Now you try! Draw the arrows to represent the motion described, then calculate the distance
and displacement for the object.
A car drives 12 meters North and 15 meters South What is the car's distance and
displacement​

Answer:

C. Both Techs A and B

Explanation:

For voltage drop to be measured in the circuit, then  there must be a voltage in the circuit. Once there is a voltage across the circuit, there will be current flowing through the the circuit, hence technician A is correct. Voltage drop is usually measured across components in the circuit. Components in a circuit are consumptive in the circuit, hence their is usually a voltage drop when current flows through them in a circuit.  Technician B is correct.

Answer:

C

Explanation:

Answer:

Probeware and computer

Explanation:

Computers are more powerful and better than a graphing calculator for this situation.

are the tools he must use.

Answer:

The  current is [tex]I = 0.5425 \ A[/tex]

Explanation:

From the question we are told that

   The  magnetic field is  [tex]B = 1.2 \ T[/tex]

   The first length is  [tex]a = 8 \ cm = 0.08 \ m[/tex]

    The  second length is  [tex]a^* = 16 \ cm = 0.16 \ m[/tex]

    The first width is  [tex]b = 17 \ cm = 0.17 \ m[/tex]

     The second  width is  [tex]b^* = 22 \ cm = 0.22 \ m[/tex]

    The time interval  is  [tex]dt = 0.04 \ s[/tex]

     The resistance is  [tex]R = 1.2 \ \Omega[/tex]

Generally the first area is

     [tex]A = a * b[/tex]

=>    [tex]A = 0.08 * 0.17[/tex]

=>     [tex]A = 0.0136 \ m^2[/tex]

The second area is  

      [tex]A^* = a^* * b^*[/tex]

=>   [tex]A^* = 0.16 * 0.22[/tex]

=>     [tex]A^* = 0.0352 \ m^2[/tex]

Generally the induced emf is mathematically represented as

       [tex]\epsilon = - \frac{ B * [A^* - A]}{dt}[/tex]

This negative show that it is moving in the opposite direction of the motion producing it

=>   [tex]|\epsilon | = \frac{ 1.2 * [ 0.0352-0.0135]}{0.04}[/tex]

=>    [tex]|\epsilon | = 0.651 \ V[/tex]

The induced current is

     [tex]I = \frac{|\epsilon|}{R}[/tex]

=>   [tex]I = \frac{ 0.651}{1.2}[/tex]

=>   [tex]I = 0.5425 \ A[/tex]

Answer:

I = 6 mA

Explanation:

Given that,

Number of strands are 125

Resistance of each strand is 2.65 mΩ

The same potential difference is applied between the ends of all the strands and results in a total current of 0.750 A.

We need to find the current in each strand.

Total current is 0.75 A

Number of strands are 125

So, current in each strand :

[tex]I=\dfrac{0.75}{125}\\\\I=0.006\ A\\\\I=6\ mA[/tex]

So, 6 mA of current flows in each strand.

Explanation:

At the highest point, the tension force is 0, so the only force acting on the sphere is gravity.  Sum of forces on the sphere in the centripetal direction:

∑F = ma

mg = mv²/r

v = √(gr)

v = √(9.8 m/s² × 1 m)

v = 3.13 m/s

If the speed is constant, then the linear speed at the lowest point is also 3.13 m/s.  Otherwise, we would need to know the tension in the string at that point.

Answer:

Acceleration, [tex]a=0.69\ m/s^2[/tex]

Explanation:

Given that,

Initial speed of the motorboat, u = 0

Final speed off the motorboat, v = 9 m/s

Time, t = 13 s

We need to find the boat's average acceleration. It is equal to the change in velocity divided by time taken. SO,

[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{9-0}{13}\\\\a=0.69\ m/s^2[/tex]

So, the acceleration of the boat is [tex]0.69\ m/s^2[/tex].

Answer:

[tex]t=15.3s[/tex]

Explanation:

Hello,

In this case, since the speed is defined in terms of the distance over time:

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

We can easily solve for the time with the given speed and distance:

[tex]t=\frac{x}{V}=\frac{1500m}{98m/s}\\ \\t=15.3s[/tex]

Regards.

Answer:

[tex]\mathtt{Q_{sh} = 600.75 \ vars}[/tex]

Explanation:

Given that:

A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA

Here:

the initial power factor  i.e cos θ₁ = 0.7 lag

θ₁ = cos⁻¹ (0.7)

θ₁ = 45.573°

Active power P = 1500 watts

Apparent power S = 2100 VA

What amount of vars must be added to bring the pf to 0.85

i.e the required power factor here is cos θ₂ = 0.85 lag

θ₂ =  cos⁻¹   (0.85)

θ₂ = 31.788°

However; the initial reactive power [tex]Q_1[/tex] = P×tanθ₁

the initial reactive power [tex]Q_1[/tex] = 1500 × tan(45.573)

the initial reactive power [tex]Q_1[/tex] = 1500 × 1.0202

the initial reactive power [tex]Q_1[/tex] =  1530.3 vars

The amount of vars that must therefore be added to bring the pf to 0.85

can be calculated as:

[tex]Q_{sh} = P( tan \theta_1 - tan \theta_2)[/tex]

[tex]Q_{sh} = 1500( tan \ 45.573 - tan \ 31.788)[/tex]

[tex]Q_{sh} = 1500( 1.0202 - 0.6197)[/tex]

[tex]Q_{sh} = 1500( 0.4005)[/tex]

[tex]\mathtt{Q_{sh} = 600.75 \ vars}[/tex]

Answer:

The different types of fluid flow are: Steady and Unsteady Flow. Uniform and Non-Uniform Flow. ... Compressible and Incompressible Flow. Rotational and Irrotational Flow.

Answer:

The current moves in the terminal.

Answer:

D. zero

Explanation:

For momentum of an isolated or closed system to be conserved (initial momentum must equal final momentum), the net external force acting on the system must be zero.

There is always external forces acting on a system, for this system’s momentum to remain constant, all the external forces acting on the system must cancel out, so that the net external force on the system is zero.

[tex]F_{ext} = 0[/tex]

Therefore, the correct option is "D"

D. zero

Answer:

r = 4.747 cm and h = 9.4925 cm

Explanation:

We know that volume of a cylinder is given as:

V = πr²h

Also, surface area is given as;

S = 2πr² + 2πrh

Where r is radius and h is height

Now, we are told that the volume is 672 cm³

Thus, πr²h = 672

Making h the subject gives;

h = 672/πr²

Putting 672/πr² for h in the surface area equation gives;

S = 2πr² + 2πr(672/πr²)

Factorizing gives;

S = 2π[r² + 672/πr]

Differentiating to get first derivative gives;

S' = 2π[2r - (672/πr²)]

Equating to zero gives;

2π[2r - (672/πr²)] = 0

4πr - 1344/r² = 0

4πr = 1344/r²

r³ = 1344/4π

r³ = 106.95212175775

r = ∛106.95212175775

r = 4.747 cm

So, since h = 672/πr²

Then, h = 672/π(4.747)²

h = 9.4925 cm

Answer:

Posterior of the body

Explanation:

Gluteal region is located at the proximal end of the femur and posterior to the pelvic girdle.  The gluteal muscles help to move the lower limb at the hip joint. The gluteal region is divided into two groups: Deep lateral rotators and superficial abductors and extenders.  

The lumbar is the lower region of the spine commonly known as lower back, it has five vertebrates.  The lumbar contain tissue and nerves that control communication between legs and brain. In anatomical terms they are located inferior to the rib cage, at the bottom section of the vertebral column and superior to sacrum and pelvis.

Answer: 12

Explanation:

Given: VF=Vi+AT

-------------------------

In this case, substitute all the given values into the equation

VF=Vi+AT

VF=0+(3)(4)

VF=0+12

VF=12

Hope this helps!! :)

Answer:

[tex]\huge \boxed{V_f=12}[/tex]

Explanation:

[tex]V_f=V_i+AT[/tex]

This is the formula for final velocity.

The values are given for initial velocity, acceleration, and time elapsed.

[tex]V_i=0, \ A=3, \ T=4[/tex]

Solve for [tex]V_f[/tex].

[tex]V_f=0+(3)(4)[/tex]

Evaluate.

[tex]V_f=12[/tex]

Answer: P = 4.86 × 10⁻²

Therefore, the particle's quantum number is 4.86 × 10⁻²

Explanation:

The expression of wave function for a particle in one dimensional box is given as;

φ(x) = ( √2/L ) sin ( nπx/L )

now we input our given figures, the limit of the particle to find it within the center of the box is

xₓ = L/2 + 20% of L/2

xₓ = L/2 + (0.2)L/2

xₓ = 3L/5

And the lower limit is,

x₁ = L/2 - 20% of L/2

x₁ = L/2 - (0.2) L/2

x₁ = 2L / 5

The expression for the probability of finding the particle within the center of the box is

P = ∫ˣˣₓ₁ ║φ(x)║² dx

P = ∫ ³L/⁵ ₂L/₅║(√2/L) sin ( nπx/L)║²dx

= 2/L ( ∫ ³L/⁵ ₂L/₅║sin ( nπx/L)║²dx

= 2/L ( ∫ ³L/⁵ ₂L/₅ (( 1 - cos ( 2πnx/L)/2) dx)

The particle is in its first excited state, then

n =2

Then calculate  the particle's quantum number as follows;

= 2/L ( ∫ ³L/⁵ ₂L/₅ (( 1 - cos ( 2π(2)x/L)/(2)) dx)

= 1/L ( ∫ ³L/⁵ ₂L/₅ (( 1 - cos ( 4πx/L)/2) dx)

= 1/L ( x - (L/4π)sin (4πx/L)) ³L/⁵ ₂L/₅

= 1/L ((3L/5) - (L/4π) sin (( 4π(3L/5)/L)) - (( 2L/5) - (L/4π)sin  ( 4π(2L/5)/L)))

= 1/L ( L/5 + L/4π (sin(8π/5) - sin ( 12π/5)))

Use the trigonometric formula to solve the above equation

sinA - sinB = 2sin ( A-B/2) cos (A+B/2)

Calculate the particle's quantum number as follows

P = 1/L ( L/5 + L/4π (sin(8π/5) - sin ( 12π/5)))

= 1/5 + 1/4π ( 2sin( (8π/5 -12π/5 ) / 2 ) cos ( (8π/5 + 12π/5) / 2 ))

= 1/5 + 1/2π ( -sin(2π/5) cos2π

= 1/5 - 1/2π ( sin (2π/5)(1))

= 0.0486 (10⁻²)(10²)

= 4.86 × 10⁻²

Therefore, the particle's quantum number is 4.86 × 10⁻²

Answer:

Explanation:

We can use the expression B=μ₀*n*I-------1 for the magnetic field that enters a coil  and

n= N/L (number of turns per unit length)

Given data

The number of turns n= 1200 turns

length L= 0.42 m

magnetic field B= 1*10^-4 T

μ₀= [tex]4\pi*10^-^7 T.m/A[/tex]

Applying the equation  B=μ₀*n*I

I= B/μ₀*n

I= B*L/μ₀*n

[tex]I= \frac{1*10^-^4*0.42}{4\pi*10^-^7*1.2*10^3 }[/tex]

[tex]I= 2.65*10^-^2 mA[/tex]

Answer:

The  value is  [tex]V_n = 2.2498 \ m^3[/tex]

Explanation:

From the question we are told that

   The volume of  liquid nitrogen is  [tex]V_n = 3.6 \ L= 3.6 *10^{-3} \ m^3[/tex]

   The  density of  nitrogen at gaseous form   is  [tex]\rho_n = 1.2929 \ kg/m^3[/tex]  =  The dry air at sea level

Generally the density of nitrogen at liquid form is  

         [tex]\rho _l = 808 \ kg/m^3[/tex]

And this is mathematically represented as

      [tex]\rho_l = \frac{m}{V_l }[/tex]

=>   [tex]m = \rho_l * V_l[/tex]

Now the density of  gaseous nitrogen is

       [tex]\rho_n = \frac{m}{V_n }[/tex]

=>   [tex]m = \rho_n * V_n[/tex]

Given that the mass is constant

       [tex]\rho_n * V_n = \rho_l * V_l[/tex]

        [tex]1.2929* V_n = 808 * 3.6*10^{-3}[/tex]

=>   [tex]V_n = 2.2498 \ m^3[/tex]

Answer:

a) Tantalum

b) 1.93 V

Explanation:

The energy of the incident photon= hc/λ

h= Plank's constant=6.63×10^-34 Is

c= speed of light = 3×10^8 ms-1

λ= wavelength of incident photon

E= 6.63×10^-34 × 3×10^8/ 200×10^-9

E= 0.099×10^-17

E= 9.9×10^-19 J

The kinetic energy of the electron = eV

Where;

e= electronic charge = 1.6×10^-19 C

V= 1.93 V

KE= 1.6×10^-19 C × 1.93 V

KE= 3.1 ×10^-19 J

From Einstein's photoelectric equation;

KE= E -Wo

Wo= E -KE

Wo=9.9×10^-19 J - 3.1 ×10^-19 J

Wo= 6.8×10^-19 J

Wo= 6.8×10^-19 J/1.6×10^-19

Wo= 4.25 ev

The metal is Tantalum

b) the stopping potential remains 1.93 V because intensity of incident photon has no effect on the stopping potential.

Answer:

The greater of the two currents is 0.692 A

Explanation:

Given;

distance between the two parallel wires; r = 6 mm = 6 x 10⁻³ m

let the current in the first wire = I₁

then, the current in the second wire = 2I₁

length of the wires, L = 3.0 m

magnitude of force on the wires, F = 8 μN = 8 x 10⁻⁶ N

The magnitude of force on the two parallel wires is given by;

[tex]F = \frac{\mu_o I_1(2I_1)}{2\pi r}\\\\F = \frac{\mu_o 2I_1^2}{2\pi r}\\\\I_1^2 = \frac{F*2\pi r}{2\mu_o} \\\\I_1^2 = \frac{8*10^{-6}*2\pi (6*10^{-3})}{2(4\pi*10^{-7})}\\\\I_1^2 = 0.12\\\\I_1 = \sqrt{0.12}\\\\ I_1 =0.346 \ A[/tex]

the current in the second wire = 2I₁ = 2 x 0.346 A = 0.692 A

Therefore, the greater of the two currents is 0.692 A

Answer:

a) How many parsecs are there in one astronomical unit?

[tex]4.85x10^{-6}pc[/tex]

(b) How many meters are in a parsec?

[tex]3.081x10^{16}m[/tex]

(c) How many meters in a light-year?

[tex]9.46x10^{15}m[/tex]

(d) How many astronomical units in a light-year?

[tex]63325AU[/tex]

(e) How many light-years in a parsec?

3.26ly

Explanation:

The parallax angle can be used to find out the distance using triangulation. Making a triangle between the nearby star, the Sun and the Earth, knowing that the distance between the Earth and the Sun ([tex]1.496x10^{11} m[/tex]) is defined as 1 astronomical unit:

[tex]\tan{p} = \frac{1AU}{d}[/tex]

Where d is the distance to the star.

Since p is small it can be represent as:

[tex]p(rad) = \frac{1AU}{d}[/tex]  (1)

Where p(rad) is the value of in radians

However, it is better to express small angles in arcseconds

[tex]p('') = p(rad)\frac{180^\circ}{\pi rad}.\frac{60'}{1^\circ}.\frac{60''}{1'}[/tex]

[tex]p('') = 2.06x10^5 p(rad)[/tex]

[tex]p(rad) = \frac{p('')}{2.06x10^5}[/tex] (2)

Then, equation 2 can be replace in equation 1:

[tex]\frac{p('')}{2.06x10^5} = \frac{1AU}{d}[/tex]  

[tex]\frac{d}{1AU} = \frac{2.06x10^5}{p('')}[/tex]  (3)

From equation 3 it can be see that [tex]1pc = 2.06x10^5 AU[/tex]

a) How many parsecs are there in one astronomical unit?

[tex]1AU . \frac{1pc}{2.06x10^5AU}[/tex] ⇒ [tex]4.85x10^{-6}pc[/tex]

(b) How many meters are in a parsec?

[tex]2.06x10^{5}AU . \frac{1.496x10^{11}m}{1AU}[/tex] ⇒ [tex]3.081x10^{16}m[/tex]

(c) How many meters in a light-year?

To determine the number of meters in a light-year it is necessary to use the next equation:

[tex]x = c.t[/tex]

Where c is the speed of light ([tex]c = 3x10^{8}m/s[/tex]) and x is the distance that light travels in 1 year.

In 1 year they are 31536000 seconds

[tex]x = (3x10^{8}m/s)(31536000s)[/tex]

[tex]x = 9.46x10^{15}m[/tex]

(d) How many astronomical units in a light-year?

[tex]9.46x10^{15}m . \frac{1AU}{1.496x10^{11}m}[/tex] ⇒ [tex]63325AU[/tex]

(e) How many light-years in a parsec?

[tex]2.06x10^{5}AU . \frac{1ly}{63235AU}[/tex] ⇒ [tex]3.26ly[/tex]

Answer:

0.153 m/s

Explanation:

The flowrate Q = 0.625 x 10-3 m^3-/s

The diameter of the nozzle d = 5.19 x 10^-3 m

the velocity V = ?

The cross-sectional area of the flow A = [tex]\pi d^{2}/4[/tex]

==> (3.142 x 5.19 x 10^-3)/4 = 4.077 x 10^-3 m^2

From the continuity equation,

Q = AV

V = Q/A = (0.625 x 10-3)/(4.077 x 10^-3) = 0.153 m/s

Answer:

P = 198.118 kPa

Explanation:

Given:

Atmospheric pressure = P[tex]_{atm\\}[/tex] = 1.01×10⁵

depth = h = 9.91 m

To find:

Absolute pressure P[tex]_{abs}[/tex]

Solution:

Density of water = ρ = 1.000x10 ³kg/m ³

acceleration due to gravity = ρ = 9.8 m/s²

P[tex]_{abs}[/tex] = P[tex]_{atm\\}[/tex] + ρgh

      = 1.01×10⁵ + 1.000x10 ³x 9.8 x 9.91

      = 101000 + 1000(9.8)(9.91)

      = 101000 + 97118

      = 198118 Pa

      = 198.118 kPa

P[tex]_{abs}[/tex] = 198.118 kPa

The absolute pressure at a depth below the surface of this deep lake is 198.118 kPa.

Given the following data:

Scientific data:

To calculate the absolute pressure at a depth below the surface of a deep lake:

Mathematically, absolute pressure is given by this formula:

[tex]P_{abs} = P + \rho gh[/tex]

Substituting the given parameters into the formula, we have;

[tex]P_{abs} = 1.01 \times 10^5 + 1000 \times 9.8 \times 9.91 \\\\ P_{abs} = 101000+ 97118 \\\\[/tex]

Absolute pressure = 198118 Pa

Note: 1 kPa = 1000 Pa

Absolute pressure = 198.118 kPa

Read more on absolute pressure here: https://brainly.com/question/10013312

Answer:

  Q₁ = 2.4 10⁻⁴ C

Explanation:

We have a circuit with several capacitors, let's find the equivalent capacitor of the parallel

    [tex]C_{eq1}[/tex] = C₂ + C₃

    C_{eq1} = (10 +30) 10⁻⁶

    C_{eq1} = 40 10⁻⁶ F

There remains a series system between C₁ and C_{eq1}, let's find the equivalent capacitor

     1/C_{eq2} = 1 / C₁ + 1 / C_{eq1}

     1 /C_{eq2} = 1 / 20 10⁻⁶ + 1/40 10⁻⁶

     1 / C_{eq2} = 0.075 10⁶

     C_{eq2} = 13.33 10⁻⁶ F

let's use the relationship

        V = Q / C_{eq2}

        Q = V C_{eq2}

        Q = 18  13.33 10⁻⁶

        Q = 2.4 10⁻⁴ C

In a combination of capacitors in series the charge is constant, so the charge on C₁ is the same

        Q₁ = 2.4 10⁻⁴ C

Answer:

70 N

Explanation:

Draw a free body diagram of the boy.  There are four forces:

Weight force mg pulling down,

A 300 N normal force pushing up,

A 75 N applied force pulling right,

and a 5 N friction force pushing left.

The boy's acceleration in the y direction is 0, so the net force in the y direction is 0.

The net force in the x direction is 75 N − 5 N = 70 N.

Answer:

Answer:

43.4J

Explanation:

We know that

Work done = total heat energy

But work done is force x distance

=> F = 19 x8.6 = 163.4 J

So the total heat. Will be Heat of cube + heat of floor = 163.4J

So that heat of floor will now be

floor = 163.4 J - 120 J = 43.4 Joules

Explanation:

Answer:

The speed must a ball be thrown vertically from ground level to rise to a maximum height is 28.35 m/s.

Explanation:

Given;

maximum vertical height of the throw, H = 41 m

Apply the following kinematic equation;

V² = U² + 2gH

where;

V is the final speed with which the ball will rise to a maximum height

U is the initial speed of the ball = 0

g is acceleration due to gravity = 0

V² = U² + 2gH

V² = 0² + 2gH

V² =  2gH

V = √2gH

V = √(2 x 9.8 x 41)

V = 28.35 m/s

Therefore, the speed must a ball be thrown vertically from ground level to rise to a maximum height is 28.35 m/s.

Answer:

Explanation:

Without wearing glasses , her near point is 20 cm .

for correction of eye

u = infinity ,

v = - 150 cm

f = ?

[tex]\frac{1}{v} -\frac{1}{u} = \frac{1}{f }[/tex]

[tex]\frac{1}{-150} -0 = \frac{1}{f }\\[/tex]

f = - 150 cm

He must be wearing glass of focal length of 150 cm .

If near point be x after wearing glass ,

u = x

v = - 20 cm

f = - 150 cm

[tex]\frac{1}{v} -\frac{1}{u} = \frac{1}{f }[/tex]

[tex]\frac{1}{-20} -\frac{1}{x} = \frac{1}{-150 }[/tex]

[tex]\frac{1}{-20} + \frac{1}{150 }= \frac{1}{x}[/tex]

x = 23 cm .

While wearing the glasses, Yang's near point will be 23.08 cm.

Given information:

Yang can focus on objects 150 cm away with a relaxed eye.

With full accommodation, she can focus on objects 20 cm away.

For correction, we have to use a concave lens such that it can make the image of a distant object at 150 cm.

So, the object distance will be infinity, and the image distance will be [tex]v=-150[/tex] cm.

So, the focal length of the lens can be calculated by lens formula as,

[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\dfrac{1}{-150}-\dfrac{1}{\infty}=\dfrac{1}{f}\\f=-150\rm\;cm[/tex]

Now, after using the lens, the image distance will be [tex]v=-20[/tex] cm. Let u be the near point.

The near point, after correction, can be calculated as,

[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\dfrac{1}{-20}-\dfrac{1}{u}=\dfrac{1}{-150}\\\dfrac{1}{u}=\dfrac{1}{150}-\dfrac{1}{20}\\u=23.08\rm\; cm[/tex]

Therefore, while wearing the glasses, Yang's near point will be 23.08 cm.

For more details, refer to the ink:

https://brainly.com/question/4419161

Answer:

Velocity of moon = 586.23 km/h

Explanation:

We are given;

Distance of moon from the Earth = 384403 km

Time taken to orbit earth;t = 27.32166 days

24 hours make 1 day, thus 27.32166 days = 27.32166 × 24 = 655.72 hours

Formula for velocity is distance/time

Thus,

Velocity of moon = distance from moon to earth/time taken to orbit the earth

Velocity of moon = 384403/655.72 = 586.23 km/h

Answer:

E) 80 N/m

Explanation:

Given;

mass of the block, m = 4.8 kg

displacement of the block, x = -0.5 m

velocity of the block, v = -0.8 m/s

acceleration of the block, a = 8.3 m/s²

From Newton's second law of motion;

F = ma

Also, from Hook's law;

F = -Kx

where;

k is the force constant

Thus, ma = -kx

k = -ma/x

k = -(4.8 x 8.3) / (-0.5)

k = 79.7 N/m

k ≅ 80 N/m

Therefore, the force constant of the spring is closest to 80 N/m

Answer:

Hey mate ,

Area of rectangle = l×b

1.82×1.5

2.73cm2