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weqwewe [10]
3 years ago
5

a car travels 50 kilometers West and 10 minutes. after reaching the destination the car troubles back to the starting point, aga

in taking 5 minutes. What is the average velocity of the car
Physics
1 answer:
Gre4nikov [31]3 years ago
6 0
SPEED is

               (distance covered) / (time to cover the distance) .

The car's average SPEED is

                            (100 km)  /  (15min)

                      =     6-2/3 km per minute.


VELOCITY is  

             (distance and direction between start-point and end-point)
divided by
             (time to travel from start-point to end-point).

The car ended up at the starting point, so the distance between
stating-point and end-point is zero.

      Velocity  =  (zero) / (time to travel from start-point to end-point)

                    =  Zero .
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What is the wavelength of the radio waves from an FM station operating at a frequency of 99.5 MHz
ser-zykov [4K]

Answer:

Wavelength, \lambda=3.01\ m

Explanation:

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4 years ago
A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a sp
dusya [7]

from the question you can see that some detail is missing, using search engines i was able to get a similar question on "https://www.slader.com/discussion/question/a-student-throws-a-water-balloon-vertically-downward-from-the-top-of-a-building-the-balloon-leaves-t/"

here is the question : A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 60.0m/s. Air resistance may be ignored,so the water balloon is in free fall after it leaves the throwers hand. a) What is its speed after falling for 2.00s? b) How far does it fall in 2.00s? c) What is the magnitude of its velocity after falling 10.0m?

Answer:

(A) 26 m/s

(B) 32.4 m

(C) v = 15.4 m/s

Explanation:

initial speed (u) = 6.4 m/s

acceleration due to gravity (a) = 9.9 m/s^[2}

time (t) = 2 s

(A)   What is its speed after falling for 2.00s?

  from the equation of motion v = u + at we can get the speed

v = 6.4 + (9.8 x 2) = 26 m/s

(B) How far does it fall in 2.00s?

  from the equation of motion s=ut+0.5at^{2} we can get the distance covered

s = (6.4 x 2) + (0.5 x 9.8 x 2 x 2)

s = 12.8 + 19.6 = 32.4 m

c) What is the magnitude of its velocity after falling 10.0m?

from the equation of motion below we can get the velocity

v^{2} = u^{2} + 2as\\v^{2} = 6.4^{2} + (2x9.8x10)\\V^{2} = 236.96\\v = \sqrt{236.96}

v = 15.4 m/s

7 0
3 years ago
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