Which word describes a statement that has been accepted, tested, and supported by multiple sets of evidence? A. Hypothesis B. Theory C. Procedure D. Data

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: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:

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

motion is the action of moving one place to another place

Answer:

We’re imagining imagining imagining an imagination...

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.

Find out more information about charge here:

https://brainly.com/question/25922783

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.

Answer:

1.    Kicking a soccer ball

2.      Playing tug of war  

3.         Bouncing a Ball

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.

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°

learn more about corresponding angles

https://brainly.com/question/12521848?referrer=searchResults

#SPJ3

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:

the intercept is the orgin (0,0)

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]

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

Explanation:

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]

Learn more here:https://brainly.com/question/17132341

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

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:

The energy of the wave is 1.435 x 10⁻⁴ J

Explanation:

Given;

area of the window, A = 0.5 m²

the rms value of the field, E = 0.06 V/m

The peak value of electric field is given by;

[tex]E_o = \sqrt{2} *E_{rms}\\\\E_o = \sqrt{2}*0.06\\\\E_o = 0.0849 \ V/m[/tex]

The average intensity of the wave is given by;

[tex]I_{avg} = \frac{c \epsilon_o E_o^2 }{2}\\\\I_{avg} = \frac{(3*10^8)( 8.85*10^{-12}) (0.0849)^2 }{2}\\\\I_{avg} = 9.569*10^{-6} \ W/m^2[/tex]

The average power of the wave is given by;

P = I x A

P = (9.569 x 10⁻⁶ W/m²) (0.5 m²)

P = 4.784 x 10⁻⁶ W

The energy of the wave is given by;

E = P x t

E = (4.784 x 10⁻⁶ W)(30 s)

E = 1.435 x 10⁻⁴ J

Therefore, the energy of the wave is 1.435 x 10⁻⁴ J

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:

yes bc they are falling :]

Explanation:

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

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:

With changing speed and/or in a circle

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.

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:

Increase in velocity propagation would lead to a longer sinusoidal pattern.

Explanation:

The velocity of propagation has a relationship with the tension in a standing wave and is given by;

v = √(T/μ)

where:

μ is mass per unit length.

T is tension in string

For the sinusoidal pattern on the string to be longer or shorter, it means that the wavelength will be longer or shorter.

Now, relationship between wavelength and velocity and tension is;

v = λ/T

Where V is velocity of propagation, λ is wavelength and T is Tension.

So, λ = vT

From the earlier standing wave equation, we will see that if we increase the velocity, it means the Tension of the spring will increase as well.

Thus, from this wavelength equation, an increase in velocity means an increase in tension and which will in turn mean an increase in wavelength.

Therefore, an increase in velocity propagation means a longer sinusoidal pattern.

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) an expression for the car's speed is given as

v = u + at

where

v is the car's speed

u is the initial speed of the car

a is the car's acceleration

t is the time spent accelerating

2) The car does not hit the tree limb

Explanation:

The initial velocity of the car = 0 m/s  (since it accelerates from rest)

acceleration of the car = 1.76 m/s

time spent accelerating = 20 s

For the car's speed just before the driver begins braking, we use the expression

v = u + at

where v is the final speed of the car just before the driver begins braking

u is the initial velocity with which the car starts moving

a is the acceleration of the car

t is the time spent accelerating from u to v

substituting values, we have

v = 0 + 1.76(20)

v = 0 + 35.2

the car's speed v = 35.2 m/s

In this time the car accelerates, the car moves a distance given by

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

where s is the distance covered in this time

u is the initial speed of the vehicle

a is the acceleration

t is the time taken

substituting, we have

s = 0(20) + [tex]\frac{1}{2}[/tex](1.76)[tex]20^{2}[/tex]

s = 0 + 352

distance s = 352 m

When the driver brakes, we have

time spent braking = 5 s

acceleration = -5.93 m/s

and the distance to the limb = 550 m from where the car begun

to get the distance covered in this period, we use the expression

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

where s is the distance traveled at this time

u is the speed of the car before it starts braking = 35.2 m/s

a is the acceleration at this point

t is the time taken to decelerate to a stop

substituting values, we have

s = 35.2(5) + [tex]\frac{1}{2}[/tex](-5.93 x [tex]5^2[/tex])

s = 176 - 74.125

s = 101.88 m

Total distance moved by the car = 352 m + 101.88 m = 453.88 m

Since the total distance traveled by the car is less than the distance from the starting point to the place where the tree limb is, the car does not hit the tree limb.

Answer:

no more than 10 pills a day depending on what type it is

Explanation: