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marissa [1.9K]
4 years ago
10

NEED HELP!! WILL GIVE BRAINIEST!! When a driver presses the brake pedal, his car can stop with an acceleration of -5.4 meters pe

r second squared. how far will the car travel while coming to a complete stop if its initial speed was 25 meters per second?
Physics
2 answers:
nexus9112 [7]4 years ago
7 0

Answer:

d = 57.87 m

Explanation:

It is given that,

Acceleration of car due to applied brake is -5.4\ m/s^2

Its initial speed was 25 m/s

Its final speed was 0 as its comes to rest.

Leat d is the distance covered by the car. It can be calculated using third equation of kinematics as :

v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-5.4)}\\\\d=57.87\ m

So, the distance covered by the car is 57.87 m.

8_murik_8 [283]4 years ago
3 0

Answer:

d = 57.87 m

Explanation:  have a nice day :)

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The difference in frequency of the two signals is 1.33 \times 10^{10} \ kHz.

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E = hf\\\\E_1 = h \frac{c}{\lambda} \\\\E_1 =  \frac{(6.626 \times 10^{-34})\times 3\times 10^8}{20 \times 10^{-6}} \\\\E_1 = 9.94 \times 10^{-21} \ J

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The difference in frequency of the two signals is calculated as follows;

E_1- E_2 = hf_1 - hf_2\\\\E_1 - E_2 = h(f_1 - f_2)\\\\f_1 - f_2 = \frac{E_1 - E_2 }{h} \\\\f_1 - f_2 = \frac{(9.94\times 10^{-21}) - (1.11 \times 10^{-21})}{6.626\times 10^{-34}} \\\\f_1 - f_2 = 1.33 \times 10^{13} \ Hz\\\\f_1 - f_2 = 1.33\times 10^{10} \ kHz

Thus, the difference in frequency of the two signals is 1.33 \times 10^{10} \ kHz.

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