Answer:
maximum speed of the bananas is 18.8183 m/s
Explanation:
Given data
amplitude A = 23.195 cm
spring constant K = 15.2676 N/m
mass of the bananas m = 56.9816 kg
to find out
maximum speed of the bananas
solution
we know that radial oscillation frequency formula that is = √(K/A)
radial oscillation frequency = √(15.2676/23.195)
radial oscillation frequency is 0.8113125 rad/s
so maximum speed of the bananas = radial oscillation frequency × amplitude
maximum speed of the bananas = 0.8113125 × 23.195
maximum speed of the bananas is 18.8183 m/s
Answer:
600,000,000 degree C
Explanation:
This stage is the last stage and is refereed to as supernova. In the beginning of this stage, gravity pulls the inner core and crush it, due to which fusion of atoms starts. Carbon and Oxygen fuse together and the temperature is about of 600,000,000 degree C.
The most heavier atom that can be formed out of this fusion is the iron. The moment all the atoms becomes of iron, no further fusion is possible hence that body emits radiation of high intensity and collapse causing a big supernova.
Answer:
Given,
Frame rate = 25 frames per second
To find,
Time interval between one frame and the next.
Solution,
We can simply solve this numerical problem by using the following process.
Now,
Number of frames = 25
Total time taken to display the given number of frames (ie. 25 frames) = 1 second
To calculate the time interval between one frame and next, we need to divide the time taken to display total number of frames by total number of frames.
So,
Time interval between one frame and next :
= Time taken to display total number of frames / Total frames
= 1/25
= 0.04 second
Hence, time interval between one frame and next is 0.04 second.
Answer:
a) about 20.4 meters high
b) about 4.08 seconds
Explanation:
Part a)
To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.
In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:
Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:
Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)
To solve for "t" in this quadratic equation, we can factor it out as shown:
Therefore there are two possible solutions when each of the two factors equals zero:
1) t= 0 (which is not representative of our case) , and
2) the expression in parenthesis is zero:
