{6}(1.28)(2.02)^3 + (4.27)(2.02) +0.79=11.2 m[/tex]

The question deals with calculating the time-dependent acceleration of a bus using the equation ax(t)=αt, applicable to Physics (specifically mechanics), and is suitable for a high school level.

The question pertains to the acceleration of an object, specifically a bus, which makes this a Physics problem. Acceleration is defined as the change in velocity over time and is a vector quantity having both magnitude and direction. Given the equation ax(t)=αt, where α is a constant equal to 1.28 m/s3, we can recognize that the acceleration is linearly increasing over time because for each second, the acceleration increases by 1.28 m/s2. This kind of problem typically appears in high school Physics.

For instance, to calculate the acceleration of a bus at a specific time point, we simply multiply the constant α by the time t at which we want to know the acceleration. So if we wanted to know the acceleration at t=15 s, we would calculate it as ax(15s) = 1.28 m/s3 × 15 s = 19.2 m/s2. This demonstrates that as time progresses, the acceleration value increases.

Answer:

airplane speed 135mph windspeed 45 mph

Explanation:

This information helps us to write down a system of linear equations

When going head wind, the speed of the wind is substracted from that of the airplane and on the return trip it is added, then:

A:=Airlplane speed

W:= Wind speed

(A+W)*1h=180mi (1)

(A-W)*2h=180mi (2)

then from (1) A=180-W (3), replacing this in (2) we get (180-W-W)*2h =180mi, then

360-4W=180, or 180=4W, then W=45 mph. Replacing this in (3) we have that A=180-45=135 mph.

The problem was solved by setting up equations based on the given distances and times. By solving these equations, it was determined that the plane's speed in still air is 105 miles per hour, and the wind's speed is 15 miles per hour.

To solve this problem, we can set up two equations using the relationship between distance, speed, and time. We will denote the plane's speed in still air as P and the wind's speed as W.

When flying against the wind, the plane's effective speed is P - W, and the time taken to cover the 180 miles is 2 hours. We can represent this with the equation:
180 = 2(P - W) ... (1)

On the return trip, with a tailwind, the plane's effective speed is P + W, and the time taken is 1.5 hours. This gives us a

second equation:
180 = 1.5(P + W) ... (2)

By solving these two equations simultaneously, we can find the values of P and W. Multiplying equation (2) by 2 to eliminate the fractions, we get:
360 = 3(P + W) ... (2')

Now, we subtract equation (1) from equation (2'):
360 - 180 = 3(P + W) - 2(P - W)
180 = 3P + 3W - 2P + 2W
180 = P + 5W

Using equation (1), we express P in terms of W:
180 = 2P - 2W
P = 90 + W ... (3)

Substituting (3) into the equation we got after subtracting:
180 = (90 + W) + 5W
180 = 90 + 6W
90 = 6W
W = 15 miles per hour

Now, we substitute the value of W back into equation (3) to find P:
P = 90 + 15
P = 105 miles per hour

The plane's speed in still air is 105 miles per hour and the wind's speed is 15 miles per hour.

Recrystallization is a temperature-induced process that can vary among materials. Metal parts may recrystallize at different temperatures, and recrystallization can happen unevenly within a single part.

Recrystallization is a process where a material undergoes structural changes due to increased temperature. Metallic glasses can have different recrystallization temperatures based on the composition of the metals. Recrystallization can occur non-uniformly in a part, with some regions undergoing the process before others due to variations in temperature or cooling rates.

Answer:

magnitude=34.45 m

direction=[tex]55.52\°[/tex]

Explanation:

Assuming the initial point P1 of this vector is at the origin:

P1=(X1,Y1)=(0,0)

And knowing the other point is P2=(X2,Y2)=(19.5,28.4)

We can find the magnitude and direction of this vector, taking into account a vector has a initial and a final point, with an x-component and a y-component.

For the magnitude we will use the formula to calculate the distance [tex]d[/tex] between two points:

[tex]d=\sqrt{{(Y2-Y1)}^{2} +{(X2-X1)}^{2}}[/tex] (1)

[tex]d=\sqrt{{(28.4 m - 0 m)}^{2} +{(19.5 m - 0m)}^{2}}[/tex] (2)

[tex]d=\sqrt{1186.81 m^{2}}[/tex] (3)

[tex]d=34.45 m[/tex] (4) This is the magnitude of the vector

For the direction, which is the measure of the angle the vector makes with a horizontal line, we will use the following formula:

[tex]tan \theta=\frac{Y2-Y1}{X2-X1}[/tex]  (5)

[tex]tan \theta=\frac{24.8 m - 0m}{19.5 m - 0m}[/tex]  (6)

[tex]tan \theta=\frac{24.8}{19.5}[/tex]  (7)

Finding [tex]\theta[/tex]:

[tex]\theta= tan^{-1}(\frac{24.8}{19.5})[/tex]  (8)

[tex]\theta= 55.52\°[/tex]  (9) This is the direction of the vector

Answer:

The distance traveled by the swan is 60.35 meters.

Explanation:

Given that,

A swan accelerate from rest (u = 0) to 6.5 m/s to take off.

Acceleration of the swan, [tex]a=0.35\ m/s^2[/tex]

We need to find the distance Δx it travel before becoming airborne. From the third equation of motion as :

[tex]\Delta x=\dfrac{v^2-u^2}{2a}[/tex]

[tex]\Delta x=\dfrac{(6.5)^2}{2\times 0.35}[/tex]

[tex]\Delta x=60.35\ m[/tex]

So, the distance traveled by the swan is 60.35 meters. Hence, this is the required solution.

Answer:

The answer to your question is:

a) t = 3.64 s

b) vox = 31.59 m/s

c) vy = 35.71 m/s

d) v = 47.67 m/s

Explanation:

a) To calculate the time, we know that voy = 0 m/s so we this this formula

   h = voy + 1/2(gt²)

   65  = 0 + 1/2(9.81)(t²)

   65 = 4.905t²

   t² = 65/4.905 = 13.25

   t = 3.64 s

b) To calculate vox  we use this formula vox = d/t ; vox is constant

   vox = 115/3.64 = 31.59 m/s

c) To calculate voy we use the formula    vy = voy + gt

but voy = 0

               vy = gt = 9.81 x 3.64 = 35.71 m/s

d) To calculate v  we use the pythagorean theorem

            c2 = a2 + b2

            c2 = 31.59² + 35.71² = 997.92 + 1275.20

            c2 = 2273.12

              c = 47.67 m/s

Answer:

A.[tex]a=203.14\ \frac{m}{s^2}[/tex]

B.s=397.6 m

Explanation:

Given that

speed  u= 284.4 m/s

time t = 1.4 s

here he want to reduce the velocity from 284.4 m/s to 0 m/s.

So the final speed v= 0 m/s

We know that

v= u + at

So now by putting the values

0 = 284.4 -a x 1.4     (here we take negative sign because this is the case of de acceleration)

[tex]a=203.14\ \frac{m}{s^2}[/tex]

So the acceleration  while stopping will be [tex]a=203.14\ \frac{m}{s^2}[/tex].

Lets take distance travel before come top rest is s

We know that

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

[tex]0=284.4^2-2\times 203.14\times s[/tex]

s=397.6 m

So the distance travel while stopping is 397.6 m.

Answer:

[tex]a) acceleration = -203.14 m/s^2\\\\b) distance = 199.1 m[/tex]

Explanation:

Given

initial speed u = 284.4 m/s

final speed v = 0 m/s

duration t = 1.4 s

Solution

a)

Acceleration

[tex]a = \frac{v - u}{t} \\\\a = \frac{0-284.4}{1.4} \\\\a = -203.14 m/s^2[/tex]

b)

Distance

[tex]v^2  - u^2 = 2as\\\\0^2 - 284.4^2 = 2 \times  (-203.14) \times S\\\\S = 199.1 m[/tex]

Answer:

0.006 N

Explanation:

First find the mass of the horse which is the only unknown variable.

Then use it with the new data. Working out is in the attachment.

Answer:

[tex]\Delta S = 2.11 J/K[/tex]

Explanation:

As we know that entropy change for phase conversion is given as

[tex]\Delta S = \frac{\Delta Q}{T}[/tex]

Here we know that heat required to change the phase of the water is given as

[tex]\Delta Q = mL[/tex]

here we have

[tex]m = 0.349 g = 3.49\times 10^{-4} kg[/tex]

L = 2250000 J/kg

now we have

[tex]\Delta Q = (3.49 \times 10^{-4})\times 2250000[/tex]

[tex]\Delta Q = 788.9 J[/tex]

also we know that temperature is approximately same as boiling temperature

so we have

[tex]T = 373 k[/tex]

so here we have

[tex]\Delta S = \frac{788.9}{373}[/tex]

[tex]\Delta S = 2.11 J/K[/tex]

Answer:

x = +7, x = -7

Explanation:

The equation to solve is:

[tex]\left|\frac{x}{7}\right|=1[/tex]

Which means that the absolute value of the fraction [tex]\frac{x}{7}[/tex] must be equal to 1. Since we have an absolute value, we have basically two equations to solve:

1) [tex]\frac{x}{7}=+1[/tex]

Solving this one, we find

[tex]\frac{x}{7}\cdot 7 = 1\cdot 7 \rightarrow x = +7[/tex]

2) [tex]\frac{x}{7}=-1[/tex]

Solving this one, we find

[tex]\frac{x}{7}\cdot 7 = -1\cdot 7 \rightarrow x = -7[/tex]

So the two solutions are +7 and -7.

Final answer:

A person can produce approximately 85 W of useful electric power using a bicycle-powered generator when factoring in the typical 25% body efficiency and an 85% efficient generator.

Explanation:

A typical person can maintain a steady energy expenditure of 400 W on a bicycle. Assuming a typical efficiency for the body and a generator that is 85% efficient, the useful electric power produced can be calculated as follows:

First, consider the efficiency of the human body. If we assume that the body is 25% efficient (which means that only 25% of the energy consumed by the body is converted to mechanical work, while the rest is lost as heat), then the actual mechanical work a person can produce is 25% of their energy expenditure.

Therefore, the mechanical work output would be:

Next, we account for the efficiency of the generator:

To conclude, a person can produce approximately 85 W of useful electric power with a bicycle-powered generator when considering typical human body efficiency and an 85% efficient generator.

Answer:

For cardboard = 29.4 g/cm²

For aluminium = 113.4 g/cm²

For lead = 193.8 g/cm²

Explanation:

Given:  

Density of the cardboard, d₁ = 0.6 g/cm³

Density of the aluminium, d₂ = 2.7 g/cm³

Density of the lead, d₃ = 11.4 g/cm³

Length of the cardboard,  L₁ = 49 cm

Length of the aluminium, L₂ = 42 cm

Length of the lead, L₃ = 17 cm

Now,

The absorber thickness is calculated as:

= Density × Length

therefore,

For cardboard = d₁ × L₁ = 0.6 × 49 = 29.4 g/cm²

For aluminium = d₂ × L₂ = 2.7 × 42 = 113.4 g/cm²

For lead = d₃ × L₃ = 11.4 × 17 = 193.8 g/cm²

Answer:

Normal force: 167.48 N

Explanation:

        ∑[tex]F_{x}[/tex]:  [tex]F{x}-20 = 0[/tex]  

        ∑[tex]F_{y}[/tex]: [tex]N -W+F_{y}=0[/tex]

       [tex]N = W - Fy[/tex]

        [tex]N = mg -Fsin(\theta)[/tex]

        [tex]Fcos(\theta) -20 = 0[/tex]   ⇒   [tex]Fcos(\theta) = 20[/tex][tex]\theta                  

        [tex]\theta=cos^{-1}(20/F)[/tex]

       [tex]N = 20 x 9.81 - 35sin(55.15)[/tex]  ⇒ [tex]N = 167.48 N[/tex]

Answer:

314.92 m

Explanation:

Acceleration of the helicopter = 5.4 m/s² = a

Time taken by the helicopter to reach maximum height = 10.8 s = t

Initial velocity = 0 = u

Final velocity = v

Displacement = s

Equation of motion

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times 10.8+\frac{1}{2}5.4\times 10.8^2\\\Rightarrow s=314.92\ m[/tex]

Maximum height above ground reached by the helicopter is 314.92 m

Answer:

Explanation:

[tex]\overrightarrow{A} = 3\widehat{i}+3\widehat{j}[/tex]

[tex]\overrightarrow{B} = \widehat{i}-4\widehat{j}[/tex]

[tex]\overrightarrow{C} = -2\widehat{i}+5\widehat{j}[/tex]

(a)

[tex]\overrightarrow{D} =\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}[/tex]

[tex]\overrightarrow{D} =\left ( 3+1-2 \right )\widehat{i} +\left ( 3-4+5 \right )\widehat{j}[/tex]

[tex]\overrightarrow{D} =\left 2\widehat{i} +4\widehat{j}[/tex]

Magnitude of [tex]\overrightarrow{D}[/tex] = [tex]\sqrt{2^{2}+4^{2}}[/tex]

                                                                     = 4.47 m

Let θ be the direction of vector D

[tex]tan\theta =\frac{4}{2}[/tex]

θ = 63.44°

(b)

[tex]\overrightarrow{E} =

- \overrightarrow{A}-\overrightarrow{B}+\overrightarrow{C}[/tex]

[tex]\overrightarrow{E} =\left ( - 3- 1 -2 \right )\widehat{i} +\left ( - 3 + 4+5 \right )\widehat{j}[/tex]

[tex]\overrightarrow{E} =- \left 6\widehat{i} +6\widehat{j}[/tex]

Magnitude of [tex]\overrightarrow{E}[/tex] = [tex]\sqrt{6^{2}+6^{2}}[/tex]

                                                                     = 8.485 m

Let θ be the direction of vector D

[tex]tan\theta =\frac{6}{-6}[/tex]

θ = 135°

a) For vector [tex]\( \mathbf{D} \)[/tex]: Magnitude: [tex]\( 4 \, \text{meters} \)[/tex] ,Direction: [tex]\( 90^\circ \)[/tex]

b) For vector [tex]\( \mathbf{E} \)[/tex]: Magnitude: [tex]\( 7.21 \, \text{meters} \)[/tex] ,Direction: [tex]\( 123.69^\circ \)[/tex]

To find the resultant vectors [tex]\( \mathbf{D} = \mathbf{A} + \mathbf{B} + \mathbf{C} \) and \( \mathbf{E} = -\mathbf{A} - \mathbf{B} + \mathbf{C} \)[/tex] using the component method, we need to break each vector into its [tex]\( i \)[/tex] and [tex]\( j \)[/tex] components, sum these components, and then find the magnitude and direction of the resulting vectors.

Given vectors:

[tex]\[ \mathbf{A} = 3\mathbf{i} + 3\mathbf{j} \][/tex]

[tex]\[ \mathbf{B} = -\mathbf{i} - 4\mathbf{j} \][/tex]

[tex]\[ \mathbf{C} = -2\mathbf{i} + 5\mathbf{j} \][/tex]

a) Vector [tex]\(\mathbf{D} = \mathbf{A} + \mathbf{B} + \mathbf{C}\)[/tex]

1. Sum the components:

[tex]\[ \mathbf{D} = (3\mathbf{i} + 3\mathbf{j}) + (-\mathbf{i} - 4\mathbf{j}) + (-2\mathbf{i} + 5\mathbf{j}) \][/tex]

Combine like terms:

 [tex]\[ D_i = 3 - 1 - 2 = 0 \][/tex]

  [tex]\[ D_j = 3 - 4 + 5 = 4 \][/tex]

  So, the components of [tex]\( \mathbf{D} \)[/tex] are:

  [tex]\[ \mathbf{D} = 0\mathbf{i} + 4\mathbf{j} \][/tex]

2. Magnitude of [tex]\( \mathbf{D} \)[/tex]:

 [tex]\[ |\mathbf{D}| = \sqrt{D_i^2 + D_j^2} = \sqrt{0^2 + 4^2} = 4 \, \text{meters} \][/tex]

3. Direction of [tex]\( \mathbf{D} \)[/tex]:

  The angle [tex]\( \theta_D \)[/tex] from the positive x-axis is:

 [tex]\[ \theta_D = \tan^{-1}\left(\frac{D_j}{D_i}\right) = \tan^{-1}\left(\frac{4}{0}\right) = 90^\circ \][/tex]

b) Vector [tex]\(\mathbf{E} = -\mathbf{A} - \mathbf{B} + \mathbf{C}\)[/tex]

1. Sum the components:

 [tex]\[ \mathbf{E} = - (3\mathbf{i} + 3\mathbf{j}) - (-\mathbf{i} - 4\mathbf{j}) + (-2\mathbf{i} + 5\mathbf{j}) \][/tex]

  Simplify the negative signs:

  [tex]\[ \mathbf{E} = (-3\mathbf{i} - 3\mathbf{j}) + (\mathbf{i} + 4\mathbf{j}) + (-2\mathbf{i} + 5\mathbf{j}) \][/tex]

  Combine like terms:

 [tex]\[ E_i = -3 + 1 - 2 = -4 \][/tex]

[tex]\[ E_j = -3 + 4 + 5 = 6 \][/tex]

So, the components of [tex]\( \mathbf{E} \)[/tex] are:

  [tex]\[ \mathbf{E} = -4\mathbf{i} + 6\mathbf{j} \][/tex]

2. Magnitude of [tex]\( \mathbf{E} \)[/tex]:

 [tex]\[ |\mathbf{E}| = \sqrt{E_i^2 + E_j^2} = \sqrt{(-4)^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} \approx 7.21 \, \text{meters} \][/tex]

3. Direction of [tex]\( \mathbf{E} \)[/tex]:

  The angle [tex]\( \theta_E \)[/tex] from the positive x-axis is:

  [tex]\[ \theta_E = \tan^{-1}\left(\frac{E_j}{E_i}\right) = \tan^{-1}\left(\frac{6}{-4}\right) = \tan^{-1}\left(-1.5\right) \][/tex]

Since [tex]\( E_i \)[/tex] is negative and [tex]\( E_j \)[/tex] is positive, [tex]\( \theta_E \)[/tex] is in the second quadrant:

 [tex]\[ \theta_E = 180^\circ - \tan^{-1}(1.5) \approx 180^\circ - 56.31^\circ = 123.69^\circ \][/tex]

Answer:

20.7 s

Explanation:

The equation to calculate the velocity for a uniform acceleration a, time t and initial velocity v₀:

v = a*t + v₀

Solve for t:

t = (v - v₀)/a

The work function of the metal is approximately 3.67 eV.

The work function, or binding energy, of a metal is the minimum energy required to eject an electron from the metal surface. In the photoelectric effect, the maximum kinetic energy of ejected electrons (photoelectrons) is given by the equation KE = hf - BE, where hf is the photon energy and BE is the work function. To find the work function W0 of the metal, we can use Equation 6.16, which states that the threshold wavelength for observing the photoelectric effect is given by λ = (hc)/(W0), where h is Planck's constant, c is the speed of light, and W0 is the work function.

Given that the maximum wavelength that can still eject electrons is 542 nm, we can rearrange the equation to solve for W0:

W0 = (hc)/λ

Plugging in the values, we have:

W0 = (6.626 × 10^-34 J·s)(3 × 10^8 m/s)/(542 × 10^-9 m)

Simplifying the calculation, we find that the work function W0 of the metal is approximately 3.67 eV.

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Final answer:

The maximum wavelength for ejecting electrons from the metal surface is 542 nm. Using the energy equation for photons, we calculate the work function (W0) to be approximately 2.29 electron volts.

Explanation:

The question relates to the photoelectric effect and the calculation of the work function of a metal when electromagnetic radiation of a certain wavelength is incident upon it. The work function (W0) is the minimum energy needed to eject an electron from the metal surface. According to the question, the maximum wavelength that can eject an electron is 542 nm.

To find the work function in electron volts, we can use the equation:

E(photon) = hc / λ

Where:

E(photon) is the energy of the photon

h is Planck's constant (4.135667696 × [tex]10^-^1^2[/tex] eV·s)

c is the speed of light (3 × 108 m/s)

λ is the wavelength of the photon in meters (542 nm = 542 × [tex]10^-^9[/tex] m)

First, convert the energy into electron volts:

E(photon) = (4.135667696 × [tex]10^-^1^2[/tex] eV·s × 3 × 108 m/s) / (542 × [tex]10^-^9[/tex] m)

After calculation, we find that E(photon) is approximately 2.29 eV. For a photon to eject an electron, its energy must be equal to or greater than the work function of the metal. Therefore, the work function of the metal is also 2.29 eV.

it takes approximately [tex]\(1.3863 \times 10^{-4}\)[/tex] seconds for the copper rod to cool to an average temperature of 25°C when exposed to an air stream at 20°C with a heat transfer coefficient of [tex]\(20,000 \, \text{W/m}^2 \cdot \text{K}\)[/tex].

The rate at which the copper rod loses heat can be described by Newton's Law of Cooling, which is given by the formula:

[tex]\[ Q(t) = Q_0 e^{-ht} \][/tex]

where:

[tex]\( Q(t) \)[/tex] is the heat at time[tex]\( t \)\\[/tex],

[tex]\( Q_0 \)[/tex]is the initial heat content,

[tex]\( h \)\\[/tex] is the heat transfer coefficient,

[tex]\( t \)[/tex] is the time.

The heat transfer coefficient [tex]\( h \)[/tex] can be calculated using the formula:

[tex]\[ h = \frac{k}{D} \][/tex]

where:

[tex]\( k \)[/tex] is the thermal conductivity of copper,

[tex]\( D \)[/tex] is the diameter of the rod.

The time ( t ) it takes for the rod to cool to a certain temperature can be determined by solving for ( t) when [tex]\( Q(t) \)[/tex]reaches a specific value. In this case, we want to find ( t ) when the average temperature of the rod is 25°C.

Let's proceed with the calculations:

1. **Calculate [tex]\( h \)[/tex]:**

  The thermal conductivity of copper[tex](\( k \)) is approximately \( 400 \, \text{W/m} \cdot \text{K} \)[/tex].

[tex]\[ h = \frac{k}{D} \][/tex]

 [tex]\[ h = \frac{400 \, \text{W/m} \cdot \text{K}}{0.02 \, \text{m}} \] \[ h = 20,000 \, \text{W/m}^2 \cdot \text{K} \][/tex]

2. **Plug in values and solve for ( t ):**

  The initial temperature of the rod[tex](\( T_0 \))[/tex] is 100°C, the ambient temperature [tex](\( T_{\text{air}} \))[/tex] is 20°C, and the average temperature[tex](\( T \))[/tex] is 25°C. The temperature difference[tex]\( \Delta T \) is \( T_0 - T_{\text{air}} \)[/tex].

[tex]\[ \Delta T = T_0 - T_{\text{air}} = 100°C - 20°C = 80°C \][/tex]

 Now, [tex]\( h \) and \( \Delta T \) are known, and we can rearrange the formula \( Q(t) = Q_0 e^{-ht} \) to solve for \( t \)[/tex]:

[tex]\[ t = -\frac{1}{h} \ln\left(\frac{T - T_{\text{air}}}{T_0 - T_{\text{air}}}\right) \][/tex]

Substituting the values:

[tex]\[ t = -\frac{1}{20,000 \, \text{W/m}^2 \cdot \text{K}} \ln\left(\frac{25°C - 20°C}{100°C - 20°C}\right) \][/tex]

Calculate ( t ) using the natural logarithm function, and you'll get the time it takes for the copper rod to cool to an average temperature of 25°C.

let's calculate the time it takes for the copper rod to cool to an average temperature of 25°C.

[tex]\[ t = -\frac{1}{20,000 \, \text{W/m}^2 \cdot \text{K}} \ln\left(\frac{25°C - 20°C}{100°C - 20°C}\right) \][/tex]

[tex]\[ t = -\frac{1}{20,000 \, \text{W/m}^2 \cdot \text{K}} \ln\left(\frac{5}{80}\right) \][/tex]

Now, calculate the natural logarithm:

[tex]\[ t = -\frac{1}{20,000 \, \text{W/m}^2 \cdot \text{K}} \ln\left(0.0625\right) \][/tex]

[tex]\[ t \approx -\frac{1}{20,000 \, \text{W/m}^2 \cdot \text{K}} \times (-2.7726) \][/tex]

[tex]\[ t \approx \frac{2.7726}{20,000 \, \text{W/m}^2 \cdot \text{K}} \][/tex]

[tex]\[ t \approx 1.3863 \times 10^{-4} \, \text{s} \][/tex]

So, it takes approximately [tex]\(1.3863 \times 10^{-4}\)[/tex] seconds for the copper rod to cool to an average temperature of 25°C when exposed to an air stream at 20°C with a heat transfer coefficient of [tex]\(20,000 \, \text{W/m}^2 \cdot \text{K}\)[/tex].

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The time it would take for the copper rod to cool to an average temperature of 25°C is approximately 22.17 minutes.

To solve this problem, we will use the concept of Newton's law of cooling, which states that the rate of heat loss of a body is directly proportional to the difference in the temperatures between the body and its surroundings. The formula for the heat transfer rate is given by:

[tex]\[ \frac{dQ}{dt} = hA(T_s - T_{\infty}) \][/tex]

where:

[tex]\( \frac{dQ}{dt} \)[/tex] is the rate of heat transfer (W),

h is the heat transfer coefficient [tex](W/m^2\·K)[/tex],

A is the surface area of the object [tex](m^2)[/tex],

[tex]\( T_s \)[/tex] is the surface temperature of the object (°C), and

[tex]\( T_{\infty} \)[/tex] is the temperature of the surrounding air (°C).

The copper rod is a cylinder, so its surface area A can be calculated using the formula for the surface area of a cylinder:

[tex]\[ A = 2\pi rL \][/tex]

where:

r is the radius of the rod, and

L is the length of the rod.

Given that the diameter of the rod is 2.0 cm, the radius r is half of that, which is 1.0 cm or 0.01 m. The length L of the rod is not given, so we will assume it to be L meters.

The initial temperature of the rod is [tex]\( T_{initial} = 100 \°C \)[/tex], and the final temperature we want to reach is [tex]\( T_{final} = 25 \°C \)[/tex]. The temperature difference between the rod and the surrounding air is [tex]\( T_s - T_{\infty} \)[/tex].

The heat lost Q by the rod as it cools from [tex]\( T_{initial} \)[/tex] to [tex]\( T_{final} \)[/tex] can be found using the specific heat capacity of copper (c) and the mass (m) of the rod:

[tex]\[ Q = mc(T_{initial} - T_{final}) \][/tex]

The mass (m) of the rod can be calculated from its density \rho and volume V:

[tex]\[ m = \rho V \][/tex]

[tex]\[ V = \pi r^2 L \][/tex]

The density of copper [tex]\( \rho \)[/tex] is approximately [tex]\( 8930 \) kg/m^3[/tex].

Now, we can equate the heat lost to the heat transfer over time:

[tex]\[ mc(T_{initial} - T_{final}) = hA(T_s - T_{\infty}) \cdot \Delta t \][/tex]

Solving for [tex]\( \Delta t \)[/tex], we get:

[tex]\[ \Delta t = \frac{mc(T_{initial} - T_{final})}{hA(T_s - T_{\infty})} \][/tex]

The specific heat capacity of copper (c) is approximately 385 J/kg·K.

Substituting the expressions for (m) and (A) into the equation for [tex]\( \Delta t \)[/tex], we have:

[tex]\[ \Delta t = \frac{\rho \pi r^2 L c(T_{initial} - T_{final})}{h(2\pi rL)(T_s - T_{\infty})} \][/tex]

Simplifying, we get:

[tex]\[ \Delta t = \frac{\rho r c(T_{initial} - T_{final})}{2h(T_s - T_{\infty})} \][/tex]

Now we can plug in the values:

[tex]\( \rho = 8930 \) kg/m^3[/tex],

r = 0.01 m,

c = 385 J/kg·K,

[tex]T_{initial}[/tex] = 100°C ,

[tex]T_{final}[/tex] = 25°C,

[tex]h = 200 W/m^2\·K[/tex],

[tex]T_{\infty}[/tex] = 20°C.

[tex]\[ \Delta t = \frac{8930 \times 0.01 \times 385 \times (100 - 25)}{2 \times 200 \times (25 - 20)} \][/tex]

[tex]\[ \Delta t = \frac{8930 \times 0.01 \times 385 \times 75}{2 \times 200 \times 5} \][/tex]

[tex]\[ \Delta t = \frac{8930 \times 0.01 \times 385 \times 15}{2 \times 200} \][/tex]

[tex]\[ \Delta t = \frac{8930 \times 0.01 \times 385 \times 15}{400} \][/tex]

[tex]\[ \Delta t = \frac{532095}{400} \][/tex]

[tex]\[ \Delta t = 1330.2375 \text{ seconds} \][/tex]

To express this in minutes, we divide by 60:

[tex]\[ \Delta t = \frac{1330.2375}{60} \text{ minutes} \][/tex]

[tex]\[ \Delta t \approx 22.170625 \text{ minutes} \][/tex]

Answer:

0.000000113 or 1.13*[tex]10^{-7}[/tex] meters

Explanation:

One nanometer is [tex]10^{-9}[/tex] meters. So 113 nanometers would be 113*[tex]10^{-9}[/tex], or 1.13*[tex]10^{-7}[/tex] meters. That's expessed on "cientific notation." On the "usual" notation, it will be 0.000000113 meters.

Answer:

75 N

Explanation:

t = Time taken = 2 seconds

u = Initial velocity

v = Final velocity

s = Displacement = 30 m

a = Acceleration

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 30=0\times 2+\frac{1}{2}\times a\times 2^2\\\Rightarrow a=\frac{30\times 2}{2^2}\\\Rightarrow a=15\ m/s^2[/tex]

The acceleration due to gravity on the planet is 15 m/s²

Force

F = ma

[tex]F=5\times 15\\\Rightarrow F=75\ N[/tex]

The gravitational force exerted on the object near the planet’s surface is 75 N

We have that for the Question it can be said that he gravitational force exerted on the 5kg object near the planet’s surface

F=75N

From the question we are told

A 5kg object is released from rest near the surface of a planet such that its gravitational field is considered to be constant. The mass of the planet is unknown. After 2s, the object has fallen 30m.

Air resistance is considered to be negligible. What is the gravitational force exerted on the 5kg object near the planet’s surface?

Generally the equation for the Motion   is mathematically given as

[tex]s=ut+1/2at^2\\\\Therefore\\\\30=0+1/2(a)(2)^2\\\\a=15m/s^2[/tex]

Therefore

F=ma

F=5*15

F=75N

Hence, the gravitational force exerted on the 5kg object near the planet’s surface

F=75N

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