The final step of the decision-making model is: A-describe the situation B-weigh the pros and cons C-take action D-review your decision

Answer:

(a) 1, average velocity = -65.6 m/s

   2, average velocity = -64.8 m/s

   3, average velocity = -64.16 m/s

(b) The instantaneous velocity is -96 m/s

Explanation:

(a)

Average velocity is given  by;

[tex]y(t_2,t_1) = \frac{y(t_2) - y(t_1)}{t_2-t_1}[/tex]

(1)

[tex]y(2.1,2) = \frac{(235-16*2.1^2) - (235-16*2^2)}{2.1-2}\\\\ y(2.1,2) = -65.6 \ m/s[/tex]

(2)

[tex]y(2.05,2) = \frac{(235-16*2.05^2) - (235-16*2^2)}{2.05-2}\\\\ y(2.05,2) = -64.8 \ m/s[/tex]

(3)

[tex]y(2.01,2) = \frac{(235-16*2.01^2) - (235-16*2^2)}{2.01-2}\\\\ y(2.01,2) = -64.16 \ m/s[/tex]

b. y = 235 - 16t²

The instantaneous velocity is given by;

v = dy /dt

dy / dt = -32t

when t = 3 s

v = -32(3)

v = -96 m/s

(a)The average velocity for the 0.1 sec,0.05 sec,0.01 sec will be  -65.6 m/s, -64.8 m/s and -64.16 m/s respectively.

The total displacement traveled by an object divided by the total time taken is the average velocity.

Average velocity is given by;

[tex]\rm y(t_2,t_1)=\frac{y(t_2)-y(t_1)}{t_2-t_1} \\\\[/tex]

The average velocity for case 1;

[tex]\rm y(2.1,2)=\frac{(235-16\times 2\times 1^2)-(235-16 \times 2^2)}{2.1-2} \\\\ \rm y(2.1,2)=\-65.6 \ m/sec[/tex]

The average velocity for case 2;

[tex]\rm y(2.015,2)=\frac{(235-16\times 2.05^2)-(235-16 \times 2^2)}{2.05-2} \\\\ \rm y(2.1,2)=\-64.8 \ m/sec[/tex]

The average velocity for case 3;

[tex]\rm y(2.01,2)=\frac{(235-16\times 2.01^2)-(235-16 \times 2^2)}{2.01-2} \\\\ \rm y(2.1,2)=\-64.16 \ m/sec[/tex]

Hence the average velocity for the 0.1 sec,0.05 sec,0.01 sec will be  -65.6 m/s, -64.8 m/s and -64.16 m/s respectively.

(b) The instantaneous velocity will be -96 m/s.

The given equation in the problem is;

[tex]\rm y = 235 - 16t^2[/tex]

The instantaneous velocity is given as;

[tex]v = \frac{dy}{dt} \\\\ \frac{dy}{dt} = -32t\\\\ t = 3 s\\\\ v = -32\times 3\\\\ v = -96 m/s[/tex]

Hence the instantaneous velocity will be -96 m/s.

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Answer:

The new charge on the second object is q₂ - q₁

Explanation:

Given;

charge on the first object,  q₁

charge on the second object, q₂

Since there is no external force on isolated system, the total charge in the system is given by;

Q = q₁ + q₂

q₂ = Q - q₁

When the charge on object 1 is doubled, q₁' = 2q₁

let q₂' be the new charge on object 2

q₂' = Q - q₁'

q₂' = (q₁ + q₂) - 2q₁

q₂' = q₁ + q₂ - 2q₁

q₂' = q₂ - q₁

Therefore, the new charge on the second object is q₂ - q₁

The charge on object 2 will be "q₂ - q₁".

The charge's quantity on such an item represents the degree of imbalance between its electrons as well as protons.

Just to calculate the overall charge of a positively (+) charged item, divide the total no. of valence electrons by the overall number of protons.

According to the question,

Object 1's charge = q₁

Object 2's charge = q₂

Let,

New charge of object 2 be "q₂'".

The total charge in the system will be:

Q = q₁ + q₂

or,

q₂ = Q - q₁

hence,

The charge on object 2 be:

→ q₂' = Q - q₁'

By substituting the values,

        = (q₁ + q₂) - 2q₁

        = q₁ + q₂ - 2q₁

        = q₂ - q₁                

Thus the answer above is appropriate.

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Answer:

μN/ (kg ns) = 10³ N / (kg s)

Explanation:

In this exercise they ask us if the notation is correct. Let's write the different terms in the SI systems

force is N

time is in seconds

the unit given for the force is 1 N = 10⁶ μN

the unit of time is 1 s = 10⁹ ns

the correct way to give the answer should be: N / (kg s)

so the notation should be changed

        μN /kg ns = μN / (kg ns) (1N / 10⁶ μN) (10⁹ ns / 1 s) =

        μN/ (kg ns) = 10³ N / (kg s)

Answer:

1.    Kicking a soccer ball

2.      Playing tug of war  

3.         Bouncing a Ball

Explanation:

Answer:

B, and D

Explanation:

The two benefits of this model is it specify a process that is very complex. It can represent changes that occur very slowly. Thus option B and D is correct.

The water cycle is defined as a cycle of events that involves precipitation as rain and snow, drainage in streams and rivers, and return to the atmosphere by evaporation and transpiration. Water moves through this cycle between the earth's oceans, atmosphere, and land.

It can also be defined as the route all water takes as it travels around Earth in various conditions.

There are basically six stages of water cycle.

Thus, the two benefits of this model is it specify a process that is very complex. It can represent changes that occur very slowly. Thus option B and D is correct.

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Answer: The amount of matter is higher in a bus than a ball.

Explanation:

Answer:

Explanation:

A. We know that 1microsecond= 10^-6s

So = 9.12*10^-6=9.12*10^-6s

B 1km= 1000m

So 3.42km= 3240m

C.1ms= 10^-3s

And 1cm= 10^-2m

So 44*10^-2m/10^-3s=440m/s

D. 80km/hr= 80*1000/3600= 22.2m/s

Because 1km= 1000m

1hr= 3600s

Answer:

With changing speed and/or in a circle

Answer:

force is the amount of work or pressure given to an object

motion is the action of moving one place to another place

Answer:

yes bc they are falling :]

Explanation:

Answer:

Zero.

Explanation:

Acceleration is defined as change in velocity per unit time. For constant, the acceleration is zero.

as [tex]\delta[/tex][tex]v\\[/tex] is zero.

Answer:

The new water temperature is 26.4 °C

Explanation:

Given;

mass of copper, [tex]M_{cu}[/tex] = 26 g = 0.026 kg

temperature of copper, t = 300 °C

volume of water, V = 120 mL = 0.12 L

temperature of water, t = 21 °C

density of water, ρ = 1 kg/L

mass of water = density x volume

mass of water = (1 kg/L) x 0.12 L = 0.12 kg

heat lost by copper = heat gained by water

Both copper and water reach final temperature, T

Heat gained by water, [tex]Q_w[/tex] = [tex]m_w[/tex]cΔθ = [tex]m_w C(T - t)[/tex]

[tex]Q_w = m_w C(T - t)\\\\Q_w = 0.12*4200(T-21)\\\\Q_w = 504(T-21)[/tex]

Heat lost by copper is given by;

[tex]Q_{cu} = m_{cu}C(300-T)\\\\Q_{cu} = 0.026*385(300-T)\\\\Q_{cu} = 10.01(300 - T)[/tex]

[tex]Q_{cu} = Q_w[/tex]

504(T- 21) = 10.01(300 - T)

504 T - 10584 = 3003 - 10.01 T

504 T + 10.01 T= 3003 + 10584

514.01 T = 13587

T = (13587) / 514.01

T = 26.4 °C

Therefore, the new water temperature is 26.4 °C

The new water temperature using the given parameters is;

26.49°C

We are given;

Mass of copper; m_cu = 26 g = 0.026 kg

Temperature; T_cu = 300 °C

Volume of water; V = 120 mL = 0.12 L

Temperature of water, T_w = 21 °C

Density of water; ρ = 1 kg/L

Let us find the mass of water from the formula;

m_w = ρ × V

m_w = 1 × 0.12

m_w = 0.12 kg

Now, from the principle of conservation of energy, we can say that the total heat lost by a hot body is equal to the total heat gained by a cold body.

The hot body here is copper while the cold body is water. Thus;

Heat lost by copper = Heat gained by water

Formula for heat lost by copper is;

Q_cu = m_cu * c_cu * (T_f - T_cu)

Formula for heat gained by water is;

Q_w = m_w * c_w * (T_f - T_w)

Where;

T_f is final temperature reached by copper and water

c_cu is specific heat capacity of copper = 385 J/kg.°C

c_w is specific heat capacity of water = 4200 J/kg.°C

Thus;

Q_cu = 0.026 × 385 × (300 - T_f)

Q_cu = 3030 - 10.01T_f

similarly;

Q_w = 0.12 × 4200 × (T_f - 21)

Q_w = 504T_f - 10584

Thus;

3030 - 10.01T_f = 504T_f - 10584

3030 + 10584 = 504T_f + 10.01T_f

13614 = 514.01T_f

T_f = 13614/514.01

T_f = 26.49°C

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Answer:

Explanation:

Given the position of a particle  expressed by the equation x = (2t^3 - 6t^2 + 12)m, where t is in seconds, the acceleration function can be gotten by taking the second derivative of the function with respect to t as shown;

a = d/dt(dx/dt)

First let us get dx/dt

dx/dt = 3(2)t³⁻¹-2(6)t²⁻¹+0

dx/dt = 6t²-12t

a = d/dt(dx/dt)

a = d/dx(6t²-12t)

a = 2(6)t²⁻¹-12t¹⁻¹

a = 12t - 12t⁰

a = 12t-12

If the acceleration is zero, then;

12t-12 = 0

add 12 to both sides

12t-12+12 = 0+12

12t = 12

t = 12/12

t = 1sec

Hence the time when acceleration is zero is 1sec

Displacement is a vector magnitude that depends on the position of the body which is individualistic for the trajectory.

While, Distance is a scalar magnitude that measures over the trajectory.

Answer:

When conducting research, scientists use the scientific method to collect measurable, empirical evidence in an experiment related to a hypothesis (often in the form of an if/then statement), the results aiming to support or contradict a theory.

Use the formula,

[tex]\Delta x=v_it+\dfrac12at^2[/tex]

where [tex]\Delta x[/tex] is the cart's displacement (from the origin), [tex]v_i[/tex] is its initial speed, [tex]a[/tex] is its acceleration, and [tex]t[/tex] is time.

[tex]\Delta x=\left(2.0\dfrac{\rm m}{\rm s}\right)(5.0\,\mathrm s)+\dfrac12\left(1.6\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)^2[/tex]

[tex]\implies\boxed{\Delta x=30\,\mathrm m}[/tex]

Alternatively, since acceleration is constant, we have

[tex]\dfrac{v_f+v_i}2=\dfrac{\Delta x}t[/tex]

That is, we have these two equivalent expressions for average velocity, where [tex]v_f[/tex] is the cart's final velocity. Solve for [tex]\Delta x[/tex]:

[tex]\dfrac{10\frac{\rm m}{\rm s}+2.0\frac{\rm m}{\rm s}}2=\dfrac{\Delta x}{5.0\,\mathrm s}[/tex]

[tex]\implies\boxed{\Delta x=30\,\mathrm m}[/tex]

Answer:

Going by EM SPECTRUM WE HAVE

radio waves, microwaves, infrared, VISIBLE LIGHT, ultraviolet, X-rays, GAMMA RAYS

Explanation:

BECAUSE

V= WAVELENGTH/ FREQUENCY

AS FREQUENCY INCREASES WAVELENGTH DECREASE AN VICE VERSA

Given :

The acceleration of a piston is given by :

[tex]a=r\omega^2cos\ (\omega t-\dfrac{\pi}{4})[/tex]

To Find :

The amplitude of the acceleration of piston .

Solution :

Now , acceleration is :

[tex]a=r\omega^2cos\ (\omega t-\dfrac{\pi}{4})[/tex]

Now , amplitude is maximum of any function .

So , maximum value of [tex]cos\ \theta[/tex] is equal to 1 .

Therefore , amplitude of the acceleration of piston is :

[tex]A=r\omega^2[/tex]

Hence , this is the required solution .

Answer:

SI unit is used in most places around the world, so our use of it allows scientists from disparate regions to use a single standard in communicating scientific data without vocabulary confusion.

Explanation:

hope it helps

Si unit is used almost all around the world, so our use of it allows

Scientist from disparate regions to use a single standered in communicating scientific data without vocabulary confusion

Answer:

The true statements are:

c) Temperature differing by 25 on the Fahrenheit scale must differ by 45 on the Celsius scale.

d) Temperatures which differ by 10 on the Celsius scale must differ by 18 on the Fahrenheit scale.

Explanation:

option a is false, 90°C is equal to 194°F which is not warmer than 202°F

option b is false, 40K is equal to -233°C

option e is false 0°F corresponds to -17.8°C.

Answer:

The order of magnitude of the distance from the sun to Earth is 10⁸ km.

Explanation:

The order of magnitude of the distance from the sun to Earth can be calculated as follows:

[tex] c = \frac{x}{t} [/tex]

Where:

c: is the speed of light = 3x10⁸ m/s

t: is the time = 8 min

Hence, the distance is:

[tex] x = c*t = 3 \cdot 10^{8} m/s*8 min*\frac{60 s}{1 min} = 1.44 \cdot 10^{11} m = 1.44 \cdot 10^{8} km [/tex]

Therefore, the order of magnitude of the distance from the sun to Earth is 10⁸ km.

I hope it helps you!

Answer:

 F = 0.111015 N

Explanation:

For this exercise the force is given by Coulomb's law

        F = k q₁q₂ / r₂₁²

we calculate the electric force of the other two particles on the charge q1

Charges q₁ and q₂

the distance between them is

          r₁₂ = y₁ -y₂

          r₁₂ = 0.30 + 0.30

          r₁₂ = 0.60 m

let's calculate

          F₁₂ = 9 10⁹ 2 10⁻⁶ 2 10⁻⁶ / 0.60 2

          F₁₂ = 1 10⁻¹ N

directed towards the positive side of the y-axis

Charges 1 and 3

Let's find the distance using the Pythagorean Theorem

             r₁₃ = RA [(0.40-0) 2 + (0-0.30) 2]

             r₁₃ = 0.50 m

            F₁₃ = 9 10⁹ 2 10⁻⁶ 4 10⁻⁶ / 0.50²

            F₁₃ = 1.697 10⁻² N

The direction of this force is on the line that joins the two charges (1 and 3), let's use trigonometry to find the components of this force

           tan θ = y / x

           θ = tan⁻¹ y / x

          θ = tan⁻¹ 0.3 / 0.4

           tea = 36.87º

    The angle from the positive side of the x-axis is

         θ ’= 180 - θ

        θ ’= 180 - 36.87

        θ ’= 143.13º

       sin143.13 = F_13y / F₁₃

           F_13y = F₁₃ sin 143.13

           F{13y} = 1.697 10⁻² sin 143.13

           F_13y = 1.0183 10⁻² N

            cos 143.13 = F_13x / F₁₃

           F₁₃ₓ = F₁₃ cos 143.13

           F₁₃ₓ = 1.697 10⁻² cos 143.13

           F₁₃ₓ = -1.357 10-2 N

Now we can find the components of the resultant force

          Fx = F13x + F12x

          Fx = -1,357 10-2 +0

          Fx = -1.357 10-2 N

          Fy = F13y + F12y

         Fy = 1.0183 10-2 + ​​1 10-1

          Fy = 0.110183 N

We use the Pythagorean theorem to find the modulus

         F = Ra (Fx2 + Fy2)

         F = RA [(1.357 10-2) 2 + 0.110183 2]

         F = 0.111015 N

Let's use trigonometry for the angles

         tan tea = Fy / Fx

          tea = tan-1 (0.110183 / -0.01357)

          tea = 1,448 rad

to find the angle about the positive side of the + x axis

           tea '= pi - 1,448

           Tea = 1.6936 rad

The magnitude of the net force on q1 exerted by the other two charges is 0.357 N.

The direction of the net force on q1 is 50⁰.

The given parameters;

The force on q1 due to q2 occurs only in y-direction and can be calculated using Coulomb's law as shown below;

[tex]F_1_2 = \frac{kq^2}{r^2}j = \frac{(9\times 10^9) \times (2\times 10^{-6})^2 }{(0.3 +0.3)^2} j \\\\\F_{12} = (0.1)j[/tex]

The force on q1 due to q3 occurs both in x-direction and y-direction, and it is calculated as follows;

[tex]distance \ between \ q1 \ and \ q3\ , r_{13} = \sqrt{0.3^2 \ + \ 0.4^2} = 0.5 \ m\\\\F_{13} = \frac{kq^2}{r_{13}^2} (\frac{0.4i}{0.5} \ + \ \frac{0.3j}{0.5} )\\\\F_{13} = \frac{kq^2}{r_{13}^2} (0.8i + 0.6j)\\\\F_{13} = \frac{9\times 10^9 \times (2\times 10^{-6}) \times (4\times 10^{-6})}{0.5^2} (0.8i + 0.6j)\\\\F_{13} = 0.288(0.8i + 0.6j)\\\\F_{13} = 0.23i + 0.173j[/tex]

The net force is calculated as follows;

[tex]F_{net} = F_{12} \ + \ F_{13}\\\\F_{net} = (0.1j) \ + \ (0.23i + 0.173j)\\\\F_{net} = (0.23i + 0.273j)[/tex]

The magnitude of the net force on q1 is calculated as follows;

[tex]|F| = \sqrt{(0.23^2) \ + \ (0.273^2)} \\\\|F| = 0.357 \ N[/tex]

The direction of the net force on q1 is calculated as follows;

[tex]tan(\theta )= \frac{F_y}{F_x} \\\\\theta = tan^{-1} (\frac{0.273}{0.23} )\\\\\theta = 50^0[/tex]

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Answer:

C An ideal isotropic antenna is a theoretical antenna that does not exist in practice, but is useful in explaining power density and unguided EM signal attenuation

Explanation:

Because practically speaking there is no ideal isotropic antenna it is only imaginary radiating in all directions and use as an arbitrary point for antenna gain

Answer:no

Explanation:the way to see air is by steaming something sometimes you might not see the air but you can see the shadow

Answer:

23.5 degree North

Explanation:

The latitude of the vertical (direct) rays of the sun is 23.5 degree North. This is however due to the sun's vertical rays being located directly above 23.5 degree South which is the position of the Tropic of Capricorn.

The sun ray is highest at this pap tion and it occurs during Summer solstice which happens between June 21 / 22 of every year.

Explanation:

Hey there!!

a = 40° { corresponding angles on a parallel lines are equal}.

As A and B are parallel, a and 40° are corresponding angles.

Hope it helps...

The value of angle "a" will be 40°

The angles which are on the same side of one of two lines cut by a transversal and are on the same side of the transversal is called corresponding angles.

since , angle a and angle B = 40° (given) are on the same relative position at the intersection where a straight line crosses two others and since both the lines are parallel to each other hence ,both the angles are said to be corresponding angles .This implies both the angles must be equal .

angle a = 40°

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Answer:

ANSWER

A = 5 cm = 0.05 m

T = 0.2 s

ω=2π/T=2π/0.2=10πrad/s

plz give brainlist

hope this helped

Answer:

[tex]here \: amplitude = 5cm (convert \: to \: si \: unit) = .05m \\ t = .2sec \\ omega = \frac{2\pi}{t} \\ = \frac{2\pi}{.2 } = 10\pi \frac{rad}{s} \\ we \: want \: find \: a \: and \: v \\ we \: know \: that \: a = - {omega}^{2} x \\ v = omega \sqrt{ {r}^{2} - {x}^{2} } \\(1)x = .05m \\ a = - {10\pi}^{2} \times .05 = - 5 {\pi}^{2} \frac{m}{ {s}^{2} } \\ v = 10\pi \sqrt{ {.05}^{2} - {.05}^{2} } = 0 \\ 2)x = 3cm = .03m \\ a = {(10\pi)}^{2} \times .03 = - 3 {\pi}^{2} \frac{m}{ {s}^{2} } \\ v = 10\pi \sqrt{ {.05}^{2} - {.03}^{2} } = 10\pi \times .04 = .4\pi \frac{m}{s} \\ 3)x = 0 \\ a = - {(10\pi)}^{2} \times 0 = 0 \\ v = 10\pi \times \sqrt{ {.05}^{2} - {0}^{2} } = - 10\pi \times .05 = .5\pi \frac{m}{s} \\ thank \: you[/tex]

Answer:

Part A

[tex]x = 0.33 \ m[/tex]

Part B

[tex]y = 23.49 \ m/s[/tex]

Explanation:

From the question we are told to convert

    13 inches to meters

Now

     1 in   =   0.0254 \  m

      13 \  in  =  x  m

=>  [tex]x = \frac{13 * 0.0254 }{1}[/tex]

=>     [tex]x = 0.33 \ m[/tex]

For  part B  we are told to covert 77ft/s  to meters /srconds

   So  

         1 ft/s  =  0.305 \ m/s  

         77 ft/s  =  y

=>      [tex]y = \frac{77 * 0.305 }{1}[/tex]

=>     [tex]y = 23.49 \ m/s[/tex]