A hungry bat finds the mother lode of flying insects between the headlights of two bicycles cruising towards each other through the Herrick lake forest preserve. The bikes are 100 meters apart at the start and one has a speed of 8 m/s and the other a speed of 10 m/s. The bat has a speed of 20 m/s. Assume that the bat flies from the front tire of one bike straight to the front tire of the other bike, then turns around with no delay and heads towards the other bike. This is repeated until both bikes and the bat are at the same place. Determine how far the bat flies during this flying feast.​

Answers

Answer 1

Answer:

111.1 meters flown from bat

Explanation:

So one bike is going at 8m/s and the other bike is going at 10m/s in the direction towards the other bike. So we can make one bikes velocity =0 and the other we add the velocities together to get V of bike a = 18m/s while velocity of bike b =0m/s. Then we use the distance apart to find the time. so we take our average velocity formula and rearrange it to time = displacement/average velocity. This gives t = 100/18=5.55secs. We then take the bats velocity to find to the distance traveled. 5.55secs = x/20m/s. We just cross multiply to get 111.1 repeating which we just say is 111.1m traveled from the bat.

Answer 2

The distance traveled by the bat before the two bikes meet is 1,000 m.

Time taken for for the two bikes to be in the same place

The time taken for the bikes to meet or intersect each other is calculated as follows;

(Va - Vb)t = d

where;

t is the time taken for the bikes to meet each otherd is the distance between the bikesVa is the velocity of the first bikeVb is the velocity of the second bike

(10 m/s - 8 m/s)t = 100 m

2 m/s (t) = 100 m

t = 100 m / 2 m/s

t = 50 seconds

Distance traveled by the bat within this period

Distance = average speed x time

Distance = 20 m/s x 50 s

Distance = 1,000 m

Thus, the distance traveled by the bat before the two bikes meet is 1,000 m.

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Related Questions

with what force in the rightward direction, fx , should marcel push? if your expression would give a negative result (using actual values) that just means the force should be toward the left.

Answers

The force in the rightward direction, fx , should Marcel push is this force leads to counterclockwise torque. Fx = - (w L2 + w L3 - W Lend) / h.

A force in physics is an effect that changes the velocity, or acceleration, of a mass-moving object. It is a vector quantity since it can be a push or a pull and always has magnitude and direction.

Viewing equilibrium of torque:

W Lend + Fx × h = w L2 + wL3

Fx = (w L2 + w L3 - WLend) / h

In the rightward direction, Fx,

Fx = - (w L2 + w L3 - W Lend) / h

Thus, this force leads to counterclockwise torque.

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The given question is incomplete, so the most probable complete question is,

With what force in the rightward direction, fx , should Marcel push? if your expression would give a negative result (using actual values) that just means the force should be toward the left.

The image of the question is attached below.

with what force in the rightward direction, fx , should marcel push? if your expression would give a

Glider‌ ‌A‌ ‌of‌ ‌mass‌ ‌0.355‌ ‌kg‌ ‌moves‌ ‌along‌ ‌a‌ ‌frictionless‌ ‌air‌ ‌track‌ ‌with‌ ‌a‌ ‌velocity‌ ‌of‌ ‌0.095‌ ‌m/s.‌ ‌It‌ ‌collides‌ ‌with‌ ‌glider‌ ‌B‌ ‌of‌ ‌mass‌ ‌0.710‌ ‌kg‌ ‌moving‌ ‌in‌ ‌the‌ ‌same‌ ‌direction‌ ‌at‌ ‌a‌ ‌speed‌ ‌of‌ ‌0.045‌ ‌m/s.‌ ‌After‌ ‌the‌ ‌collision,‌ ‌glider‌ ‌A‌ ‌continues‌ ‌in‌ ‌the‌ ‌same‌ ‌direction‌ ‌with‌ ‌a‌ ‌velocity‌ ‌of‌ ‌0.035‌ ‌m/s.‌ ‌What‌ ‌is‌ ‌the‌ ‌velocity‌ ‌of‌ ‌glider‌ ‌B‌ ‌after‌ ‌the‌ ‌collision?

Answers

Answer:

vB' = 0.075[m/s]

Explanation:

We can solve this problem using the principle of linear momentum conservation, which tells us that momentum is preserved before and after the collision.

Now we have to come up with an equation that involves both bodies, before and after the collision. To the left of the equal sign are taken the bodies before the collision and to the right after the collision.

\((m_{A}*v_{A})+(m_{B}*v_{B})=(m_{A}*v_{A'})+(m_{B}*v_{B'})\)

where:

mA = 0.355 [kg]

vA = 0.095 [m/s] before the collision

mB = 0.710 [kg]

vB = 0.045 [m/s] before the collision

vA' = 0.035 [m/s] after the collision

vB' [m/s] after the collison.

The signs in the equation remain positive since before and after the collision, both bodies continue to move in the same direction.

\((0.355*0.095)+(0.710*0.045)=(0.355*0.035)+(0.710*v_{B'})\\v_{B'}=0.075[m/s]\)

Calculate force a father will push a 20.0 kg baby carriage with
if he goes 4.5 m while he does 2920 J of work.

Answers

Answer:

649 N

Explanation:

W = force X distance

2920 = force * 4.5

2920/4.5 = force = 649 N

What is the frequency of light with a wavelength of 7.9 x 10^-9 m? ( the speed of light is 3.00 x 10^8)

A.) 3.80 x 10^16
B.) 2.37 Hz
C.) 2.37 m
D.) 3.80 x 10^16 Hz

Answers

using the formula v=f times lambada
then v=the speed of light.
and f=what’s we’re looking for
and lambada=the wavelength.

so then you sub what you have (v and lambada) in the formula.
then multiply the frequency(f) by the given wavelength and then solve for f
What is the frequency of light with a wavelength of 7.9 x 10^-9 m? ( the speed of light is 3.00 x 10^8)A.)

what was the dwarf planet discovered in AZ? ​

Answers

Answer:

Pluto.

Explanation:

an observary in Flagstaff, Arizona, discorvered it.

what is the correct answer?

what is the correct answer?

Answers

Answer:

(B) 39

Explanation:

Use synthetic division:

-2 | 3  0  -1    3  1

____-6 12 -22 38

     3 -6 11  -19 | 39

This means that the remainder is 39

The remainder is 39.
what is the correct answer?

a particle is projected from a point a with a velocity of 21m/s at an angle of 53.1 degees to the horizontal. one second later another particle is projected at a point 0.3m below a with initial velocity of 31.5m/s at an angle of 36.9 degrees to the horizontal.
(i) prove that the particles collide
(ii) find the time of collision
(iii) find the distance of collision from point of projection

Answers

Answer:

(i) Please see graph of the motion of the particles created with MS Excel and the calculations in the following sections

(ii) The time of collision is approximately 1.0 seconds

(iii) The common horizontal distance of point collision from the point of projection is approximately 25.2 meters

Explanation:

The velocity with which the first projectile was fired, v₁ = 21 m/s

The angle to the horizontal the particle is launched = 53.1°

The time at which the other particle was launched = 1 second later

The location from which the other particle was projected = 0.3 m below the first particle

The initial velocity of the second particle = 31.5 m/s

The angle to the horizontal at which the second particle was projected, θ = 36.9°

(i) The height reached, by each of the particle is given as follows;

y = u·t - 1/2·g·t²

For the first projectile, we have;

y = 21·(t₁+1)×sin(53.1°) - 9.81·(t₁+1)²/2 + 0.3

For the second projectile, we have;

y= 31.5·(t₁)×sin(36.9°) - 4.905·(t₁)²

If the two projectiles collide, we get;

21·(t₁+1)×sin(53.1°) - 9.81·(t₁+1)²/2 + 0.3  = 31.5·(t₁)×sin(36.9°) - 4.905·(t₁)²

Using a graphing calculator for simplifying, we get;

-11.93·t₁ + 12.2 = 0

t₁ = 12.2/11.93 ≈ 1.02

Therefore, at time t₁ = 1.02 seconds, after the launch of the second particle, the two particle will be at the same vertical height

However, whereby at the time, t₂, the particles collide, the horizontal distance travelled, 'x', will be equal;

We have;

x = u·cos(θ)·t₁

For the first particle, we have;

x₁₁ = 21 × cos (53.1°) × (t₂ + 1)

For the second particle, we have;

x₂₂ = 31.5 × cos (36.9°) × t₂

At the point of collision, we have;

x₁ = x₂

∴ 21 × cos (53.1°) × (t₂ + 1) = 31.5 × cos (36.9°) × t₂

31.5 × cos (36.9°) × t₂ - 21 × cos (53.1°) × t₂ = 21 × cos (53.1°)

t₂ = 21 × cos (53.1°)/(31.5 × cos (36.9°)  - 21 × cos (53.1°) ) = 1.00219236871

t₂ ≈ 1.0 seconds

Given that t₁ ≈ t₂, the particles reach the same height and the same horizontal distance at the same time, t₂ ≈ 1.0 and therefore, they collide.

(ii) The time of collision is found above as t₁ ≈ t₂ ≈ 1.0 seconds

(iii) The horizontal distance of the point of collision from the starting point, 'x', is given as follows;

x = 21 × cos (53.1°) × (1.0 + 1) ≈ 25.2

The horizontal distance of the point of collision from the starting point, x ≈ 25.2 meters

The vertical distance of the point of collision from the starting point of the second particle, 'y', is given as follows;

y = 21 × (1+1)×sin(53.1°) - 9.81 × (1+1)²/2 + 0.3 ≈ 14

The vertical distance of the point of collision from the starting point of the second particle, y ≈ 14 meters

The magnitude of the distance from the starting point of the second particle, r = √(25.2² + 14²) ≈ 28.8

The magnitude of the distance from the starting point of the second particle, r ≈ 28.4 meters.

a particle is projected from a point a with a velocity of 21m/s at an angle of 53.1 degees to the horizontal.

Make a claim about the time interval between lunar phases.
I will mark as BRAINLIEST ✨
just pls help ASAP
due by 4:00pm
plss

Answers

Answer:

The Lunar month is the time period for the moon to complete ONE revolution around Earth. SIDEREAL MONTH: 27.3 days SYNODIC MONTH: 29.5 days

I will give 14 points & make you the brainiest
Terrestrial planets are primarily composed of what two materials?

a. hydrogen and helium

b. silicate rocks and metals

c. frozen water and methane

d. helium and iron

Answers

Answer:

B. silicate rocks and metals

Terrestrial planets, such as Earth, are primarily composed of silicate rocks and metals. The correct answer is b. silicate rocks and metals.

Silicate rocks, which are made up of silicon and oxygen, are the dominant building blocks of these planets. They include minerals like quartz, feldspar, and mica. Terrestrial planets contain significant amounts of metals such as iron, nickel, and other heavy elements. These metals are found in the planet's core, providing structural strength and contributing to the planet's magnetic field.

This composition differentiates terrestrial planets from gas giants like Jupiter and Saturn, which are predominantly composed of hydrogen and helium. The combination of silicate rocks and metals forms the solid, rocky surfaces that characterize terrestrial planets.

Therefore, The correct answer is b. silicate rocks and metals.

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Hint?
In which of these media will sound waves travel the fastest in?
a Oil
b Steel
c Vacuum
d Water
e Air

Answers

The sound wave travels the fastest in Steel. The correct option is b.

Sound waves require a medium to travel through, as they are mechanical waves that propagate through the vibration of particles in a substance. The speed of sound depends on the properties of the medium, specifically its density and elasticity.

a) Oil: Sound waves generally travel slower in liquids compared to solids because liquids are denser and less elastic. Therefore, oil is not the medium in which sound waves will travel the fastest.

c) Vacuum: Sound waves cannot travel through a vacuum because a vacuum is a complete absence of matter. Since sound waves require particles to propagate, they cannot travel through a vacuum. Therefore, a vacuum is not the medium in which sound waves will travel the fastest.

d) Water: Sound waves travel faster in water compared to gases (like air) because water is denser than air. However, steel is denser and more elastic than water, making it a better medium for transmitting sound waves. Therefore, water is not the medium in which sound waves will travel the fastest.

e) Air: Sound waves travel slower in gases, such as air, compared to solids and liquids because gases are less dense and less elastic. Although sound waves can travel through air and are commonly experienced in our everyday lives, they propagate slower in air compared to steel. Therefore, air is not the medium in which sound waves will travel the fastest.

Therefore, b) Steel: Sound waves travel fastest in solids, particularly in materials that are both dense and elastic. Steel is a solid material that is dense and has high elasticity, allowing sound waves to travel through it at higher speeds compared to other options listed. Therefore, steel is the medium in which sound waves will travel the fastest among the given options.

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page:
+ At what height from surface of earth accelaration
due to gravity is half of the surface​

Answers

Answer:

(R=6. 4×106m)

Explanation:

The height above the surface of earth at which acceleration due to gravity is half the acceleration due to gravity at surface of earth

Explanation:

acceleration due to gravity at surface of earth is (R=6. 4×106m)

hope this answer helps you


A firework with a mass of 1.5 kg shoots up into the air with a velocity of 25 m/s. It then explodes. Use this information to work out the total momentum of all the pieces after the firework explodes

Answers

Answer:

37.5 kg⋅m/s

Explanation:

p = mv

p = (1.5)(25) = 37.5 kg⋅m/s

Momentum is conserved in all collisions and explosions!

A horizontal bar 1.5 metres long and of small cross-section rotates about vertical axis through one end. It accelerates uniformly from 1200 r.P.M. To 1500 r.P.M. In an interval of 5 seconds. What is the linear velocity at the beginning and end of the interval ? What are the normal and tangential components of the acceleration of the mid-point of the bar after 5 seconds after the acceleration begins ?

Answers

Answer:

a. i. 188.49 m/s. ii. 235.62 m/s. b. i. 117.81 m/s² ii. 4.71 m/s²

Explanation:

a. Since the linear velocity v = rω where r = radius of path = length of horizontal bar = 1.5 m and the initial angular speed ω₀ = 1200 r.P.M = 1200 × 2π/60 = 125.66 rad/s and final angular speed after the 5 seconds ω₁ = 1500 r.P.M = 1500 × 2π/60 = 157.08 rad/s

i. So, the initial linear velocity v₀ = rω₀ = 1.5 m × 125.66 rad/s = 188.49 m/s.

ii. The final linear velocity v₁ = rω₁ = 1.5 m × 157.08 rad/s = 235.62 m/s.

b. The mid-point of the bar is at a distance of one-half from the end of the bar. So its radius r' = r/2 = 1.5 m/2 = 0.75 m

i. Its normal acceleration a after the 5 seconds is a = r'ω₁² = 0.75 m × 157.08 rad/s = 117.81 m/s²

ii. Its tangential acceleration is given by a' = r'α where α = tangential acceleration

α = (ω₁ - ω₀)/t  where t = 5 seconds

α = (157.08 rad/s - 125.66 rad/s)/5 s

= 31.42 rad/s ÷ 5 s

= 6.28 rad/s²

a' = r'α

= 0.75 m × 6.28 rad/s²

= 4.71 m/s²

a diffraction grating with 145 slits per centimeter is used to measure the wavelengths emitted by hydrogen gas. at what angles in the third-order spectrum would you expect to find the two violet lines of wavelength 434 nm and of wavelength 410 nm? (angles in radians)

Answers

The two violet lines of hydrogen gas with wavelengths 434 nm and 410 nm in the third-order spectrum of a diffraction grating with 145 slits per centimetre would be expected at angles of approximately 1.09 radians and 1.22 radians, respectively.

Diffraction gratings are used to disperse light into its constituent wavelengths and measure their spectra. The number of slits per centimetre on the grating determines the angular spacing between the diffracted wavelengths. In this case, a diffraction grating with 145 slits per centimetre is used to measure the spectrum of hydrogen gas, which emits violet lines at wavelengths 434 nm and 410 nm. The third-order spectrum corresponds to diffracted wavelengths that are three times the spacing between the slits. Using the equation for diffraction grating, the angles at which these violet lines are expected to appear in the third-order spectrum can be calculated as approximately 1.09 radians for the 434 nm line and 1.22 radians for the 410 nm line.

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Each large pizza can feed three teenage boys. How many pizzas would be needed to feed the
football team? (56 teenage boys).

Answers

Answer:

19 pizzas

Explanation:

1 pizza = 3 boys

=> 56 ÷ 3 = 18.666666666666.. or about 19 pizzas

18.666666666666666 or round to 19

How does changing the charge’s value influence the force between the two charges?

As the value of charge 2 __________, the value of the force ___________.


decreases; decreases

increases; decreases

increases; does not change

decreases; does not change

Answers

The strength of the repulsion or attraction weakens and drops to one-fourth of its initial value when the gap between the two electrons is doubled.

What exactly is physics' law of attraction?

According to the Law of Attraction, when we alter our emotions and thoughts, we alter our electromagnetic field, which attracts a new reality from of the Quantum Field. We must hold that condition of being long sufficient to allow the manifestations to take place after the thought and the sensation have to coincide.

What exactly is physics' law of attraction?

According to the Law of Attraction, when we alter our opinions and feelings, we alter our electromagnetic field, which attracts a new reality from Quantum Field. The thought must coincide with the feeling, and we must then maintain that condition of being for long enough to allow the manifestation to take place.

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In a wire, when elongation is 4 cm energy stored is E. if it is stretched by 4 cm,
then what amount of elastic potential energy will be stored in it?
plz don't spam
answer fast in imp

Answers

Answer:

4E

Explanation:

From the question given above, the following data were obtained:

Initial elongation (e₁) = 4 cm = 4/100 = 0.04 m

Initial energy (E₁) = E

Final elongation (e₂) = 0.04 + 0.04 = 0.08 m

Final energy (E₂) =?

The energy stored in a s spring is given by:

E = ½Ke²

Where

E => is the energy

K => is the spring constant

e => is the elongation

From:

E = ½Ke²

Energy is directly proportional to the elongation. Thus,

E₁/e₁² = E₂/e₂²

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

Initial elongation (e₁) = 0.04 m

Initial energy (E₁) = E

Final elongation (e₂) = 0.08 m

Final energy (E₂) =?

E₁/e₁² = E₂/e₂²

E / 0.04² = E₂ / 0.08²

E / 0.0016 = E₂ / 0.0064

Cross multiply

0.0016 × E₂ = 0.0064E

Divide both side by 0.0016

E₂ = 0.0064E / 0.0016

E₂ = 4E

Therefore, the final energy is 4 times the initial energy i.e 4E

Which wave interaction results in a change in the direction of the wave as it passes through one medium to another medium? absorption diffraction reflection refraction.

Answers

Answer: Reflection

Explanation:

The incident wave is the wave coming from the source. When an incident wave hits an interface, it splits into two parts. There is the reflected wave and the transmitted wave. Assuming that the incident wave hits the interface directly, and not at an angle, the reflected wave will reflect back to the source, and the transmitted wave will pass through the interface and continue on its path. The reflected wave will be "upside down" as it gets sent back.

Answer:refraction

Explanation:

right on edge 2023

BRAINLIEST WILL BE REWARDED TO THE CORRECT ANSWER
How does the Half-life setting affect how quickly the simulated substance decays?


The greater the half-life is, the faster the rate of decay will be.

The greater the half-life is, the slower the rate of decay will be.

The smaller the half-life is, the slower the rate of decay will be.

The half-life does not affect rate of decay.

Answers

Answer:

the greater the half life is the slower the rate of decay will be.

Explanation:

half life of a radioactive substance is inversely proportional to the radioactive decay.

Answer:

B. The greater the half life is, the slower the rate of decay will be

Explanation: Most living things contain carbon-14, an unstable isotope of carbon that has a half-life of around 5,000 years. That means that when scientists dig up fossil bones, they can figure out how old they are by measuring the amount of carbon-14 remaining in the bones.

Scientific notation 47,000

Answers

47,000 written as 4.7 x 104 in s scientific notation

Yep that is right 4.7 times 104

Find the velocity and distance travelled of a car that is accelerating at 3 m/s2 for 4 seconds.

Answers

Explanation:

Given:

v₀ = 0 m/s

a = 3 m/s²

t = 4 s

Find: Δx and v

Δx = v₀ t + ½ at²

Δx = (0 m/s) (4 s) + ½ (3 m/s²) (4 s)²

Δx = 24 m

v = at + v₀

v = (3 m/s²) (4 s) + 0 m/s

v = 12 m/s

What is the work done by a force equivalent to the weight of 100kg that moves an object 200km in 7.0min?
Please guys can somebody helps me :(

Answers

The work done by a force equivalent to the weight of 100 kg that moves an object 200 km in 7.0 min would be 196000 kJ.

What is work done?

The total amount of energy transferred when a force is applied to move an object through some distance

The work done is the multiplication of applied force with displacement.

Work Done = Force ×  Displacement

As given in the problem we have to calculate the work done by a force equivalent to the weight of 100 kg that moves an object 200 km in 7.0 min.

The weight of the 100 kg = 100 × 9.8

                                          = 980 N

The work done = 980 × 200 × 1000

                          = 196000000

                          = 196000 kJ

Thus, the work done by a force equivalent to the weight of 100 kg that moves an object 200 km in 7.0 min would be 196000 kJ.

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Select the correct location on the image.
The lines on the position-time graph show the velocities of different vehicles. Which line represents a vehicle moving at constant velocity?

Answers

Answer: Green line

Explanation:

When an object is moving at constant velocity or speed rather, the line on a graph will be represented by a straight line.

They would be maintaining the same speed over a period of time so on a graph with time like the above, their journey would appear to be horizontal which would make the green line correct.

Select the correct location on the image.The lines on the position-time graph show the velocities of

Answer:

Hope this helps!!!

Explanation:

Select the correct location on the image.The lines on the position-time graph show the velocities of

if light consisted of classical particles and was sent through a double slit, the pattern on the wall would be which of the following? a single bright fringe dependent on the size of the slit an interference pattern of light and dark fringes a large round dot a bright blob with no distinct shape if light is actually a wave that only behaves like a particle in certain situations then, when light passes through a double slit, the pattern on the wall would be which of the following? a single bright fringe dependent on the size of the slit an interference pattern of light and dark fringes a large round dot two lines proportional to the shape of the two slits

Answers

When light passes through a double slit, the pattern on the wall would be an interference pattern of light and dark fringes.

If light consisted of classical particles and was sent through a double slit, the pattern on the wall would be a bright blob with no distinct shape. If light is actually a wave that only behaves like a particle in certain situations then, when light passes through a double slit, the pattern on the wall would be an interference pattern of light and dark fringes.What is a double slit?A double-slit experiment is an experiment that demonstrates the wave-like nature of light. Light passes through two small slits that are positioned close together in a double-slit experiment. Two waves emerge from the two slits and interact with each other, producing an interference pattern on a screen. The pattern will consist of a series of alternating bright and dark fringes, known as interference fringes.

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the picture below shows a centrifuge used to test astronauts

Which vectors shows the direction of the centripetal acceleration at this point?
A. A
B. B
C. C
D. D
[PLEASE SOMEONE HELP! ASAP]

the picture below shows a centrifuge used to test astronauts Which vectors shows the direction of the

Answers

Answer: B

Explanation: just did the quiz

Vector B shows the direction of the centripetal acceleration at this point.

What is centripetal acceleration?

Centripetal acceleration is the acceleration of a frame traversing a circular path. because the pace is a vector quantity (this is, it has both a magnitude, the rate, and a course), when a frame travels on a circular course, its route continuously adjustments, and accordingly its velocity changes.

What reasons for centripetal acceleration?

Centripetal forces motivate centripetal accelerations. within the special case of the Earth's circular motion around the sun – or any satellite's ground movement around any celestial body – the centripetal force causing the movement is the end result of the gravitational enchantment among them.

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I need help
what force is needed to make a ball fall on earth if it had a mass of 13kg

I'd appreciate if someone could help me thank you ​

Answers

Newton's second law allows finding the answer for the force that attracts the body is 127.4N

Newton's second law indicates that the force that is the interaction between two bodies is directly proportional to the product of the mass and the acceleration.

            F = m a

Where the bold letters indicate vectors, F is the force, m the mass, and the acceleration of the body.

When a body is close to the Earth there is an interaction between the body and the planet, we call this interaction weight, it is given by the relationship

           W = m g

Where W is the force called weight, m ​​the mass of the body and g the acceleration of the body, which in this case is called the gravity acceleration   (g = 9.8 m / s²)

They indicate that the mass of the body is m = 13 kg, let's calculate the weight

           W = 13  9.8

           W = 127.4 N

In conclusion using Newton's second law we can find the answer for the force that attracts the body is 127.4N

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The table above shows the data that Mendel collected about the offspring of
his second set of crosses. What is the median of the number of trials he
performed?

The table above shows the data that Mendel collected about the offspring ofhis second set of crosses.

Answers

The table above shows the data that Mendel collected about the offspring of his second set of crosses therefore the median of the number of trials he performed is 1,181 and is therefore denoted as option A.

What is Median?

This is referred to as the middle number in a sorted, ascending or descending list of numbers.

In this scenario there were five trials which are 1064, 7324, 8003, 1181 and 929. When it is arranged in an ascending order we have:

929 ,  1064 , 1181 , 7324 , 8003

Therefore the median number is 1181.

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A moving object has O speed.O Momentum.O energy. O velocity. O all of the above

Answers

According to the given information A moving object has all of the above.

What is momentum ?

Speed is a scalar quantity that represents the magnitude of the velocity of an object, and it is measured in units such as meters per second (m/s).

Momentum is a vector quantity that represents the product of an object's mass and velocity, and it is measured in units such as kilogram-meters per second (kg·m/s).

Energy is a scalar quantity that represents the ability of an object to do work, and it is measured in units such as Joules (J).

Velocity is a vector quantity that represents both the magnitude and direction of an object's motion, and it is measured in units such as meters per second (m/s).

A moving object has all of the above properties.

It's important to note that all of these properties are related and they can be converted to each other, but they are not the same thing.

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A 20μC point charge is placed 10cm from a 5μC point charge.
1. Calculate the force experienced by the 5μC charge.
2. What is the force on the 20μC charge?
3. What is the electric field strength located at 15cm from the 20μC charge?
4. Draw the direction of the electric field line at the 15cm mark from the 20μC charge

Answers

1. The force experienced by the 5μC charge is 0.45 N.

2. The force on the 20μC charge is -0.45 N.

3. The electric field strength located at 15cm from the 20μC charge is 12 N/C.

4. The electric field lines will point away from the 20μC charge, radially outward.

1. To calculate the force experienced by the 5μC charge, we can use Coulomb's law. Coulomb's law states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, the formula is given by:

\(F = k * (q1 * q2) / r^2\)

Where F is the force, k is the electrostatic constant (9 x 10^9 Nm²/C²), q1 and q2 are the charges, and r is the distance between them.

Plugging in the values:

\(F = (9 x 10^9 Nm^2/C^2) * ((5 x 10^-6 C) * (20 x 10^-6 C)) / (0.1 m)^2\)

= 0.45 N

Therefore, the force experienced by the 5μC charge is 0.45 N.

2. By Newton's third law of motion, the force on the 20μC charge is equal in magnitude but opposite in direction to the force experienced by the 5μC charge. Hence, the force on the 20μC charge is -0.45 N.

3. To calculate the electric field strength at a point, we can use the formula:

\(E = k * (q / r^2)\)

Where E is the electric field strength, k is the electrostatic constant, q is the charge, and r is the distance from the charge.

Plugging in the values:

\(E = (9 x 10^9 Nm^2/C^2) * (20 x 10^-6 C) / (0.15 m)^2\)

= 12 N/C

Therefore, the electric field strength located at 15cm from the 20μC charge is 12 N/C.

4. Electric field lines depict the direction of the electric field. Since the charge is positive (20μC), the electric field lines will point away from it, radially outward. Hence, at the 15cm mark from the 20μC charge, the electric field lines will extend outward from the charge in all directions.

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A plane moves at a certain velocity then accelerates at the rate of 52 m/s2 for a distance of 2300 m to reach a velocity of 620 m/s. How fast was the plane going before it began to accelerate?

Answers

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