A plane is flying horizontally with speed 167 m/s at a height 4040 m above the ground, when a package is dropped from the plane. The acceleration of gravity is 9.8 m/s 2 . Neglecting air resistance, when the package hits the ground, the plane will be:

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:

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:

d = 39.7 km

Explanation:

initial position of the boat is 45 km away at an angle of 15 degree East of North

so we will have

[tex]r_1 = 45 sin15 \hat i + 45 cos15 \hat j[/tex]

[tex]r_1 = 11.64 \hat i + 43.46\hat j[/tex]

after some time the final position of the boat is found at 30 km at 15 Degree North of East

so we have

[tex]r_2 = 30 cos15\hat i + 30 sin15 \hat j[/tex]

[tex]r_2 = 28.98\hat i + 7.76 \hat j[/tex]

now the displacement of the boat is given as

[tex]d = r_2 - r_1[/tex]

[tex]d = (28.98\hat i + 7.76 \hat j) - (11.64 \hat i + 43.46\hat j)[/tex]

[tex]d = 17.34 \hat i - 35.7 \hat j[/tex]

so the magnitude is given as

[tex]d = \sqrt{17.34^2 + 35.7^2}[/tex]

[tex]d = 39.7 km[/tex]

Answer:

The planetary system is 9 billion years old

Explanation:

In 4.5 billion years, the meteorite was half uranium and half lead. For this amount of uranium to be reduced by half again (reaching 1/4 uranium and 3/4 lead), another 4.5 billion years must have passed. Then, the planetary system should be 9 (2 x 4.5) billion years old.

Answer:

T = g μ_s ( M+m )

78.4 N

Explanation:

When both of them move with the same acceleration , small box will not slip over the bigger one. When we apply force on the lower box, it starts moving with respect to lower box. So a frictional force arises on the lower box which helps it too to go ahead . The maximum value that this force can attain is mg μ_s . As a reaction of this force, another force acts on the lower box in opposite direction .

Net force on the lower box

= T - mg μ_s = M a    ( a is the acceleration created by net force in M )

Considering force on the upper box

mg μ_s = ma

a = g μ_s

Put this value of a in the equation above

T - m gμ_s = M g μ_s

T = mg μ_s + M g μ_s

=  g μ_s ( M+m )

2 )

Largest tension required

T = 9.8 x  .50 x ( 10+6 )

= 78.4 N

The maximum tension at which the small box doesn't slip off the large box can be calculated using the formula T_max = μs * mg. Substituting the given values into the formula, the largest tension is found to be 29.4 N for the small box not to slip off a 10 kg wood sled on frictionless snow.

The maximum tension, T_max, for which the small box remains stationary with respect to the larger box can be found using the concept of static friction. This static friction, f_s, can be expressed as the product of the static friction coefficient, μs, and the normal force on the small box, which is its weight, mg. So, f_s = μs * mg.

Given that the maximum tension, T_max, equals f_s, we can derive the formula as T_max = μs * mg. This is an expression for the maximum tension in terms of the given variables.

For the numerical part of the question, given that the mass of the box m = 6 kg, and the coefficient of kinetic friction μk = 0.50, we can substitute these values into the derived formula: T_max = μk * m * g = 0.50 * 6.0 kg * 9.8 m/s^2 = 29.4 N. This is the largest tension force for which the box doesn't slip.

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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:

(a) Magnitude = [tex]1.14\times 10^5\ N/C[/tex]

Direction = toward the center of the disc

(b) Magnitude = [tex]8.90\times 10^4\ N/C[/tex]

Direction = toward the center of the disc

(c) Magnitude = [tex]1.46\times 10^5\ N/C[/tex]

Direction = toward the center of the disc

(d) See explanation

(e) See explanation.

Explanation:

Given:

Assume:

Part(a):

Using the formula for the electric field due to the uniform charged disc at a point P on its axis, we have

[tex]E = \dfrac{2 kQ}{r^2}\left ( 1-\dfrac{x}{\sqrt{x^2+r^2}} \right )\\\Rightarrow E = \dfrac{2 \times 9\times 10^9\times (-6.50\times 10^{-9})}{(0.0125)^2}\left ( 1-\dfrac{0.02}{\sqrt{(0.02)^2+(0.0125)^2}} \right )\\\Rightarrow E =-1.14\times 10^5\ N/C\\[/tex]

Hence, the magnitude of electric field at the given distance from the disc along its axis is [tex]1.14\times 10^5\ N/C[/tex] in the direction toward the center of the disc along the axis.

Part(b):

When all the charges on the disc are pushed away from the center of the disc to get uniformly distributed to its outer rim, the disc behaves like a uniformly charged ring.

Using the formula for the electric field due to the uniform charged ring at a point P on its axis, we have

[tex]E= \dfrac{kQx }{\left ( \sqrt{x^2+r^2} \right )^3}\\\Rightarrow E = \dfrac{ 9\times 10^9\times (-6.50\times 10^{-9})\times 0.02}{\left ( \sqrt{(0.02)^2+(0.0125)^2} \right )^3}\\&=-8.90\times 10^4\ N/C[/tex]

Hence, the magnitude of electric field at the given point from the rim along its axis is [tex]8.90\times 10^4\ N/C[/tex], and it is also toward the center of the along the axis.

Part (c):

When all the charge on the disc is pushed to the center of the disc, then the disc behaves as a point charge.

Using the formula of electric field due to a point charge at a fixed point, we have

[tex]E = \dfrac{kQ}{x^2}\\\Rightarrow E = \dfrac{9\times 10^9\times (-6.50\times 10^{-9})}{(0.02)^2}\\\Rightarrow E = -1.46\times 10^5[/tex]

Hence, the magnitude of electric field due to the point charge at the point P on the axis is [tex]1.46\times 10^5\ N/C[/tex]. which also points toward the center of the disc along the axis.

The direction of the electric field is found toward the negative x-axis because the electric field due to a point negative charge is radially inward.

Part (d):

The magnitude of electric field in part (a) is stronger than that in part (b) because the disc in part (a) is composed of infinite number of rings having radius varying from 0 cm to 2.5 cm whose components along the axis of electric field are added together which is greater than the electric field produced by a single ring in part (b).

Hence, the electric field in part (a) is greater than the electric field in part (b).  

Part (e):

The electric field in part (c) is the strongest of the three fields. This is because the field due to a point charge on the axis has the only component along the axis. However, the field components in part (a) and part (b) are distributed along two axes whose axial component is considerable and the other component gets canceled out. This causes the field strength to be weaker in part (a) and part (c).

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:

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

Final answer:

Richard will save approximately 13.68 minutes by driving at 73 mph for 135 miles instead of at the speed limit of 65 mph. The mathematical concept used involves calculating travel time at different speeds and finding the difference between these times.

Explanation:

The student's question involves calculating how much time Richard will save by driving at 73 mph instead of the speed limit of 65 mph for a distance of 135 miles. This is a mathematics problem that falls under the category of rate, time, and distance, which is often covered in high school math classes.

To solve this, we need to calculate the time taken to travel 135 miles at both speeds and then find the difference in time. At 65 mph, the time taken (T1) is 135 miles / 65 mph. At 73 mph, the time taken (T2) is 135 miles / 73 mph. The time saved is T1 - T2.

Using the formula time = distance / speed, we get:

Time at 65 mph: T1 = 135 mi / 65 mph = 2.077 hours

Time at 73 mph: T2 = 135 mi / 73 mph = 1.849 hours

The time saved is T1 - T2 = 2.077 hours - 1.849 hours = 0.228 hours, which is about 13.68 minutes.

Explanation:

Given that, the height of an object tossed upward with an initial velocity of 136 feet per second is given by the formula :

[tex]h=-16t^2+136t[/tex]

Where

h is the height in feet

t is the time in seconds

Let t is the time required for the object to return to its point of departure. At this point, h = 0

[tex]-16t^2+136t=0[/tex]

t = 8.5 seconds

So, the time required for the object to return to its point of departure is 8.5 seconds.

Answer:

The lowest frequency is 95.6 Hz

Explanation:

The standing waves that can be formed in this system must meet some conditions, such as until this is fixed at the bottom here there must be a node (point without oscillation) and being free at its top at this point there should be maximum elongation (antinode)

For the lowest frequency we have a node at the bottom point and a maximum at the top point, this corresponds to ¼ of the wavelength, so the full wave has

      λ = 4L

As the speed any wave is equal to the product of its frequency by the wavelength

      v = f λ

      f = v / λ    

      f = v / 4L

      f = 2730 / (4 7.14)

      f=  95.6 1 / s = 95.6 Hz

The lowest frequency of the standing wave that can be formed on the flagpole is 95.59 Hz.

The velocity, frequency and wavelength of a wave are related according to the following equation:

Velocity (v) = wavelength (λ) × frequency (f)

v = λf

With the above formula, we can obtain the frequency as follow:

v = λf

2730 = 28.56 × f

Divide both side by 28..56

f = 2730 / 28.56

f = 95.59 Hz

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Answer:

the equation is dimensionally consistent

Explanation:

To verify that the formula is dimensionally consistent, we must verify the two terms of the sum and verify that they are units of force. We achieve this by putting the units of each dimensional term of the equation and verifying that the answer is in units of force

μ=viscosity=Ns/m^2

D=diameter

V=velocity

ρ=density=Kg/m^3π

First term

3πμDV=[tex]\frac{N.s.m.m}{m^{2}.s }[/tex]=N

the first term is  dimensionally consistent

second term

(9π/16)ρV^2D^2=[tex]\frac{kg.m^2.m^2}{m^3.s^2} =\frac{kg.m}{s^2} =N[/tex]

.

as the two terms are in Newtons it means that the equation is dimensionally consistent

Answer:

[tex]v_{f} =4.9\frac{m}{s}[/tex]  :  velocity of the crate after the person has pushed the crate a distance of 6 meters

Explanation:

Crate kinetics

Crate moves with uniformly accelerated movement

v f²=v₀²+2a*d (formula 1)

d:displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Known data

v₀=0 The speed of the crate is equal to zero because part of the rest.

a= 2m/s²

d= 6m

Distance calculating

We replace data in the Formula (1)

v f²=0+2*2*d

v f²=2*2*6

v f²=24

[tex]v_{f} =\sqrt{24}[/tex]

[tex]v_{f} =4.9\frac{m}{s}[/tex]

To find the final velocity of a 50-kg crate after being pushed for 6 meters with an acceleration of 2.0 m/s², the kinematic equation v² = u² + 2as is used, resulting in a final velocity of 4.9 m/s.

To calculate the velocity of a 50-kg crate after being pushed a distance of 6 meters with an acceleration of 2.0 m/s2 from rest, we use the kinematic equation v2 = u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance moved. Since the crate starts from rest, the initial velocity, u, is 0 m/s. Substituting the given values we get:

v2 = 0 m/s + 2(2.0 m/s2)(6 m),

v2 = 24 m2/s2,

v = √(24 m2/s2),

v = 4.9 m/s.

Therefore, the velocity of the crate after being pushed 6 meters will be 4.9 m/s.

Answer:

2. Change in velocity

Explanation:

The units that acceleration use are distance/time^2 [tex](m/s^2)[/tex] (in metric units).

The acceleration is the change of velocity over time, so

[tex]\frac{dv}{dt} =\frac{dx}{dt}*\frac{1}{dt}[/tex]

In another way, if the velocity is constant, the acceleration is zero, but if the velocity change over the time (as a car do when the red light change), the acceleration takes value!.

Velocity changes in displacement divided by time. Acceleration is the change in velocity divided by time.  

Answer: Option 2

Explanation:

          Acceleration is defined as the rate of velocity changes of an object in a given time period. This can express in the equation as follows,  

                                [tex]\bold{a = \frac{v}{t} m / s^{2}}[/tex]

          There two types in it as linear and radial acceleration. When velocity changes in the opposite direction to an object then it is called deceleration. But these are the same as a change in velocity.

Final answer:

The statement regarding receiving credit for correctly positioned endpoints using an endpoint mover is typically true. Credit is awarded for each endpoint positioned correctly in a digital math exercise or tool, which may involve placing endpoints according to specific criteria.

Explanation:

The statement "When using the endpoint mover, you will receive credit for each endpoint that is positioned correctly" is generally True. In the context of mathematical tools or interactive digital platforms, an endpoint mover likely refers to a feature that allows users to manipulate the endpoints of a line segment or other geometric object. When students are tasked with positioning these endpoints correctly according to given criteria, they would typically receive credit or points for each endpoint that is placed accurately.

For example, in a digital math exercise, if you're asked to position the endpoints of a line segment so that one end is at the point (3,2) and the other end is at the point (-1,5), and you succeed in placing them correctly, you would receive credit for both endpoints. However, details may vary based on the specific platform or educational program, so it's essential to follow the instructions provided.

Answer:

option (a)

Explanation:

the angular velocity of the carousel is same througout the motion, so the angular velocity of all the horses is same, but the linear velocity is different for different horses.

As the angular displacement of all the horses are same in the same time so the angular velocity is same.

The relation between the linear velocity and the angular velocity is given by

v = r ω

where, v is linear velocity and r be the distance between the horse and axis of rotation and ω be the angular velocity.

So, the angular velocity of Alice horse is same as the angular velocity of Bob horse.

ωA = ωB

Thus, option (a) is true.

The correct option is Option A ( ω A = ω B). Both Alice and Bob's horses on the carousel have the same angular velocity, regardless of their distance from the axis. Hence, ω A == ω B.

Thus, the correct answer is:
a. ω A = ω B

Answer:

T = 92.8 min

Explanation:

Given:

The altitude of the International Space Station t minutes after its perigee (closest point), in kilometers, is given by:

                               [tex]A(t) = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})[/tex]

Find:

- How long does the International Space Station take to orbit the earth? Give an exact answer.

Solution:

- Using the the expression given we can extract the angular speed of the International Space Station orbit:

                                 [tex]A(t) = 415 - sin({\frac{2*\pi*t }{92.8} + \frac{23.2*2*\pi }{92.8} )[/tex]

- Where the coefficient of t is angular speed of orbit w = 2*p / 92.8

- We know that the relation between angular speed w and time period T of an orbit is related by:

                                T = 2*p / w

                                T = 2*p / (2*p / 92.8)

Hence,                     T = 92.8 min

The time of the International Space Station  is mathematically given as

T = 5568sec

Generally, the equation for the altitude of the International Space Station  is mathematically given as

[tex]A(t) \left = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})[/tex]

Therefore

Using the equation

[tex]A(t) \left = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})[/tex] we decipher time to be

     T = 2*p / w

In conclusion

T = 2*p / (2*p / 92.8)

T = 5568sec

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Answer:

[tex]V = 12.85 cm^3[/tex]

Explanation:

As we know that initially the air bubble is at depth 110 m

so the pressure of the air bubble is given as

[tex]P = P_o + \rho gh[/tex]

[tex]P = 1.01 \times 10^5 + (1000)(9.81)(110)[/tex]

[tex]P = 1.18 \times 10^6 Pa[/tex]

initial volume of the bubble is given as

[tex]V = 1.1 cm^3[/tex]

now we know that here temperature of air bubble is constant

so we have

[tex]P_1 V_1 = P_2 V_2[/tex]

[tex](1.18 \times 10^6)(1.1 cm^3) = (1.01 \times 10^5) V[/tex]

[tex]V = 12.85 cm^3[/tex]

Answer:

You're supposed to measure the distance from X to the end of vector 5 using the appropriate scale, and measure the angle (counterclockwise from X) using a protractor.  

Mathematically:  

x = [83cos90+55cos141+69cos229+41cos281+61co... cm = -27 cm  

y = [83sin90+55sin141+69sin229+41sin281+61si... cm = 55 cm  

so d = √(x² + y²) = 61 cm  

and Θ = arctan(55/-27) = -64º +180º (to get into QII) = 116º

Explanation:

To calculate the pinball's total displacement, you sum up the x and y components of each vector separately and find the resultant vector's magnitude and direction, which is approximately 96.60 cm at an angle of 88.7 degrees from the horizontal.

To find the total displacement of the pinball, we need to add the vectors given by their magnitude and direction. We first break down each vector into its horizontal (x) and vertical (y) components using trigonometric functions:

To find the resultant vector, sum the x and y components separately:

The magnitude of the resultant displacement vector (S) is calculated using the Pythagorean theorem:

S = √(x² + y²) = √(1.97² + 96.13²) = √(9330.56) = 96.60 cm approximately

The angle made with the horizontal (θ) is found using the tangent function:

θ = arctan(y/x) = arctan(96.13/1.97) ≈ 88.7 degrees

Answer: [tex]p_{SO_2}=0.017atm[/tex]

Explanation:

We are given:

[tex]K_c=1.7\times 10^8[/tex]

Relation of [tex]K_p[/tex] with [tex]K_c[/tex] is given by the formula:

[tex]K_p=K_c(RT)^{\Delta ng}[/tex]

Where,

[tex]K_p[/tex] = equilibrium constant in terms of partial pressure = ?

[tex]K_c[/tex] = equilibrium constant in terms of concentration

R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature = [tex]600K[/tex]

[tex]2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)[/tex]

[tex]\Delta ng[/tex] = change in number of moles of gas particles = [tex]n_{products}-n_{reactants}=2-3=-1[/tex]

Putting values in above equation, we get:

[tex]K_p=1.7\times 10^8\times (0.0821\times 600)^{-1}\\\\K_p=3.4\times 10^6[/tex]

The chemical reaction follows the equation:

                 [tex]2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)[/tex]

The expression for [tex]K_p[/tex] for the given reaction follows:

[tex]K_p=\frac{(p_{SO_3})^2}{ p_{O_2}\times {(p_{SO_2})^2}}[/tex]

We are given:

[tex]p_{SO_3}=300atm[/tex]   [tex]p_{O_2}=100atm[/tex]

Putting values in above equation, we get:

[tex]3.4\times 10^6=\frac{(300)^2}{100\times {(p_{SO_2})^2}}[/tex]

[tex]p_{SO_2}=0.017atm[/tex]

Hence, the partial pressure of the [tex]SO_2[/tex] at equilibrium is 0.017 atm.

Answer:

weight  is 442.10 kg

Explanation:

given data

weight = 1400 kg

length = 6 m

to find out

how many kilogram 19 m beam support

solution

we know by inverse variation formula of weight  that is

weight = [tex]\frac{k}{l}[/tex]   .............1

here k is constant and l is length

so  k will be

1400 =  [tex]\frac{k}{6}[/tex]

k = 8400

so for 19 m weight will be by equation 1

weight = [tex]\frac{k}{l}[/tex]

weight = [tex]\frac{8400}{19}[/tex]

weight = 442.10 kg

The weight that a horizontal beam can support varies inversely as the length of the beam. Using this relationship, we can find the weight a 19-meter beam can support given that a 6-meter beam can support 1400 kg.

The weight W that a horizontal beam can support varies inversely as the length L of the beam. This can be expressed mathematically as W = k/L, where k is a constant. In this case, we are given that a 6-meter beam can support 1400 kg. Let's use this information to find the value of k.

Using the given information, we can write the equation as 1400 = k/6. To solve for k, we can multiply both sides of the equation by 6, giving us k = 1400 x 6 = 8400.

Now that we have the value of k, we can use it to find how many kilograms a 19-meter beam can support. Using the equation W = k/L, we can plug in the values: W = 8400/19. Solving this equation gives us W = 442.105 kg (rounded to three decimal places). Therefore, a 19-meter beam can support approximately 442.105 kilograms.

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.

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:

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

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:

F=<0,-29.4,0>N

D = <0,-10,0>m

Work=F·D=294 Joules

Explanation:

Force:

The gravity has a negative direction in the axis Y in our coordinate system:

F=<0,-mg,0>=<0,-3*9.8,0>=<0,-29.4,0>N

Displacement:

The initial position A is <0, 33, 0>m

The final position B is <0, 23,0>m

The displacement vector is D=B-A = <0,-10,0>m

Gravitational Work:

The work is the scalar product between the force and the displacement

Work=F·D=(-29.4)*(-10)=294 Joules

Answer:

Increase of ten yields 10 dB increase

Explanation:

The decibel is a logarithmic scale that shortens a large range of numbers into smaller and more manageable numbers. The decibel is found using:

[tex]dB = 10log(r )[/tex]

Where r represents a ratio of a measured magnitude (pressure or intensity level in sound) to a reference pressure or intensity value. However, in our problem we only need to represent an increase factor of ten:

[tex]dB = 10log(\frac{10}{1} )\\\\dB = 10*1=10 dB[/tex]

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

Answer:

a) x component  = -31.25 km/hr

b) y component = 46.64 km/hr

Explanation:

Given data:

A position is 4km north and 2.5 km east to B

Ship A velocity = 22 km/hr

ship B velocity = 40 km/hr

A velocity wrt to velocity of B

[tex]\vec{V_{AB}} =\vec{V_A} - \vec{V_B}[/tex]

[tex]\vec{V_A} = 22 km/hr[/tex]

[tex]\vec{V_B} = 40 cos38\hat{i} + 40sin 38 \hat{j}[/tex]

                 [tex]= 31.52\hat{i} + 24.62 \hat{j}[/tex]

putting respective value to get velocity of  A with respect to B

[tex]\vec{V_{AB}} = -22 \hat {j} - (31.52\hat{i} + 24.62 \hat{j})[/tex]

[tex]\vec{V_{AB}} = -31.52\hat{i} - 46.62\hat{j}[/tex]

a) x component  = -31.25 km/hr

b) y component = 46.64 km/hr