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jekas [21]
2 years ago
6

The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

r (mph). At full power, the car can accelerate from zero to 29.0mph in time 1.50s .
Part A

At full power, how long would it take for the car to accelerate from 0 to 58.0mph ? Neglect friction and air resistance. THE ANSWER IS "6 seconds" according the mastering physics.. BUT WHY???? Please show how you derive that answer...

Part B

A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a constant force (in contrast to the constant power in Part A). If such a sports car went from zero to 29.0mph in time 1.50s , how long would it take to go from zero to 58.0mph ?

Express your answer numerically, in seconds THE ANSWER IS 3 seconds. Please show how you derrive this answer!

Express your answer in seconds.
Physics
1 answer:
kirill115 [55]2 years ago
6 0

Answer:

Part A

it would take 6 sec

it would take 3 sec

Explanation:

We are told that the power supplied to the wheel is constant which means that the sport car is gaining energy i.e

                           power \ \alpha  \ energy

Hence if power is constantly supplied energy constantly increase

From the formula of the Kinetic energy

                       KE  = \frac{1}{2} mv^2

we can see that as the speed doubles from 29 mph  to 58 mph  the energy needed is 2^2 = 4 times of the energy from the formula

   Also the time needed would also be 4 times because energy i directly proportional to time

       Hence to reach 58mph the time that it would take is

                          = 4* 1.5sec = 6sec

     

We are told that the ground pushes the car  with a constant force and

                 F = ma

this means that the acceleration is also constant

             now from newtons law

     v = u +at  

 Looking at it we see that final velocity is directly proportional with time

hence it would take twice the time to reach twice the final velocity

        Time to reach 58mph = 3 s

        since time to reach 29 mph(\frac{1}{2} \ of \ [58mph]) =( \frac{1}{2} \ of \ 3sec )1.5 s

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sertanlavr [38]

Answer:

lambda = 343 m/s divided by 340 Hz = 1.009 seconds

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4 0
2 years ago
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In Niels Bohr's 1913 model of the hydrogen atom, the single electron is in a circular orbit of radius 5.29×10⁻¹¹m and its speed
Svet_ta [14]

The magnitude of the magnetic moment due to the electron's motion is 87.87 * 10^{-37}.

<h3>What is magnetic moment?</h3>

The magnetic pull and direction of a magnet or other object that produces a magnetic field are referred to as the magnetic moment in electromagnetism. Things that have magnetic moments include electromagnets, permanent magnets, various compounds, elementary particles like electrons, and a number of celestial objects (such as many planets, some moons, stars, etc).

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Calculations:

radius= 5.29 * 10^{-11} m\\

velocity=2.9* 10^{6} m/s

Working formula, M=N/A

I=\frac{charge flow }{time taken} =\frac{e}{time taken\\}

T= \frac{2xr}{v} =\frac{2xx * 5.29 * 10^{-11} }{2.9* 10^{6} }

   =15.16 * 10^{-5} s

I= \frac{1.6 * 10^{-19} }{15.16 * 10^{-5} }= 0.10 * 10^{-14}

                     =1 * 10^{-15} C

M=1x (1* 10^{-15} * (5.29 * 10^{-11} )^{2}

  =87.87 * 10^{-37}

To learn more about magnetic moment ,visit:

brainly.com/question/14298729

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4 0
10 months ago
Question 24 1 pts Find the voltage in an extension cord having a 0.0600 22 resistance and through which a 5.00A current is flowi
Neporo4naja [7]

The voltage in the extension cord is 30 V.

The problem above can be solved using ohm's law

⇒ Formula:

V = IR.................. Equation 1

⇒ Where:

  • V = Voltage in the extension cord
  • I = Current flowing through the extension cord
  • R = Resistance of the extension cord.

From the question, I think there was a slight error in the value of the current given it suppose to be 500 A, and not 5.00 A

⇒ Given:

  • I = 500 A
  • R = 0.06 ohms

⇒ Substitute these values into equation 1

  • V = 500(0.06)
  • V = 30 V

Hence the voltage in the extension cord is 30 V

Learn more about voltage here: brainly.com/question/4429782

6 0
2 years ago
Problem 1: Three beads are placed along a thin rod. The first bead, of mass m1 = 24 g, is placed a distance d1 = 1.1 cm from the
Svet_ta [14]

Answer:

b)  x_{cm} = 4.88 cm , c) x_{cm}’= 1/M  (m₁ d₁ + m₃ d₃) and d)

x_{cm}’= 1.88 cm

Explanation:

The definition of mass center is

    x_{cm} = 1/M ∑ xi mi

Where mi, xi are the mass and distance from an origin for each mass and M is the total mass of the object.

Part b

Apply this equation to our case.

Body 1

They give us the mass (m₁ = 24 g) and the distance (d₁ = 1.1 cm) from the origin at the far left

Body 2

They give us the mass (m₂ = 12.g) and the distance relative to the distance of the body 1, let's look for the distance from the left end (origin)

    D₂ = d₁ + d₂

    D₂ = 1.1 + 1.9

    D₂ = 3.0 cm

Body 3

Give the mass (m₃ = 56 g) and the position relative to body 2, let's find the distance relative to the origin

    D₃ = D₂ + d₂

    D₃ = 3.0 + 3.9

    D₃ = 6.9 cm

With this data we substitute and calculate the center of mass

    M = m₁ + m₂ + m₃

    M = 24 + 12 + 56

    M = 92 g

    x_{cm} = 1/92 (1.1 24 + 3.0 12 + 6.9 56)

    x_{cm} = 1/92 (448.8)

    x_{cm} = 4,878 cm

    x_{cm} = 4.88 cm

This distance is from the left end of the bar

Par c)

In this case we are asked for the same calculation, but the reference system is in the center marble, we have to rewrite the distance with the reference system in this marble.

Body 1

It is at   d1 = -1.9 cm

It is negative for being on the left and the value is the relative distance of 1 to 2

Body 2

d2 = 0 cm

The reference system for her

Body 3

d3 = 3.9 cm

Positive because that is to the left of the reference system and is the relative distance between 2 and 3

Let's write the new center of mass (xcm')

    x_{cm} ’= 1/M  (m₁ d₁ + m₂ d₂ + m₃ d₃)

   

   x_{cm}’= 1/M  (m₁ d₁ + m₃ d₃)

Part d) Let's calculate the value of the center of mass

    x_{cm}’= 1/92 ((24 (-1.9) +56 3.9)

    x_{cm}’= 1/92 (172.8)

    x_{cm}’= 1.88 cm

This distance is to the right of the central marble

3 0
2 years ago
A wave of amplitude 20mm has intensity Ix. Another wave of the same frequency but of amplitude 5mm has an intensity Iy.
Alexeev081 [22]

Answer:

(C) 16

Explanation:

Given:

The amplitude of first wave (s₁) = 20 mm

The amplitude of second wave (s₂) = 5 mm

Intensity of first wave = Iₓ

Intensity of second wave = I_y

The intensity associated with a wave depends on the amplitude of the wave.

The intensity (I) is directly proportional to the square of the amplitude (s) of the wave and is expressed as:

I=ks^2\\Where\ k\to constant\ of\ proportionality

Now, the intensities of the two waves are given as:

I_x=ks_1^2=k(20)^2\\\\I_y=ks_2^2=k(5)^2

Dividing both the intensities, we get:

\frac{I_x}{I_y}=\frac{k(20)^2}{k(5)^2}\\\\\frac{I_x}{I_y}=\frac{400}{25}\\\\\frac{I_x}{I_y}=16

Therefore, the option (C) is correct.

5 0
2 years ago
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