If you are driving a car with a velocity of -25 m/s and you have an acceleration of -2 m/s^2, are you speeding up or slowing down? Why?

Answers

Answer 1

Answer:

Hmmm...

This is a bit tricky

Ok...

Negative Velocity means you're Moving in the Opposite direction....

Negative Acceleration (deceleration) means you're slowing down.

Deceleration would mean slowing down if you were Moving with a Positive velocity.

But In this case...

You're Moving with negative velocity and Negative acceleration...

This simply means that the acceleration and velocity vector are in the same direction....

Its means that...

"YOU'RE SPEEDING UP"

Just that you're doing it in the opposite direction.

Hope this helps.


Related Questions

g you hang an object of mass m on a spring with spring constant k and find that it has a period of T. If you change the spring to one that has a spring constant of 2 k, the new period is

Answers

Answer:

a)   T = 2π [tex]\sqrt{\frac{m}{k} }[/tex],  b)  T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]

Explanation:

a) A system formed by a mass and a spring has a simple harmonic motion with angular velocity

          w² = k / m

angular velocity and period are related

          w = 2π /T

     

we substitute

          4π²/ T² = k / m

           T = 2π [tex]\sqrt{\frac{m}{k} }[/tex]

b) We change the spring for another with k ’= 2 k, let's find the period

           T ’= 2π [tex]\sqrt{\frac{m}{k'} }[/tex]

           T ’= 2π [tex]\sqrt{ \frac{m}{2k} }[/tex]

           T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]

A camera lens with focal length f = 50 mm and maximum aperture f>2
forms an image of an object 9.0 m away. (a) If the resolution is limited
by diffraction, what is the minimum distance between two points on the
object that are barely resolved? What is the corresponding distance
between image points? (b) How does the situation change if the lens is
“stopped down” to f>16? Use λ= 500 nm in both cases

Answers

Answer:

The minimum distance between two points on the  object that are barely resolved is 0.26 mm

The corresponding distance between the  image points = 0.0015 m

Explanation:

Given  

focal length f = 50 mm and maximum aperture f>2

s =  9.0 m

aperture = 25 mm = 25 *10^-3 m

Sin a = 1.22 *wavelength /D  

Substituting the given values, we get –  

Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m

Sin a = 2.93 * 10 ^-5 rad

Now  

Y/9.0 m = 2.93 * 10 ^-5

Y = 2.64 *10^-4 m = 0.26 mm

Y’/50 *10^-3 = 2.93 * 10 ^-5  

Y’ = 0.0015 m

Which describes farsightedness? O Distant objects are blurry. O Concave lenses can correct it. O Objects appear larger when wearing corrective glasses. O Corrective glasses do not change apparent the size of objects.​

Answers

Answer:

O Distant objects are blurry. describes farsightedness.

Explanation:

Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry. The degree of your farsightedness influences your focusing ability.Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry.

examples of frictional forces grade 8​

Answers

Answer:

Dragging a Chair, Climbing, Rubbing Hands

A police car in hot pursuit goes speeding past you. While the siren is approaching, the frequency of the sound you hear is 5500 Hz. When the siren is receding away from you, the frequency of the sound is 4500 Hz. Use the Doppler formula to determine the velocity of the police car. Use vsound=330 m/s.
What is the velocity v of the police car ?

Answers

Vs = 34m/s
I don’t have an explanation my apologies.

When a police car in hot pursuit goes speeding past you, the velocity v of the police car is 33 m/s.

What is the Doppler formula?

The formula is used when there exists a Doppler shift. The Doppler shift is due to the relative motion of sound waves between the source and observer.

The frequency increase by the Doppler effect is represented by the formula

f' = [tex]\dfrac{v-v_{o} }{v-v_{s} }[/tex]× f

Given the frequency of source f' is 5500 Hz . Velocity of the observer v₀  is 0.

Substituting the value into the equation will give us the velocity of the police car.

[tex]5500 = \dfrac{330}{330-v} \times f[/tex]...........(1)

When the car is receding, the frequency of the receiving signal f = 4500 Hz.

[tex]4500 = \dfrac{330}{330+v} \times f[/tex]..........(2)

Solving both equation, we get the velocity of a police car.

v = 33 m/s

Therefore, the velocity v of the police car is 33 m/s.

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what is entrapersonal environment​

Answers

Answer:

The interpersonal environment is considered to be a subset of the organizational environment – defined as the employee’s perception of the practices, policies, and processes of an organization

Explanation:

An object with mass m = 0.56 kg is attached to a string of length r = 0.72 m and is rotating with an angular velocity ω = 1.155 rad/s. What is the centripetal force acting in the object?

Answers

Answer:

The centripetal force is 0.54 N.

Explanation:

mass, m = 0.56 kg

radius, r = 0.72 m

angular speed, w = 1.155 rad/s

The centripetal force is given by

[tex]F = m r w^2\\\\F =0.56\times 0.72\times 1.155\times 1.155\\\\F = 0.54 N[/tex]

help me with this question ​

Answers

Explanation:

Let's set the x-axis to be parallel to the and positive up the plane. Likewise, the y-axis will be positive upwards and perpendicular to the plane. As the problem stated, we are going to assume that m1 will move downwards so its acceleration is negative while m2 moves up so its acceleration is positive. There are two weight components pointing down the plane, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex] and two others pointing up the plane, the two tensions T along the strings. There is a normal force N pointing up from the plane and two pointing down, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex]. Now let's apply Newton's 2nd law to this problem:

x-axis:

[tex]m1:\:\:\:\displaystyle \sum_i F_i = T - m_1g \sin \theta = - m_1a\:\:\:\:(1)[/tex]

[tex]m2:\:\:\:\displaystyle \sum_i F_i = T - m_2g \sin \theta = m_2a\:\:\:\:(2)[/tex]

y-axis:

[tex]\:\:\:\displaystyle \sum_i F_i = N - m_1g \cos \theta - m_2g \cos \theta = 0[/tex]

Use Eqn 1 to solve for T,

[tex]T = m_1(g \sin \theta - a)[/tex]

Substitute this expression for T into Eqn 2,

[tex]m_1g \sin \theta - m_1a - m_2g \sin \theta = m_2a[/tex]

Collecting all similar terms, we get

[tex](m_1 + m_2)a = (m_1 - m_2)g \sin \theta[/tex]

or

[tex]a = \left(\dfrac{m_1 - m_2}{m_1 + m_2} \right)g \sin \theta[/tex]

A 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy

Answers

Answer:

3.6 KJ

Explanation: Given that a 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy

The workdone = the energy.

There are two different energies in the scenario - the potential energy (P.E ) and the kinetic energy ( K.E )

P.E = mgh

P.E = 70 × 9.8 × 1.6

P.E = 1097.6 J

P.E = 1.098 KJ

K.E = 1/2mv^2

K.E = 1/2 × 70 × 8.5^2

K.E = 2528.75 J

K.E = 2.529 KJ

The non conservative workdone = K.E + P.E

Work done = 1.098 + 2.529

Work done = 3.63 KJ

Therefore, the non conservative workdone is 3.6 KJ approximately

Lauren pushed a crate of mass 28.9 kg a distance of 2.7 meters along a horizontal surface. On that part of the surface, the crate could slide with negligible friction. Lauren exerted a constant force of 121 newtons for 2.7 meters. The crate then slid down an inclined plane of height 1.8 meters, also with negligible friction. As the crate slid down the plane, the only significant forces on it were the normal force from the plane and gravity (with g = 9.81 meters per second squared). At the bottom of the incline, the crate began sliding along a horizontal surface with ordinary kinetic friction. The coefficient of friction between this surface and the crate was 0.41. This surface ended at a vertical wall after a distance of d2 = 5.2 meters but the crate did not slide that far. Attached to the vertical wall was a long ideal spring with length dy = 3.4 meters and a spring constant of 154 newtons per meter.

Required:
Calculate the distance that the crate compressed the spring before coming to rest.

Answers

Answer:

3.034 m

Explanation:

From the law of conservation of energy, the energy at the top of the incline equals the energy at the bottom of the incline since at the top of the incline, the horizontal surface is frictionless and along the incline there is no friction.

The work done in moving the crate a distance, d = 2.7 m with a force of F = 121 N to the top of the incline is W = Fd = 121 N × 2.7 m = 326.7 J.

From work-kinetic energy principles, this work W = kinetic energy of the crate at the top of the incline, K₁.

Now, the total mechanical energy at the top of the incline, E equals the total mechanical energy at the bottom of the incline E' since there is no friction along the incline.

So, E = E'

U₁ + K₁ = U₂ + K₂ where U₁ = potential energy of crate at top of incline = mgh where m = mass of crate = 28.9 kg, g = acceleration due to gravity = 9.8 m/s², h = height of incline = 1.8 m, K₁ = kinetic energy of crate at top of incline = 326.7 J, U₂ = potential energy of crate at bottom of incline = 0 J(since it is at an elevation h = 0) and K₂ = kinetic energy of crate at bottom of incline

So, substituting the values of the variables into the equation, we have

U₁ + K₁ = U₂ + K₂

mgh + K₁ = U₂ + K₂

28.9 kg × 9.8 m/s² × 1.8 m + 326.7 J = 0 J + K₂

509.796 J + 326.7 J = K₂

K₂ = 836.496 J

K₂ ≅ 836.5 J

Now since the vertical wall is a distance d2 away and the long ideal spring has a length dy = 3.4 m, let x be the compression of the spring. So, the distance moved by the crate is thus D = d2 - dy - x.

Now, the change in kinetic energy of the crate ΔK equals the work done by friction and that done by the spring W.

So ΔK = -W (from work-kinetic energy principles)

Let W' = work done by friction = μmgD where  μ = coefficient of kinetic friction between surface and crate = 0.41, m = mass of crate = 28.9 kg, g = acceleration due to gravity = 9.8 m/s² and D = distance moved by crate = D = d2 - dy - x = 5.2 m - 3.4 m - x = 1.8 - x

So, W' = μmgD

W' = 0.41 × 28.9 kg × 9.8 m/s² (1.8 - x)

W' = 116.12(1.8 - x)

W' = 2090.16 - 116.12x

The work done by the spring W" = 1/2k(x₀² - x²) where k = spring constant = 154 N/m, x₀ = initial spring length = dy = 3.4 m and x = final spring compression.

So,  W" = 1/2k(x₀² - x²)

W" = 1/2 × 154 N/m[(3.4 m)² - x²]

W" = 77 N/m[11.56 m² - x²]

W" = 890.12  - 77x²

So, W = W' + W"

W = 2090.16 - 116.12x + 890.12  - 77x²

W = 2980.28 - 116.12x - 77x²

Since the crate stops, final kinetic energy K₃ = 0. So, ΔK = K₃ - K₂ = 0 - 836.5 J = -836.5 J

Also, ΔK = -W

-836.5 = -(2980.28 - 116.12x - 77x²)

836.5 = 2980.28 - 116.12x - 77x²

77x² + 116.12 -2980.28 + 836.5 = 0

77x² + 116.12x -2143.78 = 0

dividing through by 77, we have

x² + 1.508x -27.841 = 0

Using the quadratic formula to find x, we have

[tex]x = \frac{-1.508 +/-\sqrt{1.508^{2} - 4 X 1 X (-27.841)} }{2 X 1.508} \\x = \frac{-1.508 +/-\sqrt{2.274064 + 111.364} }{3.016} \\x = \frac{-1.508 +/-\sqrt{113.638064} }{3.016} \\x = \frac{-1.508 +/- 10.66}{3.016} \\x = \frac{-1.508 - 10.66}{3.016} or x = \frac{-1.508 + 10.66}{3.016} \\x = \frac{-12.168}{3.016} or x = \frac{9.152}{3.016} \\x = -4.03 or 3.034[/tex]

x = -4.03 or 3.034

Since the compression of the spring is positive, we choose x = 3.034

So, the crate compresses the spring 3.034 m

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool. We assume the upward direction to be positive, and the downward direction to be negative.
(a) How long are her feet in the air?(b) What is her highest point above the board?(c) What is her velocity when her feet hit the water?

Answers

Answer:

(a) t = 1.14 s

(b) h = 0.82 m

(c) vf = 7.17 m/s

Explanation:

(b)

Considering the upward motion, we apply the third equation of motion:

[tex]2gh = v_f^2 - v_i^2[/tex]

where,

g = - 9.8 m/s² (-ve sign for upward motion)

h = max height reached = ?

vf = final speed = 0 m/s

vi = initial speed = 4 m/s

Therefore,

[tex](2)(9.8\ m/s^2)h = (0\ m/s)^2-(4\ m/s)^2\\[/tex]

h = 0.82 m

Now, for the time in air during upward motion we use first equation of motion:

[tex]v_f = v_i + gt_1\\0\ m/s = 4\ m/s + (-9.8\ m/s^2)t_1\\t_1 = 0.41\ s[/tex]

(c)

Now we will consider the downward motion and use the third equation of motion:

[tex]2gh = v_f^2-v_i^2[/tex]

where,

h = total height = 0.82 m + 1.8 m = 2.62 m

vi = initial speed = 0 m/s

g = 9.8 m/s²

vf = final speed = ?

Therefore,

[tex]2(9.8\ m/s^2)(2.62\ m) = v_f^2 - (0\ m/s)^2\\[/tex]

vf = 7.17 m/s

Now, for the time in air during downward motion we use the first equation of motion:

[tex]v_f = v_i + gt_1\\7.17\ m/s = 0\ m/s + (9.8\ m/s^2)t_2\\t_2 = 0.73\ s[/tex]

(a)

Total Time of Flight = t = t₁ + t₂

t = 0.41 s + 0.73 s

t = 1.14 s

During which radioactive decay process does a neutron change into a proton?
A. Alpha decay
B. Gamma decay
C. Beta decay (positron)
D. Beta decay (electron)

Answers

A neutron changes into a proton during a radioactive decay process called beta decay (positron), which is option D.

What is beta decay?

A beta decay in physics is a nuclear reaction in which a beta particle (electron or positron) is emitted.

A positron is an electron with a positive charge.

During a beta decay, a neutron in the nucleus of the radioactive material suddenly changes into a proton, causing an increase in the atomic number of an element.

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Who stated that man is an animal

Answers

aristotle is the answer to this question

Increasing the surfactant concentration above the critical micellar concentration
will result in: Select one:
1.An increase in surface tension
2. A decrease in surface tension
3. No change in surface tension
4.None of the above​

Answers

Answer:

Explanation:no change in surface tension

An increase in the surfactant concentration above the critical micellar concentration will result in no change in surface tension.

In water-gas interface, surfactant reduces the surface tension of water by adsorbing at the liquid–gas interface.

Also, in oil-water interface, surfactant reduces the interfacial tension between oil and water by adsorbing at the oil-water interface.

The concentration of the surfactant can increase to a level called critical micellar concentration, which is an important characteristic of a surfactant.

As the concentration of the surfactant increases before critical micellar concentration, the surface tension changes strongly with an increase in the concentration of the surfactant. After reaching the critical micellar concentration, any further increase in the concentration will result in no change of the surface tension, that is the surface tension will be constant.

Thus, increasing the surfactant concentration above the critical micellar concentration will result in no change in surface tension.

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who is the biggest man in the world​

Answers

Answer:

Sultan Kösen

here is a pic

what change occurs to the mass of an object when a unbalanced

Answers

Answer:

The mass decreases

Explanation:

Just smart

A helicopter is ascending vertically witha speed of 5.40 m/s. At a height of 105 m above the earth a package is dropped from the helicopter. How much time does is take for the package to reach the ground

Answers

Answer: 5.21 s

Explanation:

Given

Helicopter ascends vertically with [tex]u=5.4\ m/s[/tex]

Height of helicopter [tex]h=105\ m[/tex]

When the package leaves the helicopter, it will have the same vertical velocity

Using equation of motion

[tex]\Rightarrow h=ut+\dfrac{1}{2}at^2\\\\\Rightarrow 105=-5.4t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-5.4t-105=0\\\\\Rightarrow t=\dfrac{5.4\pm \sqrt{5.4^2+4\times 4.9\times 105}}{2\times 4.9}\\\\\Rightarrow t=\dfrac{5.4\pm 45.68}{9.8}\\\\\Rightarrow t=5.21\ s\quad \text{Neglect negative value}[/tex]

So, package will take 5.21 s to reach the ground

A child throws a ball vertically upward to a friend on a balcony 28 m above him. The friend misses the ball on its upward flight but catches it as it is falling back to earth. If the friend catches the ball 3.0 s after it is thrown, at what time did it pass him on its upward flight

Answers

Answer:

[tex]t=1.9 sec[/tex]

Explanation:

From the question we are told that:

Height [tex]h=28m[/tex]

Time [tex]t=3s[/tex]

Generally the Newton's equation for Initial velocity upward is mathematically given by

 [tex]s=ut+\frtac{1}{2}at^2[/tex]

 [tex]28=3u-\frac{1}{2}*9.8*3^2[/tex]

 [tex]u=24.03m/s[/tex]

Generally the velocity at  elevation and depression occurs  as ball arrives and passes through S=28

 [tex]v=\sqrt{24.03-2*9.8*28}[/tex]

 [tex]v=5.35m/s and -5.35m/s[/tex]

Generally the Newton's equation for time to reach initial velocity  is mathematically given by

 [tex]v=u+at[/tex]

 [tex]5.35=24.03-9.8t[/tex]

 [tex]t=\frac{28.03-5.35}{9.8}[/tex]

 [tex]t=1.9 sec[/tex]

What does it mean when work is positive?
O Velocity is greater than kinetic energy.
O Kinetic energy is greater than velocity.
O The environment did work on an object.
O An object did work on the environment.

Answers

Answer:

O The environment did work on an object

Explanation:

LC-circuit of the radio receiver consists of variable capacitor (Cmin= 1 pF, Cmax=10 pF) and inductor
with inductance 1 µH. Determine the wavelength range of this radio receiver.

Answers

Answer:

the radio can tune wavelengths between 1.88 and 5.97 m

Explanation:

The signal that can be received is the one that is in resonance as the impedance of the LC circuit.

         X = X_c - X_L

         X = 1 / wC - w L

at the point of resonance the two impedance are equal so their sum is zero

         X_c = X_L

         1 / wC = w L

         w² = 1 / CL

         w = [tex]\sqrt{\frac{1}{CL} }[/tex]

let's look for the extreme values

C = 1  10⁻¹² F

         w = [tex]\sqrt{\frac{1}{ 1 \ 10^{-12} \ 1 \ 10^{-6}} }[/tex]

         w = [tex]\sqrt{1 \ 10^{18}}[/tex]

         w = 10⁹ rad / s

C = 10 10⁻¹² F

         w = [tex]\sqrt{\frac{1}{10 \ 10^{-12} \ 1 \ 10^{-6}} }[/tex]Ra 1/10 10-12 1 10-6

         w = [tex]\sqrt{0.1 \ 10^{18}}[/tex]Ra 0.1 1018

         w = 0.316 10⁹ rad / s

Now the angular velocity and the frequency are related

           w = 2π f

           f = w / 2π

the light velocity  is

           c = λ f

           λ = c / f

we substitute

          λ = c 2π/w

               

we calculate the two values

 C = 1 pF

          λ₁ = 3 10⁸ 2π / 10⁹

          λ₁= 18.849 10⁻¹ m

          λ₁ = 1.88 m

C = 10 pF

           λ₂ = 3 10⁸ 2π / 0.316 10⁹

           λ₂ = 59.65 10⁻¹ m

           λ₂ = 5.97 m

so the radio can tune wavelengths between 1.88 and 5.97 m

A pilot drops a package from a plane flying horizontally at a constant speed. Neglecting air resistance, when the package hits the ground the horizontal location of the plane will Group of answer choices be behind the package. be over the package. be in front of the package depend of the speed of the plane when the package was released.

Answers

Answer:

The location of helicopter is behind the packet.

Explanation:

As the packet also have same horizontal velocity as same as the helicopter, and also it has some vertical velocity as it hits the ground.

The horizontal velocity remains same as there is no force in the horizontal direction. The vertical  velocity goes on increasing as acceleration due to gravity acts.

So, the helicopter is behind the packet.

plz answer the question

Answers

Answer:

Ray A - incident ray

Ray B - reflected ray

Determine the tension in the string that connects M2 and M3.

Answers

therefore mass m1=4.8 kg and the tension in the horizontal spring T2=10N.

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A 5.0-kg mass is placed at (3.0, 4.0) m, and a 6.0-kg mass is placed at (3.0, -4.0) m. What is the moment of inertia of this system of masses about the y-axis?

Answers

Answer:

the moment of inertia of this system of masses about the y-axis is 99 kgm²

Explanation:

Given the data in the question;

mass m₁ = 5.0 kg at point ( 3.0, 4.0 )

mass m₂ = 6.0 kg at point ( 3.0, -4.0 )

Now, Moment of inertia [tex]I[/tex] of this system of masses about the y-axis will be;

Moment of inertia [tex]I[/tex]ₓ = mixi²

Moment of inertia [tex]I[/tex] = m₁x₁² + m₂x₂²

we substitute

Moment of inertia [tex]I[/tex] = [ 5.0 × ( 3 )² ]  + [ 6.0 × ( 3 )² ]

Moment of inertia [tex]I[/tex] = [ 5.0 × 9 ]  + [ 6.0 × 9 ]

Moment of inertia [tex]I[/tex] = 45 + 54

Moment of inertia [tex]I[/tex] = 99 kgm²

Therefore, the moment of inertia of this system of masses about the y-axis is 99 kgm²

Select the correct answer.
What are the directions of an object's velocity and acceleration vectors when the object moves in a circular path with a constant speed?
OA. The question is meanimgless, since the acceleration is zero.
ов.
The vectors point in opposite directions.
Oc.
Both vectors point in the same direction.
OD
The vectors are perpendicular,

Answers

Answer:

A

Explanation:

If the object is moving at a constant speed, the object isn't accelerating as the velocity doesn't change.

Answer: C.

Explanation:  plato users

Your car breaks down in the middle of nowhere. A tow truck weighing 4000 lbs. comes along and agrees to tow your car, which weighs 2000 lbs., to the nearest town. The driver of the truck attaches his cable to your car at an angle of 20 degrees to horizontal. He tells you that his cable has a strength of 500 lbs. He plans to take 10 secs to tow your car at a constant acceleration from rest in a straight line along a flat road until he reaches the maximum speed of 45 m.p.h. Can the driver carry out the plan

Answers

Answer:

F = 1010 Lb

the tension on the cable is greater than its resistance, which is why the plan is not viable

Explanation:

For this exercise we can use the kinematic relations to find the acceleration and with Newton's second law find the force to which the cable is subjected.

          v = v₀ + a t

how the car comes out of rest v₀ = 0

          a = v / t

let's reduce to the english system

          v = 45 mph (5280 ft / 1 mile) (1h / 3600) = 66 ft / s

let's calculate

          a = 66/10

          a = 6.6 ft / s²

now let's write Newton's second law

X axis

         Fₓ = ma

with trigonometry

         cos 20 = Fₓ / F

         Fₓ = F cos 20

we substitute

          F cos 20 = m a

          F = m a / cos20

          W = mg

          F = [tex]\frac{W}{g} \ \frac{a}{cos 20}[/tex]

let's calculate

          F = [tex]\frac{2000}{32} \ \frac{6.6 }{cos20}[/tex](2000/32) 6.6 / cos 20

          F = 1010 Lb

Under these conditions, the tension on the cable is greater than its resistance, which is why the plan is not viable.

Kulsum’s TV uses 45 W. How much does it cost her to watch TV for one month (30 days). She watches TV for 4 hours/day during mid-peak time (10.4 cents/kWh).

Answers

Answer:

Total cost = 56.16 cents

Explanation:

Given the following data;

Power = 45 Watts

Time = 4 hours

Number of days = 30 days

Cost = 10.4 cents

To find how much does it cost her to watch TV for one month;

First of all, we would determine the energy consumption of the TV;

Energy = power * time

Energy = 45 * 4

Energy = 180 Watt-hour = 180/1000 = 0.18 Kwh (1 Kilowatts is equal to 1000 watts).

Energy consumption = 0.18 Kwh

Next, we find the total cost;

Total cost = energy * number of days * cost

Total cost = 0.18 * 30 * 10.4

Total cost = 56.16 cents

which of the following is a correct statement. a. In dc steady state conditions, the voltages across the capacitors are constant and the currents through the capacitance are zero. The current through the inductors are constant and the voltage across the inductances are constant. b. In dc steady state conditions, the voltages across the capacitors are zero and the currents through the capacitance are constant. The current through the inductors are constant and the voltage across the inductances are zero. c. In dc steady state conditions, the voltages across the capacitors are constant and the currents through the capacitance are zero. The current through the inductors are zero and the voltage across the inductances are constant. d. WIn dc steady state conditions, the voltages across the capacitors are constant and the currents through the capacitance are zero. The current through the inductors are constant and the voltage across the inductances are zero.

Answers

Answer:

d. In dc steady state conditions, the voltages across the capacitors are constant and the currents through the capacitance are zero. The current through the inductors are constant and the voltage across the inductances are zero.

Explanation:

The current through a capacitor is given by i = CdV/dt where C = capacitance of capacitor and V = voltage across capacitor. At steady state dV/dt = 0 and V = constant. So, i = CdV/dt = C × 0 = 0.

So, in dc steady state, the voltage across a capacitor is constant and the current zero.

The voltage across an inductor is given by V = Ldi/dt where L = inductance of inductor and i = current through inductor. At steady state di/dt = 0 and V = constant. So, V = Ldi/dt = L × 0 = 0.

So, in dc steady state, the voltage across an inductor is zero and the current constant.

So, In dc steady state conditions, the voltages across the capacitors are constant and the currents through the capacitance are zero. The current through the inductors are constant and the voltage across the inductances are zero.

The answer is d.

why are you teachers regarded as professionals​

Answers

Answer:

coz teaching is their profession.

Ahmed is pushing a 4 Kg box to the right and Rashid is Pushing it to the right as well with a force of 12 N , the box accelerates by 5 m/s^2. What is the Force that is applied by Ahmed

Answers

Answer:

8 N

Explanation:

Applying,

(F'+F) = ma............... Equation 1

Where F' = Amhed's force, F = Rashid's force, m = mass of the box, a = acceleration of the box.

From the question,

Given: F = 12 N, m = 4 kg, a = 5 m/s²

Substitute these values into equation 1

(F'+12) = 4×5

(F'+12) = 20

F' = 20-12

F' = 8 N.

Hence Ahmed's force is 8 N

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