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Karo-lina-s [1.5K]
3 years ago
14

Two racecars are driving at constant speeds around a circular track. Both cars are the same distance away from the center of the

track, but car 2 is driving twice as fast as car 1. The acceleration of car 2 is ___________ the acceleration of car 1. Two racecars are driving at constant speeds around a circular track. Both cars are the same distance away from the center of the track, but car 2 is driving twice as fast as car 1. The acceleration of car 2 is ___________ the acceleration of car 1. twice half of equal to four times a quarter of
Physics
1 answer:
Alenkasestr [34]3 years ago
4 0

Answer:

The acceleration of car 2 is four times of the acceleration of car 1.

Explanation:

The centripetal acceleration of the object is possessed when it moves in a circular path. It is given by :

a=\dfrac{v^2}{r}

In this case, two race cars are driving at constant speeds around a circular track. Both cars are the same distance away from the center of the track, but car 2 is driving twice as fast as car 1.

So,

\dfrac{a_1}{a_2}=\dfrac{v_1^2}{v_2^2}

1 and 2 represent car 1 and car 2 respectively.

v_2=2v_1

So,

\dfrac{a_1}{a_2}=\dfrac{v_1^2}{(2v_1)^2}\\\\\dfrac{a_1}{a_2}=\dfrac{v_1^2}{4v_1^2}\\\\\dfrac{a_1}{a_2}=\dfrac{1}{4}\\\\a_2=4\times a_1

So, the acceleration of car 2 is four times of the acceleration of car 1.

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A thin rod of length 0.75 m and mass 0.42 kg is suspended
MrRissso [65]

Answer:

a)  K = 0.63 J, b)  h = 0.153 m

Explanation:

a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is

         w² = \frac{m g d}{I}

where d is the distance from the pivot point to the center of mass and I is the moment of inertia.

The rod is a homogeneous body so its center of mass is at the geometric center of the rod.

              d = L / 2

the moment of inertia of the rod is the moment of a rod supported at one end

              I = ⅓ m L²

we substitute

            w = \sqrt{\frac{mgL}{2}  \ \frac{1}{\frac{1}{3} mL^2} }

            w = \sqrt{\frac{3}{2}  \ \frac{g}{L} }

            w = \sqrt{ \frac{3}{2} \ \frac{9.8}{0.75}  }

            w = 4.427 rad / s

an oscillatory system is described by the expression

              θ = θ₀ cos (wt + Φ)

the angular velocity is

             w = dθ /dt

             w = - θ₀ w sin (wt + Ф)

In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1

In the exercise it is indicated that at the lowest point the angular velocity is

           w = 4.0 rad / s

the kinetic energy is

           K = ½ I w²

           K = ½ (⅓ m L²) w²

           K = 1/6 m L² w²

           K = 1/6 0.42 0.75² 4.0²

           K = 0.63 J

b) for this part let's use conservation of energy

starting point. Lowest point

             Em₀ = K = ½ I w²

final point. Highest point

             Em_f = U = m g h

energy is conserved

             Em₀ = Em_f

             ½ I w² = m g h

             ½ (⅓ m L²) w² = m g h

             h = 1/6 L² w² / g

             h = 1/6 0.75² 4.0² / 9.8

             h = 0.153 m

5 0
2 years ago
Jake calculates that the frequency of a wave is 500 hertz and that the wave is moving at 1,250 m/s. What is the wavelength of th
Neko [114]
Frequency (f) = 500 hz (SI)
Velocity (V) = 1250 m/s (SI)
Wavelength (Lambda) = ? meters

v =  \lambda \times f
1250 =  \lambda \times 500 \\ \lambda = 1250 \div 500 \\ \lambda = 2.5 \: meters
6 0
3 years ago
4. With a diameter that's 11 times larger than Earth's, _______ is the largest planet.
Komok [63]

The answer is B.Jupiter

7 0
3 years ago
A toy rocket launcher can project a toy rocket at a speed as high as 35.0 m/s.
Anestetic [448]

Answer:

(a) 62.5 m

(b) 7.14 s

Explanation:

initial speed, u = 35 m/s

g = 9.8 m/s^2

(a) Let the rocket raises upto height h and at maximum height the speed is zero.

Use third equation of motion

v^{2}=u^{2}+2as

0^{2}=35^{2}- 2 \times 9.8 \times h

h = 62.5 m

Thus, the rocket goes upto a height of 62.5 m.

(b) Let the rocket takes time t to reach to maximum height.

By use of first equation of motion

v = u + at

0 = 35 - 9.8 t

t = 3.57 s

The total time spent by the rocket in air = 2 t = 2 x 3.57 = 7.14 second.

8 0
3 years ago
A nuclear explosion may release tremendous amounts of energy in the form of noise, heat, visible light, radiation and an atmosph
posledela

light and radiation is able to experience by an observer watching the explosion from the vacuum.

<u>Explanation:</u>

  • Basically in the nuclear explosion, there will be an enormous of energy released as noise, heat, visible light, radiation and atmospheric wave.
  • Usually sound and wave propagate through medium but in vacuum, there will be no medium to transfer this type of energy. so there is no chance of sound and wave transfer.  
  • Light and radiation travel in vacuum because they didn't need the medium to transfer .  

7 0
3 years ago
Read 2 more answers
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