unit of speed is derived unit..why??
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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.

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.

Answer:

The correct statement must be: both balls hit the floor at the same time

Explanation:

This is a kinematics exercise. The ball thrown horizontally does not have vertical speed and the ball that is released does not have vertical speed, therefore both take the same time to reach the ground, if we neglect the air resistance

The correct statement must be: both balls hit the floor at the same time

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.

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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:

No there won't be.

Explanation:

An insulator does not conduct electricity readily. An insulator that is charged at one end will have the charges remain at that end, since it does not allow the free flow of charges through it. If you touch the other end of the insulator that is void of electric charges, there would be no charge transfer between you and the insulator, and hence no electric shock.

Answer: 125

Explanation: use the equation d=vt to find the first distance, 50. Then use the equation

X=Vot + 1/2at to find the second distance. Then add both together.

Vo=5

V=5

T=10

A=5

Answer:

√82/9 m/s

Explanation:

Let's say x²+1²= y²(y length of rope)

Using Pythagoras

With repect to t diff.

So 2x dx / dt= 2y dy / dt

But dy/ dt = 1.4m/s

X = 9 so from Pythagoras y = √82

So 9dx/dt = √82

So the boat will approved de dock at √82/9 m/s

Answer:

100%

Explanation:

The accuracy level can be measured by the percent of error which is defined as the average accepted 100%.

Accuracy is related to the proximity of the measurement. It is called the true value of the accuracy where as the precision is referred to the ability at the same time to be reproduced in measurement .

The instrument should be proper and produce the reliable and accurate measurement tools.

Thus here the statistical quantity that is used by the to indicate the accuracy level is purely 100%.

Answer:

C.

Explanation:

The element that is most likely to be shiny is calcium. The correct option is 3.

Calcium (Ca) is the most likely to be sparkly of the elements shown. Calcium is a metallic element with a glossy or shining look when newly cut or polished.

It belongs to the alkaline earth metals group, which have metallic characteristics and a distinctive lustre.

Sulphur (S), boron (B), and carbon (C), on the other hand, are non-metallic elements with a dull look.

They come in a variety of forms, including powders, crystals, and amorphous solids, although they are not noted for their metallic lustre.

Thus, the answer is 3. calcium.

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#SPJ7

Your question seems incomplete, the probable complete question is:

Which element is most likely to be shiny?

Explanation:

4/2.5 = 1.6 miles/hour

Answer:

1.4 m/s

Explanation:

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

Initial Displacement (d1) = 0.9 m

Final Displacement (d2) = 1.6 m

Initial time (t1) = 1.5 secs

Final time (t2) = 2 secs

Velocity (v) =..?

The velocity of an object can be defined as the rate of change of the displacement of the object with time. Mathematically, it can be expressed as follow:

Velocity = change of displacement /time

v = Δd / Δt

Thus, with the above formula, we can obtain the velocity of the car as follow:

Initial Displacement (d1) = 0.9 m

Final Displacement (d2) = 1.6 m

Change in displacement (Δd) = d2 – d1 = 1.6 – 0.9

= 0.7 m

Initial time (t1) = 1.5 secs

Final time (t2) = 2 secs

Change in time (Δt) = t2 – t1

= 2 – 1.5

= 0.5 s

Velocity (v) =..?

v = Δd / Δt

v = 0.7/0.5

v = 1.4 m/s

Therefore, the velocity of the car is 1.4 m/s

Answer:

14.7 m

Explanation:

It would fall 14.7 m in 3 s . You multiply 4.9 by 3.

Answer:

It's d on edg

Explanation:

Answer:

a

[tex]N_e  =  906 \  electrons[/tex]

b

[tex]M =  9.43 *10^{-25} \  kg[/tex]

Explanation:

From the question we are told that

   The total number of particles is  n =  1470 particles

    The  the total amount of charge on the particles is  [tex]Q =  - 5.456*10^{-17} \  C[/tex]

Generally this total number of particles can be mathematically represented as

      [tex]n  =  N_p + N_e[/tex]

Where  [tex]N_p \  and  \  N_e[/tex] represent the number of proton and electron respectively

=>    [tex]N_p  =  1470  -  N_e[/tex]

Also the total charge of these particles can be mathematically represented as

       [tex]Q = (N_p  - N_e ) e[/tex]

Here e is the charge on a single electron or proton with value  

The  negative sign is due to the fact the electrons are negative signed

     [tex]e =  1.60 *10^{-19} \  C[/tex]

[tex]-5.456*10^{-17} = ( 1470 -2N_e )* 1.60 *10^{-19}  [/tex]

    [tex]N_e = 906[/tex]

Thus  

      [tex]N_p  =  1470 - 906[/tex]

      [tex]N_p  =  564[/tex]

Generally the mass of the system is mathematically represented as

     [tex]M  =  N_e  * M_e  +  N_p  *  M_p[/tex]

Here  [tex]M_e  \  and  \  M_p  \  are  \  mass  \  of  \  electron \  and \  proton \  with \  values \\   M_e = 9.1 *10^{-31} \  kg \  and  \  M_p  =  1.67 *10^{-27} \  kg  \ respectively[/tex]

So

     [tex]M  =  906 *  9.1 *10^{-31}  +  564  *  1.67 *10^{-27}[/tex]

      [tex]M =  9.43 *10^{-25} \  kg[/tex]

Answer:

a

[tex]I  =  6637 \  W/m^2[/tex]

b

[tex]E_{max} =  500 \ N/m[/tex]

And

[tex]B_{max} =  1.67*10^{-6} \  T[/tex]

Explanation:

From the question we are told that  

   The  power is  [tex]P =  75 \  W[/tex]

   The  diameter is  [tex]d =  6.0 \ cm =  0.06 \  m [/tex]

Generally the radius is mathematically represented as

      [tex]r =  \frac{d}{2}[/tex]

=>   [tex]r =  \frac{ 0.06}{2}[/tex]

=>   [tex]r =  0.03 \  m[/tex]

Generally the area of the sphere is mathematically evaluated as

          [tex]A = 4 \pi r^2[/tex]

=>      [tex]A = 4 * 3.142 *  (0.03)^2[/tex]

=>       [tex]A =  0.0113 \ m^2[/tex]

Generally the total  Intensity of the  incandescent light bulb is mathematically represented as

     [tex]I= \frac{P}{A}[/tex]

=>    [tex]I  =  \frac{75}{ 0.0113}[/tex]

=>   [tex]I  =  6637 \  W/m^2[/tex]

Given that 5%  of the energy goes to visible  light

Then the intensity that goes visible light is  

    [tex]I_v  =  0.05 * 6637[/tex]

    [tex]I_v  =  332 \ W/m^2[/tex]

The amplitude of the electric field at the surface is mathematically represented as

      [tex]E_{max} =  \sqrt{\frac{2 * I_v}{\epsilon_o *  c } }[/tex]

=>   [tex]E_{max} =  \sqrt{\frac{2 * 332}{ 8.85*10^{-12} *  3.0*10^8} }[/tex]

=>     [tex]E_{max} =  500 \ N/m[/tex]

The amplitude of the magnetic  field at the surface is mathematically represented as

     [tex]B_{max} =  \frac{E_{max}}{c}[/tex]

=>     [tex]B_{max} =  \frac{ 500}{3.0*10^8}[/tex]

=>   [tex]B_{max} =  1.67*10^{-6} \  T[/tex]

Answer:

The answer is false

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:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

[tex]v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt[/tex]

[tex]v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt[/tex]

Where:

[tex]v_{a}[/tex], [tex]v_{b}[/tex] - Initial and final velocities, measured in meters per second.

[tex]t_{a}[/tex], [tex]t_{b}[/tex] - Initial and final times, measured in seconds.

[tex]a(t)[/tex] - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

[tex]v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt[/tex]

[tex]v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)[/tex]

[tex]v_{4} = -6\,\frac{m}{s}[/tex]

Region II (t = 4 s to t = 6 s)

[tex]v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt[/tex]

[tex]v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)[/tex]

[tex]v_{6} = -4\,\frac{m}{s}[/tex]

Region III (t = 6 s to t = 10 s)

[tex]v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt[/tex]

[tex]v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)[/tex]

[tex]v_{10} = 4\,\frac{m}{s}[/tex]

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

The velocity of a body under constant acceleration increases steadily with time

Please find attached the required velocity graph for values of t that are separated by Δt = 2s

The reasons the attached graph is correct are given as follows:

The initial velocity of the object at t = 0 is v = 2.0 m/s

The first point on the graph is (0, 2.0)

The acceleration from t = 0, to t = 4, a₁ = -2 m/s²

The velocity at t = 4 s, v₂ = 2.0 + (-2)×4 = -6

Therefore, the next point on the graph is (4, -6)

From t = 4 to t = 6, the acceleration, a₂ = 1 m/s²

Therefore, v₃ = -6 + 1 × 2 = -4

The third point on the velocity graph is (6, -4)

From t = 6 s to t = 10 s, we have, the acceleration, a₃ = 2 m/s²

The velocity, v₄ = -4 + 4 × 2 = 4

Therefore, the fourth point on the velocity graph is (10, 4)

With the above points, the velocity graph can be plotted using MS Excel

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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:

Explanation:

a) the speed increment of the hammer as it drops past the first window, is greater than that of the speed of the hammer as it drops past the second window. This can also be translated as saying that the hammer spent more time at the second window.

b) III

The best answer would be answer III, The hammer spends more time dropping past window 1, which I had already included in my explanation in (a) above.

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:

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:

Explanation:

Males measuring;

161 - 165 = 2

166 - 170 = 1

171 - 175 = 7

176 - 180 = 1

181 - 185 = 6

186 - 190 = 3

Males taller than 175 cm = 1 + 6 + 3 = 10

Total number of Males = 2 + 1 + 7 + 1 + 6 + 3  = 20

Percentage of total over 175 cm

= 10 / 20

= 50%

Answer:

Explanation:

T = 2π √l/g

The dimension for l = m

The dimension for g = m/s²

The dimension for 2π is nothing. Since it's a constant, it is dimensionless.

Now we proceed ahead. Since we are not using the 2π, for the sake of this proving, our formula will temporarily be written as

T = √l/g

Inputting the dimensions, we have

T = √(m) / (m/s²)

T = √(m * s²/m)

T = √s²

T = s

Since the unit of period itself is in s, we can adjudge that the equation is dimensionally constant.

Answer:

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

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.