Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant (and quite modest) acceleration. A train travels through a congested part of town at 4.0 m/s . Once free of this area, it speeds up to 11 m/s in 8.0 s. At the edge of town, the driver again accelerates, with the same acceleration, for another 16 s to reach a higher cruising speed. Part A What is the final speed

Answer:

I hear points of low volume sound and points of high volume of sound.

Explanation:

This is because, since the two sources of sound have the same frequency and are separated by a distance, d = 10 mm, there would be successive points of constructive and destructive interference.

Since their frequencies are similar, we should have beats of high and low frequency.

So, at points of low frequency, the amplitude of the wave is smallest and there is destructive interference. The frequency at this point is the difference between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, f - f' = 400 Hz - 400 Hz = 0 Hz. So, the volume of the sound is low(zero) at these points.

Also, at points of high frequency, the amplitude of the wave is highest and there is constructive interference. The frequency at this point is the sum between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, (f + f') = 400 Hz + 400 Hz = 800 Hz. So, the volume of the sound is high at these points.

So, as you wander around the room, I should hear points of high and low sound across the room.

Answer:

10 J.

Explanation:

Given that,

Net force acting on the rake, F = 2 N

Distance moved by the rake, d = 5 m

We need to find the kinetic energy gained by the rake. We know that,

Kinetic energy = work done

So,

K = F×d

K = 2 N × 5 m

K = 10 J

So, 10 J of kinetic energy is gained by the rake.

Violet pulls a rake horizontally on a frictionless driveway with a net force of 2.0 N for 5.0 m.

How much kinetic energy does the rake gain?

Answer: 10 J

Answer:W = m*g*h

19*9.8*32.4 = 6,032.9 rounded

honestly, I do not know if this is correct so please don't come back at me

hopefully this helps

Explanation: [do the following, if you think I am wrong]

just pick a formula,

plug in the number to the mass, gravity, and height

than multiply

get your answer, but don't forget to round to the nearest tenth

Answer:

- time t taken for car to travel is 64.57 s

- distance travelled between A and B is 1.4887 km

Explanation:

Given the data in the question;

[tex]U_{BC}[/tex] =  83 km/h = ( 83×1000 / 60×60) =  23.0555 m/s

[tex]U_{CD}[/tex] = 41 km/h = ( 41×1000 / 60×60) =  11.3888 m/s

now, we calculate the acceleration;

a =  (  [tex]U_{BC}[/tex] -  [tex]U_{CD}[/tex] ) / t

we substitute

a =  ( 23.0555 -  11.3888 ) / 4.4

a = 11.6667 / 4.4

a = 2.6515 m/s²

Now equation for displacement from BC

[tex]S_{BC}[/tex] =  [tex]U_{BC}[/tex]t + 1/2.at²

we substitute

[tex]S_{BC}[/tex] =  23.0555×4.4 + 1/2×a×(4.4)²

we substitute -2.6515m/s² for a

[tex]S_{BC}[/tex] =  23.0555×4.4 + 1/2×(-2.6515)×(4.4)²

= 101.4442 - 25.6665

[tex]S_{BC}[/tex] = 75.7792 m

Now, for total distance covered = 2.3km = ( 2.3×1000) = 2300m

so

[tex]S_{AB}[/tex] +  [tex]S_{BC}[/tex] +  [tex]S_{CD}[/tex]  =  2300 m

we substitute substitute

[tex]S_{AB}[/tex] +  75.7792 m +  [tex]S_{CD}[/tex]  =  2300 m

[tex]S_{AB}[/tex] +  [tex]S_{CD}[/tex] = 2300 - 75.7792

[tex]S_{AB}[/tex] +  [tex]S_{CD}[/tex]  = 2224.2208 m

so we substitute 23.0555t for [tex]S_{AB}[/tex]  and 11.3888t for  [tex]S_{CD}[/tex]  

23.0555t + 11.3888t  = 2224.2208

34.4443t = 2224.2208

t = 2224.2208 / 34.4443

t = 64.57 s

Therefore, time t taken for car to travel is 64.57 s

Distance Between A to B

[tex]S_{AB}[/tex]  = t ×  [tex]U_{AB}[/tex]

we substitute

[tex]S_{AB}[/tex]  = 64.57 s × 23.0555

[tex]S_{AB}[/tex]  = 1488.69 m

[tex]S_{AB}[/tex]  = 1.4887 km

Therefore, distance travelled between A and B is 1.4887 km

Answer:

The answer is B .........number 2

Explanation:

Answer:

I do believe her actions were justified.

Explanation:

Due to the school charging extra fee from her father who makes a modest amount as a teacher. There was sum of money involved that could change how he lived and her.

I do not believe her actions where justified

She had a lot going for her. She could have skipped the hardship of helping grace and pass. She could have easily have gotten a good job with a degree and paid back all the debts owed. Alot of troubles could have been avoided just by doing her own thing.

Answer:

124.51 m

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 49.4 m/s

Final velocity (v) = 0 m/s (at maximum height)

Maximum height (h) =?

NOTE: Acceleration due to gravity (g) = 9.8 m/s²

The maximum height to which the cannon ball attained before falling back can be obtained as illustrated below:

v² = u² – 2gh ( since the ball is going against gravity)

0² = 49.4² – (2 × 9.8 × h)

0 = 2440.36 – 19.6h

Collect like terms

0 – 2440.36 = –19.6h

–2440.36 = –19.6h

Divide both side by –19.6

h = –2440.36 / –19.6

h = 124.51 m

Therefore, maximum height to which the cannon ball attained before falling back is 124.51 m

Answer:

b i think so because it makes senes

Answer: When there is another magnet close by.

Explanation: The needle of a compass is itself a magnet, and thus the north pole of the magnet always points north, except when it is near a strong magnet. ... When you take the compass away from the bar magnet, it again points north. So, we can conclude that the north end of a compass is attracted to the south end of a magnet.

Answer:

2.75 m.

Explanation:

From the question given above, the following data were obtained:

Potential energy (PE) = 41772.5 J

Mass (m) of object = 1550 kg

Height (h) =?

Potential energy is the energy possess by an object due to its location. Mathematically, potential energy is expressed as shown below:

PE = mgh

Where

PE => potential energy

m => mass of the object

g => acceleration due to gravity

h => height to which the object is located.

With the above formula, we can obtain the height to which the object is located as follow:

Potential energy (PE) = 41772.5 J

Mass (m) of object = 1550 kg

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) =?

PE = mgh

41772.5 = 1550 × 9.8 × h

41772.5 = 15190 × h

Divide both side by 15190

h = 41772.5 / 15190

h = 2.75 m

Thus, the object is located at 2.75 m above the ground.

Answer:

the time for the car to reach the final velocity is 0.56 s.

Explanation:

Given;

acceleration of the car, a = 50 m/s²

final velocity of the car, v = 100 km/h = 27.778 m/s

the initial velocity of the car, u = 0

The time for the car to reach the final velocity is calculated as;

v = u + at

27.778 = 0 + 50t

27.778 =  50t

t = 27.778 / 50

t = 0.56 s

Therefore, the time for the car to reach the final velocity is 0.56 s.

Answer:

-6,329.5Joules

Explanation:

Complete question:

Sally applies a horizontal force of 462N with a rope to drag a wooden crate across a floor with a constant speed the rope tied to the crate is pulled at an angle of 56.0degree and sally moves the crate 24.5m. What work is done by the floor through the force of friction between the floor and crate.

Work done = Fd cos theta

F is the horizontal force

d is the distance covered

theta angle of inclination

Substituting into the formula

Workdone  = 462(24.5)cos 56

Workdone = 11,319(0.5592)

Workdone = 6,329.5Joules

Hence the workdone by sally is 6,329.5Joules

The work done by friction will be opposite the work done by sally, hence work done by the floor through force of friction between the floor and the crate is -6,329.5Joules

Answer:

mas

Explanation:

mass is the amount of space something occupies.

Answer:

4363.3231 feets²

Explanation:

Given that :

Distance, r = 50 ft

θ = 200°

The arc length of area covered :

Arc length = θ/360° * πr²

Arc length = (200/360) * 50 ft ^2 * π

Arc length = 0.5555555 * 2500 * π

Arc length = 4363.3231 feets²

Answer:

Explanation:

The electric field inside of a conductor is 0 because the conduction electrons are pushed to the outer edges of the conductor. The surface of the conductor still has charge.

Answer:

d = 106.41 m

Explanation:

Given that,

Initial speed of the car, u = 60 mph = 26.9 m/s

The deceleration in the car, a = -3.4 m/s²

The average reaction time, t = 0.218 m/s

It finally stops, final velocity, v = 0

We need to find the distance covered by the car as it come to a stop.

Using third equation of motion to find.

[tex]v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-26.9^2}{2\times (-3.4)}\\\\d=106.41 m[/tex]

So, the car will cover 106.41 m as it comes to a stop.

Answer:

13.9 m/s.

Explanation:

Since the vertical velocity of the skydiver is constant at v = 7.0 m/s, we find the time, t it takes him to drop from a height of h = 70 m.

So, distance = velocity time

h = vt

t = h/v = 70 m/7 m/s = 10 s

This is also the time it takes him to move horizontally a distance of d = 120 m to the target.

So, his horizontal velocity is v' = distance/time = d/t = 120m/10 s = 12 m/s.

Since both vertical and horizontal velocities are perpendicular, we add them vectorially to obtain the skydivers total speed, V.

So, V = √(v² + v'²)

= √((7.0 m/s)² + (12.0 m/s)'²)

= √(49 m²/s² + 144 m²/s²)

= √(193 m²/s²)

= 13.9 m/s.

The direction of this velocity is Ф = tan⁻¹(v/v')

= tan⁻¹(7 m/s/12 m/s)

= tan⁻¹(0.5833)

= 30.3°

Answer:

McNair graduated as valedictorian of Carver High School in 1967. In 1971, he received a Bachelor of Science degree in engineering physics, magna cu.m laude, from the North Carolina Agricultural and Technical State University in Greensboro, North Carolina.

Answer:

a) v₀ₓ = 62.76 m / s, b)   θ₁ = 17.6º,   θ₂ = 67.0º

Explanation:

We can solve this exercise using the projectile launch ratios

a) Let's find the time it takes for the bullet to reach the water level

       y = y₀ + v_{oy} t - ½ g t²

when it reaches the water its height is zero y = 0, as the bullet is fired horizontally its initial vertical velocity is zero

       0 = y₀ + 0 - ½ g t²

       t =[tex]\sqrt{2y_o/g}[/tex]

       t = [tex]\sqrt{2 \ 7 /9.8}[/tex]          

       t = 1,195 s

now we can calculate the speed with the horizontal movement

        x = v₀ₓ t

        v₀ₓ = x / t

        v₀ₓ = 75.0 / 1.195

        v₀ₓ = 62.76 m / s

b) if the speed of the bullets is half of that found

         v₀ = 62.76 / 2 = 31.38 m / s

let's write the expressions for the distance

          x = v₀ cos θ t

          y = y₀ + v_{oy} sin θ t - ½ g t²

          t = [tex]\frac{x}{v_o \ cos \theta}[/tex]

we substitute

          [tex]0 = y_o + v_o sin \theta \ \frac{x}{v_o \cos \thetay} - 1/2 g \ (\frac{x}{v_o \ cos \theta})^2[/tex]

          [tex]0 = y_o + x tan \theta - \frac{1}{2} g \ \frac{x^2}{ v_o^2 \ cos^2 \theta}[/tex]    

let's use the identified trigonometry

          sec² θ = 1 + tan² θ

         sec θ = 1 / cos θ

we substitute

          [tex]0 = y_o + x tan \theta - \frac{g x^2}{2 v_o^2} ( 1 + tan^2 \theta)[/tex]

          [tex]\frac{g x^2}{2v_o^2} tan^2 \theta - x tan \theta + \frac{gx^2}{2v_o^2} - y_o = 0[/tex]

we change variable

         tan θ = H

         [tex]\frac{gx^2}{2 v_o^2 } H^2 - x H + \frac{gx^2}{2v_o^2}-y_o =0[/tex]

we subtitle the values

         [tex]\frac{9.8 \ 75^2}{2 \ 31.38^2} H^2 - 75 H + \frac{9.8 \ 75^2}{2 \ 31.38^2}-7 =0[/tex]

         27.99 H² - 75 H + 20.99 = 0

         H² - 2.679 H + 0.75 = 0

we solve the quadratic equation

         H = [2.679 ± [tex]\sqrt{2.679^2 - 4 0.75}[/tex]] / 2

         H = [2,679 ± 2,044] / 2

         H₁ = 0.3175

         H₂ = 2.3615

now we can find the angles

          H₁ = tan θ₁

          θ₁ = tan⁻¹ H₁

          θ₁ = tan⁻¹ 0.3175

          θ₁ = 17.6º

          θ₂ = 67.0º

for these two angles the bullet hits the boat

Answer:

Explanation:

Let n be number of total number of nucleons ( protons + neutrons )

Total mass inside nucleus =  n x 1.67 x 10⁻²⁷ Kg

volume of nucleus = 4/3 π r³

= 1.33 x 3.14 x (10⁻¹⁵)³ m

= 4.17 x 10⁻⁴⁵ m³

Density = mass / volume

=  n x 1.67 x 10⁻²⁷ / 4.17 x 10⁻⁴⁵

= .4 n x 10¹⁸ kg / m³

or of the order of 10¹⁸ kg/m³

b )

Density of iron = 7900 kg / m³

or of the order of 10⁴ kg / m³

So nucleus of a matter  is about 10¹⁴ times denser than iron .

Answer:150M

Explanation:

Answer:

1. positive acceleration represents an object speeding up; negative acceleration represents an object slowing down

Explanation:

Acceleration is clearly defined as the rate of change of velocity with time. When are body is speeding up as we say, it is accelerating. When a body is coming to rest, it is decelerating.

Positive acceleration occurs when the speed of a moving continues to increase.

Negative acceleration is when the speed of a moving body reduces drastically.

Answer:

WE NEED TO ADD ALL 40+2.38 +75+5

Explanation:

PLSE GIVE SOME POINTS DUDE

Answer:

 t = 14.53 s

Explanation:

The speed of a wave is constant and is given by

         v = [tex]\sqrt{ \frac{B}{ \rho} }[/tex]

in this exercise they indicate that we assume the constant velocity, therefore we can use the uniform motion relations

          v = x / t

           t = x / v

in this case the sound pulse leaves the ship and must return so the distance is

          x = 2d

where d is the ocean depth d = 10900m and the speed of sound in seawater is v = 1500 m / s

let's calculate

           t = 2 10900/1500

           t = 14.53 s

C

It is right because I took this and I got this answer correct

Answer:i think it is c

Explanation:

Answer:

Explanation: i think its c to try it

Answer:

Friction can stop or slow down the motion of an object.

Explanation:

The slowing force of friction always acts in the direction opposite to the force causing the motion.

Answer:

energy= MGH

2*9.8*h=180

h=180/19.6

h=9.32 m

The height of the object above the ground would be equal to 9.18 m.

When an object of mass (m) is moved from infinity to a certain point inside the gravitational influence, the amount of work done in displacing it is stored in the form of potential energy and is known as gravitational potential energy.

The mathematical equation for gravitational potential energy can be written as:

Gravitational potential energy = m⋅g⋅h

Where m is the mass, g is the gravitational acceleration and h is the height above the ground.

Given, the mass of the given object, m = 2 Kg

The gravitational potential energy = 180 J

[tex]GPE = m\times g\times h[/tex]

180 = 2 × 9.8 × h

h = 9.18 m

Therefore, the object should be at a height of 9.18 meters in order to have 180 J of gravitational potential energy.

Learn more about gravitational potential energy, here:

https://brainly.com/question/3884855

#SPJ2

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated ([tex]Q[/tex]), measured in joules, by the cylindrical resistor is the power multiplied by operation time ([tex]\Delta t[/tex]), measured in hours. That is:

[tex]Q = \dot Q \cdot \Delta t[/tex] (1)

If we know that [tex]\dot Q = 1.2\,W[/tex] and [tex]\Delta t = 86400\,s[/tex], then the amount of heat dissipated by the resistor is:

[tex]Q = (1.2\,W)\cdot (86400\,s)[/tex]

[tex]Q = 103680\,J[/tex]

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux ([tex]Q'[/tex]), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder ([tex]A[/tex]), measured in square meters:

[tex]Q' = \frac{\dot Q}{A}[/tex] (2)

[tex]Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h }[/tex] (3)

Where:

[tex]D[/tex] - Diameter, measured in meters.

[tex]h[/tex] - Length, measured in meters.

If we know that [tex]\dot Q = 1.2\,W[/tex], [tex]D = 4\times 10^{-3}\,m[/tex] and [tex]h = 2\times 10^{-2}\,m[/tex], the heat flux of the resistor is:

[tex]Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }[/tex]

[tex]Q' \approx 4340.589\,\frac{W}{m^{2}}[/tex]

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces ([tex]r[/tex]), no unit, is the ratio of the top and bottom surfaces to total surface:

[tex]r = \frac{\frac{\pi}{2}\cdot D^{2}}{A}[/tex] (3)

If we know that [tex]A \approx 2.765\times 10^{-4}\,m^{2}[/tex] and [tex]D = 4\times 10^{-3}\,m[/tex], then the fraction is:

[tex]r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}[/tex]

[tex]r = 0.045[/tex]

The fraction of heat dissipated from the top and bottom surfaces is 0.045.