Explain How can you
use a distance-time graph
to calculate average
speed?

Answer:

A = air resistance

B = lift

C = thrust

D = gravity

1.00 N, 2.86° above horizontal

Explanation:

Gravity pulls down on the bird.  The thrust of the birds wings push it forward, and the lift pushes it up.  Air resistance pushes it back.

The net force in the y direction is:

∑Fᵧ = lift − gravity

∑Fᵧ = 0.20 N − 0.15 N

∑Fᵧ = 0.05 N

The net force in the x direction is:

∑Fₓ = thrust − air resistance

∑Fₓ = 1.50 N − 0.50 N

∑Fₓ = 1.00 N

The magnitude of the resultant force is:

F² = Fₓ² + Fᵧ²

F² = (1.00 N)² + (0.05 N)²

F ≈ 1.00 N

The direction of the resultant force is:

tan θ = Fᵧ / Fₓ

tan θ = (0.05 N) / (1.00 N)

θ = 2.86° above horizontal

Answer:

Alright the other day.

Answer:

Normalization is a systematic approach of decomposing tables to eliminate data redundancy. It is a multi-step process that puts data into tabular form, removing duplicated data from the relation tables.

Explanation:

Answer:

the process of bringing or returning something to a normal condition or state.

Explanation:

In "your own words"- Normalization is when you change something back into it's average form.

Answer:

Pluto & Mercury

Explanation:

Pluto's eccentricity is 0.248

Mercury's eccentricity is 0.206

Answer:

a: To provide organized information, for the development of the organization's internal analysis, mission, vision, values, and strategy.

Explanation:

Environmental analysis is a process in business wherein the internal and external factors that can affect a business are examined so as to help improve the organization's goals and strategies. The internal analysis serves as a guide to checkmate the organization and to see if they have adopted the right strategies that suit their external environment.

Competitors in the external environment can have a huge impact on the business. So the internal workings of the organization should align with the external factors. The organized information would also be useful in forecasting the growth of the business.          

Answer:

experimental group

Explanation:

please mark me as a brainlist..

Answer:

Experemtial group

Explanation:

The placebo is just sugar pills tested is ones already done and control is a mix of the two

Answer:

high frequent sound

Explanation:

because if its low than its slower.

Answer:

the mass of the air in the classroom = 2322 kg

Explanation:

given:

A classroom is about 3 meters high, 20 meters wide and 30 meters long.

If the density of air is 1.29 kg/m3

find:

what is the mass of the air in the classroom?

density = mass / volume

where mass (m) = 1.29 kg/m³

volume = 3m x 20m x 30m = 1800 m³

plugin values into the formula

  1.29 kg/m³   =        mass    

                             1800 m³

mass =  1.29 kg/m³  ( 1800 m³ )

mass = 2322 kg

therefore,

the mass of the air in the classroom = 2322 kg

If A classroom is about 3 meters high, 20 meters wide and 30 meters long. If the density of air is 1.29 kg/m3, then the mass of the air in the classroom is 2322Kg.

Density is the ratio of mass to volume. it tells how much mass a body is having for its unit volume. for example egg yolk has 1027kg/m³ of density, means if we collect numbers of egg yolk and keep it in a container having volume 1 m³ then total amount of mass it is having will be 1027kg. Density is a scalar quantity. when we add egg yolk into the water, egg yolk has greater density than water( 997 kg/m³), because of higher density of egg yolk it contains higher mass in same volume as water. hence due to higher mass higher gravitational force is acting on the egg yolk therefore it goes down on the inside the water. water will float upon the egg yolk. same situation we have seen when we spread oil in the water. ( in that case water has higher density than oil. thats why oil floats on the water).

Given,

Height = 3 m

Width = 20 m

length= 30 m

Density of air = 1.29kg/m³

The  volume of the room = 3×20×30 m³

Volume V = 1800m³

By formula,

Density = Mass/Volume

1.29kg/m³ = Mass/1800m³

Mass of the air = 1.29×1800 = 2322 Kg

The mass of the air is classroom is 2322Kg.

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Answer:

  3.0 m/s

Explanation:

The equation of motion for constant acceleration (a) is ...

  [tex]x(t)=\dfrac{1}{2}at^2 +v_0t+x_0[/tex]

The problem statement tells us v₀ = 1, and we read from the graph that x₀=1. We also read from the graph that x(10) = 21. Filling these values into the equation, we can find a and x'(10).

  [tex]21 = \dfrac{1}{2}a(10^2)+1(10)+1\\\\10=50a\qquad\text{subtract 11}\\\\a=\dfrac{1}{5}\\\\x'(t)=at+v_0\\\\x'(10)=\dfrac{1}{5}\cdot 10+1=3.0[/tex]

The final velocity of the object at t=10 s is about 3.0 m/s.

_____

Comment on the graph

We note this graph better represents increasing acceleration than it does constant acceleration. x(2) = 3.4 per the equation. It is graphed as about 4.

Answer:

thanks for de points

Explanation:

Answer: ummm arent you in my physical fundations class lol

Explanation:

are you or am i tripping? and sorry i dont have the answer :(

Common reasons people give for not being active include not having enough time, finding physical activity inconvenient, lacking self-motivation, etc.

Any consensual bodily mobilization generated by skeletal muscles that necessitate caloric expenditure is characterized as physical activity.

Physical activity includes all activities of any frequency, at any moment of day or night. It incorporates both strength training and incidental activity into the daily routine.

Physical activity can improve your brain health, help you manage your weight, lower your risk of disease, strengthen your bones and muscles, and improve your ability to do everyday tasks.

Adults who sit less and engage in moderate-to-vigorous physical activity reap health benefits.

Internal barriers were classified into three types namely, a lack of energy, a lack of motivation, and a lack of self-efficacy.

External barriers were also classified into three categories namely a lack of resources, a lack of social support, and a lack of time.

Thus, these are some reasons that may not prepare to be safe and comfortable when they participate in physical activity

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Answer:

2.11 m/s

Explanation:

Take north to be positive and south to be negative.

Average velocity = displacement / time

v = (82 m + -44 m) / (14 s + 4 s)

v = 2.11 m/s

The velocity is positive, so it is 2.11 m/s north.  The magnitude of the velocity is 2.11 m/s.

Answer:

The wavelength of the wave is 1 m

Explanation:

Given;

mass of the string, m = 20 g = 0.02 kg

length of the string, L = 3.2 m

tension on the string, T = 2.5 N

the frequency of the wave, f = 20 Hz

The velocity of the wave is given by;

[tex]v = \sqrt\frac{T}{\mu} {}[/tex]

where;

μ is mass per unit length = 0.02 kg / 3.2 m

μ = 6.25 x 10⁻³ kg/m

[tex]v = \sqrt{\frac{T}{\mu} } \\\\v = \sqrt{\frac{2.5}{6.25*10^{-3}} } \\\\v = 20 \ m/s[/tex]

The wavelength of the wave is given by;

λ = v / f

λ = (20 m/s )/ (20 Hz)

λ = 1 m

Therefore, the wavelength of the wave is 1 m

Answer:

[tex] \boxed{\sf Distance \ travelled = 12 \ m} [/tex]

Given:

Initial speed (u) = 0 m/s (Accelerates from rest)

Final speed (v) = 8 m/s

Time taken (t) = 3 seconds

To Find:

Distance travelled by cyclist (s)

Explanation:

From equation of motion of object moving with uniform acceleration in straight line we have:

[tex] \boxed{ \bold{s = (\frac{v + u}{2} )t}}[/tex]

By substituting value of v, u & t in the equation we get:

[tex] \sf \implies s = ( \frac{8 + 0}{2} ) \times 3 \\ \\ \sf \implies s = \frac{8}{2} \times 3 \\ \\ \sf \implies s = 4 \times 3 \\ \\ \sf \implies s = 12 \: m[/tex]

[tex] \therefore[/tex]

Distance travelled by cyclist (s) = 12 m

Answer:

s(distance) =36m

Explanation:

u(initial velocity) =0 m/s

a =8 m/s^2

t=3s

s=ut+1/2at^2

s=1/2(8)(3)^2

s=1/2(8)(3)(3)

s=4(9)

s=36m

Answer:

As we know that,

Now, come to the question -

[tex] \\ \implies \sf \: 2536 \times 1.66667E-5 m/s \\ \\ \\ \implies \sf \blue{0.0422666667 \:m/s } \\ [/tex]

Result : 2536 mm/min = 0.0422666667 m/s.

Answer:

Control groups let the one who is expermenting compare  the effect of the varibles in the expermental group.

Explanation:

Given:-

To Find: Speed of it.

We know,

v = s/t

where,

Putting the values,

v = (20 m)/(4 s)

→ v = 5 m/s (Ans.)

The speed of an object will be "5 m/s".

The given values are:

Distance,

Time,

As we know, the formula:

→ [tex]Distance (d)= Speed (v)\times Time (t)[/tex]

or,

→ [tex]Speed (v) = \frac{Distance (d)}{Time (t)}[/tex]

By putting the values, we get

→                [tex]= \frac{20}{4}[/tex]

→                [tex]= 5 \ m/s[/tex]

Thus the above answer is correct.

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Answer:

vector

Explanation:

The position of a particle is vector

Answer:

The value is [tex]\theta = 34.8 ^o[/tex]

Explanation:

Generally the time it takes the ball to move from maximum height to the ground is

[tex]t = \frac{v - u}{g}[/tex]

Here u= 0 m/s and v = 15 m/s

So

[tex]t = \frac{15 - 0}{9.8}[/tex]

=> [tex]t = 1.53 \ s [/tex]

Generally the maximum height traveled in time t is mathematically represented as

[tex]s = u * t + \frac{1}{2} * g * t^2[/tex]

=> [tex]s = 0 * 1.53 + \frac{1}{2} * 9.8 * 1.53^2[/tex]

=> [tex]s = 11.5 \ m [/tex]

The interval in distance between the maximum height and the initial position of the ball is

[tex]d = s - k[/tex]

Here k = 5 m (given)

[tex]d = 11.5 - 5[/tex]

[tex]d = 6.5 \ m [/tex]

Generally from the kinematic equation we have that

[tex]d = ut_1 + \frac{1}{2} g * t_1 ^2[/tex]

Here [tex] t_1 [/tex] is the time taken to travel through d

[tex]t_1 = \sqrt{ \frac{2 * d }{g} }[/tex]

=> [tex]t_1 = \sqrt{ \frac{2 * 6.5 }{9.8} }[/tex]

=> [tex]t_1 = 1.15 \ s [/tex]

Generally the total time for this fight is mathematically represented as

[tex]t_f = t + t_1[/tex]

=> [tex]t_f = 1.15 + 1.53[/tex]

=> [tex]t_f = 2.68 \ s [/tex]

Generally the vertical force exerted on the cube is mathematically represented as

[tex]F = m * g * sin (\theta )[/tex]

=> [tex] m * a = m * g * sin (\theta )[/tex]

=> [tex] a = g * sin (\theta )[/tex]

Now a is mathematically represented as

[tex]a = \frac{v - u }{t_f}[/tex]

=> [tex]a = \frac{15 - u }{2.68}[/tex]

=> [tex]a = 5.597 \ m/s^2[/tex]

So

[tex] 5.597 = 9.8 * sin (\theta )[/tex]

[tex]\theta = sin^{-1} [\frac{5.597}{9.8} ][/tex]

[tex]\theta = 34.8 ^o[/tex]

Answer:

A. Using

E=V/d

= 600/4.49*10^-3

= 1.336 x10^5 N/C

b) F = E*q = 1.33610^5 x 1.6*10^-19

= 2.17 x 10^-14 N

c) Work = Fs distance = 2.17 x 10^-14 N (4.49-3.14)*.001= 1.35 x 10^-17 J

Answer:

[tex]Velocity = \frac{s(8)}{8}[/tex]

Explanation:

Given

[tex]0 \leq t \leq 12[/tex]

Required

Determine the velocity when t = 8

This type of velocity is referred to as an instantaneous velocity.

In this case, it is calculated using

[tex]Velocity = \frac{Distance\ at\ 8 second}{t = 8}[/tex]

Given that s(t) models the distance;

s(8) = distance at 8 seconds

So;

[tex]Velocity = \frac{s(8)}{8}[/tex]

Option B answers the question

Answer: Synthetic polymers are lightweight.

Explanation:

A  benefit of using synthetic polymers is the fact that synthetic polymers are lightweight.

A polymer is a molecule composed of many repeating subunits.

Synthetic polymers are artificial polymers created by humans.

Most of the synthetic polymers are not biodegradable (unlike natural fibers such as cotton).

Synthetic polymers are classified according to their use into plastics, elastomers and synthetic fibers.

The advantages of synthetic polymers include: hard to break, being lightweight, and they last for a long time.

In conclusion, a benefit of using synthetic polymers is the fact that synthetic polymers are lightweight.

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Answer:

D) Humans

Explanation:

Photosynthesis is a useless ability without some way of exposing yourself to as much of the Sun's energy as possible. That requires a large surface area, relative to their volume. Plants achieve that with large, horizontal, light-capturing surfaces – leaves.

they get their energy from photosynthesis, so the answer would be (B) the sun

Answer: I HOPE IT HELPS

Use this Speed Calculator to calculate the speed you have travelled based ... (so add three 0s to your kilometres to get metres) and a mile is 1609 metres. ... You will have distance per time such as kilometres per hour or metres per second. ... If you want help with that, click on hyperlink for metres or for minutes and seconds.

Explanation:

Answer:

(a)  the frequency of the lowest note the piccolo can sound is 571.7 Hz

(b) the distance between adjacent antinodes is 5.72 cm

Explanation:

(a)

Given;

length of piccolo, L = 30 cm = 0.3 m

the speed of sound in air is 343 m/s

The wavelength of a pipe open at both ends, for the first harmonic is given;

L = A → N + N → A

L = λ / 4 +  λ / 4

L = λ / 2

λ = 2L

λ = 2 x 0.3 = 0.6 m

The fundamental frequency (lowest frequency) is given  by;

f₀ = v / λ

f₀ = (343 / 0.6)

f₀ = 571.7 Hz

(b)

Given;

highest note, f = 3000 Hz

the distance between adjacent antinodes is given by;

[tex]d = \frac{v}{2f}\\\\ d = \frac{343}{2*3000}\\\\ d = 0.0572 \ m\\\\d = 5.72 \ cm[/tex]

Answer:

v =  57.2 m/s

Explanation:

The average velocity of the train can be defined as the total distance covered by the train divided by the time taken by the train to cover that distance. Therefore, we will use the following formula to find the average velocity of the train:

v = s/t

where,

s = distance covered = 460 km = (460 km)(1000 m/1 km) = 4.6 x 10⁵ m

t = time taken to cover the distance = 2 h 14 min

Now, we convert it into minutes:

t = (2 h)(60 min/1 h) + 14 min

t = 120 min + 14 min = (134 min)(60 s/1 min)

t = 8040 s

Therefore, the value of velocity will be:

v = (4.6 x 10⁵ m)/8040 s

v =  57.2 m/s

Answer:

Have you tried google

Explanation:

Answer:

protect against whisplasg injuries

Answer:

Straight line in the direction of the tangential velocity the ball had at the moment the string broke

Explanation:

After the string breaks, the ball now disconnected from the centripetal force that was exerted via the string, continues its travel in a straight line in the direction of the tangential velocity it had at the moment the string broke.

Answer:

B

Explanation:

just took the test :)

Answer:

1.) 11 km/s

2.) 9.03 × 10^-5 metres

Explanation:

Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.

Electron q = 1.6×10^-19 C

Electron mass = 9.11×10^-31 Kg

(a) What is the speed of the electron 1.3 ns after entering this region?

E = F/q

F = Eq

Ma = Eq

M × V/t = Eq

Substitute all the parameters into the formula

9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19

V = 7.68×10^-18 /7.0×10^-22

V = 10971.43 m/s

V = 11 Km/s approximately

(b) How far does the electron travel during the 1.3 ns interval?

The initial velocity U = 64 km/s

S = ut + 1/2at^2

S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2

S =8.32×10^-5 + 7.13×10^-6

S = 9.03 × 10^-5 metres