The x-component and y-component of two vectors A & B are Ax = 9, Ay = 12,Bx =
15 & By = 20. Find:
/A+B/​

Answer: false

Explanation:

Answer:

false

Explanation:

According to Newton's First Law of motion, an object remains in the same state of motion unless a resultant force acts on it.

Answer:

first value+2nd +3rd

Explanation:

thug life and there

Answer:

See explanation below

Explanation:

This is a typical exercise of free falling. In this case a rock thrown straight down from the bridge, and we are asked to determine the final velocity of the rock and it's displacement at those given times.

First, just for this problem, as the rock is going straight down, we'll say that the downward direction is positive, therefore, the following expressions to calculate velocity and speed will be:

V = V₀ + gt   (1)

X = V₀t + gt²/2   (2)

In this case, g = 9.8 m/s²

Now, let's see the displacement and velocity for each given time:

a) For  t = 0.5 s

V = 14 + (9.8)*0.5

V = 18.9 m/s

X = (14*0.5) + (9.8)(0.5)²/2

X = 7 + 1.225

X = 8.225 m

b) For t = 1.00 s

V = 14 + (9.8)*1

V = 23.8 m/s

X = (14*1) + (9.8)(1)²/2

X = 14 + 4.9

X = 18.9 m

c) For t = 1.5 s

V = 14 + (9.8)*1.5

V = 28.7 m/s

X = (14*1.5) + (9.8)(1.5)²/2

X = 21 + 11.025

X = 32.025 m

d) For t = 2 s

V = 14 + (9.8)*2

V = 33.6 m/s

X = (14*2) + (9.8)(2)²/2

X = 28 + 19.6

X = 47.6 m

e) For t = 2.50 s

V = 14 + (9.8)*2.5

V = 38.5 m/s

X = (14*2.5) + (9.8)(2.5)²/2

X = 35 + 30.625

X = 65.625 m

Hope this helps

Answer:

a) ΔV = 25.59 V, b)  ΔV = 25.59 V,  c)  v = 7 10⁴ m / s,  v/c= 2.33 10⁻⁴ ,

v/c% = 2.33 10⁻²

Explanation:

a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy

starting point. Where the electrons come out

          Em₀ = U = e DV

final point. Where they hit the target

          Em_f = K = ½ m v2

energy is conserved

          Em₀ = Em_f

         e ΔV = ½ m v²

         ΔV = [tex]\frac{1}{2}[/tex] mv²/e     (1)

If the speed of light is c and this is 100% then 1% is

         v = 1% c = c / 100

         v = 3 10⁸/100 = 3 10⁶6 m/ s

let's calculate

         ΔV = [tex]\frac{1}{2} \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }[/tex]

         ΔV = 25.59 V

b) Ask for the potential difference for protons with the same kinetic energy as electrons

             [tex]K_e = K_p[/tex]

              K_p = ½ m v_e²

              K_p = [tex]\frac{1}{2}[/tex]  9.1 10⁻³¹ (3 10⁶)²

              K_p = 40.95 10⁻¹⁹ J

we substitute in equation 1

              ΔV = Kp / M

              ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹

              ΔV = 25.59 V

notice that these protons go much slower than electrons because their mass is greater

c) The speed of the protons is

             e ΔV = ½ M v²

             v² = 2 e ΔV / M

             v² = [tex]\frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }[/tex]

              v² = 49,035 10⁸

               v = 7 10⁴ m / s

Relation

        v/c = [tex]\frac{7 \ 10^4 }{ 3 \ 10^8}[/tex]

        v/c= 2.33 10⁻⁴

Answer:

a)    F = 21.16 N,  b)     a = 3.17 10²⁸ m / s

Explanation:

a) The outside between the alpha particles is the electric force, given by Coulomb's law

          F = [tex]k \frac{ q_1 q_2}{r^2}[/tex]

in that case the two charges are of equal magnitude

          q₁ = q₂ = 2q

let's calculate

         F = [tex]9 \ 10^9 \ \frac{ (2 \ 1.6 \ 10^{-19} )^2 }{ (6.60 \ 10^{-15} )^2 }[/tex]

         F = 21.16 N

this force is repulsive because the charges are of the same sign

b) what is the initial acceleration

         F = ma

         a = F / m

         a = [tex]\frac{21.16}{4.0026 \ 1.67 \ 10^{-27} }[/tex]21.16 / 4.0025 1.67 10-27

         a = 3.17 10²⁸ m / s

this acceleration is in the direction of moving away the alpha particles

Answer:

The answer is False......

Answer:

true

it is equal but opposite

Answer:

d=20m/sx60s=1200m=1200/1000Km=1.2km

Explanation:

Air masses from the tropics and the equator are warm as they form over lower latitudes. The major causes for moving air masses North America exists jet storm.

An air mass is a volume of air that in meteorology is identified by its temperature and humidity. Many hundreds or thousands of square miles are covered by air masses, which adjust to the properties of the land underneath them. Latitude and their continental or maritime source regions are used to categories them.

Warmer air masses are referred to as tropical, whilst colder air masses are referred to as polar or arctic. Superior and maritime air masses are moist, whereas continental and superior air masses are dry. Air masses with various densities are divided by weather fronts. Once an air mass has left its original location, nearby plants and bodies of water can quickly change the way it behaves. Classification systems address both the properties and modification of an air mass.

Air masses from the tropics and the equator are warm as they form over lower latitudes. They move poleward along the southern edge of the subtropical ridge and are drier and hotter than those that originate over seas. Trade air masses are another name for tropical maritime air masses. The Caribbean Sea, southern Gulf of Mexico, and tropical Atlantic Oceans, east of Florida via the Bahamas, are the origins of maritime tropical air masses that have an impact on the United States.

Monsoon air masses are moist and unstable. Rarely do dry superior air masses touch the ground. A trade wind inversion, which is a warmer and drier layer over the more moderately moist air mass below, is typically created over maritime tropical air masses when they are located above them.

Therefore, the correct answer is option C. jet storm.

To learn more about Air mass refer to:

https://brainly.com/question/19626802

#SPJ2

Answer:

a)   f ’’ = f₀ [tex]\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }[/tex] , b)   Δf = 2 f₀ [tex]\frac{v}{c}[/tex]

Explanation:

a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.

Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source

                    f ’= fo[tex]\frac{c+v}{c}[/tex]

This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.

                   f ’’ = f’ [tex]\frac{c}{ c-v}[/tex]

where c represents the sound velocity in stationary blood

therefore the received frequency is

                 f ’’ = f₀   [tex]\frac{c}{c-v}[/tex]

let's simplify the expression

                f ’’ = f₀ \frac{c+v}{c-v}

                f ’’ = f₀ [tex]\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }[/tex]

b) At the low speed limit v <c, we can expand the quantity

                 (1 -x)ⁿ = 1 - x + n (n-1) x² + ...

                 [tex]( 1- \frac{v}{c} ) ^{-1} = 1 + \frac{v}{c}[/tex]

                f ’’ = fo [tex]( 1+ \frac{v}{c}) ( 1 + \frac{v}{c} )[/tex]

                f ’’ = fo [tex]( 1 + 2 \frac{v}{c} + \frac{v^2}{ c^2} )[/tex]

leave the linear term

               f ’’ = f₀ + f₀ 2[tex]\frac{v}{c}[/tex]

the sound difference

               f ’’ -f₀ = 2f₀ v/c

               Δf = 2 f₀ [tex]\frac{v}{c}[/tex]

The coaster starts at rest, so the kinetic energy (KE) at point A is 0. It is situated 33 m above ground, so its potential energy (PE) at A is

mgh = (3000 kg) (9.80 m/s²) (33 m) = 970,200 J

The total energy is the same, 970,200 J.

Assuming no energy is lost to friction or sound etc, energy is conserved throughout the coaster's motion, so the total energy should be the same at each point.

At point B, the coaster has dropped to a height of 10 m, so it has PE

mgh = (3000 kg) (9.80 m/s²) (10 m) = 294,000 J

which means it must have KE

970,200 J = KE + 294,000 J   →   KE = 676,200 J

which gives the coast a speed v at point B of

1/2 mv ² = 1/2 (3000 kg) v ² = 676,200 J   →   v ≈ 21.2 m/s

At point C, the coaster has a speed of 16.0 m/s, so it has KE

1/2 mv ² = 1/2 (3000 kg) (16.0 m/s)² = 384,000 J

and hence PE

970,200 J = 384,000 J + PE   →   PE = 586,200 J

This lets us determine the height h at C:

mgh = (3000 kg) (9.80 m/s²) h = 586,200 J   →   h ≈ 19.939 m

which means the loop has diameter h - 10 m ≈ 9.94 m.

At point D, the coaster is 15 m above the ground so its PE at D is

mgh = (3000 kg) (9.80 m/s²) (15 m) = 441,000 J

and so its KE is

970,200 J = KE + 441,000 J   →   KE = 529,200 J

and hence has speed v at D

1/2 mv ² = 1/2 (3000 kg) v ² = 529,200 J   →   v ≈ 18.9 m/s

Answer:

the power generation potential is 2.705 x 10⁶ J/s.

Explanation:

Given;

height below the free surface of a large water reservoir, h = 120 m

mass flow rate of the water, m' = 2300 kg/s

The power generation potential is calculated as;

[tex]Power = \frac{Energy}{time} = \frac{F\times h}{t} = \frac{(mg) \times h}{t} = \frac{m}{t}\times gh = m' \times gh\\\\Power = m' \times gh\\\\Power = 2300 \ kg/s \ \times \ 9.8 \ m/s^2 \ \times \ 120 \ m\\\\Power = 2.705 \times 10^6 \ J/s[/tex]

Therefore, the power generation potential is 2.705 x 10⁶ J/s.

Answer:

Explanation:

Gravitational potential energy = mgh where m is mass , g is acceleration due to gravity and h is height from the ground .

In the first case mass m = volume x density

= 1 x 1000 = 1000 kg

height h = 30 m

potential energy = 1000 x 30 x 9.8 = 294000 J = 294 kJ .

When height decreases by 10 m , potential decreases as follows .

Decrease in potential energy

= mass x gravitational energy x decrease in height

= 1000 x 9.8 x 10

= 98000 J

= 98 kJ .

Answer:

dgfggddhdbxbxjxddhsnsxnc

Answer:

a= 0

Explanation:

       [tex]F_{net} = m*a = 0 (1)[/tex]

      ⇒ a = 0

Answer:

0.143 m

Explanation:

Since

d = 1/N = 1/520 = 1.92 * 10^-3 mm

For red light;

θ = sin^-1  (1 * λred/d) =  sin^-1  (1 * 656 * 10^-9/1.92 * 10^-6) = 19.98°

L = 1.4 * (tan 19.98) = 0.509 m

For blue light;

θ = sin^-1  (1 * λblue/d) =  sin^-1  (1 * 486 * 10^-9/1.92 * 10^-6) = 14.66°

L = 1.4 * (tan 14.66°) = 0.366 m

Distance between the first-order red and blue fringes= 0.509 m - 0.366 m = 0.143 m

Answer:

20 C

Explanation:

To do this, is pretty easy, we just need to do a little reasoning of what is happening.

When any charge  called q is placed inside this metallic shell which is spheric, all the opposite and even equal charges are induced on the inner and outer surface of the shell. Hence, we can say that if in the inner shell we have +q, in the outer will be -q.

Now, here we have the shell with 5 C, and when the charge of -11 C is placed inside the shell we can have the following changes on the inner surface and the outer surface:

Inner surface: +11 C

Outer surface: 9 + 11 = 20 C

Hope this helps

Answer:

The answer to the given points can be defined as follows:

Explanation:

In point 1:

[tex]\bold{v_f^2= v_i^2+2as}\\\\\to v_f=0\\\\\to v_i=20 \frac{m}{s}\\\\\to s= 4.50\ m -3.60 \ m \\\\[/tex]

      [tex]=0.9 \ m \\[/tex]

put the value in the above formula:

[tex]\to 0= 20^2+2 \times a \times 0.9\\\\\to -1.8\ a=400\\\\\to -a= \frac{400}{1.8} \\\\ \to a= -222.22\ \ \frac{m}{s^2}[/tex]

 [tex]\bold{v_f=v_i+at}\\\\\to 0=20+ (-222.22)t\\\\\to 222.22t=20\\\\\to t=\frac{20}{222.22}\\\\\to t= 0.0900 \ s\\\\\to v_{avg}=\frac{s}{t}=\frac{0.9}{t}= 10\ \frac{m}{s}[/tex]

for point 2:

[tex]t= 0.0900 \ s -\text{found above}[/tex]

for point 3:

[tex]\to |a| = 222.22 \frac{m}{s^2} \text{found above}\\\\\to \bold{|F| = m \cdot |a|}\\\\[/tex]

         [tex]=1600 \ kg \times 222.22 \ \frac{m}{s^2} \\\\= 3.55\times 10^{5} \ N[/tex]

Answer:

you could put the iron filings on the peace of paper and hover a magnet over top of the paper and the iron filings would stand up, or even stick to the magnet

Explanation:

Answer: B

Explanation:

The vertical component of a vector such as velocity is the magnitude of the vector multiplied by the sine of the angle.

[tex]V_y=48*sin(53)=38.3m/s[/tex]

Answer:

a) the new angular speed of the student is 0.9642 rad/s

b)

the kinetic energy of the student before the objects are pulled in is 1.9119 J

the kinetic energy of the student after the objects are pulled in is 2.4252 J

Explanation:

Given that;

mass of each object m = 1 kg

distance of objects from axis of rotation r = 0.9 m

Moment of inertia of each object initially [tex]I_{oi}[/tex]

[tex]I_{oi}[/tex] = mr² = 1kg ×(0.9m)² = 1 kg × 0.81 m²  = 0.81 kg.m²

moment of inertia of each object finally [tex]I_{of}[/tex]

[tex]I_{of}[/tex]  = mr² = 1kg × (0.33 m)² = 0.1089 kg.m²

Now

moment of inertia of student plus stool  [tex]I_{}[/tex] = 5 kg.m²

initial angular speed ω₀ = 0.76 rad/sec

final angular speed ω = ?

Now using conservation of angular momentum;

([tex]I_{}[/tex] + 2 [tex]I_{oi}[/tex] )ω₀ = ([tex]I_{}[/tex] + 2 [tex]I_{of}[/tex] )ω

so we substitute

(5 + 2 (0.81) )0.76 = (5 + 2 (0.1089) )ω

5.0312 = 5.2178 ω

ω =  5.0312 / 5.2178

ω  = 0.9642 rad/s

Therefore, the new angular speed of the student is 0.9642 rad/s

b)

K.E of student before = (0.5) ([tex]I_{}[/tex] + 2 [tex]I_{oi}[/tex] )ω₀²

= (0.5) (5 + 2 (0.81) )(0.76)²

= 0.5 × 6.62 × 0.5776

= 1.9119 J

Therefore, the kinetic energy of the student before the objects are pulled in is 1.9119 J

KE of student finally = (0.5) ([tex]I_{}[/tex] + 2 [tex]I_{of}[/tex] )ω²

= (0.5) (5 + 2 (0.1089) ) (0.9642)²

= 0.5 × 5.2178 × 0.9296

= 2.4252 J

Therefore, the kinetic energy of the student after the objects are pulled in is 2.4252 J

Answer:

The acceleration produced by the 100 N net force will be two times greater than the acceleration produced by 50 N net force.

Explanation:

Given;

first net force, F₁ = 100 N

second net force, F₂ = 50 N

If we consider equal mass for the two net forces, and apply Newton's second law of motion, the acceleration produced by the 100 N net force will be two times greater than the acceleration produced by 50 N net force.

Let a₁ be the acceleration produced by the first net force

then, a₂ be the acceleration produced by the second net force

Thus, a₁ = 2a₂

Answer:

:To find the Voltage, ( V ) [ V = I x R ] V (volts) = I (amps) x R (Ω)

To find the Current, ( I ) [ I = V ÷ R ] I (amps) = V (volts) ÷ R (Ω)

To find the Resistance, ( R ) [ R = V ÷ I ] R (Ω) = V (volts) ÷ I (amps)

To find the Power (P) [ P = V x I ] P (watts) = V (volts) x I (amps)

Answer:

B, D and E, not all matter can be filled with air

Answer:

Explanation:

The only force acting on the cart is a component of its weight parallel to ramp downwards . No other force acts parallel to the ramp .

Even when the cart is moving up after the initial push , its weight is acting downwards so acceleration is acting downwards .

When the cart is stationary at the top position , its weight is acting downwards so acceleration is downwards at that moment also . When the cart is going downwards , still its weight is acting down so acceleration is acting downwards .

After this push, as the car moves up the ramp, the direction of the acceleration of the cart is _down _______ the ramp. After the reaches its highest point, turns around, and begins moving down the ramp, the direction of the acceleration of the cart is _down ________ the ramp. At the highest point the cart reaches on the ramp, when the cart momentarily comes to rest, the magnitude of the acceleration of the cart is _downwards ______.

Answer:

the answer would be A. electricity don't specify the direction of any cardinal points the flow of charges moves.

Answer:

A

Explanation:

I did the test on ap3x

Answer:

   F = 196 N

Explanation:

For this exercise we will use Newton's second law,  we define a reference system with the x axis in the direction of movement of the stones and the y axis vertically

Y axis  

       N- W = 0

       N = mg

X axis

       F -fr = ma

In this case, they ask us for the force to keep moving, so the stones go at constant speed, which implies that the acceleration is zero.

       F- fr = 0

       F = fr

the friction force has the equation

       fr = μ N

       fr = μ mg

we substitute

        F = μ mg

let's calculate

         F = 0.80 9.8 25

         F = 196 N

Answer:

What am I suppose to solve

Explanation: