Determine the slope of this graph from zero seconds to five seconds.

Answer:

observe o gráfico abaixo que mostra o número de internação em um hospital no município no período de 1960

Answer:

provides a first hand account of an event or time period and are considered to be authoritative

Answer:

main source

Explanation:

Answer:

It will take 15 seconds

Complete Question

A field mouse trying to escape a hawk runs east for 5.0m, darts southeast for 3.0m, then drops 1.0m down a hole into its burrow. What is the magnitude of the net displacement of the mouse?

Answer:

The  values is  [tex]s =  7.49 \  m[/tex]

Explanation:

From the question we are told that

   The  distance it travels eastward is  [tex]x =  5.0 \ m[/tex]

   The distance it travel towards the southeast  is [tex]l  =  3.0\  m[/tex]

   The distance it travel towards the south is  [tex]z =  1 \  m[/tex]

Let x-axis  be east

      y-axis  south

       z-axis into the ground

The angle made between east and south is  [tex]\theta  =  45^o[/tex]

The displacement toward x-axis is

       [tex]x =  5 +  3cos(45)[/tex]

       [tex]x =  7.12[/tex]

 The  displacement toward the y-axis is  

     [tex]y  =  3 *  sin (45)[/tex]

      [tex]y =  2.123[/tex]

Now the overall displacement of the rat is mathematically evaluated as

        [tex]s =  \sqrt{7.12^2 +  2.12^2 +  1^2}[/tex]

      [tex]s =  7.49 \  m[/tex]

All we need to do is divide the momentum and weight to find the velocity:

v = m / w

Solve using the values we are given.

v = 952/68 = 14 m/s

Best of Luck!

Answer:

2.52 s

Explanation:

We can use equation of motion to solve this.

[tex]s \: = ut + \frac{1}{2} a {t}^{2} [/tex]

Assuming initial velocity to be 0 and no air resistance,

[tex] s \: = \frac{1}{2} a {t}^{2} \\ 31.16 = \frac{1}{2} (9.81) {t}^{2} \\ t = 2.52s[/tex]

The time of flight is 3.54 s.

There are three equation of motions that can be used when motion of the object is under constant acceleration and on a straight path.

They are listed below as:

[tex]v = u + at\\\\s = ut + \dfrac{1}{2} at^2\\\\v^2 = u^2 + 2as[/tex]

where the symbols have following meanings:

u = initial velocity of the considered object

v = final velocity of the object

a = acceleration of the object

s = distance traveled by the object in 't' time.

It is given that A woman leaning over the railing of a tall bridge accidentally drops her cell phone. If the bridge is 31.16 m high, then  how long will it take the phone to hit the river below the bridge.

We can use equation of motion to solve this;

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

where:

s = 31.16 m is the vertical displacement of the phone

u = 0 is the initial vertical velocity

Assuming initial velocity to be 0 and no air resistance,

[tex]s = 1/2 (9.8)(t)^2\\31.16 = 1/2 (9.8)(t)^2\\t = 2.52 s[/tex]

Hence, The time of flight is 3.54 s.

Learn more about free fall here:

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#SPJ2

Answer:

A)i) 1. constant,  2. constant,  3. constant,  4. decrease

   ii)  frecuency increase

  iii)   L = n /2f   √T/μ  

B)   L_b = 0.534 m

Explanation:

We can approximate the violin string as a system of a fixed string at its two ends, therefore we have a node at each end and a maximum in the central part for a fundamental vibration,

         λ = 2L / n

where n is an integer

The wavelength and frequency are related

          v = λ  f

and the speed of the wave is given by

          v = √T /μ

with these expressions we can analyze the questions

A)

i) In this case the woman decreases the length of the rope L = L₂

      therefore the wavelength changes

        λ₂ = 2 (L₂) / n

as L₂ <L₀ the wavelength is

         λ₂ < λ₀

The tension of the string is given by the force of the plug as it has not moved, the tension must not change and the density of the string is a constant that does not depend on the length of the string, therefore the speed of the string wave in the string should not change.

ii) how we analyze if the speed of the wave does not change

         v = λ  f

as the wavelength decreases, the frequency must increase so that the speed remains constant

     fy> fx

iii) It is asked to find the length of the chord

let's use the initial equations

            λ  = 2L / n

            v = λ  f

            v = 2L / n f

            v = √ T /μ

we substitute

           2 L / n f = √ T /μ

           L = n /2f   √T/μ

this is the length the string should be for each resonance

b) in this part they ask to calculate the frequency

         f = n / 2L √ T /μ

the linear density is

         μ = m / L

         μ = 2.00 10⁻³ / 60.0 10⁻²

         μ = 3.33 10⁻³ kg / m

we assume that the length is adequate to produce a fundamental frequency in each case

f_{a} = 440Hz

        λ = 2La / n

        λ = 2 0.60 / 1

        λ = 1.20 m

        v = λ f

        v = 1.20 440

        v = 528 m / s

        v² = T /μ

       T = v² μ

       T = 528² 3.33 10⁻³

       T = 9.28 10² N

Let's find the length of the chord for fb

f_{b} = 494 hz

        L_b = 1 /(2 494)  √(9.28 10² / 3.33 10⁻³)

        L_b = 0.534 m

Answer:

D

Explanation:

The Angular Position: The angular position is also called the angular displacement, and it is the angle through which a point revolves around the centre line that has been rotated in a particular manner about a given axis.

The Angular Velocity: The angular velocity of an object is the measure of the speed, or say, how fast an object rotates at one point with respect to a point.

The Angular Acceleration: This is very simple as it's a derivative of angular velocity. They both share the same relationship velocity and acceleration share. Angular acceleration is the rate of change of angular velocity.

The only options that are equal at all points on the rod are;

1. Angular position.

2. Angular velocity.

3. Angular acceleration.

Let us look at each of the options;

1) Angular Position: This is also referred to as angular displacement. In centripetal motion, it is defined as the angle through which a a body has been rotated from the initial position or reference position. This must always be equal at all points on the rod.

2) Angular Velocity: This is defined as the speed with which an object is moving in a circular motion. This must always be equal at all points on the rod.

3) Angular Acceleration: This is simply defined as the rate of change of angular velocity with time. Now, since angular velocity must be equal at all points on the rod, then angular acceleration must also be equal at all points on the rod.

4) Tangential acceleration; This is defined as the rate of change of tangential velocity with time. Tangential velocity is the linear component of the speed and it is not equal at all points on the rod and as such the tangential acceleration will also not be equal at all points on the rod.

Read more at; https://brainly.com/question/13014974

Answer:

Explanation:

A way to see this is that the definite integral of the acceleration is the change in velocity (i.e. the final velocity minus the initial velocity), and the change in velocity divided by the length of the time interval is the average acceleration on the interval.

Answer:

6?

Explanation:

Answer:

did you get the answer bc I think its -3

Explanation:

Answer:

smooth

go far

Answer:

[tex]\huge\boxed{\sf Frequency = 10\ Hz}[/tex]

[tex]\huge\boxed{\sf Time\ Period = 0.1\ secs}[/tex]

Explanation:

Given:

No. of times = 600 times

Time = 60 seconds

Required:

Frequency = f = ?

Time Period = T = ?

Solution:

Frequency = 600 / 60

Frequency = 10 Hz

Time Period = 1 / 10

Time Period = 0.1 secs

Answer:

The correct option is D

Explanation:

This question is incomplete because of the absence of the setup which as been attached below. The setup shows/determines/tests the friction of wood (which is a block material), since Jerry wants to test the friction between different types of materials, he will have to replace the wooden block with another type of block material of choice so as to determine the friction of that also.

In order to have a comprehensive experiment, Jerry can use 4-5 different types of block material in the course of the experiment.

Answer:

D

Explanation: i did it and my teacher said it was right

Answer:

It's d on edg

Explanation:

Answer:50  miles per hour 50/1hr

Explanation:100 divided by 2 is 50, divide 2 by 2 thats 1

Answer:

50 miles/hr

Explanation:

[tex]Distance = 100\: miles\\Time = 2\:hours\\\\Speed = \frac{Distance}{Time} \\\\Speed = \frac{100\:miles}{2\:hours}\\\\ Speed = 50\:miles/hour[/tex]

Answer:

Jun 29, 2016 · Tectonic plates are large segments of the Earth’s crust that move slowly. Suppose that one such plate has an average speed of 4.0 cm/year. (a) What distance does it move in 1.0 s at this speed? (b) What is its speed in kilometers per million years?

Explanation:

Answer:

Speed, [tex]v=1.6\times 10^8\ m/s[/tex]

Explanation:

Given that,

Distance covered by the electron, d = 32 cm = 0.32 m

Time, t = 2 ns

We need to find the speed of an electron. Speed is equal to distance covered divided by time. So,

[tex]v=\dfrac{0.32}{2\times 10^{-9}}\\\\v=1.6\times 10^8\ m/s[/tex]

So, the speed of the electron is [tex]1.6\times 10^8\ m/s[/tex].

Answer:

A ) [tex]5.197 \times 10 ^{-9} inches[/tex]

b) There are approximately 100,000,000 atoms on the 1.32 cm lines

Explanation:

To ensure the accuracy of our answer, we will have to make sure we work in the appropriate units.

A

To convert    [tex]1.32 \times 10^{-10} m[/tex] to inches, we can use this factor.

1 m = 39.3701 inches

Therefore [tex]1.32 \times 10^{-10} m[/tex] =  

B

To find the number of atoms on a 1.32 cm line, we will first of all need to convert 1.32 cm to meters.

1.32 cm = 0.0132 metres (divided 1.32 by 100 to convert to metres)

Assuming all the atoms are arranged side by side, with their edges touching on the 0.0132m line, the number of atoms present will be

[tex]0.0132/1.32\times 10^{-10}[/tex] = 100,000,000 atoms

Answer:

If the earth stops rotating the apparent weight will increase at most of the places because the upward component of the centrifugal force due to rotation will disappear.

Answer:

Car first should be 30.13 m behind the second car.

Explanation:

Given that,

Constant speed = 108.5 Km/hr

Time = 1 sec

Let the distance covered by second car S and by first car S'

We need to calculate the distance covered by second car

Using equation of motion

[tex]S=ut-\dfrac{1}{2}at^2[/tex].....(I)

The distance covered by first car to avoid collision

[tex]S''=S'+S[/tex]

Put the value into the formula

[tex]S'+S=ut+ut-\dfrac{1}{2}at^2[/tex]...(II)

We need to calculate the distance covered by first car

Using equation (I) and (II)

[tex]S'=ut[/tex]

Put the value into the formula

[tex]S'=108.5\times\dfrac{5}{18}\times1[/tex]

[tex]S'=30.13\ m[/tex]

Hence, Car first should be 30.13 m behind the second car.

Explanation:

4/2.5 = 1.6 miles/hour

Explanation:

Given that,

Initial speed, u = 0

Final speed, v = 8.9 m/s

Time, t = 9.5 s

Let a is the acceleration of the snowboarder. It can be calculated as follows :

[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{8.9-0}{9.5}\\\\a=0.936\ m/s^2[/tex]

The distance traveled by the snowboarder in this time is calculated using third equation of motion as follows :

[tex]v^2=u^2=2as\\\\s=\dfrac{v^2}{2a}\\\\s=\dfrac{(8.9)^2}{2\times 0.936}\\\\s=42.31\ m[/tex]

So, 42.31 m of distance is traveled by snowboarder in this time.

Answer:

The answer is false

Answer:

Clockwise direction

Explanation:

In a case of a wire carrying a current, the right hand rule is used.

The thumb in the direction of current while the finger curl around in the direction of the magnetic field.

The right hand rule applies to a current in a straight line wire.

If the direction a wire carrying a current perpendicularly into this page, the direction of the magnetic field will be in a clockwise direction .

Answer:

[tex]1.0\; \rm m \cdot s^{-1}[/tex].

Explanation:

Consider a time period of duration [tex]\Delta t[/tex]. Let change in the position of an object during that period of time be denoted as [tex]\Delta x[/tex]. The average velocity of that object during that period would be:

[tex]\displaystyle \bar{v} = \frac{\Delta x}{\Delta t}[/tex].

For the toy car in this question, the time interval has a duration of [tex]2.0 \; \rm s - 0.5\; \rm s = 1.5\; \rm s[/tex]. That is: [tex]\Delta t = 1.5\; \rm s[/tex]. (One decimal place, two significant figures.)

On the other hand, what would be the change in the position of this toy car during that [tex]1.5\; \rm s[/tex]?

Note, that from readings on the snapshot in the diagram:

Therefore, the change to the position of this toy car over that time period would be [tex]\Delta x = 1.6\; \rm m - 0.1\; \rm m = 1.5\; \rm m[/tex]. (One decimal place, two significant figures.)

The average velocity of this car over this period of time would thus be:

[tex]\displaystyle \bar{v} = \frac{\Delta x}{\Delta t} = \frac{1.5\; \rm m}{1.5\; \rm s} = 1.0\; \rm m \cdot s^{-1}[/tex]. (Two significant figures.)

Answer:

-5%

Explanation:

We know that volume is mass/density

So to find percentage

Let first mass be = M1

Second be 0.94m( 1-6%)

Second density be=0.99

First density be d

So finlq volume v2= 0.94m/0.99p=

V2= (0.94/0.99)V1

V2= 0.95v1

So volume decrease will be 1-0.95= 0.05

= -5%

Answer:

The field of view is reduced.

Explanation:

Given that,

The field of view for every resultant magnification like you change objectives from 4 to 10 to 43.

We know that,

Field of view :

When the view is observed at a point in a defined field then these field called field of view.

The normal angle of  field of view is 90°.

The formula of field of view is define as,  

[tex]field\ of\ view = \dfrac{field\ number}{magnification}[/tex]

We can say that,

The field of view is inversely proportional to the magnification.

When magnification is low then field of view will be large.

When magnification is higher then field of view will be small .

According to question,  

When the magnification adjust from 4 to 10 to 43, the field of view is reduced.    

Hence, The field of view is reduced.