show that the centre of a rod of mass M and length L lies midway between the ends assuming that the rod has uniform cross section per unit length​

Answer:

a) The assumption that error theorists claim is at the heart of morality is that sincere moral judgments are always in error, therefore there is no moral knowledge possible.

Explanation:

Answer:

SOUND ENERGY is the energy which gets released or formed.

Explanation:

Remember , Energy is ALWAYS conserved. It can neither be destroyed nor be created(made).

∴

Energy gets converted significantly to sound energy - from the following options.

Final answer:

The type of energy made by a door slamming is sound energy, which is a result of the door's mechanical energy being partially transformed into pressure waves in the air.

Explanation:

When a door slams, it produces sound energy as a result of the forceful closure that creates pressure waves in the air we perceive as sound. While the door itself experiences mechanical energy, which is necessary for the motion of slamming, the type of energy being asked about in this case is the one that is made as a consequence of that action, which is sound energy. Mechanical energy is the sum of kinetic and potential energy within a system and is responsible for physical movement, as when a person pushes a door causing it to slam. Upon impact, part of this mechanical energy is transformed into sound energy.

The velocity of the wave is 40 m/s

Explanation:

The relationship between the velocity of a wave and its frequency and wavelength is given by

[tex]v=f \lambda[/tex]

where

v is the velocity

f is the frequency

[tex]\lambda[/tex] is the wavelength

For the ocean wave in this problem, we have:

[tex]\lambda = 10 m[/tex] (wavelength)

f = 4.0 Hz (frequency)

Therefore, its velocity is

[tex]v=(4.0)(10)=40 m/s[/tex]

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The velocity of an ocean wave with a wavelength of 10 m and a frequency of 4.0 Hz is 40 m/s.

The subject of this question is Physics, and more specifically, it deals with the topic of wave motion and wave velocity. The velocity of a wave can be calculated using the equation v = f × lambda, where v is the velocity of the wave, f is its frequency, and lambda is its wavelength.

Given that the wavelength lambda is 10 m and the frequency f is 4.0 Hz, we can substitute these values into the equation to find the wave velocity:

v = f × lambda = 4.0 Hz × 10 m = 40 m/s.

Therefore, the velocity of the ocean wave is 40 meters per second (m/s).

The acceleration of an object in circular motion is calculated by the formula ac = v²/r. Given that in case a, both the speed and radius are twice that of case b, the acceleration in case a is twice that of case b. This shows the squared relationship between speed and acceleration in circular motion.

In physics, the acceleration of an object moving in a circular path, also known as centripetal acceleration, is calculated by the formula ac = v²/r where v is the velocity or speed of the object and r is the radius of the circle.

In case a, the speed is twice that of case b and the circle's radius is also two times of case b. If we plug these values into our formula, we can see that the acceleration in case a (let's call it acA) could be represented as acA = (2v)² / 2r = 4v²/2r = 2v²/r. In case b, the acceleration (acB) would be represented as acB = v² / r.

Hence, comparing both, you can see the acceleration in case a is twice that in case b. Therefore, even if the speed and the radius are increased by the same factor, the acceleration increases by a factor of that increase due to the squared relationship between velocity and acceleration in circular motion.

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The acceleration in case A is twice that of case B.Given the conditions that the speed and radius of circle in case A are twice those in case B.

Let's analyze the relationship between acceleration, speed, and radius in circular motion. We know the acceleration of an object in uniform circular motion can be given by the equation:

|a| = |v|²/r

From the problem, we have two cases:

[tex]Case A: Speed (|v_A|) is \twice \ that \ of \ case \ B (|v_B|).\\\\Case A: Radius (|r_A|) \ is \ also \ two\ times \ that\ of \ case\ B (|r_B|).[/tex]

Let's denote the acceleration in case A as [tex]|a_A|[/tex] and in case B as [tex]|a_B|[/tex]

For case A:

[tex]|a_A| = |v_A|^2 / |r_A|[/tex]

For case B:

[tex]|a_B| = |v_B|^2 / |r_B|[/tex]

Given that [tex]|v_A| = 2|v_B|[/tex] and [tex]|r_A| = 2|r_B|[/tex], substituting these into the equations gives:

[tex]|a_A| = (2|v_B|)^2 / (2|r_B|)[/tex]

Simplifying this, we get:

[tex]|a_A| = 4|v_B|^2 / 2|r_B| = 2(|v_B|^2 / |r_B|) = 2|a_B|[/tex]

Therefore, the acceleration in case A is twice that of case B.

Answer: Infrared radiation

Explanation:

Infrared radiation is not visible to the human eye, since its wavelengths are outside the visible spectrum (between 700 nm and 1 mm), however this type of electromagnetic radiation is detected by the human body as heat.

These waves can be divided into:

- Near infrared or long wave infrared: it is the least sensitive to color and is easily absorbed by water.  

- Medium or medium wave infrared: it is also insensitive to color and easily absorbed by water and many types of plastics and paints.  

- Far infrared or short wave infrared: it is more penetrating than the long wave and is good for heating metals, these waves also can pass through clear materials.  This light has many uses, including heating lamps in physiotherapy and medical treatments, heat sensing devices, among others.

a) The horizontal component of the velocity is 7.4 m/s

b) The vertical component of the velocity is 3.8 m/s

c) The balloon reaches the highest point after 0.39 s

d) The maximum height is 0.74 m

e) The total time of flight is 0.78 s

f) The range of the balloon is 5.77 m

Explanation:

a)

The motion of the balloon is the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

The horizontal component of the velocity (which is constant) is given by

[tex]v_x = u cos \theta[/tex]

where

u = 8.3 m/s is the initial velocity of the balloon

[tex]\theta=27^{\circ}[/tex] is the angle of projection

Substituting,

[tex]v_x = (8.3)(cos 27^{\circ})=7.4 m/s[/tex]

b)

The vertical component of the initial velocity of a projectile is given by

[tex]u_y = u sin \theta[/tex]

where

u is the initial velocity

[tex]\theta[/tex] is the angle of projection

Here we have

u = 8.3 m/s

[tex]\theta=27^{\circ}[/tex]

Substituting,

[tex]u_y = (8.3)(sin 27^{\circ})=3.8 m/s[/tex]

c)

The vertical component of the velocity of the balloon follows the suvat equation

[tex]v_y = u_y - gt[/tex]

where

[tex]v_y[/tex] is the vertical velocity at time t

[tex]u_y = 3.8 m/s[/tex] is the initial vertical velocity

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

The balloon reaches the maximum height when the vertical velocity becomes zero:

[tex]v_y = 0[/tex]

So we get:

[tex]0=u_y -gt\\t=\frac{u_y}{g}=\frac{3.8}{9.8}=0.39 s[/tex]

d)

The maximum height of the balloon can be calculated using the suvat equation:

[tex]s=u_y t - \frac{1}{2}gt^2[/tex]

where

[tex]u_y = 3.8 m/s[/tex] is the initial vertical velocity

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

t = 0.39 s is the time at which the highest point is reached

Substituting,

[tex]s=(3.8)(0.39)-\frac{1}{2}(9.8)(0.39)^2=0.74 m[/tex]

e)

The total time of flight of a projectile is twice the time needed to reach the maximum height, and it is given by

[tex]t=\frac{2u_y}{g}[/tex]

where

[tex]u_y[/tex] is the initial vertical velocity

[tex]g[/tex] is the acceleration of gravity

Here we have

[tex]u_y = 3.8 m/s[/tex]

[tex]g=9.8 m/s^2[/tex]

Substituting,

[tex]t=\frac{2(3.8)}{9.8}=0.78 s[/tex]

f)

The range of a projectile is the horizontal distance covered by the projectile, so it can be found by multiplying its horizontal velocity (which is constant) by the time of flight:

[tex]d=v_x t[/tex]

where

[tex]v_x[/tex] is the horizontal velocity

t is the time of flight

Here we have

[tex]v_x = 7.4 m/s[/tex]

[tex]t = 0.78 s[/tex]

Substituting,

[tex]d=(7.4)(0.78)=5.77 m[/tex]

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Answer:the answe would be joint comitee

Explanation:

Answer:

g'(10) = [tex]\frac{-1}{16}[/tex]

Explanation:

Since g is the inverse of f ,

We can write

g(f(x)) = x     (Identity)

Differentiating both sides of the equation we get,

g'(f(x)).f'(x) = 1

g'(10) = [tex]\frac{1}{f'(x)}[/tex]    --equation[1]    Where f(x) = 10

Now, we have to find x when f(x) = 10

Thus 10 = [tex]\frac{4}{x}[/tex] + 2

[tex]\frac{4}{x}[/tex] = 8

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

Since f(x) = [tex]\frac{4}{x}[/tex] + 2

f'(x) = -[tex]\frac{4}{x^{2} }[/tex]

f'([tex]\frac{1}{2}[/tex])  =  -4 × 4 = -16            

Putting it in equation 1, we get:

We get g'(10) = -[tex]\frac{1}{16}[/tex]

The angle of incidence and the angle of reflection are measured from the B) normal to the surface.

Explanation:

Reflection is a phenomenon that occurs when a wave hits a boundary between two mediums and bounces back into the original medium.

The direction of the wave after reflection changes according to the law of reflection, which states that:

In this framework, the two angles are defined as follows:

Therefore, the complete sentence is

The angle of incidence and the angle of reflection are measured from the B) normal to the surface.

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Answer: The answer is B.

Explanation: i got it correct on Usatestprep.

It means Constructive interference has occurred.

Explanation:

Waves travelling from opposite sides can collide to results into constructive interference or destructive interference.

In constructive interference, the crust of the waves merge.This results into increase in energy.In destructive interference the two waves cancel each other to cause loss of energy.

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Answer:

Constructive Interference

Explanation:

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In a collision between two 300 kg spaceships, one at rest and the other moving at 6 m/s, the final velocity after they stick together is 2 m/s.

When two objects with masses of 300 kg collide and stick together, their momentum is conserved. Utilizing the principle of conservation of momentum, we can calculate their final velocity after the collision.

In this case, since both spaceships have equal masses and one is at rest while the other moves with a known speed, the final velocity after the collision would be 2 m/s.

1) Frequency: 1.95 Hz

2) Equilibrium position: 0.138 m

3) Amplitude: 0.357 m

4) Maximum speed: 4.36 m/s

5) Maximum acceleration: [tex]53.1 m/s^2[/tex]

6) Spring constant: 15.0 N/m

7) Total mechanical energy: 0.96 J

Explanation:

1)

The frequency of a simple harmonic motion is equal to the reciprocal of the period:

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

where

f is the frequency

T is the period

For the particle in this problem, we have

T = 0.513 s (period)

So the frequency of motion is

[tex]f=\frac{1}{0.513}=1.95 Hz[/tex]

2)

In a simple harmonic motion, the object oscillates between two maximum positions [tex]+x_m[/tex] and [tex]-x_m[/tex] which are equidistance from the equilibrium position. So, the equilibrium position is the midpoint between these two positions.

For the particle in this problem, the two extreme positions are:

[tex]x_1 = -0.219 m[/tex]

[tex]x_2 = 0.495 m[/tex]

So the mid-point (the equilibrium position) is

[tex]x_{ep} = \frac{x_1 + x_2}{2}=\frac{-0.219+0.495}{2}=0.138 m[/tex]

3)

The amplitude of a simple harmonic motion is the maximum displacement of the object, measured from the equilibrium position.

This means that we can calculate the amplitude simply as the difference between one of the extreme positions and the equilibrium position.

Taking

[tex]x_2 = 0.495 m[/tex]

and

[tex]x_{ep} = 0.138 m[/tex]

We find the amplitude:

[tex]A=x_2 - x_{ep} = 0.495-0.138 =0.357 m[/tex]

4)

In a simple harmonic motion, the maximum speed is given by

[tex]v_{max}=\omega A[/tex]

where

[tex]\omega[/tex] is the angular frequency

A is the amplitude

The angular speed can be calculated from the frequency as follows:

[tex]\omega=2\pi f=2 \pi (1.95 Hz)=12.2 rad/s[/tex]

The amplitude is

A = 0.357 m

So, the maximum speed is

[tex]v_{max} = (12.2)(0.357)=4.36 m/s[/tex]

5)

The maximum acceleration in a simple harmonic motion is given by

[tex]a_{max}= \omega^2 A[/tex]

Where we already know that:

[tex]\omega=12.2 rad/s[/tex] is the angular frequency

A = 0.357 m is the amplitude of motion

Substituting, we find the maximum acceleration:

[tex]a_{max}=(12.2)^2(0.357)=53.1 m/s^2[/tex]

6)

The angular speed in a simple harmonic motion can be written as

[tex]\omega=\sqrt{\frac{k}{m}}[/tex]

where

k is the spring constant

m is the mass

Here we know that:

[tex]\omega=12.2 rad/s[/tex] is the angular speed

m = 0.101 kg is the mass of the particle

So we can solve the formula for k, the spring constant:

[tex]k=\omega^2 m =(12.2)^2(0.101)=15.0 N/m[/tex]

7)

Since the energy is conserved, the total mechanical mechanical energy is just equal to the maximum potential energy of the system, which occurs when the particle is at maximum displacement (x=A) and the speed is zero (so the kinetic energy is zero), therefore it is given by:

[tex]E=\frac{1}{2}kA^2[/tex]

where

k = 15.0 N/m is the spring constant

A = 0.357 m is the amplitude

And substituting,

[tex]E=\frac{1}{2}(15.0)(0.357)^2=0.96 J[/tex]

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The angle of the force is [tex]51.1^{\circ}[/tex]

Explanation:

To solve this problem, we can apply Newton's second law along the horizontal direction of motion of the suitcase:

[tex]\sum F_x = ma_x[/tex]

where

[tex]\sum F_x[/tex] is the net force along the x-axis

m is the mass of the suitcase

[tex]a_x[/tex] is the acceleration along the x-axis

The suitcase is moving at constant speed, so the acceleration is zero:

[tex]a_x=0[/tex]

Therefore the net force must also be zero:

[tex]\sum F_x = 0[/tex] (1)

We have two forces acting along the horizontal direction:

- The component of the push (forward) in the horizontal direction, [tex]F cos \theta[/tex], with

F = 43 N

[tex]\theta[/tex] = angle of the force with the horizontal

- The  force of friction, [tex]F_f = 27 N[/tex], backward

So the net force can be written as

[tex]\sum F_x = F cos \theta - F_f[/tex] (2)

Combining (1) and (2),

[tex]F cos \theta - F_f = 0[/tex]

And so we can find the angle:

[tex]\theta = cos^{-1}(\frac{F_f}{F})=cos^{-1}(\frac{27}{43})=51.1^{\circ}[/tex]

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Out of all given options, poor school performance  is a characteristic of sexually active teens.

Answer: Option B

Explanation:

Socio-demographic characteristics include age, gender and the highest level of education of the respondent. Most respondents had mothers with completed high school levels or diploma. Just over a quarter of respondents did not know their mother's highest education.  A low level of education and some family situations made the teen to be sexually active.

Teen sexual activity is "a fact in most African countries, as in most Western societies." There is a relationship between the sexual activity of young girls and their school resignation. Therefore, additional consequences of early sexual intercourse are women (and sometimes men) who drop out of school when they become pregnant.

Sexually active teens are more likely to come from single and step-parent families and have poor school performance.

The characteristic of sexually active teens is A and B are correct, which means they are more likely to come from single and step-parent families and have poor school performance.

While there may be studies suggesting correlations between certain factors, making broad statements about the behavior of sexually active teens based solely on family structure and school performance can be misleading.

Teenage sexual behavior is influenced by a variety of factors, including individual, family, cultural, and socioeconomic aspects. Family structure alone does not determine a teen's likelihood of engaging in sexual activity. Additionally, attributing poor school performance solely to family structure oversimplifies the complex interplay of factors influencing academic achievement.

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Answer:

a. 12 m/s² down

Explanation:

Acceleration has units of length per time squared.  Acceleration is a vector, so it also has a direction.

Answer:

iron

Explanation:

Increasing the number of coils of wire wrapped around the nail increases the strength of the electromagnet, as measured by the number of paper clips the magnet can pick up

iron

Explanation:

Increasing the number of coils of wire wrapped around the nail increases the strength of the electromagnet, as measured by the number of paper clips the magnet can pick up

Answer:

[tex]v_0[/tex]=4.761 m/s

t=0.786 sec

Explanation:

In a projectile motion (or 2D motion), the object is launched with an initial angle [tex]\theta[/tex] and an initial velocity [tex]v_0[/tex]

The components of the velocity are

[tex]v_{ox}=v_ocos\theta[/tex]

[tex]v_{oy}=v_osin\theta[/tex]

Similarly the displacement has the components

[tex]x=v_{ox}.t=v_ocos\theta.t[/tex]

[tex]y=v_osin\theta.t-\frac{gt^2}{2}[/tex]

The last formula is valid only if the object is launched at ground level, as our frog does.  

There are two times where the value of y is zero, when t=0 (at launching time) and when it lands back from the air. We need to find that time t by making y=0

[tex]0=v_osin\theta.t-\frac{gt^2}{2}[/tex]

Dividing by t (assuming t different from zero)

[tex]0=v_osin\theta-\frac{gt}{2}[/tex]

Then we find the total flight as

[tex]t=\frac{2v_osin\theta}{g}[/tex]

Replacing this time in the formula of x

[tex]x=v_ocos\theta\frac{2v_osin\theta}{g}[/tex]

We can solve for [tex]v_o[/tex]

[tex]\displaystyle v_0=\sqrt{\frac{xg}{sin2\theta}}[/tex]

Knowing that x=2.20 m and [tex]\theta=54[/tex]°

[tex]\displaystyle v_0=\sqrt{\frac{xg}{sin2\theta}}=4.761m/s[/tex]

We now compute t

[tex]t=\frac{2v_osin\theta}{g}=0.786\ sec[/tex]

Final answer:

The launch speed of the frog is 4.81 m/s, and the time the frog spends in the air is 0.737 seconds.

Explanation:

A) Using the kinematic equation for projectile motion, we can find the launch speed. The launch speed, in this case, would be 4.81 m/s.

B) To determine the time the frog spends in the air, we can use the time of flight formula for projectile motion. The time in the air for the frog is 0.737 seconds.

The maximum speed of the car is 24.3 m/s

Explanation:

For a car moving along an unbanked turn, the frictional force provides the centripetal force required to keep the car in circular motion. Therefore, we can write:

[tex]\mu mg = m\frac{v^2}{r}[/tex]

where the term on the left is the frictional force while the term on the right is the centripetal force, and where

[tex]\mu=0.40[/tex] is the coefficient of friction between the tires and the road

m is the mass of the car

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

v is the speed of the car

r = 150 m is the radius of the curve

Solving for v, we find the (maximum) speed at which the car can move along the turn:

[tex]v=\sqrt{\mu gr}=\sqrt{(0.40)(9.8)(150)}=24.3 m/s[/tex]

For speed larger than this value, the frictional force is no longer enough to keep the car along the turn.

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Final answer:

On a rainy day, the maximum speed at which a curve with a radius of 150 m can be safely negotiated is 12.43 m/s. This calculation is based on the centripetal force provided by the friction between the tires and the road surface.

Explanation:

To determine the maximum speed at which the curve can be safely negotiated, we need to consider the centripetal force acting on the car. The centripetal force is provided by the friction between the tires and the road surface. The formula for centripetal force is Fc = mv²/r, where m is the mass of the car, v is the velocity, and r is the radius of the curve.

On a rainy day, the coefficient of friction between the tires and the road is given as 0.40. The maximum frictional force is given by the equation Ff = μN, where μ is the coefficient of friction and N is the normal force.

Since the normal force is equal to the gravitational force acting on the car, we can calculate it using N = mg, where m is the mass of the car and g is the acceleration due to gravity.

To find the maximum speed, we need to equate the centripetal force and the frictional force: mv²/r = μN. We can rearrange this equation to solve for the maximum velocity v: v = sqrt(μrg).

Substituting the given values, v = sqrt(0.40 * 9.8 * 150) = 12.43 m/s.

The distance covered is 30.2 m

Explanation:

The motion of Cynthia is a uniformly accelerated motion (constant acceleration), so we can find the distance covered by using the following suvat equation:

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

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

In this problem,

u = 0 (since Cynthia started from rest)

t = 3.2 s

[tex]a=5.9 m/s^2[/tex] is her acceleration

Substituting,

[tex]s=0+\frac{1}{2}(5.9)(3.2)^2=30.2 m[/tex]

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Using the equation of motion S = ut + ½at2, we calculated that Susan traveled 30.216 meters during the initial stage of her downhill luge, where she accelerated at 5.9 m/s2 for 3.2 seconds.

To calculate how far Susan traveled during the initial stage of her downhill luge, where she accelerated from rest at 5.9 m/s2 for 3.2 seconds, we use the equation of motion for constant acceleration:

S = ut + ½at2

Here, S represents the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.

Since Susan starts from rest, her initial velocity u = 0. We plug in the acceleration a = 5.9 m/s2 and time t = 3.2 s into the equation:

S = (0)(3.2) + ½(5.9)(3.2)2

S = 0 + ½(5.9)(10.24)

S = 30.216 m

Therefore, Susan traveled 30.216 meters during the initial stage of her downhill luge.

a) The acceleration of gravity is [tex]4.96 m/s^2[/tex]

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

a)

The acceleration of gravity for an object near a planet is given by

[tex]g=\frac{GM}{(R+h)^2}[/tex]

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

[tex]M=5.98\cdot 10^{24} kg[/tex] (mass of the Earth)

[tex]R=6.37\cdot 10^6 m[/tex] (Earth's radius)

[tex]h=2.60\cdot 10^6 m[/tex] (altitude of the satellite)

Substituting,

[tex]g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2[/tex]

b)

The critical velocity for a satellite orbiting around a planet is given by

[tex]v=\sqrt{\frac{GM}{R+h}}[/tex]

where we have again:

[tex]M=5.98\cdot 10^{24} kg[/tex] (mass of the Earth)

[tex]R=6.37\cdot 10^6 m[/tex] (Earth's radius)

[tex]h=2.60\cdot 10^6 m[/tex] (altitude of the satellite)

Substituting,

[tex]v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s[/tex]

Converting into km/h,

[tex]v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h[/tex]

c)

The period of the orbit is given by the circumference of the orbit divided by the velocity:

[tex]T=\frac{2\pi (R+h)}{v}[/tex]

where

[tex]R=6.37\cdot 10^6 m[/tex]

[tex]h=2.60\cdot 10^6 m[/tex]

v = 6668 m/s

Substituting,

[tex]T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s[/tex]

d)

One day consists of:

[tex]t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s[/tex]

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

[tex]n=\frac{t}{T}=\frac{86400}{8452}=10.2[/tex]

e)

The escape velocity for an object in the gravitational field of a planet is given by

[tex]v=\sqrt{\frac{2GM}{R+h}}[/tex]

where here we have:

[tex]M=5.98\cdot 10^{24} kg[/tex]

[tex]R=6.37\cdot 10^6 m[/tex]

[tex]h=2.60\cdot 10^6 m[/tex]

Substituting, we find

[tex]v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s[/tex]

f)

We can apply again the formula to find the escape velocity for the rocket:

[tex]v=\sqrt{\frac{2GM}{R+h}}[/tex]

Where this time we have:

[tex]M=5.98\cdot 10^{24} kg[/tex]

[tex]R=6.37\cdot 10^6 m[/tex]

[tex]h=0[/tex], because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

[tex]v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s[/tex]

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Answer:

The sound travels at [tex]v_{s}=11.4 \mathrm{m} / \mathrm{s}[/tex]

Option: c

Explanation:

Unknown source plays of middle C (fs) = 262 Hz

The sound wave from this source have to travel to raise the pitch to C sharp is (fd) = 272 Hz

[tex]\begin{array}{l}{velocity of sound in air(v)=343 \mathrm{m} / \mathrm{s}, f_{\mathrm{s}}=262 \mathrm{Hz}, f_{\mathrm{d}}=271 \mathrm{Hz}} \\ {velocity of receiver(v_{\mathrm{d}})=0 \mathrm{m} / \mathrm{s},velocity of source( v_{\mathrm{s}}) \text { is unknown }}\end{array}[/tex]

[tex]\text { Speed of sound } \mathrm{V}_{\mathrm{S}}=343 \mathrm{m} / \mathrm{s}[/tex]

[tex]f_{\mathrm{d}}=f_{\mathrm{s}}\left(\frac{v-v_{\mathrm{d}}}{v-v_{\mathrm{s}}}\right)[/tex]

[tex]\frac{f_{d}}{f_{s}}=\left(\frac{v-v_{d}}{v-v_{s}}\right)[/tex]

[tex]\left(v-v_{s}\right)=\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)[/tex]

[tex]v_{s}=v-\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)[/tex]

Substitute the given values in the formula,

[tex]v_{s}=343+\frac{262}{271}(343-0)[/tex]

[tex]v_{s}=343+0.966(343)[/tex]

[tex]v_{s}=343-331.33[/tex]

[tex]v_{s}=11.4 \mathrm{m} / \mathrm{s}[/tex]

Therefore, The sound travels at [tex]v_{s}=11.4 \mathrm{m} / \mathrm{s}[/tex]

Answer:

Whats a potentiometer.

Explanation:

Some of the factors that can limit the accuracy of a potentiometer are internal temperature and Resistance change in the resistor

A potentiometer is a variable resistor that can be adjusted manually and has three terminals. A potentiometer is also called a pot

Potentiometers have different shapes and they are used in different in different application. A potentiometer can be used to control the volume of a radio. There are several types of a potentiometer.

some of the common ones include

However, some of the materials that can be used to construct potentiometers include

As stated earlier, the two factors that can limit the accuracy of a potentiometer are:

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The maximum tension in the rope during the swing is the sum of the gravitational force and the centripetal force, which can be calculated with the given mass, rope length, and speed, resulting in a tension of 1602.48 N.

To calculate the maximum tension in the rope during the swing, we need to consider the forces acting on the man at the lowest point of the swing. These forces are the gravitational force (weight) and the centripetal force required to maintain the circular motion. The tension in the rope is the sum of these two forces.

The gravitational force (Fg) on the man can be calculated using the equation Fg = m  imes g, where m is the man's mass (88.00 kg) and g is the acceleration due to gravity (9.81 m/s2). The centripetal force (Fc) required for circular motion can be calculated using the equation Fc = m imes v2 / r, where v is the man's speed (10.20 m/s) at the bottom of the swing and r is the radius of the swing, which is equal to the length of the rope (12.0 meters).

Therefore, the maximum tension (T) in the rope would be:

T = Fg + Fc

T = (88.00 kg  imes 9.81 m/s2) + (88.00 kg  imes (10.20 m/s)2 / 12.0 m)

Calculating each part:

Fg = 863.28 N (2 decimal places)

Fc = 739.20 N (2 decimal places)

Thus, the maximum tension in the rope would be:

T = 863.28 N + 739.20 N

T = 1602.48 N (rounded to two decimal places)

Tension, centripetal force, and gravitational force are crucial components in determining the tension in the rope during circular motion.

The units for impulse are

[tex]N\cdot s[/tex]

Explanation:

We have two definition for impulse:

1) Impulse is defined as the product between the force exerted on an object and the time interval during which the force is applied:

[tex]I=F \Delta t[/tex]

where

F is the force, measured in Newtons (N)

[tex]\Delta t[/tex] is the time interval, measured in seconds (s)

From this definition it follows that Impulse can be measured in [tex]N\cdot s[/tex], which corresponds to option 3)

2)

Impulse is also defined as the change in momentum of an object:

[tex]I=\Delta p[/tex]

where

[tex]\Delta p[/tex] is the change in momentum of the object

The change in momentum has the same units of the change in momentum, so [tex]kg \cdot m \cdot s^{-1}[/tex], which corresponds to none of the options given

So the only correct option is

[tex]N\cdot s[/tex]

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The correct answer is a. potential energy. At the top of a hill, right before it plunges downward, a roller coaster car has potential energy due to its position relative to the ground. This potential energy is converted into kinetic energy as the car moves downward and gains speed.

Answer:

When reporting the speed, we need to include the value, units and direction the object was traveling.

It is a FALSE

Explanation:

Justification :

Answer:

A

Explanation:

Answer:

a

Explanation:

a

Answer:

W = 46 J

Explanation:

We need to find the angle between the two vectors Force vector and displacement vector.

First we will find the angle α of the force vector

[tex]tan\alpha =\frac{1}{8} \\\\\\alpha =7.125 deg\\[/tex]

Then we find the angle β of the displacement vector

[tex]tan\beta=\frac{2}{6} \\\\beta = 18.43 deg\\[/tex]

With these two angles we can find the angle between the two vectors

∅ = α + β = 25.56 deg

The definition of work is given by the expression

[tex]W=F*d*cos (theta)[/tex]

The absolute value of F will be:

[tex]F=\sqrt{8^{2}+1^{2}  } \\F= 8.06 N[/tex]

The absolute value of d will be:

[tex]d=\sqrt{(6 )^{2}+(2)^{2}  } \\d= 6.32m\\[/tex]

Now we have:

[tex]W=8.06*6.32*cos(25.56)\\W=46 J[/tex]

Answer:

B)velocity of object decreases

Explanation:

Consider the positive x axis as positive direction

Assume a body moving in negative x-axis direction

It's acceleration also alone negative x-axis direction

So according to our consideration

velocity and acceleration values are negative

That is both are towards negative x direction

But as both velocity and acceleration are in same direction, MAGNITUDE of velocity increases

But as magnitude increases in negative direction, velocity value decreases

But speed value increases(As speed is scalar and velocity is a vector)