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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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What is the relative size and composition of the universe, a galaxy, and a solar system? Is the universe endless?
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Consider a projectile of mass 20 kg launched with a speed 9 m/s at an elevation angle of 45 degrees. Taking the launch point as
viva [34]

Answer:

a) L=0. b) L = 262 k ^   Kg m²/s and c)  L = 1020.7 k^   kg m²/s

Explanation:

It is angular momentum given by

      L = r x p

Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum

One of the easiest ways to make this vector product is with the use of determinants

{array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]

Let's apply this relationship to our case

Let's start by breaking down the speed

      v₀ₓ = v₀ cosn 45

      voy =v₀ sin 45

      v₀ₓ = 9 cos 45

      voy = 9 without 45

      v₀ₓ = 6.36 m / s

      voy = 6.36 m / s

a) at launch point r = 0 whereby L = 0

. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero

   vfy² = voy²- 2 g y

   y = voy² / 2g

   y = (6.36)²/2 9.8

   y = 2.06 m

Let's calculate the angular momentum

L= \left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]

L = -px y k ^

L = - (m vox) (2.06) k ^

L = - 20 6.36 2.06 k ^

L = 262 k ^   Kg m² / s

The angular momentum is on the z axis

c) At the point of impact, at this point the height is zero and the position on the x-axis is the range

     R = vo² sin 2θ / g

     R = 9² sin (2 45) /9.8

     R = 8.26 m

L = \left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]

L = - x py k ^

L = - x m voy

L = - 8.26 20 6.36 k ^

L = 1020.7 k^   kg m² /s

5 0
3 years ago
2.<br> 1,200 kg<br> 8.33 m/s<br> 3.3
Sliva [168]

Answer:

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Explanation:

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3 0
3 years ago
a tennis ball is thrown straight up with an initial velocity of 22.5 m/s how much total time is the ball in the air
lutik1710 [3]

Answer:

4.6s

Explanation:

v=u+at

0=22.5+(-9.8)t

-22.5=-9.8t

t=-22.5/-9.8

t=2.295 s

The total time will double

2.295×2=4.59s

=4.6s

6 0
3 years ago
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