A person walks first at a constant speed of 5.00 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 2.90 m/s. What is the average speed over the entire trip?

Answers

Answer 1

Answer:

The Average Speed over the entire trip is 3.671 meters per second.

The person is moving in a straight line at a constant Speed. Let suppose that Distance between points A and B is $r$ , then we construct and use the Kinematic formulas for each stage of the travel to determine Times and Average Speed, based on the fact that Speed is inversely proportional to Time, we derive the resulting expression:

First Stage (from point A to point B):

$t_(1) = (r)/(v_(1))$ (1)

Second Stage (from point B to point A):

$t_(2) = (r)/(v_(2))$ (2)

Average Speed:

$t_(1)+t_(2) = (2\cdot r)/(\bar v)$ (3)

Where:

$t_(1)$ - Travelling time for the first stage, in seconds.

$t_(2)$ - Travelling time for the second stage, in seconds.

$r$ - Distance from A to B, in meters.

$v_(1)$ - Speed of the person in the first stage, in meters per second.

$v_(2)$ - Speed of the person in the second stage, in meters per second.

$\bar v$ - Average speed, in meters per second.

By applying (1) and (2) in (3), we derive an expression to determine the Average Speed:

$(r)/(v_(1)) + (r)/(v_(2)) = (2\cdot r)/(\bar v)$

$(v_(2)+v_(1))/(v_(1)\cdot v_(2)) = (2)/(\bar v)$

$\bar v = (2\cdot (v_(1)\cdot v_(2)))/(v_(2)+v_(1))$ (4)

If we know that $v_(1) = 5\,(m)/(s)$ and $v_(2) = 2.90\,(m)/(s)$ , the average speed over the entire trip is:

$\bar v = (2\cdot (v_(1)\cdot v_(2)))/(v_(2)+v_(1))$

$\bar v = (2\cdot \left(5\,(m)/(s) \right)\cdot \left(2.90\,(m)/(s) \right))/(2.90\,(m)/(s) + 5\,(m)/(s) )$

$\bar v = 3.671\,(m)/(s)$

The Average Speed over the entire trip is 3.671 meters per second.

Please see this question related to Average Speed for further details: brainly.com/question/19335778

Answer 2

Answer: This problem is a very interesting problem.
we know wacth:

Speed = distance / time.

Consider that:
distance from point A to point B=distance from point B to point A=d

We calculate the time to go from point A to point B.
time=distance / speed.
T₁=d / (5 m/s)

We calculate the time to go from point B to point A.
T₂=d / (2.9 m/s)

Therefore; the total time wil be: T₁+T₂
Total time=d/(5 m/s) + d / (2.9 m/s)
Least common multiple=(5 m/s)(2.9 m / s)=14.5 m²/s²
Total time=(2.9 m/s d + 5 m/s d)/ 14.5m²/s²
Total time=[(7.9 d) m/s] / (14.5 m²/s²)
Total time=7.9 d/ (14.5 m/s)

Now, the total distance will be =d+d=2d

Therefore:
Average speed=total distance / total time
Average speed=[(2d)] / [(7.9 d)/ (14.5 m/s)]
Average speed≈3.67 m/s.

Answer: the average speed will be ≈3.67 m/s.

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A stone is thrown horizontally at 8.0 m/s from a cliff 78.4 m high. How far from the base of the cliff does the stone strike the ground?

Answers

A stone is thrown horizontally at 8.0 m/s from a cliff 78.4 m high. The distance from the base of the cliff does the stone strike the ground is 64 m.

What is distance?

Distance is defined as the entire path traveled by an object in motion, regardless of journey direction. Three sight distances are primarily taken into account in the geometric design of road construction. Things that can be measured in terms of length include the dimensions of a section or a room. Distance is the separation between two things, like two cities or two points. There will never be a negative value for either length or distance.

A projectile is an object that is launched vertically into the air and moves while being pulled by gravity. A horizontally launched object that defies gravity is referred to as a projectile.

Thus, a stone is thrown horizontally at 8.0 m/s from a cliff 78.4 m high. The distance from the base of the cliff does the stone strike the ground is 64 m.

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radiation in the ultraviolet region of the electromagnetic spectrum is quite energetic. it is this radiation that causes dyes to fade and your skin to develop a sunburn. if you are bombarded with 1.00 mol of photons with a wavelength of 370. nm, what amount of energy, in kilojoules per mole of photons, are you being subjected to?

Answers

If you are bombarded with 1.00 mol of photons with a wavelength of 370 nm, you are being subjected to 322 kJ/mol of energy.

The energy of a photon is given by the equation:

E = hc/λ

where:

h = Planck's constant = 6.626 x $10^{-34$ J s

c = speed of light = 2.998 x $10^8$ m/s

λ = wavelength of the radiation

To find the energy of a mole of photons, we need to multiply the energy of a single photon by Avogadro's number (6.022 x $10^{23$ ).

First, we need to convert the wavelength of the radiation from nanometers to meters:

370 nm = 370 x $10^{-9$ m

Now, we can use the equation to calculate the energy of a single photon:

E = hc/λ

E = (6.626 x $10^{-34$ J s) x (2.998 x $10^8$ m/s) / (370 x $10^{-9$ m)

E = 5.35 x $10^{-19$ J

Next, we can calculate the energy of a mole of photons:

(5.35 x $10^{-19$ J/photon) x (6.022 x $10^{23$ photons/mol) = 3.22 x $10^5$ J/mol

Finally, we can convert the energy to kilojoules per mole of photons:

3.22 x $10^5$ J/mol = 322 kJ/mol

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A constellation is _______ that forms a pattern resembling animals, mythological characters, or objects.A. a group of meteors
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Answers

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Explain why the car reaches a top speed even though the thrust Force remains constant at 3500N

Answers

The car reaches its top speed when the constant thrust force equals the opposing air resistance force, resulting in zero net external force and constant velocity. This balance between thrust and resistance forces determines the top speed.

Why the car reaches a top speed despite the thrust Force remaining constant

When the applied thrust force typically supplied by the engine equals the total of the opposing forces pushing on the vehicle, the vehicle reaches its maximum speed.

Air resistance, or drag, is the main opposing force and it grows as the car's speed squares. The car first accelerates because the propulsion force overcomes this resistance as it accelerates. But the air resistance force likewise increases with the car's speed until it reaches the same level as the thrust force.

The car now maintains a steady velocity when it reaches its maximum speed and the net external force acting on it zeroes. The maximum speed is also influenced by other elements like mechanical efficiency and rolling resistance.

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Air resistance will eventually equal the thrust force, so there is no resultant force meaning the car will stay at the same speed and can no longer accelerate.

Sarah throws a ball directly upward at the edge of a cliff with a starting velocity of 3.0 m/sIt lands on the ground with a final speed of 6.0 m/s
How long is the ball in the air?
Round the answer to two significant digits

Answers

Answer:

The correct answer is t = 0.92s

Explanation:

Initial velocity v0 = 3.0 m/s

Displacement Δy = ?

Acceleration a = -9.8m/s2

Final velocity v = -6.0m/s

Time t=? Target unknown

We can use the kinematic formula missing Δy to solve for the target unknown t:

V=v0+at

We can rearrange the equation to solve to t:

V-v0=at

t= v-v0/a

Substituting the known value into the kinematic formula gives:

t= (-6.0m/s)-(3.0m/s)

————————————

-9.8m/s2

= -9m/s

—————-

-9.8m/s2

=0.92s

Sarah throws a ball directly upward at the edge of a cliff with a starting velocity of 3.0 m/s. It lands on the ground with a final speed of 6.0 m/s.

How long is the ball in the air?

Answer: 0.92

*PLEASE HURRY ITS FOR A QUIZ*A motorcycle and rider, having a total mass of 875 kg, are moving with a velocity of 15 m/s due east. What is the total momentum of the rider and cycle?

A. 58 kg.m/s east

B. 60 kg.m/s east

C. 13.3 kg.m/s east

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Answers

I think it's D because momentum= mass times velocity

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