A jetliner rolls down the runway with constant acceleration from rest, it reaches its take off speed of 250 km/h in 1 min. What is its acceleration? Express in km/h^2, does this result make sense?

Answer:

Option B

Explanation:

The fundamental quantities are the basic quantities from which other quantities are derived.

There are a total of 7 fundamental quantities, namely Length, Mass, Time, Luminous Intensity, Amount of substance, Electric Current, Temperature.

From these 7 quantities, the other quantities are derived.

Weight is the product of its mass and the gravitational acceleration 'g' which acts on that mass.

Where mass is the fundamental quantity and 'g' is also derived and hence weight is a derived quantity.

Answer:

3073 m/s

Explanation:

The key is that the period of the satellite is 24 hours  because it is synchronized with the rotation of the Earth. You can use Kepler's Third law to find the radius of the orbit:

  [tex]4\pi^2 r^3 = G MT^2[/tex]

[tex]4\pi^2 r^3= 6.67\times10^{-11}\times6\times10^{24}\times86400^2[/tex]

     r =   4.225 x 10^7 meters

So  one complete orbit is a distance of

   [2πr = 2× π × 4.225 x 10^7 = 26.55 x 10^7 meters

So the speed is

  distance / time = 26.55 x 10^7 meters / 86400 seconds =

        =   3073 m/s

Answer:

A. structure of the classroom

Explanation:

According to the Creative Curriculum model, the A. structure of the classroom  sets the context for teaching and learning.

Creative curriculum model stands to enhance the skill of the students through a structured class room. Creative curriculum goes beyond the rote learning and focuses on big ideas and individual passion of each students.

Answer:

Radius of aluminium sphere which has same mass as of sphere of iron with radius 14 m is 40.745 meters.

Explanation:

Let the radius of aluminium sphere be [tex]r_{a}[/tex]

From the relation between density, mass and volume we know that

[tex]Mass=Density\times volume...............(i)[/tex]

Applying equation 'i' separately to iron and aluminium sphere we get

[tex]M_{a}=\rho _{a}\times V_{a}[/tex]

[tex]M_{ir}=\rho _{ir}\times V_{ir}[/tex]

Equating the masses of iron and aluminium spheres we get

[tex]M_{a}=M_{ir}[/tex]

[tex]\rho _{a}\times \frac{4\pi r_a^3}{3}=\rho _{ir}\times \frac{4\pi r_{ir}^3}{3}[/tex]

[tex]\rho _{a}\times r_{a}^3=\rho _{ir}\times r_{ir}^{3}\\\\\therefore r_{a}=(\frac{\rho _{ir}}{\rho _{a}})^{1/3}\cdot r_{ir}\\\\r_{a}=\frac{7.8610}{2.701}\times 14\\\\\therefore r_a=40.745m[/tex]

Answer:

The period of the pendulum is 1.816 sec.

Explanation:

Given that,

Length = 81.9 cm

Mass = 165 g

Angle = 10°

We need to calculate the period of the pendulum

Using formula of period

[tex]T = 2\pi\sqrt{\dfrac{l}{g}}[/tex]

Where, l = length

g = acceleration due to gravity

Put the value into the formula

[tex]T =2\pi\sqrt{\dfrac{81.9\times10^{-2}}{9.80}}[/tex]

[tex]T=1.816\ sec[/tex]

Hence, The period of the pendulum is 1.816 sec.

Answer:

The charge transferred equals [tex]-4.96\times 10^{-6}Columbs[/tex]

Explanation:

The charge quantizatrion principle states that the charge is always transferred in terms of product of fundamental charges.

Mathematically [tex]Q=ne[/tex]

where

'e' is the fundamental magnitude of charge

Thus charge transferred by [tex]4.4\times 10^{13}[/tex] electrons equals

[tex]Q_{1}=4.4\times 10^{13}\times -1.6\times 10^{-19}=-7.04\times 10^{-6}Columbs[/tex]

Similarly charge transferred by [tex]1.3\times 10^{13}[/tex] protons equals

[tex]Q_{2}=1.3\times 10^{13}\times 1.6\times 10^{-19}=2.08\times 10^{-6}Columbs[/tex]

Thus the net charge transferred equals

[tex]Q_{1}+Q_{2}=-7.04\times 10^{-6}+2.08\times 10^{-6}\\\\=-4.96\times 10^{-6}columbs[/tex]

Answer:

Explained

Explanation:

a) the work done will be positive since the chicken is scratching the ground. Here displacement is along the direction of force.

b) A person studying does no work in the language of physics because there is no displacement.

C) the work is done on the bucket by the crane and work is positive and here the displacement is in the direction of force.

d) Gravitational force act on the bucket in downward direction, here the work done will be negative as the force displacement are opposite to each other.

E)Negative work will be done as the force applied  by the muscle is in opposite to the displacement.

Answer:

x=22.33m

Explanation:

Kinematics equation for constant deceleration:

[tex]x =v_{o}*t - 1/2*at^{2}=8.9*6.3-1/2*1.70*6.3^{2}=22.33m[/tex]

Answer:

the magnitude of each spaceships velocity is 0.495c

Explanation:

This is a relative velocity problem, that means that the velocity is relative to the observer, in this case, there is an observer on earth, from this reference point the velocity of each spaceship are the same, but for the observer on a spaceship would be different, we can see it as a velocity difference :

[tex]V=Va-Vb\\[/tex]

because the spaceship B is going on the opposite direction from A, the velocity is negative, the observer on earth says the velocity is the same for both o them, so:

[tex]0.99c=va-(-va)\\0.99c=2va\\va=0.495c[/tex]

vb=-va

Answer:

3.54 nC.

Explanation:

Given:

The surface area of each of the electrode is given by

[tex]A=2.0\ cm \times 2.0\ cm = 4.0\ cm^2 = 4.0\times 10^{-4}\ m^2.[/tex]

The strength of the electric field inside a capacitor is given by

[tex]E = \dfrac{\sigma}{\epsilon_o}[/tex]

where,

Therefore,

[tex]E = \dfrac{q}{A\epsilon_o}\\q=EA\epsilon_o\\=1.0\times 10^6\times 4.0\times 10^{-4}\times 8.85\times 10^{-12}\\=3.54\times 10^{-9}\ C\\=3.54\ nC.[/tex]

It is the charge on the positive electrode.

The value of an electric field at a particular point is known as the strength of the electric field. the value of the charge will be 3.54 NC.

The electric field is the area around an electric charge where its impact may be felt. The force encountered by a unit positive charge put at a spot is the electric field strength at that point.

Electrode dimension = 2.0 cm×2.0 cm

Area of electrode = 4.0 cm²

seperated distance beween electrode = 2.2 mm = 2.2 × 10⁻³ m

strength of electric field = E = 1.0 × 10⁶ N/C

Electric field strength is given by

[tex]\rm E = \frac{\sigma}{\varepsilon_0}[/tex]

where,

[tex]\varepsilon_0[/tex]= electrical permittivity of free space

[tex]\sigma[/tex] = surface charge density of the electrode

q=Electrode charge

[tex]\rm \sigma =\RM{ \frac{q}{A} }[/tex]

[tex]\rm E = \frac{q}{A\varepsilon_0}[/tex]

[tex]\rm q = \sigma{A\varepsilon_0}[/tex]

[tex]\rm q =1.0\times 10^-6\times 4.0\times10^{-4}\times 8.85 \times 10^-12[/tex]

[tex]\rm q = 3.54\times 10^-9 \; C[/tex]  

[tex]\rm q = 3.54\; nC[/tex]

Hence the value of the charge will be 3.54 NC.

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Answer:

3.324 km

Explanation:

d1 = 2 km south

d2 = 1.5 km at 37° east of south

Write the displacements in vector form

[tex]\overrightarrow{d_{1}}=-2\widehat{j}[/tex]

[tex]\overrightarrow{d_{2}}=1.5\left (Sin37\widehat{i}-Cos37\widehat{j}  \right )=0.9\widehat{i}-1.2\widehat{j}[/tex]

The resultant displacement is given by

[tex]\overrightarrow{d} = \overrightarrow{d_{1}}+ \overrightarrow{d_{2}}[/tex]

[tex]\overrightarrow{d} = \left ( 0.9 \right )\widehat{i}+\left ( -2-1.2 \right )\widehat{j}[/tex]

[tex]\overrightarrow{d} = \left ( 0.9 \right )\widehat{i}+\left ( -3.2\right )\widehat{j}[/tex]

The magnitude of displacement is given by

[tex]d=\sqrt{0.9^{2}+\left ( -3.2 \right )^{2}}=3.324 km[/tex]

Thus, the bird has to travel 3.324 km in a straight line to return to its original place.

Answer:

The diameter of the piston of the players equals 55.136 cm.

Explanation:

from the principle of transmission of pressure in a hydraulic lift  we have

[tex]\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{1}}[/tex]

Since the force in the question is the weight of the individuals thus upon putting the values in the above equation we get

[tex]\frac{57.0\times 9.81}{\frac{\pi \times (19.0)^{2}}{4}}=\frac{4\times 120\times 9.81}{\frac{\pi \times D_{2}^{2}}{4}}[/tex]

Solving for [tex]D_{2}[/tex] we get

[tex]D_{2}^{2}=\frac{4\times 120}{57}\times 19^{2}\\\\\therefore D_{2}=\sqrt{\frac{4\times 120}{57}}\times 19\\\\D_{2}=55.136cm[/tex]

The diameter of the football players' piston is  [tex]\({55.4 \text{ cm}} \).[/tex]

To solve for the diameter of the football players' piston in the hydraulic lift, we start by using the principle of hydraulic systems, which states that pressure is constant throughout the fluid.

Given data:

- Mass of the cheerleader, [tex]\( m_1 = 57.0 \)[/tex] kg

- Mass of four football players, [tex]\( m_2 = 4 \times 120 \)[/tex] kg

- Height difference, [tex]\( h = 1.10 \)[/tex] m

- Diameter of the cheerleader's piston, [tex]\( D_1 = 19.0 \)[/tex] cm

First, convert the diameter of the cheerleader's piston into meters:

[tex]\[ D_1 = 19.0 \text{ cm} = 0.19 \text{ m} \][/tex]

The force exerted by the cheerleader's piston is:

[tex]\[ F_1 = m_1 g \][/tex]

The force exerted by the football players' piston is:

[tex]\[ F_2 = m_2 g \][/tex]

According to Pascal's principle:

[tex]\[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \][/tex]

where [tex]\( A_1 \)[/tex] and [tex]\( A_2 \)[/tex] are the cross-sectional areas of the cheerleader's and football players' pistons, respectively.

The area [tex]\( A \)[/tex] of a piston is given by:

[tex]\[ A = \frac{\pi D^2}{4} \][/tex]

Now, calculate [tex]\( A_1 \)[/tex]:

[tex]\[ A_1 = \frac{\pi (0.19)^2}{4} \][/tex]

[tex]\[ A_1 = \frac{\pi \times 0.0361}{4} \][/tex]

[tex]\[ A_1 \approx 0.0284 \text{ m}^2 \][/tex]

Next, find [tex]\( F_1 \)[/tex]:

[tex]\[ F_1 = m_1 g = 57.0 \times 9.8 \][/tex]

[tex]\[ F_1 \approx 558.6 \text{ N} \][/tex]

Now, find [tex]\( A_2 \)[/tex]:

[tex]\[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \][/tex]

[tex]\[ A_2 = \frac{F_2 \times A_1}{F_1} \][/tex]

Calculate [tex]\( F_2 \)[/tex]:

[tex]\[ F_2 = m_2 g = 480 \times 9.8 \][/tex]

[tex]\[ F_2 = 4704 \text{ N} \][/tex]

Now, find [tex]\( A_2 \)[/tex]:

[tex]\[ A_2 = \frac{4704 \times 0.0284}{558.6} \][/tex]

[tex]\[ A_2 \approx 0.2397 \text{ m}^2 \][/tex]

Finally, solve for [tex]\( D_2 \)[/tex]:

[tex]\[ A_2 = \frac{\pi (D_2)^2}{4} \][/tex]

[tex]\[ (D_2)^2 = \frac{4 \times 0.2397}{\pi} \][/tex]

[tex]\[ D_2 \approx \sqrt{\frac{0.9588}{\pi}} \][/tex]

[tex]\[ D_2 \approx 0.554 \text{ m} \][/tex]

Convert [tex]\( D_2 \)[/tex] from meters to centimeters:

[tex]\[ D_2 \approx 55.4 \text{ cm} \][/tex]

Answer:

9.81 × 10⁻¹⁰ C

Explanation:

Given:

Distance between the tissue and the tip of the scale, r = 6 cm = 0.06 m

Charge on the ruler, Q = - 12 μC = - 12 × 10⁻⁶ C

Mass of the tissue = 3 g = 0.003 Kg

Now,

The force required to pick the tissue, F = mg

where, g is the acceleration due to gravity

also,

The force between (F) the charges is given as:

[tex]F=\frac{kQq}{r^2}[/tex]

where,

q is the charge on the tissue

k is the Coulomb's constant = 9 × 10⁹ Nm²/C²

thus,

[tex]mg=\frac{kQq}{r^2}[/tex]

on substituting the respective values, we get

[tex]0.003\times9.81=\frac{9\times10^9\times(-12\times10^{-6})\times q}{0.06^2}[/tex]

or

q = 9.81 × 10⁻¹⁰ C

Minimum charge required to pick the tissue paper is 9.81 × 10⁻¹⁰ C

Answer:

Option (a) is correct.

Explanation:

The angle of banking of curved path is given as

tan θ[tex]=\frac{v^{2} }{rg}[/tex]

Here, v is linear velocity, r is radius of curved path, θ is bank angle and g is acceleration due to gravity.

We have,  θ= 20.0°, r = 100 m and take [tex]g=9.8 m/s^{2}[/tex]

Substituting these values in above formula, we get

[tex]tan 20 =\frac{v^{2} }{100 m*9.8m/s^{2} }[/tex]

[tex]v^{2} =356.69[/tex]

[tex]v=\sqrt{356.69} =18.9m/s[/tex]

Thus, the ideal speed is 18.9 m/s.

The ideal speed to take a 100 m radius curve banked at a 20.0° angle is 18.9 m/s. The correct Option is a.

The speed of any moving object is the ratio of the distance covered and the time taken to cover that distance.

Speed s = distance d / time t

Given a 100 m radius curve banked at a 20.0° angle.

The angle of banking of curved path is given as

tan θ= V² /2g

Here, V is linear velocity, r is radius of curved path, θ is bank angle and g is acceleration due to gravity.

Substituting the values in above formula, we get

tan 20° = V² /2x9.81

V = 18.9 m/s

Thus, the correct option is a.

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Answer: 8*10^-15 N

Explanation: In order to calculate the force applied on an electron in the middle of the two planes at 500 V we know that,  F=q*E

The electric field between  the plates is given by:

E = ΔV/d = 500 V/0.01 m=5*10^3 N/C

the force applied to the electron is: F=e*E=8*10^-15 N

Answer:

The modulation index in the amplitude modulation will be 0.384

Explanation:

We have given amplitude of adulating signal [tex]A_m=0.5volt[/tex]

Amplitude of carrier signal [tex]A_C=1.3volt[/tex]

We have to find modulation index

Modulation index is the ratio of amplitude of modulating signal and amplitude of carrier signal

So modulation index [tex]m=\frac{A_m}{A_c}=\frac{0.5}{1.3}=0.384[/tex]

So the modulation index in the amplitude modulation will be 0.384

Answer:

2.4 m

Explanation:

Plane Mirror: It is a type of mirror whose reflecting surface is plane. it produces an image of an object behind it at a distance equal to the object lying in front of the mirror.

According to the question, the person has to take the photo of himself who is looking at the mirror.

Let us assume that the person is standing in front of a plane mirror at a distance of 1.2 m.

Due to the reflection of light, the image of the person will be formed at a distance 1.2 m behind the mirror. This means the image of the person is 2.4 m apart from him.

As the person could take the picture of his own image from the classic camera in his hand. This means the image of the person will act as an object for the classic camera to which he is going to focus on which is at a distance of 2.4 m.

Hence, the person will choose a distance of 2.4 m.

Focus length is the distance of optics from the point of convergence of light rays to create a sharp image of the object.

The distance of the focus must be 2.4 meters.

Focus length is the distance of optics from the point of convergence of light rays to create a sharp image of the object. Focus length represents in the millimetres.

In the given problem, the person is standing in front of the plane mirror. A plane mirror is the mirror with a flat reflective surface.

The image formed by the plane mirror is virtual, erect and the equal to the size of the person or thing.

Given information-

The distance of person from the mirror is 1.2 m.

As the image in the plane mirror appears at the equivalent distance to the object, thus the distance of the image of the person is,

[tex]d_i=1.2 \rm m[/tex]

Thus the distance of the image of the person appears is 1.2 meters behind the mirror.

To take the photo of this image, the distance of the focus should be equal to the sum of the distance of the image inside the mirror to the distance of person front of the mirror. Thus,

[tex]d_f=d_p+d_i\\d_f=1.2+1.2\\d_f=2.4 \rm m[/tex]

Hence the distance of the focus must be 2.4 meters.

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Answer:

320 K

Explanation:

Given:

Number of moles of gas , n = 3

Dimensions of the rectangular container = 0.2 m × 0.2 m × 0.4 m

Thus,

Volume of the container =  0.2 m × 0.2 m × 0.4 m = 0.016 m³

Pressure inside the container, P = 498,840 N/m²

Now,

From the ideal gas law, we have

PV = nRT

here,

T is the temperature

R is the ideal gas constant = 8.314 J / (mol·K)

on substituting the values, we get

498,840 × 0.016 = 3 × 8.314 × T

or

T = 320 K

Hence, the pressure inside the container is 320 K

Answer:

5.37 m/s

0.98 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -9.81\times 10+15^2}\\\Rightarrow v=5.37\ m/s[/tex]

Velocity of the ball when it passes an observer sitting at a window is 5.37 m/s

[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{5.37-15}{-9.81}\\\Rightarrow t=0.98\ s[/tex]

Time taken by the ball to pass the observer sitting at a window is 0.98 seconds

Answer:

The maximum temperature is 90.06° C

Explanation:

Given that

t= 0.1 mm

Heat generation

[tex]q_g=0.3\ MW/m^3[/tex]

Heat transfer coefficient

[tex]h=500\ W/m^2K[/tex]

Here one side(left side) of the wall is insulated so the all heat will goes in to right side .

The maximum temperature will at the left side.

Lets take maximum temperature is T

Total heat flux ,q

[tex]q=q_g\times t[/tex]

[tex]q=0.3\times 1000000\times 0.1 \times 10^{-3}\ W/m^2[/tex]

[tex]q=30\ W/m^2[/tex]

So the total thermal resistance per unit area

[tex]R=\dfrac{t}{K}+\dfrac{1}{h}[/tex]

[tex]R=\dfrac{0.1\times 10^{-3}}{25}+\dfrac{1}{500}[/tex]

R=0.002 K/W

We know that

q=ΔT/R

30=(T-90)/0.002

T=90.06° C

The maximum temperature is 90.06° C

Answer:

[tex]V=\dfrac{PT}{m}\ ln\dfrac{M+m}{M}-gT[/tex]

Explanation:

Given that

Constant rate of leak =R

Mass at time T ,m=RT

At any time t

The mass = Rt

So the total mass in downward direction=(M+Rt)

Now force equation

(M+Rt) a =P- (M+Rt) g

[tex]a=\dfrac{P}{M+Rt}-g[/tex]

We know that

[tex]a=\dfrac{dV}{dt}[/tex]

[tex]\dfrac{dV}{dt}=\dfrac{P}{M+Rt}-g[/tex]

[tex]\int_{0}^{V}V=\int_0^T \left(\dfrac{P}{M+Rt}-g\right)dt[/tex]

[tex]V=\dfrac{P}{R}\ ln\dfrac{M+RT}{M}-gT[/tex]

[tex]V=\dfrac{PT}{m}\ ln\dfrac{M+m}{M}-gT[/tex]

This is the velocity of bucket at the instance when it become empty.

The velocity of the bucket when it becomes empty can be found by setting up an equation using Newton's second law and the equation for final velocity. When the bucket becomes empty, its final velocity is equal to the force exerted by the rope multiplied by the time taken for the bucket to become empty, divided by the initial mass of the bucket.

When the bucket becomes empty, the force exerted by the rope pulling the bucket up will all be used to accelerate the bucket. We can set up an equation using Newton's second law, which states that the net force on an object is equal to its mass multiplied by its acceleration: F_net = M * a. Since there are no other forces acting on the bucket except for the force exerted by the rope, we can equate the force to the force exerted by the rope: P = M * a.

Since the bucket is initially at rest, its initial velocity is zero. As the bucket becomes empty, the mass of the bucket decreases, but the force exerted by the rope remains constant. The acceleration of the bucket will therefore increase, resulting in an increasing velocity. When the bucket becomes empty, its final velocity can be determined using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken for the bucket to become empty. Rearranging the equation, we have v = 0 + Pt/M.

Answer:

a) ω1 = 18rpm    ω2 = -18rpm

b) ω1 = 102rpm     ω2 = 138rpm

c) ω1 = ω2 = 3.18rpm

Explanation:

For the first case, we know that each wheel will spin in a different direction but with the same magnitude, so:

ωr = 6rpm   This is the angular velocity of the robot

[tex]\omega = \frac{\omega r * D/2}{r_{wheel}}[/tex]  where D is 30cm and rwheel is 5cm

[tex]\omega = \frac{6 * 30/2}{5}=18rpm[/tex]  One velocity will be positive and the other will be negative:

ω1 = 18rpm    ω2 = -18rpm

For part b, the formula is the same but distances change. Rcircle=100cm:

[tex]\omega 1 = \frac{\omega r * (R_{circle} - D/2)}{r_{wheel}}[/tex]

[tex]\omega 2 = \frac{\omega r * (R_{circle} + D/2)}{r_{wheel}}[/tex]

Replacing values, we get:

[tex]\omega 1 = \frac{6 * (100 - 30/2)}{5}=102rpm[/tex]

[tex]\omega 2 = \frac{\omega r * (100 + 30/2)}{5}=138rpm[/tex]

For part c, both wheels must have the same velocity:

[tex]\omega = \frac{V_{robot}}{r_{wheel}}=20rad/min[/tex]

[tex]\omega = 20rad/min * \frac{1rev}{2*\pi rad}=3.18rpm[/tex]

Answer:

Explanation:

We know that the equation for time dilation will be:

[tex]\Delta t = \frac{\Delta t'}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

where Δt its the time difference measured from Earth, and Δt' is the time difference measured by the astronaut.

Lets work a little the equation

[tex] \sqrt{1-\frac{v^2}{c^2}} = \frac{\Delta t'}{\Delta t}[/tex]

[tex] 1-\frac{v^2}{c^2}= (\frac{\Delta t'}{\Delta t})^2[/tex]

[tex] \frac{v^2}{c^2}= 1 - (\frac{\Delta t'}{\Delta t})^2[/tex]

[tex] \frac{v}{c}= \sqrt{ 1 - (\frac{\Delta t'}{\Delta t})^2 }[/tex]

[tex] v = \sqrt{ 1 - (\frac{\Delta t'}{\Delta t})^2 } c [/tex]

So, we got our equation. Knowing that Δt=211 years and Δt'=15 years

then

[tex] v = \sqrt{ 1 - (\frac{15 \ y}{211 \ y})^2 } c [/tex]

[tex] v = 0.99747 c [/tex]

To calculate the necessary speed, we can use the time dilation formula. Given the time experienced by the astronaut and the time experienced on Earth, we can solve for the velocity using the Lorentz factor. Using the approximate trip time of 211 years, we can calculate the necessary speed with the given equation.

To calculate the speed necessary, we can use the time dilation formula:

Δt' = Δt / γ

Where Δt' is the time experienced by the astronaut, Δt is the time experienced on Earth, and γ is the Lorentz factor given by γ = 1 / √(1 - (v² / c²)), where v is the velocity of the astronaut and c is the speed of light.

Given that the astronaut wishes to age only 15 years during the trip, we can approximate the dilated trip time to be 211 years. Substituting these values, we have:

15 = 211 / γ

Simplifying the equation, we find:

γ = 211 / 15

Using this value of γ, we can calculate the velocity of the astronaut:

v = √((1 - (1 / γ²)) * c²)

Substituting the value of γ, we have:

v = √((1 - (1 / (211 / 15)²)) * c²)

Answer:

mean = 10.68 m/s

standard deviation 0.3059

[/tex]\sigma_m = 0.14[/tex]  

Explanation:

1) [tex]Mean = \frac{ 10.2+11+10.7+11+10.5}{5}[/tex]

  mean = 10.68 m/s

2 ) standard deviation is given as

[tex]\sigma = \sqrt{ \frac{1}{N} \sum( x_i -\mu)^2}[/tex]

N = 5

   [tex]\sigma =\sqrt{ \frac{1}{5} \sum{( 10.2-10.68)^2+(11-10.68)^2 + (10.7- 10.68)^2+ (11- 10.68)^2++ (10.5- 10.68)^2[/tex]

SOLVING ABOVE RELATION TO GET STANDARD DEVIATION VALUE

\sigma  = 0.3059

3) ERROR ON STANDARD DEVIATION

[tex]\sigma_m = \frac{ \sigma}{\sqrt{N}}[/tex]

               [tex]= \frac{0.31}{\sqrt{5}}[/tex]

[tex]\sigma_m = 0.14[/tex]  

Answer:

Mean =  = 10.68 m/s

Standard deviation = σ = 0.342 m/s

Error =  0.153 .  

Explanation:

The data has 5 readings.

Let each of the readings be Y

Take average and find the mean X = (10.2+11+10.7+11+10.5)/5 = 53.4/5 = 10.68 m/s.

Take the difference between the data values and the mean and square them individually.

(10.2 - 10.68)² =(-0.48)² = 0.23

(11 - 10.68)² = 0.32² = 0.102

(10.7 - 10.68)² = (-0.02)² = 0.0004

(11-10.68)² =0.32² = 0.102

(10.5-10.68)² = (-0.18)² = 0.0324

Standard deviation = [tex]\sigma = \sqrt{\frac{\sum(Y-X)^2 }{n-1}}[/tex]

                                = [tex]\sqrt{(0.23+0.102+0.0004+0.102+0.0324)/(5-1)}[/tex]

                                 = [tex]\sqrt{0.1167}[/tex] = 0.342 m/s

Error = Standard deviation / [tex]\sqrt{n}[/tex] = 0.342/5 = 0.153 .

Answer:

[tex]F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}[/tex]

Explanation:

Since all the four charges are equidistant from the position of Q

so here we can assume this charge distribution to be uniform same as that of a ring

so here electric field due to ring on its axis is given as

[tex]E = \frac{k(4q)x}{(x^2 + R^2)^{1.5}}[/tex]

here we have

x = b

and the radius of equivalent ring is given as the distance of each corner to the center of square

[tex]R = \frac{d}{\sqrt2}[/tex]

now we have

[tex]E = \frac{4kq b}{(b^2 + \frac{d^2}{2})^{1.5}}[/tex]

so the force on the charge is given as

[tex]F = QE[/tex]

[tex]F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}[/tex]

The magnitude of the force exerted by object W on object Z is F.

The magnitude of the force exerted by object W on object Z can be determined by analyzing the geometry and the charges involved. Since objects X and Z are at the midpoints of the sides of the square, they are equidistant from all four charges. As a result, the magnitude of the force exerted by each charge on object Z will be the same as the magnitude of the force exerted by each charge on object X. Therefore, the magnitude of the force exerted by object W on object Z is also F.

In a uniform electric field, the electric potential or voltage increases or decreases linearly with distance based on the direction. Given initial voltage of 80V and field strength of 50 N/C, the voltage 1.0 m west increases by 50V to 130V.

For the given scenario, we are dealing with a uniform electric field and we need to determine the voltage at a certain distance in the field.

The key relationship between the electric field (E) and voltage (V) in a uniform electric field is that E = ΔV / Δd, where, ΔV is the potential difference and Δd is distance.

Given the electric field strength E = 50 N/C, and the initial voltage V1 = 80 V, we want to find the potential V2 a distance d = 1.0 m to the west, against the direction of the field. Since the electric field is uniform and points towards decreasing potential, the potential a distance d against the direction of field would increase. Therefore, ΔV = E * d = 50 N/C * 1.0 m = 50 V. Thus, the voltage at the point 1.0 m west would be V2 = V1 + ΔV = 80V + 50V = 130V.

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Answer:

The amount of gasoline is [tex]2.105\times10^{-2}\ gallons[/tex].

Explanation:

Given that,

Energy contained in gasoline [tex]= 1.3\times10^{8}\ J[/tex]

Mass = 2000 kg

Speed = 20 m/s

Energy used propel the car[tex] E=15\%\ of 1.3\times10^{8}\ J[/tex]

[tex]E=\dfrac{15}{100}\times1.3\times10^{8}[/tex]

[tex]E=19500000 = 1.9\times10^{7}\ J[/tex]

[tex]E=1.9\times10^{7}\ J[/tex]

We need to calculate the work done by the frictional force to stop the car

Using formula of work done

[tex]W=\Delta KE[/tex]

[tex]W=\dfrac{1}{2}m(v_{f}^2-v_{0}^2)[/tex]

[tex]W=\dfrac{1}{2}\times2000\times(0-20^2)[/tex]

[tex]W=-4.0\times10^{5}\ J[/tex]

Therefore,

Work done to bring the car back to its original speed

[tex]W=4.0\times10^{5}\ J[/tex]

[tex]Amount\ of\ gasoline\ needed = \dfrac{W}{E}[/tex]

[tex]Amount\ of\ gasoline =\dfrac{4.0\times10^{5}}{1.9\times10^{7}}[/tex]

[tex]Amount\ of\ gasoline =2.105\times10^{-2}\ gallons[/tex]

Hence, The amount of gasoline is [tex]2.105\times10^{-2}\ gallons[/tex].

Final answer:

The corresponding blackbody temperature of the radiation with a wavelength of 970 um is approximately 2.725 K (Kelvin). The blackbody radiation observed in interstellar space is known as the cosmic microwave background (CMB). This temperature represents the average temperature of the universe at the time the CMB was emitted.

Explanation:

The corresponding blackbody temperature of the radiation with a wavelength of 970 um is approximately 2.725 K (Kelvin).

The blackbody radiation observed in interstellar space is known as the cosmic microwave background (CMB). It is the afterglow of the Big Bang and fills all of space. The blackbody spectrum of the CMB has a temperature of 2.725 K, as determined from observations.

This temperature represents the average temperature of the universe at the time the CMB was emitted. It provides valuable insights into the early universe and supports the idea of the expanding universe.

Answer: [tex]-0.45^0C[/tex]

Explanation:-

Depression in freezing point is given by:

[tex]\Delta T_f=i\times K_f\times m[/tex]

[tex]\Delta T_f=T_f^0-T_f=(0-T_f)^0C[/tex] = Depression in freezing point

i= vant hoff factor = 2 (for electrolyte undergoing complete dissociation, i is equal to the number of ions produced)

[tex]KCl\rightarrow K^++Cl^-[/tex]

[tex]K_f[/tex] = freezing point constant = [tex]1.86^0C/m[/tex]

m= molality

[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]

Weight of solvent (water)= 550 g = 0.55 kg  (1kg=1000g) 

Molar mass of solute (KCl) = 74.5 g/mol

Mass of solute (KCl) = 5.0 g

[tex](0-T_f)^0C=2\times 1.86\times \frac{5g}{74.5g/mol\times 0.55kg}[/tex]

[tex](0-T_f)^0C=0.45[/tex]

[tex]T_f=-0.45^0C[/tex]

Thus the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water is [tex]-0.45^0C[/tex]

The freezing point of a solution : -0.45 °C

Solution properties are the properties of a solution that don't depend on the type of solute but only on the concentration of the solute.  

Solution properties of electrolyte solutions differ from non-electrolyte solutions because electrolyte solutions contain a greater number of particles and break down into ions. So the Solution properties of electrolytes is greater than non-electrolytes

The term is used in the Solution properties

• 1. molal  

that is, the number of moles of solute in 1 kg of solvent  

[tex]\large {\boxed {\bold {m = mole. \frac {1000} {mass \: of \: solvent (in \: grams)}}}[/tex]

• 2. mole fraction  

the ratio of the number of moles of solute to the mole of solution  

[tex]\large {\boxed {\bold {Xa = \frac {na} {na + nb}}}[/tex]

a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depression constant () for water is 1.86  

• Step 1: determine molal  

molar mass KCl = 39 + 35.5 = 74.5  

mole KCl = mass: molar mass  

mole KCl = 5 gr: 74.5  

mole KCl = 0.067  

molal = m = 0.067 x (1000: 550 gr water)  

molal = 0.122  

• Step 2: determine the freezing point of the solution  

i = 1 + (n-1) a  

i = 1 + (2-1) 1  

i = 2  

[tex]\displaystyle \Delta T_f=K_f.m.i[/tex]

---> KCl is an electrolyte solution  

[tex]\Delta T_f=1.86.0.122.2[/tex]

[tex]\displaystyle \Delta T_f=0.454[/tex]

freezing point of water = 0 °C  

[tex]0.454=0-T~solution[/tex]

[tex]T_f~solution=- 0.454 ^oC[/tex]

Raoult's law  

https://brainly.com/question/10165688  

The vapor pressure of benzene  

https://brainly.com/question/11102916  

The freezing point of a solution  

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Keywords: Freezing Point Depression, Boiling Point Elevation, Solution Properties  

The wheelbarrow's final kinetic energy at the bottom of the ramp is 4500 J, found by calculating the work done on the wheelbarrow by gravity and friction and applying the work-energy theorem.

To determine the wheelbarrow's kinetic energy at the bottom of the ramp, we need to calculate the work done on the wheelbarrow by both gravity and friction as it slides down the ramp, and then use this to find the change in kinetic energy.

The work done by gravity is simply the component of the wheelbarrow's weight (500 N) that acts parallel to the ramp (i.e., 500 N * sin(20) * 50 m = 8500 J).

The work done by friction is the frictional force (80 N) times the distance (50 m), but since friction opposes the motion it’s negative, so -4000 J.

By the work-energy theorem, the total work done on the wheelbarrow equals its change in kinetic energy. The wheelbarrow starts from rest, so its initial kinetic energy is zero. Hence, its final kinetic energy (at the bottom of the ramp) is the total work done on it: 8500 J - 4000 J = 4500 J.

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