you walk from the park to your friend's house, then back to your house. (a) What is the distance traveled? (b) What is your displacement? If you walk from your house to the library, then back to your house, repeat (a) and (b).

Answer:

[tex]\theta=145[/tex]

Explanation:

The amplitude of he combined wave is:

[tex]B=2Acos(\theta/2)\\[/tex]

A, is the amplitude from the identical harmonic waves

B, is the amplitude of the resultant wave

θ, is the phase, between the waves

The amplitude of the combined wave must be 0.6A:

[tex]0.6A=2Acos(\theta/2)\\ cos(\theta/2)=0.3\\\theta/2=72.5\\\theta=145[/tex]

Answer:

a) 0 m/s

b) - 24 m/s

c)  - 68 m

Explanation:

Given:

Initial distance = - 4 m

Initial velocity, u = 4 m/s

1) acceleration, a = - 2 m/s² for time, t = 2 seconds

thus,

velocity after 2 seconds will be

from Newton's equation of motion

v = u + at

v = 4 + (-2) × 2

v = 0 m/s

2) Velocity after 2 second is the initial velocity for this case

given acceleration = - 6 m/s² for 4 seconds

thus,

final velocity, v = 0 + ( - 6 ) × 4 = - 24 m/s

here the negative sign depicts the velocity in opposite direction to the initial direction of motion

thus, velocity after 6 seconds = - 24 m/s

3) Now,

Total displacement in 6 seconds

= Displacement in 2 seconds + Displacement in 4 seconds

From Newton's equation of motion

[tex]s=ut+\frac{1}{2}at^2[/tex]

where,  

s is the distance

u is the initial speed  

a is the acceleration

t is the time

thus,

= [tex]0\times2+\frac{1}{2}\times(-2)\times2^2[/tex]  + [tex]0\times4+\frac{1}{2}\times-6\times4^2[/tex]

= - 16 - 48

= - 64 m

Hence, the final displacement = - 64 - 4 = - 68 m

Answer:

Explanation:

Given:

Length of each side of the cube, [tex]L=10\ cm[/tex]

The Elecric flux through one of the side of the cube is, [tex]\phi =-1 kNm^2/C.[/tex]

The net flux through a closed surface is defined as the total charge that lie inside the closed surface divided by [tex]\epsilon_0[/tex]

Since Flux is a scalar quantity. It can added to get total flux through the surface.

[tex]\phi_{total}=\dfrac{Q_{in}}{\epsilon_0}\\6\times {-1}\times 10^3=\dfrac{Q_{in}}{\epsilon_0}\\\\Q_{in}=-6}\times 10^3\epsilon_0\\Q_{in}=-6}\times 10^3\times8.85\times 10^{-12}\\Q_{in}=-53.1\times10^{-9}\ C[/tex]

So the the charge at the centre is calculated.

Answer:

The shell hit at a distance of 1.9 x 10² km

The time of flight of the shell was 5.3 x 10² s

Explanation:

The position of the shell is given by the vector "r":

r  = (x0 + v0 * t * cos α ; y0 + v0 * t * sin α + 1/2 g t²)

where:

x0 = initial horizontal position

v0 = magnitude of the initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration of gravity

When the shell hit, the vertical component (ry) of the vector position r is 0. See figure.

Then:

ry = 0 =  y0 + v0 * t * sin α + 1/2 g t²

Since the gun is at the center of our system of reference, y0 and x0 = 0

0 = t (v0 sin α + 1/2 g t)

t= 0 is discarded as solution

v0 sin α + 1/2 g t = 0

t = -2v0 sin α / g

t = (-2 * 2610 m/s * sin 81.9°)/ (-9.8 m/s²) = 5.3 x 10² s. This is the time of flight of the shell until it hit.

Then, the distance at which the shell hit is:

Distance = Module of r = ( x0 + v0 * t * cos α; 0) = x0 + v0 * t * cos α  

Distance = 2.61 km/s * 5.3 x 10² s * cos 81.9 = 1.9 x 10² km

Answer:

[tex]F_1+F_2= (28.26, 9.05) N[/tex]

[tex]\alpha = 17.7\º[/tex]

[tex]F = 29.67 N[/tex]

Explanation:

Hi!

In a (x, y) coordinate representation, the two forces are:

[tex]F_1=(12.7N, 0)\\F_2=(18.1N\cos(30\º), 18.1N \sin(30\º) )\\\cos(\º30)=0.86\\\sin(\º30)= 0.5[/tex]

The sum of the two forces is:

[tex]F_1 + F_2 = ( 12.7 + 0.86*18.1, 18.1*0.5) N[/tex]

[tex]F_1+F_2= (28.26, 9.05) N[/tex]

The angle to x-axis is calculated using arctan:

[tex]\alpha = \arctan(\frac{F_y}{F_x}) = \arctan(\frac{9.05}{28.26} = 17.7\º[/tex]

The magnitude is:

[tex]F = \sqrt {F_x^2 + F_y^2}= \sqrt{798.6 + 81.9} = 29.67 N[/tex]

Answer:

Average velocity

[tex]v=\frac{d}{t}\\ v=\frac{10m}{70s}\\v=.1428 \frac{m}{s}[/tex]

Average speed,

[tex]S=\frac{D}{t}\\ S=\frac{90}{70}\\ S=1.29\frac{m}{s}[/tex]

Explanation:

(a)Average velocity

We have to find the average velocity. We know that velocity is defined as the rate of change of displacement with respect to time.

To find the average velocity we have to find the total displacement.

since displacement along east direction is 50m

and displacement along west=40m

so total displacement,

[tex]d=50m-40m\\d=10m[/tex]

total time,

[tex]t=28 s+42 s\\t=70 s[/tex]

therefore, average velocity

[tex]v=\frac{d}{t}\\ v=\frac{10m}{70s}\\v=.1428 \frac{m}{s}[/tex]

(b)Average Speed:

Average speed is defined as the ratio of total distance to the total time

it means

Average speed= total distance/total time

here total distance,

[tex]D= 50m+40m\\D=90m[/tex]

and total time,

[tex]t= 28s+40s\\t=70s[/tex]

therefore,

Average speed,

[tex]S=\frac{D}{t}\\ S=\frac{90}{70}\\ S=1.29\frac{m}{s}[/tex]

To determine the magnitude of acceleration, use the formula (final velocity - initial velocity) / time. To calculate the distance traveled during the braking period, use the formula (initial velocity + final velocity) / 2 * time.

The magnitude of acceleration can be calculated using the formula:

acceleration = (final velocity - initial velocity) / time

acceleration = (50 mi/h * 1.46667 ft/s - 74 mi/h * 1.46667 ft/s) / 3 s

The distance traveled during the braking period can be calculated using the formula:

distance = (initial velocity + final velocity) / 2 * time

distance = (74 mi/h * 1.46667 ft/s + 50 mi/h * 1.46667 ft/s) / 2 * 3 s

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Answer:

The spring constant is [tex]1.09\times10^{9}\ N/m[/tex]

Explanation:

Given that,

length = 500 mm

Diameter = 2 cm

Young's modulus = 17.4 GPa

We need to calculate the young's modulus

Using formula of young's modulus

[tex]Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}[/tex]....(I)

[tex]Y=\dfrac{Fl}{\Delta l A}[/tex]

From hook's law

[tex]F=kx[/tex]

[tex]k=\dfrac{F}{x}[/tex]

[tex]F=k\times\Delta l[/tex]....(II)

Put the value of F in equation

[tex]Y=\dfrac{k\times\Delta l\times l}{\Delta l A}[/tex]

[tex]Y=\dfrac{kl}{A}[/tex]

We need to calculate the spring constant

[tex]k = \dfrac{YA}{l}[/tex]....(II)

We need to calculate the area of cylinder

Using formula of area of cylinder

[tex]A=2\pi\times r\times l[/tex]

Put the value into the formula

[tex]A=2\pi\times 1\times10^{-2}\times500\times10^{-3}[/tex]

[tex]A=0.0314\ m^2[/tex]

Put the value of A in (II)

[tex]k=\dfrac{1.74\times10^{10}\times0.0314}{500\times10^{-3}}[/tex]

[tex]k=1.09\times10^{9}\ N/m[/tex]

Hence, The spring constant is [tex]1.09\times10^{9}\ N/m[/tex]

Answer:

Average speed, v = 48.01 km/h

Explanation:

Given that,

Distance covered by the walker, d = 4 km

Tim taken to cover that distance, t = 53 min = 0.0833 hours

We need to find the average speed of the walker for this distance. It is given by :

[tex]speed=\dfrac{distance}{time}[/tex]

[tex]speed=\dfrac{4\ km}{0.0833\ h}[/tex]  

Speed, v = 48.01 km/h

So, the average speed of the walker is 48.01 km/h. Hence, this is the required solution.

Answer:

The frequency of motion is 5.8 Hz.

Explanation:

frequency of motion of any object is defined as the number of times the object repeats it's motion in 1 second.

mathematically frequency equals [tex]f=\frac{1}{T}[/tex]

where,

'T' is the time it takes for the object to complete one revolution. Since it is given that the ball completes 5.8 revolutions in 1 seconds thus the time it takes for 1 revolution equals [tex]\frac{1}{5.8}s[/tex]

Hence[tex]T=\frac{1}{5.8}s[/tex]

thus the frequency equals

[tex]\frac{1}{\frac{1}{5.8}}s^{-1}\\\\f=5.8s^{-1}=5.8Hz[/tex]

Answer:

5.8 Hz.

Explanation:

Given:

The frequency of the motion of a rotating object is defined the total number of revolutions the object makes in unit time. It is measured in units of Hertz (Hz) or per second.

It is given that the ball is making 5.8 revolutions in one second, therefore, its frequency of motion would be the same, i.e., 5.8 Hz.

Answer:

Explanation:

mass, m = 1 kg

Position (2, 3 ) m

height, h = 2 m

acceleration due to gravity, g = 9.8 m/s^2

Here, no force is acting in horizontal direction, the force of gravity is acting in vertical direction, so the work done by the gravitational force is to be calculated.

Force  mass x acceleration due to gravity

F = 1 x 9.8 = 9.8 N

Work = force x displacement x CosФ

Where, Ф be the angle between force vector and the displacement vector.

Here the value of Ф is 180° as the force acting vertically downward and the displacement is upward

So, W = 9.8 x 2 x Cos 180°

W = - 19.6 J

Thus, option (A) is correct.

Answer:

A. [tex]H=\frac{v_{0}^{2}Sin^{2}\theta }{2g}[/tex]

B. 5556.1 m

Explanation:

A.

Launch speed, vo

Angle of projection = θ

The value of vertical component of velocity at maximum height is zero. Let the maximum height is H.

Use third equation of motion in vertical direction

[tex]v_{y}^{2}=u_{y}^{2}+2a_{y}H[/tex]

[tex]0^{2}=\left (v_{0}Sin\theta  \right )^{2}-2gH[/tex]

[tex]H=\frac{v_{0}^{2}Sin^{2}\theta }{2g}[/tex]

B.

u = 330 m/s

Let it goes upto height H.

V = 0 at maximum height

Use third equation of motion in vertical direction

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

[tex]0^{2}=330^{2}-2\times 9.8\times H[/tex]

H = 5556.1 m

Answer:

a) 14×10⁹ years

b) 4.4×10¹⁷ seconds

Explanation:

1 billion years = 1000000000 years

1000000000 years = 10⁹ years

14 billion years

= 14×1000000000 years

= 14000000000 years

= 14×10⁹ years

∴ 14 billion years = 14×10⁹ years

b) 1 year = 365.25×24×60×60 seconds

14×10⁹ years = 14×10⁹×365.25×24×60×60

= 441806400×10⁹ seconds

Rounding off,  we get

= 4.4×10⁸×10⁹

= 4.4×10¹⁷ seconds

∴ 14 billion years = 4.4×10¹⁷ seconds

By installing motion sensors, the storage room would save $1.27 per day in energy costs. The simple payback period for installing motion sensors is approximately 77.17 days.

To determine the amount of energy and money that will be saved as a result of installing motion sensors, we need to compare the energy consumption and cost of the current lighting system with that of the motion sensor system. Currently, the storage room uses six fluorescent light fixtures, each containing four lamps rated at 60 W each. The lamps are on for 12 hours a day. So, the total energy consumed by the current system per day is 6 * 4 * 60 W * 12 hours = 17,280 W. With an electricity price of $0.11/kWh, the cost of energy consumed per day is 17,280 W / 1000 * $0.11 = $1.90.

Now, let's consider the motion sensor system. If the room is only used for an average of 3 hours a day, the total energy consumed by the motion sensor system per day would be 6 * 4 * 60 W * 3 hours = 5,760 W. This results in a cost of 5,760 W / 1000 * $0.11 = $0.63 per day. Therefore, by installing motion sensors, the storage room would save $1.90 - $0.63 = $1.27 per day in energy costs.

To determine the simple payback period, we need to calculate the cost of the motion sensor system and the savings achieved per day. The total cost of the motion sensor system is $32 for the purchase price + $66 for installation, which equals $98. With daily savings of $1.27, the payback period can be calculated as $98 / $1.27 = 77.17 days. Therefore, the simple payback period for installing motion sensors in the storage room is approximately 77.17 days.

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Answer:

percent abundance of the heaviest isotope is 78 %

Explanation:

given data

atomic weight = 48.68 amu

mass 1 = 47 amu

mass 2 = 48 amu

mass 3 = 49 amu

natural abundance = 10 %

to find out

percent abundance of the heaviest isotope

solution

we consider here percent abundance of the heaviest isotope is x

so here lightest isotope = 47 amu of 10 %   ..........1

and heaviest isotope = 49 amu of x       ................2

and middle isotope = 48 amu of 100 - 10 - x      ........3

so

average mass = add equation 1 + 2 + 3

average mass = 10% ( 47) + x% ( 49) + (90 - x) % (48)

48.68 = 4.7 + 0.49 x + 43.2 - 0.48 x

x = 0.78

so percent abundance of the heaviest isotope is 78 %

Answer:

The acceleration of the bullet is 212.12 m/s²

Explanation:

Given that,

Force = 1.4 N

Mass = 6.6 g

We need to calculate the acceleration

Using newton's second law

[tex]F=ma[/tex]

[tex]a=\dfrac{F}{m}[/tex]

Where, F = force

m = mass

Put the value into the formula

[tex]a=\dfrac{1.4 }{6.6\times10^{-3}}[/tex]

[tex]a=212.12\ m/s^2[/tex]

Hence, The acceleration of the bullet is 212.12 m/s²

Answer:

Distance, d = 112.5 meters

Explanation:

Initially, the bicyclist is at rest, u = 0

Final speed of the bicyclist, v = 30 m/s

Acceleration of the bicycle, [tex]a=4\ m/s^2[/tex]

Let s is the distance travelled by the bicyclist. The third equation of motion is given as :

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

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

[tex]s=\dfrac{(30)^2}{2\times 4}[/tex]

s = 112.5 meters

So, the distance travelled by the bicyclist is 112.5 meters. Hence, this is the required solution.

Answer:

a) 38.27      b) 322.5°

c) 126.99    d) 1.17°

e) 62.27     e) 139.6°

Explanation:

First of all we have to convert the coordinates into rectangular coordinates, so:

a=( 43.3 , 25)

b=( -48.3 , -12.94)

c=( 35.36 , -35.36)

Now we can do the math easier (x coordinate with x coordinate, and y coordinate with y coordinate):

1.)  a+b+c=( 30.36 , -23.3) = 38.27 < 322.5°

2.)  a-b+c=( 126.96 , 2.6) = 126.99 < 1.17°

3.)  (a+b) - (c+d)=0   Solving for d:

     d=(a+b) - c = ( -40.36 , 47.42) = 62.27 < 139.6°

Answer:

1.004 × 10⁴ kPa

Explanation:

Given data

[tex]\frac{1.024g}{cm^{3}}.\frac{1kg}{10^{3}g} .\frac{10^{6}cm^{3}}{1m^{3} } =1.024 \times 10^{3} kg/m^{3}[/tex]

In order to prevent water from entering, the air pressure must be equal to the pressure exerted by the seawater at the bottom. We can find that pressure (P) using the following expression.

P = ρ × g × h

P = (1.024 × 10³ kg/m³) × (9.806 m/s²) × 1000 m

P = 1.004 × 10⁷ Pa

P = 1.004 × 10⁷ Pa × (1 kPa/ 10³ Pa)

P = 1.004 × 10⁴ kPa

Answer:

The final velocity of the car is 30 m/s.

Explanation:

Given that,

Initial speed of the car, u = 0

Acceleration of the car, [tex]a=5\ m/s^2[/tex]

Time taken, t = 6 s

Let v is the final velocity of the car. It can be calculated using first equation of kinematics as :

[tex]v=u+at[/tex]

[tex]v=at[/tex]

[tex]v=5\ m/s^2\times 6\ s[/tex]

v = 30 m/s

So, the final velocity of the car is 30 m/s. Hence, this is the required solution.

Answer:

Explanation:

Total volume of the shell on which charge resides

= 4/3 π ( R₁³ - R₂³ )

= 4/3 X 3.14 ( 33³ - 20³) X 10⁻⁶ m³

= 117 x 10⁻³ m³

Charge inside the shell

-117 x 10⁻³ x 1.3 x 10⁻⁶

= -152.1 x 10⁻⁹ C

Charge at the center

= - 60 x 10⁻⁹ C

Total charge inside the shell

= - (152 .1 + 60 ) x 10⁻⁹ C

212.1 X 10⁻⁹C

Force between - ve charge and proton

F = k qQ / R²

k = 9 x 10⁹ .

q = 1.6 x 10⁻¹⁹ ( charge on proton )

Q = 212.1 X 10⁻⁹ ( charge on shell )

R = 33 X 10⁻² m ( outer radius )

F = [tex]\frac{9\times10^9\times1.6\times10^{-19}\times212.1\times 10^{-9}}{(33\times10^{-2})^2}[/tex]

F = 2.8 X 10⁻¹⁵ N

This force provides centripetal force for rotating proton

mv² / R = 2.8 X 10⁻¹⁵

V² = R X 2.8 X 10⁻¹⁵ / m

= 33 x 10⁻² x 2.8 x 10⁻¹⁵ /(  1.67 x 10⁻²⁷ )

[ mass of proton = 1.67 x 10⁻²⁷ kg)

= 55.33 x 10¹⁰

V = 7.44 X 10⁵ m/s

The total charge on a cloud when breakdown of the surrounding air begins is calculated using Gauss's Law. The breakdown field strength, radius of the cloud and permittivity of free space are needed. The number of excess electrons in the cloud is then found by dividing the total charge by the charge of an electron.

The charge q on a spherically symmetric object, such as our cloud, when an electric field E is induced on its surface can be calculated using Gauss's Law, which states that the total charge enclosed by a Gaussian surface is equal to the electric field times the surface area of the Gaussian surface divided by the permittivity of free space (ε0). In our case, E = Eb (the breakdown field strength), the radius of the sphere, r = 0.5km = 500m (half the diameter), and ε0 = 8.85 x 10^-12 C2/N·m2. Therefore, q = (4πr2Eb)ε0.

For question 2: To calculate the number of excess electrons in the cloud, we should recall that the charge of an electron is -1.602 x 10^-19 C. Therefore, the number of excess electrons, N, can be calculated using N = q / Charge of an electron.

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Answer:

magnitude is 2.52 m  with 66.15° north east

Explanation:

given data

displace 1 = 3.45 m north = 3.45 j

displace 2 = 1.53 m southeast = 1.53 cos(315) i + sin(315) j

displace 3 = 0.877 m southwest  = 0.877 cos(225) i- sin(225) j

to find out

displacement and direction

solution

we consider here direction i as east and direction j as north

so here

displacement = displace 1 + displace 2 + displace 3

displacement =  3.45 j + 1.53 cos(315) i + sin(315) j + 0.877 cos(225) i- sin(225) j

displacement =  3.45 j + 1.081 i -1.0818 j - 0.062 i -0.062 j

displacement = 1.019 i + 2.306 j

so magnitude

magnitude = [tex]\sqrt{1.019^{2} + 2.306^{2}}[/tex]

magnitude = 2.52

and

angle will be = arctan(2.306/1.019)

angle = 1.15470651 rad

angle is 66.15 degree

so magnitude is 2.52 m  with 66.15° north east

The relativistic factor y (gamma) quantifies the relativistic effects based on the speed of an object. In your given case with v = 0.932c, gamma would be approximately 2.61, indicating significant relativistic effects. The provided options do not include this value, suggesting some error.

The dimensionless parameter used in relativity is denoted as y (gamma), and it is referred to as the relativistic factor. It is associated with the relative speed of an object and is defined by the equation y = 1/√(1 − (v²/c²)), where v is the object's velocity and c is the speed of light.

For your case where v = 0.932c, we'll substitute this into the formula and solve for y. This results in y being approximately equal to 2.61. This isn't available in the provided options, which would indicate an error in the question or provided choices.

Without a doubt, as y approaches and surpasses 1, the relativistic effects become more noticeable in an object's behavior. Significant relativistic effects mean that the classical interpretation, where we neglect these effects, becomes increasingly inaccurate.

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Final answer:

The race car driver is approximately 7.644 km from where she started.

Explanation:

To find the distance from where the race car driver started, we can use the Pythagorean theorem. We can consider the north-south movement as one leg of a right triangle, and the east-west movement as the other leg. The distance from the starting point is equal to the square root of the sum of the squares of the two legs. In this case, the north-south leg is 6.71 km and the east-west leg is 3.67 km. Using the Pythagorean theorem, the distance from the starting point is:

d = √((6.71 km)² + (3.67 km)²)

d = √(44.9041 km² + 13.4689 km²)

d = √(58.3730 km²)

d = 7.644 km

Therefore, the race car driver is approximately 7.644 km from where she started.

Final answer:

The time it takes for the car to fall is 0.606 s and the horizontal velocity of the car is 0.497 m/s.

Explanation:

To find the time it takes for the toy car to fall, we can use the kinematic equation:

h = (1/2)gt^2

Where h is the height, g is the acceleration due to gravity, and t is the time. Rearranging the equation, we can solve for t:

t = sqrt(2h / g)

Plugging in the values, we have:

t = sqrt(2 * 1.807 / 9.8) = 0.606 s

To find the horizontal velocity of the car, we can use the equation:

v = d / t

Where v is the velocity, d is the horizontal distance, and t is the time. Plugging in the values, we have:

v = 0.3012 / 0.606 = 0.497 m/s

Answer:

a) Focal length of the lens is 8 cm which is a convex lens

b) 6 cm

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

Explanation:

u = Object distance =  4 cm

v = Image distance = -8 cm

f = Focal length

Lens Equation

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{4}+\frac{1}{-8}\\\Rightarrow \frac{1}{f}=\frac{1}{8}\\\Rightarrow f=\frac{8}{1}=-8\ cm[/tex]

a) Focal length of the lens is 8 cm which is a convex lens

Magnification

[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{-8}{4}\\\Rightarrow m=2[/tex]

b) Height of image is 2×3 = 6 cm

Since magnification is positive the image upright

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

(a) The focal length of the lens is -2.67 cm.

(b) The magnification of the image is 2 and the height is 6 cm.

(c) The imaged formed by the lens is upright, virtual and magnified.

The focal length of the lens is determined by using lens formulas as given below;

[tex]\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\\\\frac{1}{f} = \frac{-1}{8} - \frac{1}{4} \\\\\frac{1}{f} = \frac{-1 -2}{8} = \frac{-3}{8} \\\\f = -2.67 \ cm[/tex]

The magnification of the image is calculated as follows;

[tex]m = \frac{v}{u} \\\\m = \frac{8}{4} \\\\m = 2\\\\[/tex]

The height of the image is calculated as follows;

[tex]H = mu\\\\H = 2(3cm) = 6\ cm[/tex]

The imaged formed by the lens is;

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The acceleration experienced by the space travelers during their launch would be considered unrealistically large. It would exceed both the limits of human endurance and the acceleration experienced during free fall.

To calculate the acceleration experienced by the space travelers during their launch, we can use the formula:

Acceleration = (Final Velocity - Initial Velocity) / Time

In this case, the final velocity is given as 10.97 km/s, and the initial velocity is 0 m/s (since the spaceship starts from rest). The time is not provided, so we cannot calculate the exact acceleration. However, we can compare it to the acceleration that a human can withstand, which is 15g for a short time. One g is equivalent to the acceleration due to gravity, which is approximately 9.8 m/s².

So, 15g is equal to 15 * 9.8 m/s² = 147 m/s². Therefore, any acceleration larger than 147 m/s² would be unrealistically large for the space travelers during their launch.

Comparing this with the free-fall acceleration, which is approximately 9.8 m/s², we can see that the acceleration proposed by Jules Verne would be much larger than both the limits of human endurance and the acceleration experienced during free fall.

Answer:

Explanation:

a )  V = 3 cos(0.5t)

differentiating with respect to t

dv /dt = -3 x .5 sin0.5t

= -1.5 sin0.5t.

acceleration = - 1.5 sin 0.5t

when t = 3 s

acceleration = - 1.5 sin 1.5

= - 1.496 ms⁻²

v = 3 cos.5t

b )  dx/dt = 3 cos 0.5 t

dx = 3 cos 0.5 t dt

integrating on both sides

x = 3 sin .5t / .5

x = 6 sin0.5t

At t = 2 s

x = 6 sin 1

x = 5.05 m

The car strikes the tree at approximately 10.7 m/s after slowing down with a uniform acceleration of -5.75 m/s^2 for 60.0 m.

To calculate the speed with which the car strikes the tree, we'll use the kinematic equations for uniformly accelerated motion. Given the uniform acceleration of -5.75 m/s2 and the time of 4.40 s, we can use the following equation to find the initial velocity (vi) before the car started braking:

v = vi + at
Where:

By rearranging the equation to solve for the initial velocity (vi), we get:

vi = v - at

Substituting the known values:
vi = 0 - (-5.75 m/s2 * 4.40 s)
vi = 25.3 m/s

This is the speed with which the car was travelling before it hit the brakes. However, to find the speed at which the car strikes the tree, we must consider the distance of the skid marks. Using the kinematic equation for distance (d), where d equals the initial velocity times time plus half the acceleration times time squared:

d = vit +  rac{1}{2}at2

Since the distance to the tree is 60.0 m and we're looking for the speed at the end of this distance, we rearrange the equation to solve for final velocity (v). But first, we need to calculate the time it takes to stop over this distance. We can use the formula:

d =  rac{vi2 - v2}{2a}

By solving for v we find:

v = [tex]\sqrt{x}[/tex]{vi2 - 2ad}

v = [tex]\sqrt{x}[/tex]{(25.3 m/s)2 - 2(-5.75 m/s2)(60.0 m)}

v = [tex]\sqrt{x}[/tex]{(640.09) - (-690)}

v = 10.7 m/s

Therefore, the car strikes the tree at approximately 10.7 m/s.

The car strikes the tree at a speed of 0.98 m/s.

To determine the speed at which the car strikes the tree, we can utilize the kinematic equations of motion. We are given the following data:

First, we calculate the initial velocity using the equation:

Substituting the given values:

Solving this:

Now, to find the final velocity when the car strikes the tree, we use the kinematic equation:

Substituting the values:

So, the car strikes the tree at a speed of 0.98 m/s.