The earth's radius is about 4000 miles. Kampala, the capital of Uganda, and Singapore are both nearly on the equator. The distance between them is 5000 miles as measured along the earth's surface.
a. Through what angle do you turn, relative to the earth, if you fly from Kampala to Singapore? Give your answer in both radians and degrees.
b. The flight from Kampala to Singapore take 9 hours. What is the plane's angular speed relative to the earth?

Answers

Answer 1

Answer:

a) the required angle in both radian and degree is  1.25 rad and 71.6°

b) the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec

Explanation:

Given the data in the question;

a)

we know that The expression for the angle subtended by an arc of circle at the center of the circle is,

θ = Length / radius

given that Length is 5000 miles and radius is 4000 miles

we substitute

θ = 5000 miles / 4000 miles

θ = 1.25 rad

Radian to Degree

θ = 1.25 rad × ( 180° / π rad )

θ =  71.6°

Therefore, required angle in both radian and degree is  1.25 rad and 71.6°

b)

The flight from Kampala to Singapore take 9 hours.

the plane's angular speed relative to the earth = ?

we know that, the relation between angular velocity and angular displacement is;

ω = θ / t

given that θ is 1.25 rads and time t is 9 hours or ( 9 × 3600 sec ) = 32400 sec

we substitute

ω = 1.25 rad / 32400 sec

ω = 3.86 × 10⁻⁵ rad/sec

Therefore, the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec


Related Questions

It takes 130 J of work to compress a certain spring 0.10m. (a) What is the force constant of this spring? (b) To compress the spring an additional 0.10 m, does it take 130 J, more than 130 J or less than 130 J? Verify your answer with a calculation.

Answers

Explanation:

Given that,

Work done to stretch the spring, W = 130 J

Distance, x = 0.1 m

(a) We know that work done in stretching the spring is as follows :

[tex]W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m[/tex]

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m

So,

[tex]W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J[/tex]

So, the new work is more than 130 J.

A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Newtons. What is the frequency of this mode of vibration

Answers

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

[tex]F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz[/tex]

Therefore, the frequency of this mode of vibration is 138.87 Hz

Two pistons are connected to a fluid-filled reservoir. The first piston has an area of 3.002 cm2, and the second has an area of 315 cm2. If the first cylinder is pressed inward with a force of 50.0 N, what is the force that the fluid in the reservoir exerts on the second cylinder?​

Answers

Answer:

The force on the second piston is 5246.5 N .

Explanation:

Area of first piston, a = 3.002 cm^2

Area of second piston, A = 315 cm^2

Force on first piston, f = 50 N

let the force of the second piston is F.

According to the Pascal's law

[tex]\frac{f}{a} = \frac{F}{A}\\\\\frac{50}{3.002}=\frac{F}{315}\\\\F = 5246.5 N[/tex]

A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a wall at the bottom of the incline. A light string attached to the block runs over a frictionless pulley to a 60.0-g suspended mass. The suspended mass is given an initial downward speed of 1.60 m/s .
How far does it drop before coming to rest? (Assume the spring is unlimited in how far it can stretch.)
Express your answer using two significant figures.

Answers

Answer:

0.5

Explanation:

because the block is attached to the pulley of the string

Assuming the atmospheric pressure is 1 atm at sea level, determine the atmospheric pressure at Badwater (in Death Valley, California) where the elevation is 86.0 m below sea level.

Answers

Answer:

Atmospheric pressure at Badwater is 1.01022 atm

Explanation:

Data given:

1 atmospheric pressure (Pi) = 1.01 * 10[tex]^{5}[/tex] Pa

Elevation (h) = 86m

gravity (g) = 9.8 m/s2

Density of air P = 1.225 kg/m3

Therefore pressure at bad water Pb = Pi + Pgh

Pb = (1.01 * 10[tex]^{5}[/tex]) + (1.225 * 9.8 * 86)

Pb = (1.01 * 10[tex]^{5}[/tex]) + 1032.43 = 102032 Pa

hence:

Pb = 102032 /1.01 * 10[tex]^{5}[/tex] = 1.01022 atm

3. You have a variable-voltage power supply and a capacitor in the form of two metal disks of radius 0.6 m, held a distance of 1 mm apart. What is the largest voltage you can apply to the capacitor without the air becoming highly conductive

Answers

Answer:

The breakdown of air occurs at a maximum voltage of 3kV/mm.

Explanation:

The breakdown of air occurs at a maximum voltage of 3kV/mm.

At this level of voltage the air between the plates become highly ionised and breakdown occurs. Since, the distance held between the plates is 1mm , it can withstand a maximum voltage of 3 kV.

After this voltage the air will become conductive in nature and will form ions in the air between the plates and ultimately breakdown will take place with a flash.

A 0.500-kg block slides up a plane inclined at a 30° angle. If it slides 1.50 m before coming to rest while encountering a frictional force of 2 N, find (a) its acceleration, and (b) its initial velocity.

Answers

B it’s Intail velocity

A 6.0-cm-diameter horizontal pipe gradually narrows to 4.0 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 32.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow?

Answers

Answer:

a n c

Explanation:

find out the odd one and give reason (length, volume, time, mass​

Answers

Answer:

Time

Explanation:

The answer to the question is actually time. Time is not needed when you calculate the mass or volume of an object, a square, sphere, rectangle, or any other 3D shape. You must also calculate the length to know what numbers you will be multiplying by. The answer to the question is time.

A wheel rotates about a fixed axis with an initial angular velocity of 13 rad/s. During a 8-s interval the angular velocity increases to 57 rad/s. Assume that the angular acceleration was constant during this time interval. How many revolutions does the wheel turn through during this time interval

Answers

Answer:

The number of revolutions is 44.6.

Explanation:

We can find the revolutions of the wheel with the following equation:

[tex]\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}[/tex]

Where:

[tex]\omega_{0}[/tex]: is the initial angular velocity = 13 rad/s              

t: is the time = 8 s

α: is the angular acceleration

We can find the angular acceleration with the initial and final angular velocities:

[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]

Where:

[tex] \omega_{f} [/tex]: is the final angular velocity = 57 rad/s

[tex] \alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2} [/tex]

Hence, the number of revolutions is:

[tex] \theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev [/tex]

Therefore, the number of revolutions is 44.6.

       

I hope it helps you!

The density of blood is 1055 kg/m3 . If the blood at the very top of your head exerts a minimum gauge pressure of 45 mm Hg (6000 Pa), estimate the gauge pressure at your heart in pascals.

Answers

Answer:

   P = 10135.6 Pa

Explanation:

For this exercise we use that the pressure varies with the height

           P = P₀ + ρ g h

where h is the height from the head to the heart, which is approximately

h = 40 cm = 0.40m  and P₀ is the head pressure P₀ = 6000 Pa

          P = 6000 + 1055 9.8 0.40

          P = 6000 + 4135.6

          P = 10135.6 Pa

Why are objects measured?​

Answers

In order to find out how long/wide/heavy/high/dense/deep/ massive/voluminous/reflective/opaque/ tansparent/warm/cold/hard/soft/ malleable/flexible/rigid/radioactive/old/ valuable/symmetrical/flat/regular/ irregular they are.

In a way that you can easily and conveniently describe to other people.

A nerve impulse travels along a myelinated neuron at 90.1 m/s.
What is this speed in mi/h?

Answers

Answer:

201.5537 mph

Explanation:

Given the following data;

Speed = 90.1 m/s

Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.

Mathematically, speed is given by the formula;

Speed = distance/time

To convert this value into miles per hour;

Conversion;

1 meter = 0.000621 mile

90.1 meters = 90.1 * 0.000621 = 0.05595 miles

1 metre per second = 2.237 miles per hour

90.1 meters per seconds = 90.1 * 2.237 = 201.5537 miles per hour

90.1 m/s = 201.5537 mph

A light source radiates 60.0 W of single-wavelength sinusoidal light uniformly in all directions. What is the average intensity of the light from this bulb at a distance of 0.400 m from the bulb

Answers

Answer: [tex]29.85\ W/m^2[/tex]

Explanation:

Given

Power [tex]P=60\ W[/tex]

Distance from the light source [tex]r=0.4\ m[/tex]

Intensity is given by

[tex]I=\dfrac{P}{4\pi r^2}[/tex]

Inserting values

[tex]\Rightarrow I=\dfrac{60}{4\pi (0.4)^2}\\\\\Rightarrow I=\dfrac{60}{2.010}\\\\\Rightarrow I=29.85\ W/m^2[/tex]

Answer:

29.85 W/ m^2

Explanation:

The bulk modulus of water is B = 2.2 x 109 N/m2. What change in pressure ΔP (in atmospheres) is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C?

Answers

Answer:

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

Explanation:

The bulk modulus of water ([tex]B[/tex]), in newtons per square meters, can be estimated by means of the following model:

[tex]B = \rho_{o}\cdot \frac{\Delta P}{\rho_{f} - \rho_{o}}[/tex] (1)

Where:

[tex]\rho_{o}[/tex] - Water density at 10.9 °C, in kilograms per cubic meter.

[tex]\rho_{f}[/tex] - Water density at 40 °C, in kilograms per cubic meter.

[tex]\Delta P[/tex] - Pressure change, in pascals.

If we know that [tex]\rho_{o} = 999.623\,\frac{kg}{m^{3}}[/tex], [tex]\rho_{f} = 992.219\,\frac{kg}{m^{3}}[/tex] and [tex]B = 2.2\times 10^{9}\,\frac{N}{m^{2}}[/tex], then the bulk modulus of water is:

[tex]\Delta P = B\cdot \left(\frac{\rho_{f}}{\rho_{o}}-1 \right)[/tex]

[tex]\Delta P = \left(2.2\times 10^{9}\,\frac{N}{m^{3}} \right)\cdot \left(\frac{992.219\,\frac{kg}{m^{3}} }{999.623\,\frac{kg}{m^{3}} }-1 \right)[/tex]

[tex]\Delta P = -16294943.19\,Pa \,(-160.819\,atm)[/tex]

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?​

Answers

Answer:

The correct answer is "[tex]4.49\times 10^{10} \ joules[/tex]".

Explanation:

According to the question,

The work will be:

⇒ [tex]Work=-\frac{kQq}{R}[/tex]

              [tex]=-\frac{1}{4 \pi \varepsilon \times (18-30)\times 3\times 0.2}[/tex]

              [tex]=-\frac{1}{4 \pi \varepsilon \times (-12)\times 3\times 0.2}[/tex]

              [tex]=\frac{0.3978}{\varepsilon }[/tex]

              [tex]=4.49\times 10^{10} \ joules[/tex]

Thus the above is the correct answer.    

We have that the workdone  is mathematically given as

W=4.49*10e10 J

From the question we are told

A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?​

Workdone

Generally the equation for the workdone   is mathematically given as

W=-kQq/R

Therefore

0.3978/ε0 =-1/(4πε0*(18-30)*3*0.2

Hence

W=4.49*10e10 J

For more information on Charge visit

https://brainly.com/question/9383604

Which of the following statements is correct about the magnitude of the static friction force between an object and a surface?

a. Static friction depends on the mass of the object.
b. Static friction depends on the shape of the object.
c. Static friction depends on what the object is made of but not what the surface is made of.
d. None of the above is correct.

Answers

Answer:

Static friction depends on the mass of the object.

Explanation:

Friction is the force between two surfaces in contact. The force of friction between two surfaces in contact depends on;

1) nature of the object and the surface(how rough or smooth the surfaces are)

2)surface area of the object and the surface

3) mass of the object

Since;

F=μmg

Where;

μ= coefficient of static friction

m= mass of the object

g= acceleration due to gravity

Hence, as the mass of the object increases, the magnitude of static friction force between an object and a surface increases and vice versa.

PLZ help asap :-/
............................ ​

Answers

Explanation:

[16]

[tex]\underline{\boxed{\large{\bf{Option \; A!! }}}} [/tex]

Here,

[tex]\rm { R_1} [/tex] = 2Ω[tex]\rm { R_2} [/tex] = 2Ω[tex]\rm { R_3} [/tex] = 2Ω[tex]\rm { R_4} [/tex] = 2Ω

We have to find the equivalent resistance of the circuit.

Here, [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] are connected in series, so their combined resistance will be given by,

[tex]\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\ [/tex]

[tex]\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\ [/tex]

[tex]\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\ [/tex]

Now, the combined resistance of [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] is connected in parallel combination with [tex]\rm { R_3} [/tex], so their combined resistance will be given by,

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\ [/tex]

Reciprocating both sides,

[tex]\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\ [/tex]

Now, the combined resistance of [tex]\rm { R_1} [/tex], [tex]\rm { R_2} [/tex] and [tex]\rm { R_3} [/tex] is connected in series combination with [tex]\rm { R_4} [/tex]. So, equivalent resistance will be given by,

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4}{3} + 2 \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4 + 6}{3} \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{10}{3} \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \bf {R_{(1,2,3,4)}= 3.33 \; \Omega} \\ [/tex]

Henceforth, Option A is correct.

_________________________________

[17]

[tex]\underline{\boxed{\large{\bf{Option \; B!! }}}} [/tex]

Here, we have to find the amount of flow of current in the circuit. By using ohm's law,

[tex] \longrightarrow [/tex] V = IR

[tex] \longrightarrow [/tex] 3 = I × 3.33

[tex] \longrightarrow [/tex] 3 ÷ 3.33 = I

[tex] \longrightarrow [/tex] 0.90 Ampere = I

Henceforth, Option B is correct.

____________________________

[tex] \tt \purple{Hope \; it \; helps \; you, Army! \heartsuit } \\ [/tex]

a girl is moving with a uniform velocity of 1.5 m/s then mathematically find her acceleration​

Answers

Answer:

0

Explanation:

a = dv/dt

if v is constant than the slope of the v graph will be 0, so dv/dt is 0

a= 0

how can scientific method solve real world problems examples

Answers

The scientific method is nothing more than a process for discovering answers. While the name refers to “science,” this method of problem solving can be used for any type of problem

How are elastic and inelastic collisions different?


A: Elastic collisions occur when the colliding objects move separately after the collision; after inelastic collisions, the objects are connected and move together.

B: Elastic collisions occur when the objects are going the same direction when they collide; inelastic collisions occur when the objects are going in opposite directions when they collide.

C: Momentum is conserved in elastic collisions; momentum is not conserved in inelastic collisions.

D: Elastic collisions occur between objects of the same mass; inelastic collisions occur between different masses.

Answers

Answer:

a

Explanation:

Answer:

the answer is c

'

Explanation:

What are the differences among elements, compounds, and mixtures?

Answers

Answer:

Elements have a characteristic number of electrons and protons.Both Hydrogen(H) and oxygen(O) are two different elements.

••••••••••••••••

Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of .Hydrogen(H) and Oxygen(O) both qr4 naturally gases,but they react to form water(H2O),which is liquid compound.

•••••••••••••••

A mixture is made of atleast two parts》 solid,liquid or gas.The difference is that it's not a chemical substance that's bonded by other elements. 

------------------------------

Hope it helps...

Have a great day!!!

Answer: Elements have a characteristic number of electrons and protons. Both Hydrogen(H) and oxygen(O) are two different elements. Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of.Hydrogen(H) and Oxygen(O) both qr4 naturally gases, but they react to form water(H2O), which is liquid compound. A mixture is made of at least two parts solid, liquid, or gas. The difference is that it's not a chemical substance that's bonded by other elements.

A charge Q exerts a 1.2 N force on another charge q. If the distance between the charges is doubled, what is the magnitude of the force exerted on Q by q

Answers

Answer:

0.3 N

Explanation:

Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.

The magnitude of the force exerted on Q by q when the distance between them is doubled is 0.3 N

Coulomb's law equation

F = Kq₁q₂ / r²

Where

F is the force of attraction K is the electrical constant q₁ and q₂ are two point charges r is the distance apart

Data obtained from the question Initial distance apart (r₁) =  rInitial force (F₁) = 1.2 NFinal distance apart (r₂) = 2rFinal force (F₂) =?

How to determine the final force

From Coulomb's law,

F = Kq₁q₂ / r²

Cross multiply

Fr² = Kq₁q₂

Kq₁q₂ = constant

F₁r₁² = F₂r₂²

With the above formula, we can obtain the final force as follow:

F₁r₁² = F₂r₂²

1.2 × r² = F₂ × (2r)²

1.2r² = F₂ × 4r²

Divide both side by 4r²

F₂ = 1.2r² / 4r²

F₂ = 0.3 N

Learn more about Coulomb's law:

https://brainly.com/question/506926

Your cell phone typically consumes about 300 mW of power when you text a friend. If the phone is operated using a lithium-ion battery with a voltage of 3.5 V, what is the current (in A) flowing through the cell-phone circuitry under these circumstances

Answers

Answer:

I = 0.0857 A

Explanation:

Given that,

Power consumed by the cellphone, P = 300 mW

The voltage of the battery, V = 3.5 V

Let I is the current flowing through the cell-phone. We know that,

P = VI

Where

I is the current

So,

[tex]I=\dfrac{P}{V}\\\\I=\dfrac{300\times 10^{-3}}{3.5}\\\\I=0.0857\ A[/tex]

So, the current flowing the cell-phone is 0.0857 A.


If an electrical component with a resistance of 53 Q is connected to a 128-V source, how much current flows through the component?

Answers

Answer:

the current that flows through the component is 2.42 A

Explanation:

Given;

resistance of the electrical component, r = 53 Ω

the voltage of the source, V = 128 V

The current that flows through the component is calculated using Ohm's Law as demonstrated below;

[tex]V = IR\\\\I = \frac{V}{R} = \frac{128 \ V}{53 \ ohms} = 2.42 \ A[/tex]

Therefore, the current that flows through the component is 2.42 A

why is it wrong to leave our light on​

Answers

Answer:

you will get huge electricity bills ............

Four equal-value resistors are in series with a 5 V battery, and 2.23 mA are measured. What isthe value of each resistor

Answers

Answer:

560.54 Ω

Explanation:

Applying,

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

Where V = Voltage of the battery, I = currrent, R' = Total resistance of the resistors

make R' the subject of the equation

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

From the question,

Given: V = 5 V, I = 2.23 mA = 2.23×10⁻³ A

Substitute these values into equation 2

R' = 5/(2.23×10⁻³ )

R' = 2242.15 Ω

Since the fours resistor are connected in series and they are equal,

Therefore the values of each resistor is

R = R'/4

R = 2242.15/4

R = 560.54 Ω

A car accelerates at 2 meters/s/s. Assuming the car starts from rest how far will it travel in 10 seconds

Answers

Answer:

Distance = velocity x time, so 10 m/s X 10 s = 100 m

Explanation:

If you accelerate at 2 m/s^2 for 10 seconds, at the end of the 10 seconds you are moving at a rate of 20 m/s.

V(f) = V(i) + a*t

Final velocity = initial velocity + acceleration x time

Your average velocity will be half of your final, because you accelerated at a constant rate. So your average velocity is 10 m/s.

Distance = velocity x time, so 10 m/s X 10 s = 100 m

Answer:

100 m

Explanation:

Given,

Initial velocity ( u ) = 0 m/s

Acceleration ( a ) = 2 m/s^2

Time ( t ) = 10 sec s

To find : Displacement ( s ) = ?

By 2nd equation of motion,

s = ut + at^2 / 2

= ( 0 ) ( 10 ) + ( 2 ) ( 10 )^2 / 2

= 0 + ( 2 ) ( 100 ) / 2

= 200 / 2

s = 100 m

A football quarterback runs 15.0 m straight down the playing field in 3.00 s. He is then hit and pushed 3.00 m straight backward in 1.71 s. He breaks the tackle and runs straight forward another 24.0 m in 5.20 s. Calculate his average velocity (in m/s) for the entire motion. (Assume the quarterback's initial direction is positive. Indicate the direction with the sign of your answer.)

Answers

Answer:

Average Velocity = 3.63 m/s

Explanation:

First, we will calculate the total displacement of the quarterback, taking forward direction as positive:

Total Displacement = 15 m - 3 m + 24 m = 36 m

Now, we will calculate the total time taken for this displacement:

Total Time = 3 s + 1.71 s + 5.2 s = 9.91 s

Therefore, the average velocity will be:

[tex]Average\ Velocity = \frac{Total\ Displacement}{Total\ Time}\\\\Average\ Velocity = \frac{36\ m}{9.91\ s}[/tex]

Average Velocity = 3.63 m/s

The correct equation for the x component of a vector named A with an angle measured from the x axis would be which of the following?

Answers

Answer:

Acosθ

Explanation:

The x-component of a vector is defined as :

Magnitude * cosine of the angle

Maginitude * cosθ

The magnitude is represented as A

Hence, horizontal, x - component of the vector is :

Acosθ

Furthermore,

The y-component is taken as the sin of the of the angle multiplied by the magnitude

Vertical, y component : Asinθ

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Problem: The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72inches and standard deviation 3.17 inches. Compute the probability that a simple random sample of size n=10 results in a sample mean greater than 40 inches. That is, compute P(mean >40).Gestation period The length of human pregnancies is approximately normally distributed with mean u = 266days and standard deviation o = 16 days.TaggedMath1. What is the probability a randomly selected pregnancy lasts less than 260 days?2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 daysor less?3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 daysor less?4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days ofthe mean?KnowLearnBoosteV See Please help, it is a graph question Using the following image, solve for x which of the following words that that does not belong to each sound group.bean,pick,bead,peak. What theories existed that tried to explain the spread of the plague? When the dry-bulb temperature is 22C and the wet-bulb temperature is 13C, the relativehumidity is1. 10%2. 33%3. 41%4. 59% Help me find the domain and range please! 8 Find the value of x a (22)x is exponent of (22) = 64 Vasopressin works on write an expression for 15 divided by a numbershow your work Rearrange this wordVeyre Boloy what was one majort effect of the spread of railroads throughout great Britain during the industrial revolution answer please i need to pass this class A block slides down a frictionless plane that makes an angle of 24.0 with the horizontal. What is theacceleration of the block? Suppose velocity is constant, but real GDP is not independent of the money supply. If this is the case, a 10 percent increase in the money supply will: Find the value of x that will make A||B. Please help! Which function represents the graph below? Drag the operator to the correct location on the image.Which operation results in a binomial? Please help. I'm stuck on this problem Investment interest expense includes: A. interest expense from loans to purchase corporate bonds and interest expense from loans to purchase stocks. B. interest expense from loans to purchase corporate bonds. C. interest expense from loans to purchase municipal bonds. D. interest expense from loans to purchase municipal bonds and interest expense from loans to purchase corporate bonds. E. interest expense from loans to purchase stocks.