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VMariaS [17]
3 years ago
13

suppose a car manufacturer tested its cars for front end collsion by hauling them up on a crane and dropping them from a certain

height. What height corresponds to a collision at 130 km/h
Physics
1 answer:
IRINA_888 [86]3 years ago
5 0

Initial height: 66.5 m

Explanation:

The problem can be solved by using the principle of conservation of energy.

If we neglect air resistance, the total mechanical energy of the car is conserved during the fall, therefore we can write:

K_i + U_i = K_f + U_f

where :

K_i = 0 is the kinetic energy of the car at the top (it starts from rest)

U_i = mgh is the gravitational potential energy of the car at the top, with:

m = the mass of the car

g = the acceleration of gravity

h = the heigth of the car

K_f = \frac{1}{2}mv^2 is the kinetic energy of the car just before hitting the ground, with

v = 130 km/h final speed of the car

U_f = 0 is the gravitational potential energy of the car at the bottom

Re-arranging the equation,  we find

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

and we have:

g=9.8 m/s^2\\v = 130 km/h = 36.1 m/s

Solving for h, we find the initial height of the car:

h=\frac{v^2}{2g}=\frac{36.1^2}{2(9.8)}=66.5 m

Learn more about kinetic energy and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647  

brainly.com/question/10770261  

#LearnwithBrainly

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A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a
jeka94

Answer:

a)  t_l - t_r = 12.54 us

b)  (t_l - t_r) / T = 0.0157  

Explanation:

Given:

- Frequency of source f = 1250 Hz

- Distance from source to right ear d_r = 2.6 m

- Distance from source to left ear d_l = ?

- Separation between ears s = 0.15 m

Find:

a. What is the difference in the arrival time of the sound at the left ear and the right ear?

b. What is the ratio of this time difference to the period of the sound wave?

Solution:

- Apply Pythagoras theorem to calculate the distance d_l from source to left ear:

                                      d_l = sqrt ( 2.6^2 + 0.15^2)

                                      d_l = sqrt ( 6.7825 )

                                      d_l = 2.6043 m

- The time deference can be calculated from a simple distance - speed formula:

                                      t_l - t_r = (1 / v) * ( d_l - d_r)

Where, v = 343 m/s speed of sound in air:

                                      t_l - t_r = (1 / 343) * ( 2.6043 - 2.6)  

                                      t_l - t_r = ( 0.0043 / 343 )

                                      t_l - t_r = 12.54 us

- Now we compute the Time period of the sound wave:

                                      T = 1 / f

                                      T = 1 / 1250 = 8*10^-4 s

- The ratio of differential time to Time period T is:

                                      (t_l - t_r) / T = 12.54 * 10^-6 / 8*10^-4

                                      (t_l - t_r) / T = 0.0157  

3 0
3 years ago
A marble slides without friction in a vertical loop around the inside of a smooth, 28.6 cm diameter horizontal pipe. The marble'
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Answer:3.49 m/s

Explanation:

Given

Speed of marble at Bottom v=4.22 m/s

Diameter of loop d=28.6 cm

As Energy is conserved therefore Energy at top is equal to energy at bottom

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v_0^2=v^2+2gh

v^2=v_0^2-2gh

v=\sqrt{v_0^2-2gh}

v=\sqrt{4.22^2-2\times 9.8\times 0.286}

v=\sqrt{17.8084-5.6056}

v=3.49 m/s

                       

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Their Period number is common means their "Principal Quantum Numbers" are same

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