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Schach [20]
3 years ago
11

A watermelon cannon fires a watermelon vertically up into the air at a velocity of +9.5m/s, starting from an initial position of

1.2m above the ground. When the watermelon reaches the peak of its flight, what is:
a) the velocity _____

b) the acceleration _____

c) the elapsed time ______

d) the height above the ground _______
Physics
1 answer:
choli [55]3 years ago
3 0
In 1 second the watermelon would be 10.7m above the ground. 2 seconds, 21.4m, 3 seconds,32.1m so on, so forth.
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When water moves from a gas state to a liquid state, it is experiencing ______.
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A crowbar makes our work easier by multiplying effort true or false​
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True

Explanation:

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3 years ago
Suppose that a ball is released from the window of a train that is moving with constant velocity.  The path of the ball, as obse
barxatty [35]

True, the path of the ball, as observed from the train window, will be a horizontal straight line.

An object projected from a certain height has a parabolic path when observed from a fixed point.

However, if the reference point is moving at the same velocity as the object, the path of the object's motion appears to be a straight line.

When the ball is released from the window of the train, it will move at the same constant velocity as the train, and the path of the ball's motion observed from the train window will be a straight line.

Thus, we can conclude that the given statement is true. The path of the ball, as observed from the train window, will be a horizontal straight line.

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7 0
2 years ago
30 km/h is _________m/s?<br> A.)8.3<br> B.)5.6<br> C.)13.9<br> D.)11.1
Juli2301 [7.4K]

Answer:

Correct  Option :-  A

Explanation:

3 0
2 years ago
Find deacceleration An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When
earnstyle [38]

Answer:

The deceleration is  a = -0.7273 \ m/s^2  

Explanation:

From the question we are told that

  The distance of the car from the crossing is d = 360 \  m

   The speed is u  = 16 \  m/s

    The reaction time of the engineer is  t = 0.53 \ s

Generally the distance covered during the reaction time is  

      d_r = u * t

=>   d_r = 16 * 0.53

=>   d_r = 8.48 \ m

Generally distance of the car from the crossing after the  engineer reacts is

      D = d- d_r        

=> D = 360 - 8.48      

=> D = 352 \ m

Generally from kinematic equation

      v^2 = u^2 + 2as

Here v is the final velocity of the car which is  0 m/s

So

        0^2 = 16^2 + 2 * a * 352

=>    a = -0.7273 \ m/s^2  

3 0
2 years ago
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