A 15.0 kilogram cart initially traveling at 2.0 meters per second east accelerates uniformly at 0.75 meter per second squared east for 6.0
seconds. What is the speed of the cart at the end of this 6,0 second interval? Please include all work - including givens, formula, substitution,
and units for full credit. (2)

Answer:

Explanation:

The equation you need for this is a one-dimensional:

[tex]v_f^2=v_0^2 +2a[/tex]Δx (sorry that looks weird. It is all one equation, even though it's "crooked"!)

We have final velocity, acceleration, and the displacement, and we are looking for initial velocity. I see after working this out that you don't use the die-hard rules for significant digits, so I adapted to get an answer that matches exactly what one of your choices is. Filling in:

[tex](13.5)^2=v_0^2+2(.50)(49.3)[/tex] and

[tex](13.5)^2-2(1.50)(49.3)=v_0^2[/tex] and

[tex]\sqrt{(13.5)^2-2(1.50)(49.3)}=v_0[/tex] so

v₀ = 5.86 m/s

Answer:

yes! what are the questions??

Explanation:

Answer:

D.

[tex]impulse = force \times time \\ = 60 \times 10 \times 20 \\ = 12000 \: kgm {s}^{ - 1} [/tex]

Option C: 60 mass of child/kg & 10 time / s

hope this helps :)

Answer:

f = 0.365 Hz

Explanation:

The angular velocity of a simple pendulum is

           w = [tex]\sqrt{g/L}[/tex]

angular velocity and frequency are related

           w = 2π f

we substitute

           2π f = [tex]\sqrt{g/L}[/tex]

Let's find the initial length of the pendulum

            L = [tex]\frac{g}{4\pi^2 } \ f_o[/tex]

            L₀ = [tex]\frac{9.8}{4 \pi ^2} \ 0.4[/tex]

            L₀ = 0.6206 m

Indicates that the length of the chord triples

             L = 3 L₀

             L = 3 0.6206

             L = 1.8618 m

let's find the frequency

             f = [tex]\frac{1}{2\pi } \ \sqrt{\frac{9.8}{1.8618} }[/tex]

             f = 0.365 Hz

Answer:

To decrease the current;

1) Use fewer loops or number of turns

2) Use a lower speed of rotation of the coil in the magnetic field

3) Use a weaker magnetic

Explanation:

According to Faraday's Law of induction, which is the basis of the electromagnetism, electromagnetic induction and therefore the basis of the  electric generator, can be written as follows;

[tex]\epsilon = -N \cdot \dfrac{\Delta \phi}{\Delta t}[/tex]

Where;

ε = The induced voltage

N = The number of turns (loops)

ΔФ = The change in the magnetic flux

Δt = The change in the time (the duration)

Given that voltage is directly related to the current, decreasing the voltage, decreases the current

To decrease the voltage, and therefore, the current we can;

1) Reduce the number of loops in the coil

2) Increase the time change per unit change in flux by slowing down the speed of rotation of the generator

3) Decrease the amount of change in the magnetic field per turn, by using a weaker magnetic

Answer:

With the help of formula.

Explanation:

We can calculate the electric potential of any point through the formula of electric potential which is given below.

Electric potential =  Coulomb constant x charge/ distance of separation.

Symbolically it can be written as,  V = k q/ r where

V = electric potential  

k = Coulomb constant

q = charge

r = distance of separation

If we have all these data, we can simply put the data in the formula and we will get the value of electric potential.

Explanation:

Please, check the question properly

Answer:

When the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off, this is an example of resistance, which provides light and heat.  

Explanation:

Opening or closing a circuit.

Answer:

Explanation:

Kinetic Energy has a formula of

[tex]KE=\frac{1}{2}mv^2[/tex] and

Potential Energy has a formula of

PE = mgh. The first 2 problems use the KE formula and the next 2 use the PE formula.

Q1. KE = 16 and v = 4 m/s:

[tex]16=\frac{1}{2}m(4)^2[/tex] and

[tex]\frac{2(16)}{16}=m[/tex] so

m = 2 kg

Q2. mass = 1 kg and v = 10 m/s:

[tex]KE=\frac{1}{2}(1)(10)^2[/tex] so

KE = 50 J

Q3. PE = 40, h = 2, g = 9.8:

40 = m(9.8)(2) and

[tex]\frac{40}{(9.8)(2)}=m[/tex] so

m = 2.0 kg

Q4. mass = 2 kg, g = 9.8 m/s/s, h = 20:

PE = 2(9.8)(20) so

PE = 392 J

**I completely disregarded the rules for sig fig's because your numbers were very inexact**

Answer:

The amount of heat energy received at any location on the globe is a direct effect of Sun angle on climate, as the angle at which sunlight strikes Earth varies by location, time of day, and season due to Earth's orbit around the Sun and Earth's rotation around its tilted axis.

Answer:

The earth's climate system depends entirely on the sun for its energy. Solar radiation warms the atmosphere and is fundamental to atmospheric composition.

Explanation:

hope this helps :)

The average velocity is 25 mph north.

Average velocity = displacement/time

Displacement =

(400 mi north) + (275 mi south)

Displacement = 125 mi north

Average V = (125 mi north) / 5 hours.

Average V = 25 mph north

Question

A radio station transmitted a radio wave with frequency of 25MHz the value of the frequency is equivalent to? Leave answer in Hz and in standard form.

Answer:

2.5 x 10⁷Hz

Explanation:

The measured frequency is 25MHz

To convert this to Hz, we simply change the M, which means mega, to its equivalent value which is 10⁶.

Therefore, we have;

25MHz = 25(10⁶) Hz

=> 25MHz = 25 x 10⁶ Hz  

Now express this value in standard form as follows;

25 x 10⁶ Hz = 2.5 x 10¹ x 10⁶

=> 25 x 10⁶ Hz = 2.5 x 10⁷ Hz

Therefore, the value of 25MHz is equivalent to 2.5 x 10⁷Hz

The value of the frequency of the radio wave is equivalent to 2.5 × 10⁷ Hertz.

Given the data in the question;

Frequency of radio wave; [tex]f = 25MHz[/tex]

Frequency is the number of occurrences of a repeating event per unit of time. The SI unit for frequency is the hertz (Hz)

We know that;

[tex]1MHz\ ( Megahertz) = 1000000 Hz\ ( Hertz )[/tex]

so

[tex]25MHz = x[/tex]

We cross multiply

[tex]x = \frac{25MHz\ *\ 1000000Hz}{1MHz} \\\\x = 2.5 * 10^7 Hz[/tex]

Therefore, the value of the frequency of the radio wave is equivalent to 2.5 × 10⁷ Hertz.

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Answer:

The man's total displacement is equal to 0.

Explanation:

Given that,

A man ran a 5 mile race. The race looped around a city park and back  to the starting line.

We need to find the total displacement of the man.

We know that,

Displacement = shortest path covered

Also,

Displacement = final position - initial position

As it reaches back to its starting line, it means, the displacement is equal to 0.

Hence, the man's total displacement is equal to 0.

Answer:

120,000seconds

Explanation:

400 minutes x 5= 2000minutes (2000minutes x 60)seconds =120,000seconds.

btw :

stay safe! :3

Answer:

12000

Explanation:

5 x 400 = 2000, 2000 x 60 = 12000

Answer:

The electromagnetic theory of light , that's the answer

Answer:

the electromagnetic theory of light

Explanation:

got it right on o.w

Answer:

A

Explanation:

Wavelength is represented by lambda

Answer:

44.1 m

Explanation:

Consider the displacement being x:

x = (g•t²)/2

x = (9.8•3²)/2

x = (9.8•9)/2

x = 88.2/2

x = 44.1 m

Answer:

Explanation:

that's the correct answer

Answer: A.

Explanation:

Máquina térmica tem como função, manter a temperatura de um dado fluído, então, a quantidade de calor desse mesmo fluído ao entrar na máquina, será igual á quantidade de calor desse fluido saindo.

Qin - Qout describes the work done by a heat engine. Hence option A is correct.

A heat engine is a system which converts heat into useful work. Carnot engine is  an ideal engine which has maximum efficiency than any other engines. Carnot has showed that "no engine can be more efficient than Carnot engine and the 100% efficient engine can not be existed". It takes heat from the reservoir to do some work and it discharges some amount of heat to the sink. Reservoir is know as hot body and sink is know as cold body.

Efficiency of the Carnot's heat engine is given by,

Efficiency, η = Work done / Heat input

Work done, W = Q1 – Q2

Heat input = Q1

Efficiency,

η = W / Q1

= (Q1 – Q2) / Q1

=1 – (Q2 / Q1).

To have engine with 100% efficiency, Q2 must be zero and it is not possible.

Hence work done by a heat engine is Qin - Qout.

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Answer:

Onda electromagnética

Explanation:

Las ondas mecánicas requieren un medio material para la propagación, mientras que las ondas electromagnéticas no requieren un medio material para la propagación.

Generalmente, las ondas electromagnéticas se mueven con una velocidad muy alta.

Todas las ondas electromagnéticas se denominan colectivamente luz y tienen una velocidad común de 3 × 10 ^ 8 m / s en el aire.

Ninguna onda mecánica tiene una velocidad tan alta en el aire.

Answer:

F = m a

it means as m increase force increase also and acceleration is constant

the best example of this case is free fall regardless of the mass of something the free fall acceleration is constant - 9.81 m/s^2, because as mass increase gravitational force increase also

Answer:

hlw I'm jess bregoli

your answer is here (◕ᴗ◕✿)

mass is the amount of matter..Gravity affects weight, it does not affect mass.

Masses always remain the same.

Explanation:

hope it may help you

Answer:

Explanation:

PE = mgh where m is mass, g is the pull of gravity, and h is the height to which the object can possibly fall.

PE = 3.0(-9.8)(-12.4) so

PE = 360 J, rounded to the correct number of sig figs.

The gravitational potential energy of the rock is 362.4 joules.

The gravitational potential energy of an object is given by the formula:

PE = mgh

where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity (which is approximately 9.81 m/s^2 near the surface of the Earth), and h is the height of the object above a reference point.

In this case, the mass of the rock is 3.0 kg, the height above the ground is 12.4 m, and the acceleration due to gravity is 9.81 m/s^2. Substituting these values into the formula, we get:

PE = (3.0 kg) x (9.81 m/s^2) x (12.4 m) = 362.4 J

Therefore, the gravitational potential energy of the rock is 362.4 joules.

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Answer:

KE = 308642.02469136 J

Answer:

W = 24 J

Explanation:

Given that,

Applied force, F = 60 N

Distance moved, d = 40 cm = 0.4 m

We need to find the work done by the carpenter. We know that,

Work done, W = Fd

Put all the values,

W = 60 N × 0.4 m

= 24 J

Hence, the required work done is equal to 24 J.

Answer:

the angle of incident is 40°

Explanation:

NQ is the normal to the mirror, therefore

angle NQA =90°

PQA = 50°

incident angle = NQA - PQA

90°- 50° = 40°

note that the angle of reflection is equal to the angle of incident

Answer:

Yes

Explanation:

Displacement is equal to the final position of X minus the initial position of X.

Explanation:

Sound travels from the tour guide to the other of the canyon and then back, thus hearing the echo. The time of 2.0 s represents the time it takes the sound to make a round trip, therefore it takes 1.0 s for the tour guide's voice to reach the other side of the canyon. At 20°C, the speed of sound is 343 m/s therefore the width of the canyon is

[tex]x = vt = (343\:m/s)(1.0\:s) = 343\:m[/tex]

The canyon in the given question is 340 m wide.

Echo is a reflection of sound that comes to the listener after the direct sound, but with a delay. The delay is inversely proportional to the separation between the source and the listener of the reflecting surface.

To hear the echo, the following conditions must be met:

There must be a minimum of 17 meters between the sound source and the reflecting surface, and there must be at least 0.1 seconds between the original sound and its echo.

Given parameters:

When a tour guide shouts across a canyon, her echo is heard 2.00 s later. That means, in two seconds the sound goes to other wall of canyon and reflected back.

At a temperature of 20 °C, velocity of sound is 340 m/s.

So, the wide is the canyon be = velocity of sound × time/2

= 340 m/s ×2 s/2s = 340 m.

Hence, the wide is the canyon be 340 m.

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Answer:

kinetic energy of 16 J. Determine the velocity of the first toy, and the mass of the ... distance of 12 cm into the tree. ... b) the average force exerted on the bullet while slowing down to a stop.

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